| Matric Number | Name |
|---|---|
| S2153101 | ROWENA CHOY XIN HUI |
| S2115935 | LIM JIE-YING |
| S2151909 | LIM KIM HOONG |
| S2141911 | JING SU |
| S2117541 | RONG SONG |
| Column Name | Description |
|---|---|
| 1. enrollee_id | Unique ID for every candidate |
| 2. city | City code |
| 3. city_development_index | Development index of the city (scaled) |
| 4. gender | Gender of candidate |
| 5. relevant_experience | Relevant experience of candidate |
| 6. enrolled_university | Type of University course enrolled if any |
| 7. education level | Education level of candidate |
| 8. major_discipline | Education major discipline of candidate |
| 9. experience | Candidate total experience in years |
| 10. company_size | No of employees in current employer’s company |
| 11. company_type | Type of current employer |
| 12. last_new_job | Difference in years between previous job and current job |
| 13. training_hours | Training hours completed |
| 14. target | 0 – Not looking for job change, 1 – Looking for a job change |
Predict the probability of a candidate will work for the company or leave after completed the training
Interpret the model(s) in such a way that illustrates which features will affect candidates’ decision
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionlibrary(tidyr)
library(tidyverse)## ── Attaching packages ─────────────────────────────────────── tidyverse 1.3.1 ──## ✓ ggplot2 3.3.6 ✓ purrr 0.3.4
## ✓ tibble 3.1.7 ✓ stringr 1.4.0
## ✓ readr 2.1.2 ✓ forcats 0.5.1## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()library(ggplot2)
library(formattable)
library(fastDummies)
library(caTools)
library(InformationValue)
library(e1071)
library(class)
library(randomForest)## randomForest 4.7-1.1## Type rfNews() to see new features/changes/bug fixes.##
## Attaching package: 'randomForest'## The following object is masked from 'package:ggplot2':
##
## margin## The following object is masked from 'package:dplyr':
##
## combinelibrary(C50)
library(tree)
library(rpart)
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:formattable':
##
## area## The following object is masked from 'package:dplyr':
##
## selectlibrary(ROCR)data = read.csv('https://raw.githubusercontent.com/RongSong1110/WDQ7004/main/aug_train.csv')
head(data)## enrollee_id city city_development_index gender relevent_experience
## 1 8949 city_103 0.920 Male Has relevent experience
## 2 29725 city_40 0.776 Male No relevent experience
## 3 11561 city_21 0.624 No relevent experience
## 4 33241 city_115 0.789 No relevent experience
## 5 666 city_162 0.767 Male Has relevent experience
## 6 21651 city_176 0.764 Has relevent experience
## enrolled_university education_level major_discipline experience company_size
## 1 no_enrollment Graduate STEM >20
## 2 no_enrollment Graduate STEM 15 50-99
## 3 Full time course Graduate STEM 5
## 4 Graduate Business Degree <1
## 5 no_enrollment Masters STEM >20 50-99
## 6 Part time course Graduate STEM 11
## company_type last_new_job training_hours target
## 1 1 36 1
## 2 Pvt Ltd >4 47 0
## 3 never 83 0
## 4 Pvt Ltd never 52 1
## 5 Funded Startup 4 8 0
## 6 1 24 1class(data)## [1] "data.frame"dim(data)## [1] 19158 14summary(data)## enrollee_id city city_development_index gender
## Min. : 1 Length:19158 Min. :0.4480 Length:19158
## 1st Qu.: 8554 Class :character 1st Qu.:0.7400 Class :character
## Median :16982 Mode :character Median :0.9030 Mode :character
## Mean :16875 Mean :0.8288
## 3rd Qu.:25170 3rd Qu.:0.9200
## Max. :33380 Max. :0.9490
## relevent_experience enrolled_university education_level major_discipline
## Length:19158 Length:19158 Length:19158 Length:19158
## Class :character Class :character Class :character Class :character
## Mode :character Mode :character Mode :character Mode :character
##
##
##
## experience company_size company_type last_new_job
## Length:19158 Length:19158 Length:19158 Length:19158
## Class :character Class :character Class :character Class :character
## Mode :character Mode :character Mode :character Mode :character
##
##
##
## training_hours target
## Min. : 1.00 Min. :0.0000
## 1st Qu.: 23.00 1st Qu.:0.0000
## Median : 47.00 Median :0.0000
## Mean : 65.37 Mean :0.2493
## 3rd Qu.: 88.00 3rd Qu.:0.0000
## Max. :336.00 Max. :1.0000str(data)## 'data.frame': 19158 obs. of 14 variables:
## $ enrollee_id : int 8949 29725 11561 33241 666 21651 28806 402 27107 699 ...
## $ city : chr "city_103" "city_40" "city_21" "city_115" ...
## $ city_development_index: num 0.92 0.776 0.624 0.789 0.767 0.764 0.92 0.762 0.92 0.92 ...
## $ gender : chr "Male" "Male" "" "" ...
## $ relevent_experience : chr "Has relevent experience" "No relevent experience" "No relevent experience" "No relevent experience" ...
## $ enrolled_university : chr "no_enrollment" "no_enrollment" "Full time course" "" ...
## $ education_level : chr "Graduate" "Graduate" "Graduate" "Graduate" ...
## $ major_discipline : chr "STEM" "STEM" "STEM" "Business Degree" ...
## $ experience : chr ">20" "15" "5" "<1" ...
## $ company_size : chr "" "50-99" "" "" ...
## $ company_type : chr "" "Pvt Ltd" "" "Pvt Ltd" ...
## $ last_new_job : chr "1" ">4" "never" "never" ...
## $ training_hours : int 36 47 83 52 8 24 24 18 46 123 ...
## $ target : num 1 0 0 1 0 1 0 1 1 0 ...data[,c('enrollee_id')]<-list(NULL)names(data)[names(data) == 'relevent_experience'] <- "relevant_experience"col_name<-names(data)
categorical_col_name<-names(Filter(is.character,data))
num_col_name<-setdiff(col_name,categorical_col_name)
num_col_name<-num_col_name[!num_col_name %in% 'target']
categorical_data<-data[,c(categorical_col_name)]
unique_value<-function(x){
print("Unique values of categorical variables in the dataset:")
lapply(x,unique)
}
unique_value(categorical_data)## [1] "Unique values of categorical variables in the dataset:"## $city
## [1] "city_103" "city_40" "city_21" "city_115" "city_162" "city_176"
## [7] "city_160" "city_46" "city_61" "city_114" "city_13" "city_159"
## [13] "city_102" "city_67" "city_100" "city_16" "city_71" "city_104"
## [19] "city_64" "city_101" "city_83" "city_105" "city_73" "city_75"
## [25] "city_41" "city_11" "city_93" "city_90" "city_36" "city_20"
## [31] "city_57" "city_152" "city_19" "city_65" "city_74" "city_173"
## [37] "city_136" "city_98" "city_97" "city_50" "city_138" "city_82"
## [43] "city_157" "city_89" "city_150" "city_70" "city_175" "city_94"
## [49] "city_28" "city_59" "city_165" "city_145" "city_142" "city_26"
## [55] "city_12" "city_37" "city_43" "city_116" "city_23" "city_99"
## [61] "city_149" "city_10" "city_45" "city_80" "city_128" "city_158"
## [67] "city_123" "city_7" "city_72" "city_106" "city_143" "city_78"
## [73] "city_109" "city_24" "city_134" "city_48" "city_144" "city_91"
## [79] "city_146" "city_133" "city_126" "city_118" "city_9" "city_167"
## [85] "city_27" "city_84" "city_54" "city_39" "city_79" "city_76"
## [91] "city_77" "city_81" "city_131" "city_44" "city_117" "city_155"
## [97] "city_33" "city_141" "city_127" "city_62" "city_53" "city_25"
## [103] "city_2" "city_69" "city_120" "city_111" "city_30" "city_1"
## [109] "city_140" "city_179" "city_55" "city_14" "city_42" "city_107"
## [115] "city_18" "city_139" "city_180" "city_166" "city_121" "city_129"
## [121] "city_8" "city_31" "city_171"
##
## $gender
## [1] "Male" "" "Female" "Other"
##
## $relevant_experience
## [1] "Has relevent experience" "No relevent experience"
##
## $enrolled_university
## [1] "no_enrollment" "Full time course" "" "Part time course"
##
## $education_level
## [1] "Graduate" "Masters" "High School" ""
## [5] "Phd" "Primary School"
##
## $major_discipline
## [1] "STEM" "Business Degree" "" "Arts"
## [5] "Humanities" "No Major" "Other"
##
## $experience
## [1] ">20" "15" "5" "<1" "11" "13" "7" "17" "2" "16" "1" "4"
## [13] "10" "14" "18" "19" "12" "3" "6" "9" "8" "20" ""
##
## $company_size
## [1] "" "50-99" "<10" "10000+" "5000-9999" "1000-4999"
## [7] "10/49" "100-500" "500-999"
##
## $company_type
## [1] "" "Pvt Ltd" "Funded Startup"
## [4] "Early Stage Startup" "Other" "Public Sector"
## [7] "NGO"
##
## $last_new_job
## [1] "1" ">4" "never" "4" "3" "2" ""#replace 'has relevant experience' and 'no relevant experience' with TRUE and FALSE
data$relevant_experience[data$relevant_experience=='Has relevent experience']<-'yes'
data$relevant_experience[data$relevant_experience=='No relevent experience']<-'no'
#replace the inconsistent value on company size variable
data$company_size<-replace(data$company_size,data$company_size == '10/49', '10-49')
data$company_size<-replace(data$company_size,data$company_size == '100-500', '100-499')
data$company_size<-replace(data$company_size,data$company_size == '10000+', '>9999')
data$last_new_job<-replace(data$last_new_job,data$last_new_job=='never',0)missing_value<-function(x){
print("Missing values in the dataset:")
for(i in x) {
print(paste(i,sum(data[i]==""|is.na(data[i]))))
}
}
missing_value(col_name)## [1] "Missing values in the dataset:"
## [1] "city 0"
## [1] "city_development_index 0"
## [1] "gender 4508"
## [1] "relevant_experience 0"
## [1] "enrolled_university 386"
## [1] "education_level 460"
## [1] "major_discipline 2813"
## [1] "experience 65"
## [1] "company_size 5938"
## [1] "company_type 6140"
## [1] "last_new_job 423"
## [1] "training_hours 0"
## [1] "target 0"cols_with_nan_data = data[,c('gender', 'enrolled_university', 'major_discipline', 'experience', 'company_size', 'last_new_job', 'company_type', 'education_level')]
print(head(cols_with_nan_data))## gender enrolled_university major_discipline experience company_size
## 1 Male no_enrollment STEM >20
## 2 Male no_enrollment STEM 15 50-99
## 3 Full time course STEM 5
## 4 Business Degree <1
## 5 Male no_enrollment STEM >20 50-99
## 6 Part time course STEM 11
## last_new_job company_type education_level
## 1 1 Graduate
## 2 >4 Pvt Ltd Graduate
## 3 0 Graduate
## 4 0 Pvt Ltd Graduate
## 5 4 Funded Startup Masters
## 6 1 Graduatelapply(names(cols_with_nan_data), function(col) {
ggplot(cols_with_nan_data, aes(.data[[col]], ..count..)) +
geom_bar(aes(fill = .data[[col]]), position = "dodge")
}) -> list_plots
list_plots## [[1]]
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## [[3]]
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## [[4]]
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## [[5]]
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## [[6]]
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## [[7]]
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for(i in col_name){
data[!is.na(data[i])&data[i]=="",i]<-NA
}
# Find the size of missing value to each variable
missing_value_size<-function(x){
print("The proportion of missing value")
for(i in x){
if(sum(is.na(data[i]))>0){
print(paste(i,percent((sum(is.na(data[i]))/nrow(data)))))
}
}
}
missing_value_size(col_name)## [1] "The proportion of missing value"
## [1] "gender 23.53%"
## [1] "enrolled_university 2.01%"
## [1] "education_level 2.40%"
## [1] "major_discipline 14.68%"
## [1] "experience 0.34%"
## [1] "company_size 30.99%"
## [1] "company_type 32.05%"
## [1] "last_new_job 2.21%"From the above result, we observe that the missing value for ‘experience’ is extremely small. Hence, we replace it with the mode value
data$experience<-replace(data$experience,is.na(data$experience), '>20')We will replace the rest of the missing values with a new class called ‘unknown’ as the proportion of the missing values is extremely large.
col_miss<-c('gender','enrolled_university','major_discipline','company_size','last_new_job','company_type','education_level')
for(i in col_miss){
data[is.na(data[i]),i]<-'unknown'
}
head(data)## city city_development_index gender relevant_experience
## 1 city_103 0.920 Male yes
## 2 city_40 0.776 Male no
## 3 city_21 0.624 unknown no
## 4 city_115 0.789 unknown no
## 5 city_162 0.767 Male yes
## 6 city_176 0.764 unknown yes
## enrolled_university education_level major_discipline experience company_size
## 1 no_enrollment Graduate STEM >20 unknown
## 2 no_enrollment Graduate STEM 15 50-99
## 3 Full time course Graduate STEM 5 unknown
## 4 unknown Graduate Business Degree <1 unknown
## 5 no_enrollment Masters STEM >20 50-99
## 6 Part time course Graduate STEM 11 unknown
## company_type last_new_job training_hours target
## 1 unknown 1 36 1
## 2 Pvt Ltd >4 47 0
## 3 unknown 0 83 0
## 4 Pvt Ltd 0 52 1
## 5 Funded Startup 4 8 0
## 6 unknown 1 24 1missing_value(col_name)## [1] "Missing values in the dataset:"
## [1] "city 0"
## [1] "city_development_index 0"
## [1] "gender 0"
## [1] "relevant_experience 0"
## [1] "enrolled_university 0"
## [1] "education_level 0"
## [1] "major_discipline 0"
## [1] "experience 0"
## [1] "company_size 0"
## [1] "company_type 0"
## [1] "last_new_job 0"
## [1] "training_hours 0"
## [1] "target 0"# handling the value of the variable 'city' as a numeric type
data <- separate(data, city, c('Name','city'),sep = '_')
data[,'Name']<-list(NULL)
head(data)## city city_development_index gender relevant_experience enrolled_university
## 1 103 0.920 Male yes no_enrollment
## 2 40 0.776 Male no no_enrollment
## 3 21 0.624 unknown no Full time course
## 4 115 0.789 unknown no unknown
## 5 162 0.767 Male yes no_enrollment
## 6 176 0.764 unknown yes Part time course
## education_level major_discipline experience company_size company_type
## 1 Graduate STEM >20 unknown unknown
## 2 Graduate STEM 15 50-99 Pvt Ltd
## 3 Graduate STEM 5 unknown unknown
## 4 Graduate Business Degree <1 unknown Pvt Ltd
## 5 Masters STEM >20 50-99 Funded Startup
## 6 Graduate STEM 11 unknown unknown
## last_new_job training_hours target
## 1 1 36 1
## 2 >4 47 0
## 3 0 83 0
## 4 0 52 1
## 5 4 8 0
## 6 1 24 1for(i in num_col_name){
boxplot(data[i],main="Boxplot")
print(paste(i,length(data[,i][data[,i] %in% boxplot.stats(data[,i])$out])))
}
## [1] "city_development_index 17"
## [1] "training_hours 984"From the above result, we observe that there are 17 outliers on vairable ‘city_development_index’ and 984 outliers on ‘training hours’. We will need to check the minimum outlier value and maximum value outlier for ‘training_hour’.
print(paste('minimum outlier value for training hour is:',min(data[,'training_hours'][data[,'training_hours'] %in% boxplot.stats(data[,'training_hours'])$out])))## [1] "minimum outlier value for training hour is: 188"print(paste('maximum outlier value for training hour is:',max(data[,'training_hours'][data[,'training_hours'] %in% boxplot.stats(data[,'training_hours'])$out])))## [1] "maximum outlier value for training hour is: 336"If we assume the working hour for a data scientist per day is 8 hours, the minimum outlier value for training hour of a data scientist is 23.5 days (which is almost 1 month) while the maximum outlier value for training hour of a data scientist 42 days (which is almost 2 months). The training hours seems to be quite reasonable and training hour might contribute a significant decision factors on the output, hence we decide not to replace the outlier for training_hour.
Next, we check on the outlier for ’city_development index.
print(unique(data[,'city_development_index'][data[,'city_development_index'] %in% boxplot.stats(data[,'city_development_index'])$out]))## [1] 0.448The outlier for city_development_index is 0.448. We now will need to check if is all the entry for the outlier is coming from the same city.
unique(filter(data,city_development_index==0.448)[,'city'])## [1] "33"From the above result, we observe that the city development index with 0.448 is all come from the same city with is city, 33. We now further validate if all the entry with city 33 are having a same city development index, 0.448.
unique(filter(data,city==33)[,'city_development_index'])## [1] 0.448From the above result, we can prove all the entry with city, 33 have the same city development index 0.448. Hence, we could say the city development index with 0.448 is not an outlier and is a reasonable entry.
data_preprocess_categorical<-dummy_cols(data,select_columns=categorical_col_name)
data_preprocess_categorical[,c(categorical_col_name)]<-list(NULL)
dim(data_preprocess_categorical)## [1] 19158 194# city & target
dat0 <- data.frame(table(data$city,data$target))
names(dat0) <- c("City","target","Count")
dat0<-spread(dat0,target,Count)
dat0['prob_of_stay']<-round(dat0['0']/(dat0['0']+dat0['1']),4)
dat0<-dat0[order(dat0[,"prob_of_stay"],decreasing=TRUE),]
dat0## City 0 1 prob_of_stay
## 13 111 3 0 1.0000
## 26 129 3 0 1.0000
## 35 140 1 0 1.0000
## 63 2 7 0 1.0000
## 77 39 11 0 1.0000
## 93 62 5 0 1.0000
## 109 8 4 0 1.0000
## 112 82 4 0 1.0000
## 106 77 31 1 0.9688
## 121 97 96 8 0.9231
## 71 28 177 15 0.9219
## 75 36 147 13 0.9188
## 32 138 110 10 0.9167
## 66 23 166 16 0.9121
## 7 104 273 28 0.9070
## 92 61 178 19 0.9036
## 14 114 1203 133 0.9004
## 101 72 18 2 0.9000
## 122 98 71 8 0.8987
## 104 75 274 31 0.8984
## 64 20 26 3 0.8966
## 31 136 525 61 0.8959
## 2 10 77 9 0.8953
## 86 50 125 15 0.8929
## 8 105 70 9 0.8861
## 1 1 23 3 0.8846
## 49 16 1354 179 0.8832
## 97 69 15 2 0.8824
## 27 13 42 6 0.8750
## 56 173 132 19 0.8742
## 90 57 90 13 0.8738
## 96 67 374 57 0.8677
## 110 80 13 2 0.8667
## 46 157 19 3 0.8636
## 95 65 151 24 0.8629
## 94 64 98 16 0.8596
## 83 45 97 16 0.8584
## 72 30 18 3 0.8571
## 76 37 12 2 0.8571
## 111 81 6 1 0.8571
## 100 71 227 39 0.8534
## 78 40 58 10 0.8529
## 36 141 23 4 0.8519
## 48 159 80 14 0.8511
## 19 12 11 2 0.8462
## 123 99 79 15 0.8404
## 113 83 120 23 0.8392
## 20 120 5 1 0.8333
## 5 102 252 52 0.8289
## 16 116 106 22 0.8281
## 98 7 22 5 0.8148
## 70 27 38 9 0.8085
## 119 93 21 5 0.8077
## 24 127 8 2 0.8000
## 29 133 8 2 0.8000
## 79 41 71 18 0.7978
## 69 26 19 5 0.7917
## 6 103 3427 928 0.7869
## 57 175 11 3 0.7857
## 88 54 11 3 0.7857
## 44 152 40 11 0.7843
## 52 165 64 18 0.7805
## 67 24 48 14 0.7742
## 42 149 78 24 0.7647
## 50 160 646 199 0.7645
## 3 100 210 65 0.7636
## 115 89 51 16 0.7612
## 43 150 49 16 0.7538
## 34 14 21 7 0.7500
## 53 166 3 1 0.7500
## 60 18 3 1 0.7500
## 73 31 3 1 0.7500
## 102 73 206 74 0.7357
## 22 123 58 21 0.7342
## 84 46 93 35 0.7266
## 39 144 21 8 0.7241
## 116 9 13 5 0.7222
## 30 134 31 12 0.7209
## 105 76 36 14 0.7200
## 61 180 5 2 0.7143
## 89 55 10 4 0.7143
## 51 162 91 37 0.7109
## 58 176 17 7 0.7083
## 114 84 17 7 0.7083
## 54 167 7 3 0.7000
## 91 59 7 3 0.7000
## 37 142 37 16 0.6981
## 47 158 34 15 0.6939
## 17 117 9 4 0.6923
## 87 53 18 8 0.6923
## 117 90 135 62 0.6853
## 9 106 6 3 0.6667
## 11 109 6 3 0.6667
## 18 118 18 9 0.6667
## 21 121 2 1 0.6667
## 28 131 6 3 0.6667
## 82 44 12 6 0.6667
## 118 91 29 16 0.6444
## 41 146 5 3 0.6250
## 108 79 5 3 0.6250
## 62 19 74 45 0.6218
## 120 94 16 10 0.6154
## 15 115 33 21 0.6111
## 38 143 25 16 0.6098
## 99 70 25 19 0.5682
## 68 25 2 2 0.5000
## 107 78 15 16 0.4839
## 103 74 50 54 0.4808
## 23 126 13 15 0.4643
## 85 48 6 7 0.4615
## 25 128 40 52 0.4348
## 4 101 32 43 0.4267
## 81 43 5 7 0.4167
## 40 145 26 37 0.4127
## 65 21 1105 1597 0.4090
## 12 11 100 147 0.4049
## 59 179 2 3 0.4000
## 74 33 6 11 0.3529
## 10 107 2 4 0.3333
## 80 42 4 9 0.3077
## 45 155 3 11 0.2143
## 33 139 1 4 0.2000
## 55 171 0 1 0.0000According to the analysis, those data scientist who based in city 111, 129, 140, 2, 39, 62, 8 and 82 will choose to stay in their company and don’t feel like changing a job.
filter(dat0,prob_of_stay<0.5)## City 0 1 prob_of_stay
## 1 78 15 16 0.4839
## 2 74 50 54 0.4808
## 3 126 13 15 0.4643
## 4 48 6 7 0.4615
## 5 128 40 52 0.4348
## 6 101 32 43 0.4267
## 7 43 5 7 0.4167
## 8 145 26 37 0.4127
## 9 21 1105 1597 0.4090
## 10 11 100 147 0.4049
## 11 179 2 3 0.4000
## 12 33 6 11 0.3529
## 13 107 2 4 0.3333
## 14 42 4 9 0.3077
## 15 155 3 11 0.2143
## 16 139 1 4 0.2000
## 17 171 0 1 0.0000Out of 123 cities where the data scientist live, data scientist from only 17 cities has a higher probability to change their job. The highest number of the data scientist who choose to leave is from city 21 (1105) followed by city 11 (100) and city 74 (50).
dat0['total']<-dat0['0']+dat0['1']
dat0<-dat0[order(dat0[,"total"],decreasing=TRUE),]
dat0## City 0 1 prob_of_stay total
## 6 103 3427 928 0.7869 4355
## 65 21 1105 1597 0.4090 2702
## 49 16 1354 179 0.8832 1533
## 14 114 1203 133 0.9004 1336
## 50 160 646 199 0.7645 845
## 31 136 525 61 0.8959 586
## 96 67 374 57 0.8677 431
## 104 75 274 31 0.8984 305
## 5 102 252 52 0.8289 304
## 7 104 273 28 0.9070 301
## 102 73 206 74 0.7357 280
## 3 100 210 65 0.7636 275
## 100 71 227 39 0.8534 266
## 12 11 100 147 0.4049 247
## 92 61 178 19 0.9036 197
## 117 90 135 62 0.6853 197
## 71 28 177 15 0.9219 192
## 66 23 166 16 0.9121 182
## 95 65 151 24 0.8629 175
## 75 36 147 13 0.9188 160
## 56 173 132 19 0.8742 151
## 113 83 120 23 0.8392 143
## 86 50 125 15 0.8929 140
## 16 116 106 22 0.8281 128
## 84 46 93 35 0.7266 128
## 51 162 91 37 0.7109 128
## 32 138 110 10 0.9167 120
## 62 19 74 45 0.6218 119
## 94 64 98 16 0.8596 114
## 83 45 97 16 0.8584 113
## 121 97 96 8 0.9231 104
## 103 74 50 54 0.4808 104
## 90 57 90 13 0.8738 103
## 42 149 78 24 0.7647 102
## 48 159 80 14 0.8511 94
## 123 99 79 15 0.8404 94
## 25 128 40 52 0.4348 92
## 79 41 71 18 0.7978 89
## 2 10 77 9 0.8953 86
## 52 165 64 18 0.7805 82
## 122 98 71 8 0.8987 79
## 8 105 70 9 0.8861 79
## 22 123 58 21 0.7342 79
## 4 101 32 43 0.4267 75
## 78 40 58 10 0.8529 68
## 115 89 51 16 0.7612 67
## 43 150 49 16 0.7538 65
## 40 145 26 37 0.4127 63
## 67 24 48 14 0.7742 62
## 15 115 33 21 0.6111 54
## 37 142 37 16 0.6981 53
## 44 152 40 11 0.7843 51
## 105 76 36 14 0.7200 50
## 47 158 34 15 0.6939 49
## 27 13 42 6 0.8750 48
## 70 27 38 9 0.8085 47
## 118 91 29 16 0.6444 45
## 99 70 25 19 0.5682 44
## 30 134 31 12 0.7209 43
## 38 143 25 16 0.6098 41
## 106 77 31 1 0.9688 32
## 107 78 15 16 0.4839 31
## 64 20 26 3 0.8966 29
## 39 144 21 8 0.7241 29
## 34 14 21 7 0.7500 28
## 23 126 13 15 0.4643 28
## 36 141 23 4 0.8519 27
## 98 7 22 5 0.8148 27
## 18 118 18 9 0.6667 27
## 1 1 23 3 0.8846 26
## 119 93 21 5 0.8077 26
## 87 53 18 8 0.6923 26
## 120 94 16 10 0.6154 26
## 69 26 19 5 0.7917 24
## 58 176 17 7 0.7083 24
## 114 84 17 7 0.7083 24
## 46 157 19 3 0.8636 22
## 72 30 18 3 0.8571 21
## 101 72 18 2 0.9000 20
## 116 9 13 5 0.7222 18
## 82 44 12 6 0.6667 18
## 97 69 15 2 0.8824 17
## 74 33 6 11 0.3529 17
## 110 80 13 2 0.8667 15
## 76 37 12 2 0.8571 14
## 57 175 11 3 0.7857 14
## 88 54 11 3 0.7857 14
## 89 55 10 4 0.7143 14
## 45 155 3 11 0.2143 14
## 19 12 11 2 0.8462 13
## 17 117 9 4 0.6923 13
## 85 48 6 7 0.4615 13
## 80 42 4 9 0.3077 13
## 81 43 5 7 0.4167 12
## 77 39 11 0 1.0000 11
## 24 127 8 2 0.8000 10
## 29 133 8 2 0.8000 10
## 54 167 7 3 0.7000 10
## 91 59 7 3 0.7000 10
## 9 106 6 3 0.6667 9
## 11 109 6 3 0.6667 9
## 28 131 6 3 0.6667 9
## 41 146 5 3 0.6250 8
## 108 79 5 3 0.6250 8
## 63 2 7 0 1.0000 7
## 111 81 6 1 0.8571 7
## 61 180 5 2 0.7143 7
## 20 120 5 1 0.8333 6
## 10 107 2 4 0.3333 6
## 93 62 5 0 1.0000 5
## 59 179 2 3 0.4000 5
## 33 139 1 4 0.2000 5
## 109 8 4 0 1.0000 4
## 112 82 4 0 1.0000 4
## 53 166 3 1 0.7500 4
## 60 18 3 1 0.7500 4
## 73 31 3 1 0.7500 4
## 68 25 2 2 0.5000 4
## 13 111 3 0 1.0000 3
## 26 129 3 0 1.0000 3
## 21 121 2 1 0.6667 3
## 35 140 1 0 1.0000 1
## 55 171 0 1 0.0000 1While if we sort the table, most of the data scientist are live in city 103 followed by city 65 and city 49.
# city_development_index & target
dat1 <- data.frame(table(data$city_development_index,data$target))
names(dat1) <- c("city_development_index","target","Count")
dat1<-spread(dat1,target,Count)
dat1['total']<-dat1['0']+dat1['1']
dat1<-dat1[order(dat1[,"total"],decreasing=TRUE),]
dat1## city_development_index 0 1 total
## 86 0.92 4073 1127 5200
## 15 0.624 1105 1597 2702
## 83 0.91 1354 179 1533
## 91 0.926 1203 133 1336
## 28 0.698 489 194 683
## 79 0.897 525 61 586
## 92 0.939 451 46 497
## 68 0.855 374 57 431
## 58 0.804 252 52 304
## 89 0.924 273 28 301
## 41 0.754 206 74 280
## 74 0.887 210 65 275
## 73 0.884 227 39 266
## 9 0.55 100 147 247
## 84 0.913 178 19 197
## 81 0.899 166 16 182
## 57 0.802 151 24 175
## 90 0.925 147 24 171
## 76 0.893 147 13 160
## 72 0.878 132 19 151
## 39 0.743 119 27 146
## 88 0.923 120 23 143
## 78 0.896 125 15 140
## 61 0.827 113 24 137
## 14 0.579 65 70 135
## 42 0.762 93 35 128
## 46 0.767 91 37 128
## 63 0.836 110 10 120
## 24 0.682 74 45 119
## 22 0.666 98 16 114
## 75 0.89 97 16 113
## 71 0.866 90 13 103
## 25 0.689 78 24 102
## 65 0.843 80 14 94
## 85 0.915 79 15 94
## 54 0.794 82 11 93
## 8 0.527 40 52 92
## 77 0.895 77 9 86
## 49 0.776 69 13 82
## 82 0.903 64 18 82
## 35 0.738 58 21 79
## 93 0.949 71 8 79
## 12 0.558 32 43 75
## 37 0.74 43 24 67
## 10 0.555 26 37 63
## 53 0.789 33 21 54
## 32 0.727 37 16 53
## 45 0.766 34 15 49
## 67 0.848 38 9 47
## 26 0.691 29 16 45
## 66 0.847 36 5 41
## 62 0.83 31 1 32
## 69 0.856 27 5 32
## 56 0.796 26 3 29
## 64 0.84 21 8 29
## 2 0.479 13 15 28
## 19 0.647 22 5 27
## 30 0.722 18 9 27
## 43 0.763 23 4 27
## 70 0.865 21 5 26
## 44 0.764 17 7 24
## 47 0.769 19 3 22
## 55 0.795 18 2 20
## 31 0.725 12 6 18
## 1 0.448 6 11 17
## 11 0.556 3 11 14
## 36 0.739 10 4 14
## 4 0.493 6 7 13
## 13 0.563 4 9 13
## 17 0.64 11 2 13
## 6 0.516 5 7 12
## 80 0.898 11 0 11
## 38 0.742 8 2 10
## 40 0.745 8 2 10
## 48 0.775 7 3 10
## 87 0.921 7 3 10
## 23 0.68 6 3 9
## 29 0.701 6 3 9
## 34 0.735 5 3 8
## 33 0.73 6 1 7
## 52 0.788 7 0 7
## 7 0.518 2 4 6
## 50 0.78 5 1 6
## 3 0.487 1 4 5
## 5 0.512 2 3 5
## 18 0.645 5 0 5
## 20 0.649 3 1 4
## 27 0.693 4 0 4
## 59 0.807 3 1 4
## 60 0.824 3 1 4
## 16 0.625 3 0 3
## 51 0.781 2 1 3
## 21 0.664 0 1 1plot(data$city_development_index,data$target)
From the scatter plot, it can be seen that most of the entry of this dataset is from city development index 0.8 to 0.93 except for city with city development index of 0.624, 0.754 and 0.55.
# Gender & target
dat2 <- data.frame(table(data$gender,data$target))
names(dat2) <- c("Gender","target","Count")
dat2## Gender target Count
## 1 Female 0 912
## 2 Male 0 10209
## 3 Other 0 141
## 4 unknown 0 3119
## 5 Female 1 326
## 6 Male 1 3012
## 7 Other 1 50
## 8 unknown 1 1389ggplot(data = dat2,aes(y = Gender,
x = Count, fill=target))+
geom_bar(stat="identity",position="dodge")+
labs(title = 'Number of target by gender', x= 'Number of target', y='Gender')
The bar chart illustrates that for both men and women, it is obviously that most of the data is collected from make. Overall, we can see that employees choose to stay more than leave. While look in more details, the probability of male who choose to stay in their company is higher compared to female.
# relevent_experience & target
dat3 <- data.frame(table(data$relevant_experience,data$target))
names(dat3) <- c("Relevant_experience","target","Count")
dat3## Relevant_experience target Count
## 1 no 0 3550
## 2 yes 0 10831
## 3 no 1 1816
## 4 yes 1 2961ggplot(data = dat3,aes(y = Relevant_experience,
x = Count, fill=target))+
geom_bar(stat="identity",position="dodge")+
labs(title = 'Number of targets by relevant experience', x= 'Number of target', y='Relevant experience')
It can be seen from the bar chart that, regardless of whether they have relevant work experience, the proportion of employees who choose to stay is greater than that of those who choose to leave. But for overall mobility, most of the data scientist in this dataset has relevant experience and the proportion of those who has relevant experience and choose to stay in their company is much higher than those who has no relevant experience.
# enrolled_university & target
dat4 <- data.frame(table(data$enrolled_university,data$target))
names(dat4) <- c("Enrolled_university","target","Count")
dat4## Enrolled_university target Count
## 1 Full time course 0 2326
## 2 no_enrollment 0 10896
## 3 Part time course 0 896
## 4 unknown 0 263
## 5 Full time course 1 1431
## 6 no_enrollment 1 2921
## 7 Part time course 1 302
## 8 unknown 1 123ggplot(data = dat4,aes(y = Enrolled_university,
x = Count, fill=target))+
geom_bar(stat="identity",position="dodge")+
labs(title = 'Number of targets by enrolled university', x= 'Number of target', y='Enrolled university')
The bar chart shows that employees choose to stay more than leave, regardless of whether they have enrolled in university or not. But for overall mobility ,most of the data is collected from those who do not enroll to any full time or part time univeristy couse, and they are more prone to stay the company.
# education_level & target
dat5 <- data.frame(table(data$education_level,data$target))
names(dat5) <- c("Education_level","target","Count")
dat5## Education_level target Count
## 1 Graduate 0 8353
## 2 High School 0 1623
## 3 Masters 0 3426
## 4 Phd 0 356
## 5 Primary School 0 267
## 6 unknown 0 356
## 7 Graduate 1 3245
## 8 High School 1 394
## 9 Masters 1 935
## 10 Phd 1 58
## 11 Primary School 1 41
## 12 unknown 1 104ggplot(data = dat5,aes(y = Education_level,
x = Count, fill=target))+
geom_bar(stat="identity",position="dodge")+
labs(title = 'Number of targets by education level', x= 'Number of target', y='Education level')
The bar chart shows that, regardless of education level, employees choose to stay more than they leave. For overall mobility, graduate students account for the largest proportion. We also can see from the graph that, it is very rare for a primary school education level to be in a data scientist. For high school education level, they are more likely to stay in the company while those graduate education level, they are more likely to went for a job change.
# major_discipline & target
dat6 <- data.frame(table(data$major_discipline,data$target))
names(dat6) <- c("Major_discipline","target","Count")
dat6## Major_discipline target Count
## 1 Arts 0 200
## 2 Business Degree 0 241
## 3 Humanities 0 528
## 4 No Major 0 168
## 5 Other 0 279
## 6 STEM 0 10701
## 7 unknown 0 2264
## 8 Arts 1 53
## 9 Business Degree 1 86
## 10 Humanities 1 141
## 11 No Major 1 55
## 12 Other 1 102
## 13 STEM 1 3791
## 14 unknown 1 549ggplot(data = dat6,aes(y = Major_discipline,
x = Count, fill=target))+
geom_bar(stat="identity",position="dodge")+
labs(title = 'Number of targets by major discipline', x= 'Number of target', y='Major discipline')
From the bar chart, we can observe that most of the employees worked as data scientists are majored in STEM.
# experience & target
dat7 <- data.frame(table(data$experience,data$target))
names(dat7) <- c("Experience","target","Count")
dat7<-spread(dat7,target,Count)
dat7['prob_of_stay']<-round(dat7['0']/(dat7['0']+dat7['1']),4)
dat7<-dat7[order(dat7[,"prob_of_stay"],decreasing=TRUE),]
dat7## Experience 0 1 prob_of_stay
## 10 16 436 72 0.8583
## 12 18 237 43 0.8464
## 2 >20 2825 526 0.8430
## 9 15 572 114 0.8338
## 11 17 285 57 0.8333
## 13 19 251 53 0.8257
## 8 14 479 107 0.8174
## 6 12 402 92 0.8138
## 7 13 322 77 0.8070
## 4 10 778 207 0.7898
## 22 9 767 213 0.7827
## 15 20 115 33 0.7770
## 5 11 513 151 0.7726
## 21 8 607 195 0.7569
## 19 6 873 343 0.7179
## 18 5 1018 412 0.7119
## 20 7 725 303 0.7053
## 17 4 946 457 0.6743
## 14 2 753 374 0.6681
## 16 3 876 478 0.6470
## 3 1 316 233 0.5756
## 1 <1 285 237 0.5460We can observe that the the data scientist with a lower experience are more likely to went for a job change. While the highest probability of those will stay in the company are those has more than 10 years experience.
# company_size & target
dat8 <- data.frame(table(data$company_size,data$target))
names(dat8) <- c("Company_size","target","Count")
dat8## Company_size target Count
## 1 <10 0 1084
## 2 >9999 0 1634
## 3 10-49 0 1127
## 4 100-499 0 2156
## 5 1000-4999 0 1128
## 6 50-99 0 2538
## 7 500-999 0 725
## 8 5000-9999 0 461
## 9 unknown 0 3528
## 10 <10 1 224
## 11 >9999 1 385
## 12 10-49 1 344
## 13 100-499 1 415
## 14 1000-4999 1 200
## 15 50-99 1 545
## 16 500-999 1 152
## 17 5000-9999 1 102
## 18 unknown 1 2410ggplot(data = dat8,aes(y = Company_size, x = Count, fill=target))+
geom_bar(stat="identity",position="dodge")+
labs(title = 'Number of targets by company size', x= 'Number of target', y='Company size')
Since the ‘unknown’ category represents missing value, so this category will not be considered. It can be seen from the bar chart that most of the data scientist are working in the company with size of 50-99 employees.
# company_type & target
dat9 <- data.frame(table(data$company_type,data$target))
names(dat9) <- c("Company_type","target","Count")
dat9## Company_type target Count
## 1 Early Stage Startup 0 461
## 2 Funded Startup 0 861
## 3 NGO 0 424
## 4 Other 0 92
## 5 Public Sector 0 745
## 6 Pvt Ltd 0 8042
## 7 unknown 0 3756
## 8 Early Stage Startup 1 142
## 9 Funded Startup 1 140
## 10 NGO 1 97
## 11 Other 1 29
## 12 Public Sector 1 210
## 13 Pvt Ltd 1 1775
## 14 unknown 1 2384ggplot(data = dat9,aes(y =Company_type,
x = Count, fill=target))+
geom_bar(stat="identity",position="dodge")+
labs(title = 'Number of targets by company type', x= 'Number of target', y='Company type')
We can observe that most of the data scientists are working in the company type of Pvt Ltd. The funded startup companies are more able to retain their data scientist workpower.
# last_new_job & target
dat10 <- data.frame(table(data$last_new_job,data$target))
names(dat10) <- c("Last_new_job","target","Count")
dat10## Last_new_job target Count
## 1 >4 0 2690
## 2 0 0 1713
## 3 1 0 5915
## 4 2 0 2200
## 5 3 0 793
## 6 4 0 801
## 7 unknown 0 269
## 8 >4 1 600
## 9 0 1 739
## 10 1 1 2125
## 11 2 1 700
## 12 3 1 231
## 13 4 1 228
## 14 unknown 1 154ggplot(data = dat10,aes(y = Last_new_job, x = Count, fill=target))+
geom_bar(stat="identity",position="dodge")+
labs(title = 'Number of targets by last_new_job', x= 'Number of target', y='last_new_job')
We can observe that the proportion of employees who have left their last job for about a year is the largest, followed by employees who have been more than four years and two years. While if we analyse the data by probability, the data scientist who did not change their job for more than past 4 years are more likely to continue work for their company. However, for those data scientists who has just left their previous job for less than 1 year are more likely to leave the company they are working now.
# training_hours & target
dat11 <- data.frame(table(data$training_hours,data$target))
names(dat11) <- c("training_hours","target","Count")
dat11<-spread(dat11,target,Count)
dat11['total']<-dat11['0']+dat11['1']
dat11['prob_of_stay']<-dat11['0']/dat11['total']
dat11<-dat11[order(dat11[,"total"],decreasing=TRUE),]
dat11## training_hours 0 1 total prob_of_stay
## 28 28 250 79 329 0.7598784
## 12 12 229 63 292 0.7842466
## 18 18 215 76 291 0.7388316
## 22 22 220 62 282 0.7801418
## 50 50 195 84 279 0.6989247
## 20 20 210 68 278 0.7553957
## 17 17 202 71 273 0.7399267
## 24 24 200 73 273 0.7326007
## 6 6 193 68 261 0.7394636
## 34 34 194 67 261 0.7432950
## 23 23 192 66 258 0.7441860
## 21 21 186 70 256 0.7265625
## 26 26 182 72 254 0.7165354
## 56 56 175 75 250 0.7000000
## 42 42 186 56 242 0.7685950
## 10 10 175 66 241 0.7261411
## 11 11 187 50 237 0.7890295
## 48 48 174 63 237 0.7341772
## 9 9 162 72 234 0.6923077
## 14 14 177 54 231 0.7662338
## 15 15 174 56 230 0.7565217
## 8 8 176 51 227 0.7753304
## 4 4 186 38 224 0.8303571
## 46 46 163 60 223 0.7309417
## 13 13 148 65 213 0.6948357
## 36 36 156 55 211 0.7393365
## 7 7 148 61 209 0.7081340
## 32 32 144 63 207 0.6956522
## 44 44 151 54 205 0.7365854
## 25 25 143 56 199 0.7185930
## 43 43 141 58 199 0.7085427
## 52 52 155 41 196 0.7908163
## 16 16 145 47 192 0.7552083
## 40 40 137 55 192 0.7135417
## 30 30 139 48 187 0.7433155
## 31 31 144 40 184 0.7826087
## 29 29 132 47 179 0.7374302
## 39 39 131 47 178 0.7359551
## 51 51 122 54 176 0.6931818
## 45 45 129 46 175 0.7371429
## 55 55 131 40 171 0.7660819
## 78 78 107 58 165 0.6484848
## 19 19 119 44 163 0.7300613
## 37 37 125 38 163 0.7668712
## 35 35 124 38 162 0.7654321
## 54 54 127 34 161 0.7888199
## 47 47 117 40 157 0.7452229
## 72 72 120 33 153 0.7843137
## 33 33 114 36 150 0.7600000
## 41 41 117 28 145 0.8068966
## 80 80 110 34 144 0.7638889
## 57 57 102 40 142 0.7183099
## 101 102 95 42 137 0.6934307
## 53 53 99 37 136 0.7279412
## 64 64 102 30 132 0.7727273
## 70 70 108 24 132 0.8181818
## 58 58 98 33 131 0.7480916
## 74 74 99 32 131 0.7557252
## 62 62 91 37 128 0.7109375
## 93 94 97 27 124 0.7822581
## 95 96 92 31 123 0.7479675
## 3 3 87 32 119 0.7310924
## 90 90 96 22 118 0.8135593
## 27 27 87 29 116 0.7500000
## 38 38 88 27 115 0.7652174
## 68 68 77 36 113 0.6814159
## 99 100 87 26 113 0.7699115
## 84 84 89 22 111 0.8018018
## 5 5 81 26 107 0.7570093
## 66 66 85 22 107 0.7943925
## 61 61 73 25 98 0.7448980
## 82 82 78 20 98 0.7959184
## 60 60 75 22 97 0.7731959
## 92 92 76 21 97 0.7835052
## 2 2 71 25 96 0.7395833
## 111 112 66 30 96 0.6875000
## 107 108 76 19 95 0.8000000
## 86 86 76 18 94 0.8085106
## 105 106 76 18 94 0.8085106
## 88 88 66 26 92 0.7173913
## 67 67 70 20 90 0.7777778
## 77 77 65 23 88 0.7386364
## 83 83 68 18 86 0.7906977
## 76 76 59 24 83 0.7108434
## 65 65 62 17 79 0.7848101
## 97 98 63 16 79 0.7974684
## 69 69 61 17 78 0.7820513
## 109 110 57 20 77 0.7402597
## 63 63 54 21 75 0.7200000
## 162 166 56 15 71 0.7887324
## 59 59 57 12 69 0.8260870
## 91 91 49 19 68 0.7205882
## 113 114 48 19 67 0.7164179
## 103 104 48 17 65 0.7384615
## 104 105 56 9 65 0.8615385
## 89 89 47 17 64 0.7343750
## 114 116 41 23 64 0.6406250
## 106 107 43 20 63 0.6825397
## 73 73 48 14 62 0.7741935
## 81 81 46 16 62 0.7419355
## 79 79 48 13 61 0.7868852
## 85 85 52 9 61 0.8524590
## 110 111 39 21 60 0.6500000
## 108 109 42 17 59 0.7118644
## 153 156 38 20 58 0.6551724
## 75 75 40 17 57 0.7017544
## 87 87 35 20 55 0.6363636
## 132 134 39 15 54 0.7222222
## 156 160 44 10 54 0.8148148
## 128 130 42 9 51 0.8235294
## 143 146 38 12 50 0.7600000
## 49 49 36 12 48 0.7500000
## 120 122 37 11 48 0.7708333
## 137 140 34 14 48 0.7083333
## 98 99 37 10 47 0.7872340
## 149 152 35 11 46 0.7608696
## 112 113 36 9 45 0.8000000
## 141 144 32 12 44 0.7272727
## 96 97 33 10 43 0.7674419
## 122 124 32 11 43 0.7441860
## 147 150 31 12 43 0.7209302
## 155 158 36 6 42 0.8571429
## 151 154 35 6 41 0.8536585
## 158 162 32 9 41 0.7804878
## 134 136 28 12 40 0.7000000
## 135 138 30 10 40 0.7500000
## 171 182 31 9 40 0.7750000
## 164 168 26 13 39 0.6666667
## 175 192 26 13 39 0.6666667
## 100 101 28 10 38 0.7368421
## 116 118 26 12 38 0.6842105
## 94 95 25 12 37 0.6756757
## 126 128 29 8 37 0.7837838
## 102 103 26 9 35 0.7428571
## 124 126 29 5 34 0.8529412
## 145 148 26 8 34 0.7647059
## 189 222 27 6 33 0.8181818
## 169 178 25 7 32 0.7812500
## 181 204 27 5 32 0.8437500
## 142 145 23 8 31 0.7419355
## 185 214 24 7 31 0.7741935
## 130 132 21 9 30 0.7000000
## 159 163 24 5 29 0.8275862
## 170 180 23 6 29 0.7931034
## 177 196 21 8 29 0.7241379
## 183 210 24 5 29 0.8275862
## 167 174 19 9 28 0.6785714
## 182 206 24 4 28 0.8571429
## 129 131 23 4 27 0.8518519
## 154 157 20 7 27 0.7407407
## 173 188 19 8 27 0.7037037
## 178 198 18 9 27 0.6666667
## 133 135 21 5 26 0.8076923
## 138 141 19 7 26 0.7307692
## 146 149 23 3 26 0.8846154
## 152 155 19 6 25 0.7600000
## 172 184 18 7 25 0.7200000
## 123 125 18 6 24 0.7500000
## 165 170 21 3 24 0.8750000
## 187 218 20 4 24 0.8333333
## 131 133 19 4 23 0.8260870
## 136 139 16 7 23 0.6956522
## 179 200 18 5 23 0.7826087
## 140 143 20 2 22 0.9090909
## 115 117 14 7 21 0.6666667
## 125 127 17 4 21 0.8095238
## 148 151 13 8 21 0.6190476
## 180 202 15 6 21 0.7142857
## 71 71 14 6 20 0.7000000
## 121 123 18 2 20 0.9000000
## 127 129 15 5 20 0.7500000
## 139 142 14 6 20 0.7000000
## 190 224 17 3 20 0.8500000
## 191 226 16 4 20 0.8000000
## 160 164 15 4 19 0.7894737
## 166 172 14 5 19 0.7368421
## 168 176 16 3 19 0.8421053
## 176 194 16 3 19 0.8421053
## 117 119 13 5 18 0.7222222
## 157 161 15 3 18 0.8333333
## 174 190 15 3 18 0.8333333
## 118 120 14 3 17 0.8235294
## 163 167 13 4 17 0.7647059
## 186 216 12 5 17 0.7058824
## 188 220 12 5 17 0.7058824
## 119 121 13 3 16 0.8125000
## 161 165 12 4 16 0.7500000
## 184 212 12 4 16 0.7500000
## 150 153 10 5 15 0.6666667
## 204 256 14 1 15 0.9333333
## 208 264 13 2 15 0.8666667
## 231 314 12 3 15 0.8000000
## 236 326 13 2 15 0.8666667
## 144 147 11 3 14 0.7857143
## 193 232 12 2 14 0.8571429
## 200 246 12 2 14 0.8571429
## 214 278 14 0 14 1.0000000
## 228 308 10 4 14 0.7142857
## 202 250 11 2 13 0.8461538
## 205 258 7 6 13 0.5384615
## 223 298 7 6 13 0.5384615
## 226 304 10 3 13 0.7692308
## 234 322 11 2 13 0.8461538
## 198 242 12 0 12 1.0000000
## 207 262 11 1 12 0.9166667
## 219 288 10 2 12 0.8333333
## 224 300 11 1 12 0.9166667
## 227 306 10 2 12 0.8333333
## 230 312 11 1 12 0.9166667
## 232 316 9 3 12 0.7500000
## 239 332 8 4 12 0.6666667
## 201 248 8 3 11 0.7272727
## 210 268 6 5 11 0.5454545
## 221 292 8 3 11 0.7272727
## 237 328 9 2 11 0.8181818
## 238 330 10 1 11 0.9090909
## 240 334 9 2 11 0.8181818
## 241 336 8 3 11 0.7272727
## 233 320 9 1 10 0.9000000
## 1 1 7 2 9 0.7777778
## 203 254 8 1 9 0.8888889
## 206 260 8 1 9 0.8888889
## 217 284 6 3 9 0.6666667
## 220 290 5 4 9 0.5555556
## 225 302 6 3 9 0.6666667
## 235 324 7 2 9 0.7777778
## 199 244 6 2 8 0.7500000
## 216 282 6 2 8 0.7500000
## 229 310 8 0 8 1.0000000
## 192 228 3 4 7 0.4285714
## 195 236 7 0 7 1.0000000
## 197 240 6 1 7 0.8571429
## 211 270 4 3 7 0.5714286
## 215 280 4 3 7 0.5714286
## 209 266 5 1 6 0.8333333
## 213 276 6 0 6 1.0000000
## 222 294 6 0 6 1.0000000
## 194 234 5 0 5 1.0000000
## 212 272 4 1 5 0.8000000
## 218 286 2 3 5 0.4000000
## 196 238 4 0 4 1.0000000plot(data$training_hours,data$target)
From the scatter plot, it can be seen that most of the data scientist have went for about 150 hours or below of training.
dat11<-dat11[order(dat11[,"prob_of_stay"],decreasing=TRUE),]
dat11## training_hours 0 1 total prob_of_stay
## 214 278 14 0 14 1.0000000
## 198 242 12 0 12 1.0000000
## 229 310 8 0 8 1.0000000
## 195 236 7 0 7 1.0000000
## 213 276 6 0 6 1.0000000
## 222 294 6 0 6 1.0000000
## 194 234 5 0 5 1.0000000
## 196 238 4 0 4 1.0000000
## 204 256 14 1 15 0.9333333
## 207 262 11 1 12 0.9166667
## 224 300 11 1 12 0.9166667
## 230 312 11 1 12 0.9166667
## 140 143 20 2 22 0.9090909
## 238 330 10 1 11 0.9090909
## 121 123 18 2 20 0.9000000
## 233 320 9 1 10 0.9000000
## 203 254 8 1 9 0.8888889
## 206 260 8 1 9 0.8888889
## 146 149 23 3 26 0.8846154
## 165 170 21 3 24 0.8750000
## 208 264 13 2 15 0.8666667
## 236 326 13 2 15 0.8666667
## 104 105 56 9 65 0.8615385
## 155 158 36 6 42 0.8571429
## 182 206 24 4 28 0.8571429
## 193 232 12 2 14 0.8571429
## 200 246 12 2 14 0.8571429
## 197 240 6 1 7 0.8571429
## 151 154 35 6 41 0.8536585
## 124 126 29 5 34 0.8529412
## 85 85 52 9 61 0.8524590
## 129 131 23 4 27 0.8518519
## 190 224 17 3 20 0.8500000
## 202 250 11 2 13 0.8461538
## 234 322 11 2 13 0.8461538
## 181 204 27 5 32 0.8437500
## 168 176 16 3 19 0.8421053
## 176 194 16 3 19 0.8421053
## 187 218 20 4 24 0.8333333
## 157 161 15 3 18 0.8333333
## 174 190 15 3 18 0.8333333
## 219 288 10 2 12 0.8333333
## 227 306 10 2 12 0.8333333
## 209 266 5 1 6 0.8333333
## 4 4 186 38 224 0.8303571
## 159 163 24 5 29 0.8275862
## 183 210 24 5 29 0.8275862
## 59 59 57 12 69 0.8260870
## 131 133 19 4 23 0.8260870
## 128 130 42 9 51 0.8235294
## 118 120 14 3 17 0.8235294
## 70 70 108 24 132 0.8181818
## 189 222 27 6 33 0.8181818
## 237 328 9 2 11 0.8181818
## 240 334 9 2 11 0.8181818
## 156 160 44 10 54 0.8148148
## 90 90 96 22 118 0.8135593
## 119 121 13 3 16 0.8125000
## 125 127 17 4 21 0.8095238
## 86 86 76 18 94 0.8085106
## 105 106 76 18 94 0.8085106
## 133 135 21 5 26 0.8076923
## 41 41 117 28 145 0.8068966
## 84 84 89 22 111 0.8018018
## 107 108 76 19 95 0.8000000
## 112 113 36 9 45 0.8000000
## 191 226 16 4 20 0.8000000
## 231 314 12 3 15 0.8000000
## 212 272 4 1 5 0.8000000
## 97 98 63 16 79 0.7974684
## 82 82 78 20 98 0.7959184
## 66 66 85 22 107 0.7943925
## 170 180 23 6 29 0.7931034
## 52 52 155 41 196 0.7908163
## 83 83 68 18 86 0.7906977
## 160 164 15 4 19 0.7894737
## 11 11 187 50 237 0.7890295
## 54 54 127 34 161 0.7888199
## 162 166 56 15 71 0.7887324
## 98 99 37 10 47 0.7872340
## 79 79 48 13 61 0.7868852
## 144 147 11 3 14 0.7857143
## 65 65 62 17 79 0.7848101
## 72 72 120 33 153 0.7843137
## 12 12 229 63 292 0.7842466
## 126 128 29 8 37 0.7837838
## 92 92 76 21 97 0.7835052
## 31 31 144 40 184 0.7826087
## 179 200 18 5 23 0.7826087
## 93 94 97 27 124 0.7822581
## 69 69 61 17 78 0.7820513
## 169 178 25 7 32 0.7812500
## 158 162 32 9 41 0.7804878
## 22 22 220 62 282 0.7801418
## 67 67 70 20 90 0.7777778
## 1 1 7 2 9 0.7777778
## 235 324 7 2 9 0.7777778
## 8 8 176 51 227 0.7753304
## 171 182 31 9 40 0.7750000
## 73 73 48 14 62 0.7741935
## 185 214 24 7 31 0.7741935
## 60 60 75 22 97 0.7731959
## 64 64 102 30 132 0.7727273
## 120 122 37 11 48 0.7708333
## 99 100 87 26 113 0.7699115
## 226 304 10 3 13 0.7692308
## 42 42 186 56 242 0.7685950
## 96 97 33 10 43 0.7674419
## 37 37 125 38 163 0.7668712
## 14 14 177 54 231 0.7662338
## 55 55 131 40 171 0.7660819
## 35 35 124 38 162 0.7654321
## 38 38 88 27 115 0.7652174
## 145 148 26 8 34 0.7647059
## 163 167 13 4 17 0.7647059
## 80 80 110 34 144 0.7638889
## 149 152 35 11 46 0.7608696
## 33 33 114 36 150 0.7600000
## 143 146 38 12 50 0.7600000
## 152 155 19 6 25 0.7600000
## 28 28 250 79 329 0.7598784
## 5 5 81 26 107 0.7570093
## 15 15 174 56 230 0.7565217
## 74 74 99 32 131 0.7557252
## 20 20 210 68 278 0.7553957
## 16 16 145 47 192 0.7552083
## 27 27 87 29 116 0.7500000
## 49 49 36 12 48 0.7500000
## 135 138 30 10 40 0.7500000
## 123 125 18 6 24 0.7500000
## 127 129 15 5 20 0.7500000
## 161 165 12 4 16 0.7500000
## 184 212 12 4 16 0.7500000
## 232 316 9 3 12 0.7500000
## 199 244 6 2 8 0.7500000
## 216 282 6 2 8 0.7500000
## 58 58 98 33 131 0.7480916
## 95 96 92 31 123 0.7479675
## 47 47 117 40 157 0.7452229
## 61 61 73 25 98 0.7448980
## 23 23 192 66 258 0.7441860
## 122 124 32 11 43 0.7441860
## 30 30 139 48 187 0.7433155
## 34 34 194 67 261 0.7432950
## 102 103 26 9 35 0.7428571
## 81 81 46 16 62 0.7419355
## 142 145 23 8 31 0.7419355
## 154 157 20 7 27 0.7407407
## 109 110 57 20 77 0.7402597
## 17 17 202 71 273 0.7399267
## 2 2 71 25 96 0.7395833
## 6 6 193 68 261 0.7394636
## 36 36 156 55 211 0.7393365
## 18 18 215 76 291 0.7388316
## 77 77 65 23 88 0.7386364
## 103 104 48 17 65 0.7384615
## 29 29 132 47 179 0.7374302
## 45 45 129 46 175 0.7371429
## 100 101 28 10 38 0.7368421
## 166 172 14 5 19 0.7368421
## 44 44 151 54 205 0.7365854
## 39 39 131 47 178 0.7359551
## 89 89 47 17 64 0.7343750
## 48 48 174 63 237 0.7341772
## 24 24 200 73 273 0.7326007
## 3 3 87 32 119 0.7310924
## 46 46 163 60 223 0.7309417
## 138 141 19 7 26 0.7307692
## 19 19 119 44 163 0.7300613
## 53 53 99 37 136 0.7279412
## 141 144 32 12 44 0.7272727
## 201 248 8 3 11 0.7272727
## 221 292 8 3 11 0.7272727
## 241 336 8 3 11 0.7272727
## 21 21 186 70 256 0.7265625
## 10 10 175 66 241 0.7261411
## 177 196 21 8 29 0.7241379
## 132 134 39 15 54 0.7222222
## 117 119 13 5 18 0.7222222
## 147 150 31 12 43 0.7209302
## 91 91 49 19 68 0.7205882
## 63 63 54 21 75 0.7200000
## 172 184 18 7 25 0.7200000
## 25 25 143 56 199 0.7185930
## 57 57 102 40 142 0.7183099
## 88 88 66 26 92 0.7173913
## 26 26 182 72 254 0.7165354
## 113 114 48 19 67 0.7164179
## 180 202 15 6 21 0.7142857
## 228 308 10 4 14 0.7142857
## 40 40 137 55 192 0.7135417
## 108 109 42 17 59 0.7118644
## 62 62 91 37 128 0.7109375
## 76 76 59 24 83 0.7108434
## 43 43 141 58 199 0.7085427
## 137 140 34 14 48 0.7083333
## 7 7 148 61 209 0.7081340
## 186 216 12 5 17 0.7058824
## 188 220 12 5 17 0.7058824
## 173 188 19 8 27 0.7037037
## 75 75 40 17 57 0.7017544
## 56 56 175 75 250 0.7000000
## 134 136 28 12 40 0.7000000
## 130 132 21 9 30 0.7000000
## 71 71 14 6 20 0.7000000
## 139 142 14 6 20 0.7000000
## 50 50 195 84 279 0.6989247
## 32 32 144 63 207 0.6956522
## 136 139 16 7 23 0.6956522
## 13 13 148 65 213 0.6948357
## 101 102 95 42 137 0.6934307
## 51 51 122 54 176 0.6931818
## 9 9 162 72 234 0.6923077
## 111 112 66 30 96 0.6875000
## 116 118 26 12 38 0.6842105
## 106 107 43 20 63 0.6825397
## 68 68 77 36 113 0.6814159
## 167 174 19 9 28 0.6785714
## 94 95 25 12 37 0.6756757
## 164 168 26 13 39 0.6666667
## 175 192 26 13 39 0.6666667
## 178 198 18 9 27 0.6666667
## 115 117 14 7 21 0.6666667
## 150 153 10 5 15 0.6666667
## 239 332 8 4 12 0.6666667
## 217 284 6 3 9 0.6666667
## 225 302 6 3 9 0.6666667
## 153 156 38 20 58 0.6551724
## 110 111 39 21 60 0.6500000
## 78 78 107 58 165 0.6484848
## 114 116 41 23 64 0.6406250
## 87 87 35 20 55 0.6363636
## 148 151 13 8 21 0.6190476
## 211 270 4 3 7 0.5714286
## 215 280 4 3 7 0.5714286
## 220 290 5 4 9 0.5555556
## 210 268 6 5 11 0.5454545
## 205 258 7 6 13 0.5384615
## 223 298 7 6 13 0.5384615
## 192 228 3 4 7 0.4285714
## 218 286 2 3 5 0.4000000However, data scientists who took the training more than 150 hours are more likely to choose to stay in the company.
preprocessed_data<-data_preprocess_categorical
preprocessed_data$target<-as.factor(preprocessed_data$target)
set.seed(1)
sample<-sample.split(preprocessed_data$target,SplitRatio=0.7)
train<-subset(preprocessed_data,sample==TRUE)
test<-subset(preprocessed_data,sample==FALSE)performance = function(xtab, desc=""){
cat(desc,"\n")
ACR = sum(diag(xtab))/sum(xtab)
TPR = xtab[1,1]/sum(xtab[,1]); TNR = xtab[2,2]/sum(xtab[,2])
PPV = xtab[1,1]/sum(xtab[1,]); NPV = xtab[2,2]/sum(xtab[2,])
FPR = 1 - TNR ; FNR = 1 - TPR
RandomAccuracy = (sum(xtab[,2])*sum(xtab[2,]) +
sum(xtab[,1])*sum(xtab[1,]))/(sum(xtab)^2)
Kappa = (ACR - RandomAccuracy)/(1 - RandomAccuracy)
print(xtab)
cat("\nAccuracy (ACR) :", ACR, "\n")
cat("Sensitivity(TPR) :", TPR, "\n")
cat("Specificity (TNR) :", TNR, "\n")
cat("Positive Predictive Value (PPV) :", PPV, "\n")
cat("Negative Predictive Value (NPV) :", NPV, "\n")
cat("False Positive Rate (FPR) :", FPR, "\n")
cat("False Negative Rate(FNR) :", FNR, "\n")
}
prob_gen<-function(x){
print(paste("Predicted probability of candidate will stay at the company",round(sum(x[2,])/sum(x),4)))
print(paste("Predicted probability of candidate will leave the company",round(sum(x[1,])/sum(x),4)))
}set.seed(123)
LDA_model = lda(target ~., data = train)## Warning in lda.default(x, grouping, ...): variables are collinearsummary(LDA_model)## Length Class Mode
## prior 2 -none- numeric
## counts 2 -none- numeric
## means 386 -none- numeric
## scaling 193 -none- numeric
## lev 2 -none- character
## svd 1 -none- numeric
## N 1 -none- numeric
## call 3 -none- call
## terms 3 terms call
## xlevels 0 -none- listhead(coef(LDA_model),10)## LD1
## city_development_index -2.8176411446
## training_hours -0.0008037416
## city_1 -0.9476649769
## city_2 -1.2760041362
## city_7 -0.9714193180
## city_8 -2.8664365483
## city_9 -0.0911228721
## city_10 -0.2367186872
## city_11 0.7851554560
## city_12 -0.7449616264plot(LDA_model)
LDA_pred= LDA_model %>% predict(test)
mean(LDA_pred$class == test$target)## [1] 0.7845833LDA = test$target
mean(LDA_pred$class == LDA)## [1] 0.7845833cfmat_LDA= table(LDA_pred$class,LDA)
cfmat_LDA = cfmat_LDA[2:1,2:1]
performance(cfmat_LDA, "LDA")## LDA
## LDA
## 1 0
## 1 634 439
## 0 799 3875
##
## Accuracy (ACR) : 0.7845833
## Sensitivity(TPR) : 0.4424285
## Specificity (TNR) : 0.8982383
## Positive Predictive Value (PPV) : 0.5908667
## Negative Predictive Value (NPV) : 0.8290543
## False Positive Rate (FPR) : 0.1017617
## False Negative Rate(FNR) : 0.5575715Linear Discriminant Analysis (LDA) is one of the “parametric” and generative models on the dataset. In LDA, it assumes the predictors are numeric and are drawn from multivariate gaussian/normal distribution.This is one of the LDA limitation as it only can takes in numerical predictors. In our assignment, we include all our attributes to train LDA model.From the confusion matrix stimulated above, we observed that the performance of LDA model is not operating well for data scientists job change analysis. By assuming 1 is the data scientists looking for job change and while 0 is vice-versa, the accuracy for the prediction is computed to be 78.46% which is not high and not low. Besides, the sensitivity is quite low (with 44.24%), but the PPV is higher than sensitivity with 59.09%, which means 44.24% of the predicted cases turned out to be positive cases whereas 59.09% of the positive were successfully predicted by LDA model. Apart from that, FPR (with 10.18%) has achieved an impressive performance, which means there is low probability of getting Type 1 error.
prob_gen(cfmat_LDA)## [1] "Predicted probability of candidate will stay at the company 0.8133"
## [1] "Predicted probability of candidate will leave the company 0.1867"We realized the probability of candidates who stay in the company is higher than candidates who leave the company.
set.seed(123)
log_model<-glm(target~.,data=train,family=binomial)
head(summary(log_model)$coef)## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 5.733172973 4.775930e+00 1.20043062 0.229972143
## city_development_index -9.037900075 7.635061e+00 -1.18373639 0.236517430
## training_hours -0.001075671 3.947257e-04 -2.72511003 0.006428006
## city_1 -0.322369868 2.165329e+00 -0.14887802 0.881649882
## city_2 -12.214310032 6.304742e+02 -0.01937321 0.984543380
## city_7 -0.913119562 9.205612e-01 -0.99191616 0.321238424This is the summary statistics of the logistic regression model.
logreg_prob<-log_model %>% predict(test,type="response")## Warning in predict.lm(object, newdata, se.fit, scale = 1, type = if (type == :
## prediction from a rank-deficient fit may be misleadinghead(logreg_prob)## 2 5 8 10 11 12
## 0.09507453 0.11999194 0.17029563 0.12335782 0.64942558 0.11304205hist(logreg_prob)
The above table show the probability of each data scientist to stay or leave the company.
contrasts(test$target)## 1
## 0 0
## 1 1logreg_pred<-ifelse(logreg_prob>0.5,1,0)
cfmat_LR1 = table(logreg_pred,test$target)
cfmat_LR1 = cfmat_LR1[2:1,2:1]
performance(cfmat_LR1)##
##
## logreg_pred 1 0
## 1 562 371
## 0 871 3943
##
## Accuracy (ACR) : 0.7838872
## Sensitivity(TPR) : 0.3921842
## Specificity (TNR) : 0.9140009
## Positive Predictive Value (PPV) : 0.602358
## Negative Predictive Value (NPV) : 0.8190694
## False Positive Rate (FPR) : 0.08599907
## False Negative Rate(FNR) : 0.6078158From the above result, we observe there is a 78.39% of accuracy with cut-off point = 0.5. Besides, the LR model shows that the positive class has a lower recall (sensitivity) of 39.22% and lower precision (positive predictive value) of 60.23% while the negative class has higher recall (specificity) of 91.40% and slightly higher precision (negative predictive value) of 81.91%.
prob_gen(cfmat_LR1)## [1] "Predicted probability of candidate will stay at the company 0.8377"
## [1] "Predicted probability of candidate will leave the company 0.1623"Here, the probability of candidates who stay in the company is higher than candidates who leave the company.
Next, we would like to find the optimal cut-off point for this model.
optcutoff<-optimalCutoff(test$target, logreg_prob)[1] The optimum cut-off point is 0.3883685
logreg_pred1<-ifelse(logreg_prob>optcutoff,1,0)
cfmat_LR2=table(logreg_pred1,test$target)
cfmat_LR2=cfmat_LR2[2:1,2:1]
performance(cfmat_LR2)##
##
## logreg_pred1 1 0
## 1 842 620
## 0 591 3694
##
## Accuracy (ACR) : 0.7892814
## Sensitivity(TPR) : 0.5875785
## Specificity (TNR) : 0.8562819
## Positive Predictive Value (PPV) : 0.5759234
## Negative Predictive Value (NPV) : 0.862077
## False Positive Rate (FPR) : 0.1437181
## False Negative Rate(FNR) : 0.4124215Once we obtained the optimum cut-off point (0.3883685), we then use it to train logistic regression model, we then obtained a higher accuracy which is 78.93% but the previous logistic regression model with cut-off point = 0.5 obtained an accuracy of 78.40%.
prob_gen(cfmat_LR2)## [1] "Predicted probability of candidate will stay at the company 0.7456"
## [1] "Predicted probability of candidate will leave the company 0.2544"Again, the probability of candidates who stay in the company is still higher than candidates who leave the company with optimum cut-off point.
set.seed(123)
naive_model<-naiveBayes(target~.,data=train)
naive_pred<-predict(naive_model,test)
cfmat = table(naive_pred,test$target)
cfmat = cfmat[2:1,2:1]
performance(cfmat)##
##
## naive_pred 1 0
## 1 1287 3276
## 0 146 1038
##
## Accuracy (ACR) : 0.4045589
## Sensitivity(TPR) : 0.8981158
## Specificity (TNR) : 0.240612
## Positive Predictive Value (PPV) : 0.2820513
## Negative Predictive Value (NPV) : 0.8766892
## False Positive Rate (FPR) : 0.759388
## False Negative Rate(FNR) : 0.1018842In this section, we have used Naïve Bayes theorem (without Laplace Smoothing)to build the model by using training data.Then, we carried out the prediction with testing data and the confusion matrix for evaluating the accuracy.The accuracy without laplace smoothing is 40.46% which is extremely low. This indicates that it can predict correctly the status of job change, whether looking for a job change or not, with only accuracy of 40.46%. For the positive classes, the Naïve Bayes has a high recall/sensitivity of 89.81% but with only 28.21% of precision (PPV). This means the ability of this model to detect a person is looking for a job is 89.81%, and for those that are detected, it can predict 28.21% of them correctly. For negative classes, Naïve Bayes has a very low recall of 24.10% but has high precision up to 87.67%. This reflects that although the model does not capture much of the people that is not looking for job change but it is able to perform well to predict it.
prob_gen(cfmat)## [1] "Predicted probability of candidate will stay at the company 0.206"
## [1] "Predicted probability of candidate will leave the company 0.794"We now observed the probability of candidates who stay in the company is lower than candidates who leave the company.
We now build naive bayes model with laplace smoothing.
set.seed(123)
naive_model_laplace<-naiveBayes(target~.,data=train,laplace=1)
naive_pred_laplace<-predict(naive_model_laplace,test,type="class")
cfmat_laplace=table(naive_pred_laplace,test$target)
cfmat_laplace = cfmat_laplace[2:1,2:1]
performance(cfmat_laplace)##
##
## naive_pred_laplace 1 0
## 1 1287 3276
## 0 146 1038
##
## Accuracy (ACR) : 0.4045589
## Sensitivity(TPR) : 0.8981158
## Specificity (TNR) : 0.240612
## Positive Predictive Value (PPV) : 0.2820513
## Negative Predictive Value (NPV) : 0.8766892
## False Positive Rate (FPR) : 0.759388
## False Negative Rate(FNR) : 0.1018842Besides, we also have constructed the Naïve Bayes model with Laplace Smoothing. As the algorithm used by Naïve Bayes may cause the numerator to have zero value, it will produce zero probability leading the Naïve Bayes classifier to fail to perform. Thus, Laplace Smoothing is introduced to solve the problem and then improve the prediction or accuracy of our model. After training Naive Bayes model with Laplace Smoothing. The above result indicates that Laplace smoothing does not bring much effect on the prediction of our model. Our model able to perform a well prediction without Laplace Smoothing and give rise to the same accuracy and same confusion matrix result. In other words, our model may not have any zero probabilities that affect the prediction of our model previously (without Laplace Smoothing) and this results in the same accuracy.
prob_gen(cfmat_laplace)## [1] "Predicted probability of candidate will stay at the company 0.206"
## [1] "Predicted probability of candidate will leave the company 0.794"Here, we found that the probability of candidates who stay in the company is lower than candidates who leave the company.
We first try with k=115 and k=116
set.seed(123)
knn115_model<-knn(train, test, cl=train$target, k=floor(sqrt(nrow(train))))
knn116_model<-knn(train, test,cl=train$target, k=ceiling(sqrt(nrow(train))))
cfmat_115=table(knn115_model,test$target)
cfmat_115 = cfmat_115[2:1,2:1]
performance(cfmat_115)##
##
## knn115_model 1 0
## 1 200 0
## 0 1233 4314
##
## Accuracy (ACR) : 0.7854533
## Sensitivity(TPR) : 0.1395673
## Specificity (TNR) : 1
## Positive Predictive Value (PPV) : 1
## Negative Predictive Value (NPV) : 0.7777177
## False Positive Rate (FPR) : 0
## False Negative Rate(FNR) : 0.8604327prob_gen(cfmat_115)## [1] "Predicted probability of candidate will stay at the company 0.9652"
## [1] "Predicted probability of candidate will leave the company 0.0348"cfmat_116=table(knn116_model,test$target)
cfmat_116 = cfmat_116[2:1,2:1]
performance(cfmat_116)##
##
## knn116_model 1 0
## 1 188 0
## 0 1245 4314
##
## Accuracy (ACR) : 0.7833652
## Sensitivity(TPR) : 0.1311933
## Specificity (TNR) : 1
## Positive Predictive Value (PPV) : 1
## Negative Predictive Value (NPV) : 0.7760389
## False Positive Rate (FPR) : 0
## False Negative Rate(FNR) : 0.8688067prob_gen(cfmat_116)## [1] "Predicted probability of candidate will stay at the company 0.9673"
## [1] "Predicted probability of candidate will leave the company 0.0327"From the above results, we can see the knn model with k=115 has the higher accuracy, 78.55% as compared to K=116 (78.34%). When k=115 and 116, the highest specifity is achieved at 100%. When k=115, this indicates knn model predicted person looking for a job changed is 100%, and for those that are predicted, it can predict 77.78% of them correctly.However, k=116 has also the similar confusion matrix result.The probability of candidates who leave the company is lower than candidates who stay in the company when k=115,116.
Now, We would like to find the optimum k to score a better accuracy.
k.optm=1
k_optm=c()
for (i in 1:300){
knn.mod <- knn(train, test,cl=train$target, k=i)
k.optm [i]<- 100 * sum(test$target == knn.mod)/NROW(test$target)
k_optm[i]<-k.optm[i]
}
k_optm## [1] 83.41744 83.88725 86.28850 86.53210 87.24552 86.93231 87.75013 87.57613
## [9] 87.68053 87.59353 87.45432 87.55873 87.40212 87.24552 87.22812 87.12372
## [17] 87.10632 86.89751 86.79311 86.70611 86.61911 86.35810 86.34070 86.44510
## [25] 86.21890 86.14930 86.30590 85.92309 85.88829 85.66208 85.50548 85.40108
## [33] 85.26188 85.36628 85.08787 85.01827 85.01827 84.94867 84.87907 84.49626
## [41] 84.46146 84.44406 84.42666 84.35706 84.28745 84.13085 83.88725 83.74804
## [49] 83.66104 83.59144 83.45224 83.40003 83.24343 83.19123 83.08683 82.87802
## [57] 82.77362 82.61702 82.49521 82.51262 82.40821 82.28641 82.14721 82.09501
## [65] 82.00800 81.86880 81.71220 81.66000 81.55559 81.43379 81.27719 81.15539
## [73] 80.92918 80.91178 80.98138 80.82478 80.65077 80.65077 80.58117 80.54637
## [81] 80.42457 80.42457 80.40717 80.32017 80.32017 80.26797 80.25057 80.30277
## [89] 80.25057 80.21576 80.14616 80.00696 79.97216 79.85036 79.79816 79.74595
## [97] 79.62415 79.57195 79.46755 79.46755 79.32835 79.22394 79.22394 79.22394
## [105] 79.24134 79.04994 79.04994 78.89334 78.82373 78.84113 78.70193 78.70193
## [113] 78.64973 78.58013 78.56273 78.38872 78.28432 78.21472 78.12772 78.09292
## [121] 78.17992 78.07552 78.04072 78.09292 78.09292 77.97112 77.98852 77.88411
## [129] 77.84931 77.77971 77.72751 77.65791 77.58831 77.62311 77.58831 77.50131
## [137] 77.51871 77.48390 77.46650 77.44910 77.44910 77.39690 77.27510 77.34470
## [145] 77.25770 77.29250 77.24030 77.20550 77.15330 77.04890 77.06630 77.06630
## [153] 77.06630 77.03149 77.01409 76.96189 76.97929 76.96189 76.92709 76.90969
## [161] 76.87489 76.85749 76.80529 76.80529 76.73569 76.75309 76.71829 76.71829
## [169] 76.66609 76.64869 76.64869 76.54428 76.47468 76.40508 76.42248 76.35288
## [177] 76.37028 76.31808 76.28328 76.21368 76.16148 76.12667 76.09187 76.09187
## [185] 76.09187 76.07447 76.05707 76.05707 76.05707 76.02227 75.98747 75.98747
## [193] 75.95267 75.93527 75.88307 75.86567 75.81347 75.77867 75.76127 75.76127
## [201] 75.76127 75.74387 75.70907 75.63946 75.60466 75.60466 75.60466 75.58726
## [209] 75.62206 75.62206 75.60466 75.56986 75.56986 75.55246 75.55246 75.58726
## [217] 75.50026 75.48286 75.50026 75.53506 75.50026 75.48286 75.48286 75.44806
## [225] 75.46546 75.48286 75.46546 75.44806 75.46546 75.46546 75.46546 75.46546
## [233] 75.46546 75.46546 75.41326 75.39586 75.37846 75.37846 75.37846 75.37846
## [241] 75.36106 75.36106 75.32626 75.34366 75.30886 75.32626 75.30886 75.32626
## [249] 75.32626 75.32626 75.29146 75.30886 75.30886 75.27406 75.27406 75.27406
## [257] 75.27406 75.25666 75.25666 75.27406 75.25666 75.25666 75.25666 75.22185
## [265] 75.22185 75.22185 75.20445 75.18705 75.16965 75.16965 75.16965 75.16965
## [273] 75.16965 75.16965 75.16965 75.16965 75.16965 75.15225 75.15225 75.15225
## [281] 75.15225 75.15225 75.15225 75.15225 75.13485 75.13485 75.13485 75.13485
## [289] 75.13485 75.13485 75.13485 75.13485 75.11745 75.13485 75.11745 75.11745
## [297] 75.10005 75.10005 75.10005 75.10005k_optm_pred<-which(k_optm==max(k_optm))
print(k_optm_pred)## [1] 7plot(k.optm,type="b",xlab="k-value",ylab="Accuracy level")
Hence, from the above result we will use the optimum k value, which is k=7 to generate the optimum knn model.
knn_optm_model<-knn(train, test, cl=train$target, k=k_optm_pred)
cfmat_optm=table(knn_optm_model,test$target)
cfmat_optm = cfmat_optm[2:1,2:1]
performance(cfmat_optm)##
##
## knn_optm_model 1 0
## 1 855 135
## 0 578 4179
##
## Accuracy (ACR) : 0.8759353
## Sensitivity(TPR) : 0.5966504
## Specificity (TNR) : 0.9687065
## Positive Predictive Value (PPV) : 0.8636364
## Negative Predictive Value (NPV) : 0.8784948
## False Positive Rate (FPR) : 0.03129346
## False Negative Rate(FNR) : 0.4033496With k=7, the knn model is able to generate an accuracy of 87.59%. When we compare all the knn model with different k-values, k=7 obtained the highest accuracy among all. Now, we no longer observed the sensitivity to be 100% anymore. However,the sensitivity is now 59.67% and also with higher specificity (which is 96.87%) after reducing the number of the k. Besides, with k=7, we able to note that the false negative rate is quite low. In our case, we are more concerned about the accuracy and also the false negative rate. Accuracy is the portion of true results, either true positive or true negative in a population. It measures the degree of veracity of a prediction test.False negative rate is an error in which a prediction test result improperly by indicating no presence of the person looking for a job changed (negative result), when in fact it is actually present.Therefore, both of these measures are equally important because we do not want to predict wrongly when a person who is intended to look for job changed but it is predicted as the opposite one.
prob_gen(cfmat_optm)## [1] "Predicted probability of candidate will stay at the company 0.8277"
## [1] "Predicted probability of candidate will leave the company 0.1723"The probability of candidates who leave the company is lower than candidates who stay in the company when k=7 (optimum k).
set.seed(123)
train_RF<-train
train_RF[,c('target')]<-list(NULL)
set.seed(2)
RF_model<-randomForest(x=train_RF,y=train$target,mtry=6,importance=TRUE)
RF_pred<-predict(RF_model,test,type="class")
cfmat_RF=table(RF_pred,test$target)
cfmat_RF = cfmat_RF[2:1,2:1]
performance(cfmat_RF,"\n*** Random Forest (RF) Strategy (m=6) Performance")##
## *** Random Forest (RF) Strategy (m=6) Performance
##
## RF_pred 1 0
## 1 699 506
## 0 734 3808
##
## Accuracy (ACR) : 0.7842353
## Sensitivity(TPR) : 0.4877879
## Specificity (TNR) : 0.8827075
## Positive Predictive Value (PPV) : 0.580083
## Negative Predictive Value (NPV) : 0.8383972
## False Positive Rate (FPR) : 0.1172925
## False Negative Rate(FNR) : 0.5122121prob_gen(cfmat_RF)## [1] "Predicted probability of candidate will stay at the company 0.7903"
## [1] "Predicted probability of candidate will leave the company 0.2097"The probability of candidates leave the company is lower than candidate who stay.
print(head(importance(RF_model),10))## 0 1 MeanDecreaseAccuracy
## city_development_index 22.48625289 30.8574514 28.909420
## training_hours -1.64999709 1.8695377 -0.115869
## city_1 -4.08862763 1.1247649 -2.575812
## city_2 0.00000000 0.0000000 0.000000
## city_7 -0.10053494 -2.3408929 -1.911969
## city_8 0.00000000 0.0000000 0.000000
## city_9 -3.20366875 0.4181672 -2.975831
## city_10 -2.14319873 2.8904382 -0.262619
## city_11 14.76277210 9.9920435 17.217732
## city_12 0.01345356 2.4638771 0.894456
## MeanDecreaseGini
## city_development_index 307.14523553
## training_hours 68.54850291
## city_1 0.37960444
## city_2 0.01146603
## city_7 0.87084751
## city_8 0.10958582
## city_9 0.88825253
## city_10 1.38089198
## city_11 17.30325071
## city_12 0.69658997For our selected dataset, we performed Random Forest model by setting the mtry = 6. Based on the confusion matrix performance, Random Forest has achieved the second highest performance with accuracy of 78.42%.The ranges for the accuracy, sensitivity, specificity, PPV, and NPV are from 48% to 88%. The accuracy and specificity for this model has achieved above 78% which is 78.42% and 88.27% respectively. However, the sensitivity in Random Value only achieved 48.78% which is extremely low. The positive class has a lower recall (sensitivity with 48.78%) and lower precision (positive predictive value with 58%) as they are both below 60%, while the negative class has higher recall (specificity with 88.27%) and higher precision (negative predictive value with 83.84%) when compared with the positive class. Comparing within positive class, the positive recall(sensitivity) is lower than the positive precision (positive predictive value) which means the model does not capture a lot of positive predictions but the one captures are mostly correct. The number of people who looked for job changed that are captured correctly in the random forest model is lower. Compared within the negative class, the negative recall (specificity) is higher than the negative precision (negative predictive value). This means the model captures a lot of negative predictions and the ones that are captured are mostly incorrect. The number of people who looked for another job that are captured incorrectly in the random forest model is higher.
set.seed(123)
C50_model = C5.0(x=train_RF, y=train$target)
summary(C50_model)##
## Call:
## C5.0.default(x = train_RF, y = train$target)
##
##
## C5.0 [Release 2.07 GPL Edition] Mon Jun 20 13:12:54 2022
## -------------------------------
##
## Class specified by attribute `outcome'
##
## Read 13411 cases (194 attributes) from undefined.data
##
## Decision tree:
##
## city_development_index <= 0.624:
## :...experience_15 > 0: 0 (35/13)
## : experience_15 <= 0:
## : :...education_level_High School <= 0: 1 (2154/859)
## : education_level_High School > 0:
## : :...company_type_Pvt Ltd <= 0: 1 (176/81)
## : company_type_Pvt Ltd > 0: 0 (26/7)
## city_development_index > 0.624:
## :...company_size_unknown <= 0: 0 (7667/708)
## company_size_unknown > 0:
## :...major_discipline_unknown > 0: 0 (953/163)
## major_discipline_unknown <= 0:
## :...education_level_Phd > 0: 0 (55/7)
## education_level_Phd <= 0:
## :...city_114 > 0: 0 (110/20)
## city_114 <= 0:
## :...company_type_Pvt Ltd > 0: 0 (124/25)
## company_type_Pvt Ltd <= 0:
## :...city_50 > 0: 0 (29/5)
## city_50 <= 0:
## :...city_136 > 0: 0 (86/21)
## city_136 <= 0:
## :...city_16 > 0:
## :...relevant_experience_no <= 0: 0 (111/13)
## : relevant_experience_no > 0:
## : :...education_level_Graduate <= 0: 0 (17/4)
## : education_level_Graduate > 0: 1 (52/20)
## city_16 <= 0:
## :...city_67 > 0: 0 (57/16)
## city_67 <= 0:
## :...city_104 > 0: 0 (32/9)
## city_104 <= 0:
## :...city_75 > 0: 0 (41/12)
## city_75 <= 0:
## :...experience_16 > 0: 0 (40/13)
## experience_16 <= 0:
## :...city_103 > 0: 1 (624/239)
## city_103 <= 0:
## :...experience_>20 > 0: [S1]
## experience_>20 <= 0: [S2]
##
## SubTree [S1]
##
## city_160 <= 0: 0 (137/31)
## city_160 > 0: 1 (44/13)
##
## SubTree [S2]
##
## training_hours > 157: 0 (57/15)
## training_hours <= 157:
## :...education_level_Graduate > 0: 1 (635/275)
## education_level_Graduate <= 0:
## :...experience_4 <= 0: 0 (135/53)
## experience_4 > 0: 1 (14/3)
##
##
## Evaluation on training data (13411 cases):
##
## Decision Tree
## ----------------
## Size Errors
##
## 25 2625(19.6%) <<
##
##
## (a) (b) <-classified as
## ---- ----
## 8577 1490 (a): class 0
## 1135 2209 (b): class 1
##
##
## Attribute usage:
##
## 100.00% city_development_index
## 82.17% company_size_unknown
## 25.00% major_discipline_unknown
## 18.17% company_type_Pvt Ltd
## 17.90% education_level_Phd
## 17.83% experience_15
## 17.57% education_level_High School
## 17.49% city_114
## 15.74% city_50
## 15.52% city_136
## 14.88% city_16
## 13.54% city_67
## 13.12% city_104
## 12.88% city_75
## 12.57% experience_16
## 12.27% city_103
## 7.62% experience_>20
## 6.36% education_level_Graduate
## 6.27% training_hours
## 1.35% city_160
## 1.34% relevant_experience_no
## 1.11% experience_4
##
##
## Time: 0.6 secsC50_pred=predict(C50_model, test)
cfmat_C50=table(C50_pred,test$target)
cfmat_C50 = cfmat_C50[2:1,2:1]
performance(cfmat_C50)##
##
## C50_pred 1 0
## 1 920 673
## 0 513 3641
##
## Accuracy (ACR) : 0.7936315
## Sensitivity(TPR) : 0.6420098
## Specificity (TNR) : 0.8439963
## Positive Predictive Value (PPV) : 0.5775267
## Negative Predictive Value (NPV) : 0.8765046
## False Positive Rate (FPR) : 0.1560037
## False Negative Rate(FNR) : 0.3579902In this section, we fit C50 decision tree model to our training dataset. From there, we observed the accuracy to be 79.36%. Comparing within positive classes, positive recall(sensitivity) is higher than the positive precision (positive predictive value). This means the decision tree capture a lot of positive predictions with 64.20% and the one captures correctly are only 57.75%.
prob_gen(cfmat_C50)## [1] "Predicted probability of candidate will stay at the company 0.7228"
## [1] "Predicted probability of candidate will leave the company 0.2772"The probability of candidates leave the company is lower than candidate who stay.
set.seed(123)
DT_model<-rpart(target~.,train)
DT_pred<-predict(DT_model,test,type="class")
cfmat_DT=table(DT_pred,test$target)
cfmat_DT = cfmat_DT[2:1,2:1]
performance(cfmat_DT)##
##
## DT_pred 1 0
## 1 607 429
## 0 826 3885
##
## Accuracy (ACR) : 0.7816252
## Sensitivity(TPR) : 0.4235869
## Specificity (TNR) : 0.9005563
## Positive Predictive Value (PPV) : 0.5859073
## Negative Predictive Value (NPV) : 0.8246657
## False Positive Rate (FPR) : 0.09944367
## False Negative Rate(FNR) : 0.5764131In this section, we fit rpart decision tree model to our training dataset. From there, we observed the accuracy to be lower compared to C50 decision tree model (78.16%).The main difference of C50 and rpart decision tree is that rpart is a regression tree model while c50 is a classification tree. We built two different decision models to compare and see which accuracy is higher.However, now by comparing within positive class, sensitivity is lower than the positive predicted value. This indicates the model does not capture a lot of positive predictions but the one captures are mostly correct. For negative class, specificity is still lower than negative predicted value.
prob_gen(cfmat_DT)## [1] "Predicted probability of candidate will stay at the company 0.8197"
## [1] "Predicted probability of candidate will leave the company 0.1803"The probability of candidates leave the company is lower than candidate who stay.
set.seed(123)
SVM_model = svm(target ~ ., data = train, kernel='radial', scale=F)
summary(SVM_model)##
## Call:
## svm(formula = target ~ ., data = train, kernel = "radial", scale = F)
##
##
## Parameters:
## SVM-Type: C-classification
## SVM-Kernel: radial
## cost: 1
##
## Number of Support Vectors: 7271
##
## ( 3344 3927 )
##
##
## Number of Classes: 2
##
## Levels:
## 0 1SVM_pred = predict(SVM_model, newdata = test)
cfmat_svm = table(SVM_pred, test$target)
cfmat_svm = cfmat_svm[2:1,2:1]
performance(cfmat_svm, "SVM Model")## SVM Model
##
## SVM_pred 1 0
## 1 144 105
## 0 1289 4209
##
## Accuracy (ACR) : 0.7574387
## Sensitivity(TPR) : 0.1004885
## Specificity (TNR) : 0.9756606
## Positive Predictive Value (PPV) : 0.5783133
## Negative Predictive Value (NPV) : 0.7655511
## False Positive Rate (FPR) : 0.02433936
## False Negative Rate(FNR) : 0.8995115From the confusion matrix, we can see that the SVM model for the accuracy is 75.74%. It has an accuracy which is second lowest among all machine learning algorithm. Besides, we able to observe the accuracy measures like ACR, TNR, and NPV are at least 75% and above except for sensitivity (TPR) where it is 10% only. With such low sensitivity, this means that the model can predict inaccurately the amount of people who actually looked for a job change. For example, in a total of 100 person looking for a job change, only 10 person is predicted correctly that they actually went to look for another job. Also, this model has a quite high false negative rate which is 89.95%. This means that it might mispredict on the person who did not look for a job where there will be around 90% person among 100 people who actually looked for a job change but predicted as they did not. On the other hand, the FPN is quite low which is 2.4% only.
prob_gen(cfmat_svm)## [1] "Predicted probability of candidate will stay at the company 0.9567"
## [1] "Predicted probability of candidate will leave the company 0.0433"The probability of candidates leave the company is lower than candidate who stay.
The table below provides a summarization of performance among the model built and trained.
A caption
We had achieved an accuracy of 88% using K-Nearest-Neighbor with a k value of 7, which was determined by the optimum k to score a better accuracy. This KNN model was trained and tested on our datasets effectively with a binary response such as ours. Such a KNN model is time-consuming in a way that finding the optimum K value using the looping method takes time. KNN might be very easy to implement but as the dataset grows efficiency or speed of the algorithm declines very fast under the optimum K method. Thus, we recommend that K-fold cross-validation can be used for further work as it can prevent overfitting or probably generate higher accuracy provided it can be time-consuming. Noticeably that when k = 115 or 116 for the KNN model, it obtained a 100% specificity and PPV rate. It means that the trained model is capable of predicting the true negative class as well as positive predictive value. However, the model only obtained an accuracy of approximately 78%. It may not be a good model for predicting the true positive class and negative predictive value.
On top of that, we also achieved an accuracy of 79% using a Random Forest C5.0 model. This model works by splitting our training data based on the field that provides the maximum information gain. In our training dataset, we utilized the ‘city_development_index’ as a root node to indicate whether a person will leave their job to work as a Data Scientist. While these decreases in accuracy may look small, they are crucial to reducing the time and cost associated with investing in false positive errors, such as when a Data Science company devotes 150 hours to a Data Science training hosted by a company from which no employees apply to the Data Scientist job.
One of the interesting points to be noticed is that we obtained the same result for the Naïve Bayes model with or without Laplace smoothing. It is a technique that helps to tackle the problem of zero probability in the algorithm. Thus, we can conclude that our datasets do not exist any zero-probability for the prior information.
Next, for the probability that a candidate will leave the company after the training on each model, all predicted probability for stay is higher except for Naïve Bayes model. For Naïve Bayes model, the predicted probability for leave is higher than the stay. If naïve bayes model is used to predict the job change of Data Scientist, the outcome generated will have the highest probability for leave, which means the data scientist will likely leave the company.
We discovered that the magnitude of people in developed cities is a lot higher than anywhere else, and high rates of job retention are majorly in developed cities. In comparison, people from underdeveloped cities are choosing to leave their jobs due to reason of prospect that they will thrive more in a developed city with Data Science work opportunities. Another stunning discovery is that younger professionals who just started their careers after graduating would not mind testing their potential elsewhere. In such cases, they will potentially be prone to leave their jobs as Data scientists. This will lead the employers to filter in candidates with the least experience in the field. However, this discovery was non-sensical as it might filter out potential talent in the first place.
These findings significantly enhanced the allocation of resources by the Data Science company in their outreach toward training for prospective employees. Human Resources researchers at Data Science companies should target their training to companies in the cities with lower city development indices. Moreover, they should also target those individuals with less professional experience as the models suggested above. In return, those people who have an unrealized passion for work in Data Science as their long-term career will be selected for free Data Scientist bootcamps and receive the opportunity in boosting and improving their skills and their motivation to take the leap of faith and apply for a Data Scientist position.