ANLY 530-53- B-2019 Lab #3
Student ID: 235208
2020-01-26
Part 1
Step 1:Pre-processing
Wholesale <- read.csv("~/lab3/Wholesale_customers_data.csv")
#View(Wholesale)
str(Wholesale)## 'data.frame': 440 obs. of 8 variables:
## $ Channel : int 2 2 2 1 2 2 2 2 1 2 ...
## $ Region : int 3 3 3 3 3 3 3 3 3 3 ...
## $ Fresh : int 12669 7057 6353 13265 22615 9413 12126 7579 5963 6006 ...
## $ Milk : int 9656 9810 8808 1196 5410 8259 3199 4956 3648 11093 ...
## $ Grocery : int 7561 9568 7684 4221 7198 5126 6975 9426 6192 18881 ...
## $ Frozen : int 214 1762 2405 6404 3915 666 480 1669 425 1159 ...
## $ Detergents_Paper: int 2674 3293 3516 507 1777 1795 3140 3321 1716 7425 ...
## $ Delicassen : int 1338 1776 7844 1788 5185 1451 545 2566 750 2098 ...summary(Wholesale)## Channel Region Fresh Milk
## Min. :1.000 Min. :1.000 Min. : 3 Min. : 55
## 1st Qu.:1.000 1st Qu.:2.000 1st Qu.: 3128 1st Qu.: 1533
## Median :1.000 Median :3.000 Median : 8504 Median : 3627
## Mean :1.323 Mean :2.543 Mean : 12000 Mean : 5796
## 3rd Qu.:2.000 3rd Qu.:3.000 3rd Qu.: 16934 3rd Qu.: 7190
## Max. :2.000 Max. :3.000 Max. :112151 Max. :73498
## Grocery Frozen Detergents_Paper Delicassen
## Min. : 3 Min. : 25.0 Min. : 3.0 Min. : 3.0
## 1st Qu.: 2153 1st Qu.: 742.2 1st Qu.: 256.8 1st Qu.: 408.2
## Median : 4756 Median : 1526.0 Median : 816.5 Median : 965.5
## Mean : 7951 Mean : 3071.9 Mean : 2881.5 Mean : 1524.9
## 3rd Qu.:10656 3rd Qu.: 3554.2 3rd Qu.: 3922.0 3rd Qu.: 1820.2
## Max. :92780 Max. :60869.0 Max. :40827.0 Max. :47943.0top.n.custs <- function (Wholesale,cols,n=5) { #Requires some data frame and the top N to remove
idx.to.remove <-integer(0) #Initialize a vector to hold customers being removed
for (c in cols){ # For every column in the data we passed to this function
col.order <-order(Wholesale[,c],decreasing=T) #Sort column "c" in descending order (bigger on top)
#Order returns the sorted index (e.g. row 15, 3, 7, 1, ...) rather than the actual values sorted.
idx <-head(col.order, n) #Take the first n of the sorted column C to
idx.to.remove <-union(idx.to.remove,idx) #Combine and de-duplicate the row ids that need to be removed
}
return(idx.to.remove) #Return the indexes of customers to be removed
}
#How Many Customers to be Removed?
top.custs <-top.n.custs(Wholesale,cols=3:8,n=5)
length(top.custs) #How Many Customers to be Removed?## [1] 19Wholesale[top.custs,] #Examine the customers## Channel Region Fresh Milk Grocery Frozen Detergents_Paper Delicassen
## 182 1 3 112151 29627 18148 16745 4948 8550
## 126 1 3 76237 3473 7102 16538 778 918
## 285 1 3 68951 4411 12609 8692 751 2406
## 40 1 3 56159 555 902 10002 212 2916
## 259 1 1 56083 4563 2124 6422 730 3321
## 87 2 3 22925 73498 32114 987 20070 903
## 48 2 3 44466 54259 55571 7782 24171 6465
## 86 2 3 16117 46197 92780 1026 40827 2944
## 184 1 3 36847 43950 20170 36534 239 47943
## 62 2 3 35942 38369 59598 3254 26701 2017
## 334 2 2 8565 4980 67298 131 38102 1215
## 66 2 3 85 20959 45828 36 24231 1423
## 326 1 2 32717 16784 13626 60869 1272 5609
## 94 1 3 11314 3090 2062 35009 71 2698
## 197 1 1 30624 7209 4897 18711 763 2876
## 104 1 3 56082 3504 8906 18028 1480 2498
## 24 2 3 26373 36423 22019 5154 4337 16523
## 72 1 3 18291 1266 21042 5373 4173 14472
## 88 1 3 43265 5025 8117 6312 1579 14351Step 2:Model design
Wholesale.rm.top<-Wholesale[-c(top.custs),] #Remove the Customers
set.seed(76964057) #Set the seed for reproducibility
k <-kmeans(Wholesale.rm.top[,-c(1,2)], centers=5) #Create 5 clusters, Remove columns 1 and 2
k$centers #Display cluster centers## Fresh Milk Grocery Frozen Detergents_Paper Delicassen
## 1 4189.747 7645.639 11015.277 1335.145 4750.4819 1387.1205
## 2 16470.870 3026.491 4264.741 3217.306 996.5556 1319.7593
## 3 33120.163 4896.977 5579.860 3823.372 945.4651 1620.1860
## 4 5830.214 15295.048 23449.167 1936.452 10361.6429 1912.7381
## 5 5043.434 2329.683 2786.138 2689.814 652.8276 849.8414table(k$cluster) #Give a count of data points in each cluster##
## 1 2 3 4 5
## 83 108 43 42 145Step 3: Optimum K
rng<-2:20 #K from 2 to 20
tries <-100 #Run the K Means algorithm 100 times
avg.totw.ss <-integer(length(rng)) #Set up an empty vector to hold all of points
for(v in rng){ # For each value of the range variable
v.totw.ss <-integer(tries) #Set up an empty vector to hold the 100 tries
for(i in 1:tries){
k.temp <-kmeans(Wholesale.rm.top,centers=v) #Run kmeans
v.totw.ss[i] <-k.temp$tot.withinss#Store the total withinss
}
avg.totw.ss[v-1] <-mean(v.totw.ss) #Average the 100 total withinss
}
plot(rng,avg.totw.ss,type="b", main="Total Within SS by Various K",
ylab="Average Total Within Sum of Squares",
xlab="Value of K")
### Q1-Given this is an imperfect real-world, you need to determine what you believe is the best value for “k” and write-up this portion of your lab report. Answer:
Q2-How many points do you see in each cluster?
Part II
#Read the dataset
wine <- read.csv("~/lab3/wine.csv")
str(wine)## 'data.frame': 178 obs. of 14 variables:
## $ Wine : int 1 1 1 1 1 1 1 1 1 1 ...
## $ Alcohol : num 14.2 13.2 13.2 14.4 13.2 ...
## $ Malic.acid : num 1.71 1.78 2.36 1.95 2.59 1.76 1.87 2.15 1.64 1.35 ...
## $ Ash : num 2.43 2.14 2.67 2.5 2.87 2.45 2.45 2.61 2.17 2.27 ...
## $ Acl : num 15.6 11.2 18.6 16.8 21 15.2 14.6 17.6 14 16 ...
## $ Mg : int 127 100 101 113 118 112 96 121 97 98 ...
## $ Phenols : num 2.8 2.65 2.8 3.85 2.8 3.27 2.5 2.6 2.8 2.98 ...
## $ Flavanoids : num 3.06 2.76 3.24 3.49 2.69 3.39 2.52 2.51 2.98 3.15 ...
## $ Nonflavanoid.phenols: num 0.28 0.26 0.3 0.24 0.39 0.34 0.3 0.31 0.29 0.22 ...
## $ Proanth : num 2.29 1.28 2.81 2.18 1.82 1.97 1.98 1.25 1.98 1.85 ...
## $ Color.int : num 5.64 4.38 5.68 7.8 4.32 6.75 5.25 5.05 5.2 7.22 ...
## $ Hue : num 1.04 1.05 1.03 0.86 1.04 1.05 1.02 1.06 1.08 1.01 ...
## $ OD : num 3.92 3.4 3.17 3.45 2.93 2.85 3.58 3.58 2.85 3.55 ...
## $ Proline : int 1065 1050 1185 1480 735 1450 1290 1295 1045 1045 ...#Best K value
wssplot <- function(data, nc=15, seed=1234){
wss <- (nrow(data)-1)*sum(apply(data,2,var))
for (i in 2:nc){
set.seed(seed)
wss[i] <- sum(kmeans(data, centers=i)$withinss)}
plot(1:nc, wss, type="b", xlab="Number of Clusters",
ylab="Within groups sum of squares")}
#Scale the data set
df <- scale(wine[-1])
#Optimum K
wssplot(df)
#K-means analysis
if (!require("NbClust")) {
install.packages("NbClust")
library(NbClust)
}## Loading required package: NbClust## Warning: package 'NbClust' was built under R version 3.5.2set.seed(1234)
nc <- NbClust(df, min.nc=2, max.nc = 15, method = "kmeans")
## *** : The Hubert index is a graphical method of determining the number of clusters.
## In the plot of Hubert index, we seek a significant knee that corresponds to a
## significant increase of the value of the measure i.e the significant peak in Hubert
## index second differences plot.
## 
## *** : The D index is a graphical method of determining the number of clusters.
## In the plot of D index, we seek a significant knee (the significant peak in Dindex
## second differences plot) that corresponds to a significant increase of the value of
## the measure.
##
## *******************************************************************
## * Among all indices:
## * 4 proposed 2 as the best number of clusters
## * 15 proposed 3 as the best number of clusters
## * 1 proposed 10 as the best number of clusters
## * 1 proposed 12 as the best number of clusters
## * 1 proposed 14 as the best number of clusters
## * 1 proposed 15 as the best number of clusters
##
## ***** Conclusion *****
##
## * According to the majority rule, the best number of clusters is 3
##
##
## *******************************************************************print(table(nc$Best.n[1,]))##
## 0 1 2 3 10 12 14 15
## 2 1 4 15 1 1 1 1barplot(table(nc$Best.n[1,]), xlab = "Number of Clusters", ylab = "Number of Criteria", main = "Number of Clusters Chosen by 26 Criteria")
#K-means Train and evaluation
set.seed(1234)
fit.km <- kmeans(df, 3, nstart=25)
(df_km <- table(wine$Wine, fit.km$cluster))##
## 1 2 3
## 1 59 0 0
## 2 3 65 3
## 3 0 0 48(Accuracy <- (sum(diag(df_km))/sum(df_km)*100))## [1] 96.62921#K-means Plot
if (!require("cluster")) {
install.packages("cluster")
library(cluster)
}## Loading required package: clusterclusplot(df, fit.km$cluster, main='2D representation of the Cluster solution',
color=TRUE, shade=TRUE,
labels=2, lines=0)
Part III
#Prepare the Dataset
df_rpart <- data.frame(k=fit.km$cluster, df)
#ranomize the data set
rdf <- df_rpart[sample(1:nrow(df_rpart)), ]
#Split training and testing set
train <- rdf[1:(as.integer(.8*nrow(rdf))-1), ]
test <- rdf[(as.integer(.8*nrow(rdf))):nrow(rdf), ]#Training Model
if (!require("rpart")) {
install.packages("rpart")
library(rpart)
}## Loading required package: rpart## Warning: package 'rpart' was built under R version 3.5.3if (!require("rattle")) {
install.packages("rattle")
library(rattle)
}## Loading required package: rattle## Warning: package 'rattle' was built under R version 3.5.3## Rattle: A free graphical interface for data science with R.
## Version 5.3.0 Copyright (c) 2006-2018 Togaware Pty Ltd.
## Type 'rattle()' to shake, rattle, and roll your data.##
## Attaching package: 'rattle'## The following object is masked _by_ '.GlobalEnv':
##
## winefit <- rpart(k ~ ., data=train, method="class")
fancyRpartPlot(fit)
#Model Evaluation
pred <- predict(fit, test, type="class")
(news_tbl <- table(pred, test$k))##
## pred 1 2 3
## 1 11 1 0
## 2 0 12 1
## 3 0 1 11(Accuracy <- (sum(diag(news_tbl))/sum(news_tbl)*100))## [1] 91.89189Part IV
Q3- Load the dataset of breast cancer. Do the preliminary analysis and implement a KNN (K- nearest neighbors) model for this dataset and don’t forget that whenever it is required you should use: set.seed(12345). For designing the model, use following command:
knn(train =
# prepare data
wbcd <- read.csv("~/lab3/wisc_bc_data.csv")
str(wbcd)## 'data.frame': 569 obs. of 32 variables:
## $ id : int 87139402 8910251 905520 868871 9012568 906539 925291 87880 862989 89827 ...
## $ diagnosis : Factor w/ 2 levels "B","M": 1 1 1 1 1 1 1 2 1 1 ...
## $ radius_mean : num 12.3 10.6 11 11.3 15.2 ...
## $ texture_mean : num 12.4 18.9 16.8 13.4 13.2 ...
## $ perimeter_mean : num 78.8 69.3 70.9 73 97.7 ...
## $ area_mean : num 464 346 373 385 712 ...
## $ smoothness_mean : num 0.1028 0.0969 0.1077 0.1164 0.0796 ...
## $ compactness_mean : num 0.0698 0.1147 0.078 0.1136 0.0693 ...
## $ concavity_mean : num 0.0399 0.0639 0.0305 0.0464 0.0339 ...
## $ points_mean : num 0.037 0.0264 0.0248 0.048 0.0266 ...
## $ symmetry_mean : num 0.196 0.192 0.171 0.177 0.172 ...
## $ dimension_mean : num 0.0595 0.0649 0.0634 0.0607 0.0554 ...
## $ radius_se : num 0.236 0.451 0.197 0.338 0.178 ...
## $ texture_se : num 0.666 1.197 1.387 1.343 0.412 ...
## $ perimeter_se : num 1.67 3.43 1.34 1.85 1.34 ...
## $ area_se : num 17.4 27.1 13.5 26.3 17.7 ...
## $ smoothness_se : num 0.00805 0.00747 0.00516 0.01127 0.00501 ...
## $ compactness_se : num 0.0118 0.03581 0.00936 0.03498 0.01485 ...
## $ concavity_se : num 0.0168 0.0335 0.0106 0.0219 0.0155 ...
## $ points_se : num 0.01241 0.01365 0.00748 0.01965 0.00915 ...
## $ symmetry_se : num 0.0192 0.035 0.0172 0.0158 0.0165 ...
## $ dimension_se : num 0.00225 0.00332 0.0022 0.00344 0.00177 ...
## $ radius_worst : num 13.5 11.9 12.4 11.9 16.2 ...
## $ texture_worst : num 15.6 22.9 26.4 15.8 15.7 ...
## $ perimeter_worst : num 87 78.3 79.9 76.5 104.5 ...
## $ area_worst : num 549 425 471 434 819 ...
## $ smoothness_worst : num 0.139 0.121 0.137 0.137 0.113 ...
## $ compactness_worst: num 0.127 0.252 0.148 0.182 0.174 ...
## $ concavity_worst : num 0.1242 0.1916 0.1067 0.0867 0.1362 ...
## $ points_worst : num 0.0939 0.0793 0.0743 0.0861 0.0818 ...
## $ symmetry_worst : num 0.283 0.294 0.3 0.21 0.249 ...
## $ dimension_worst : num 0.0677 0.0759 0.0788 0.0678 0.0677 ...wbcd <- wbcd[-1]
table(wbcd$diagnosis)##
## B M
## 357 212wbcd$diagnosis <- factor(wbcd$diagnosis, levels = c("B", "M"), labels = c("Benign", "Malignant"))
round(prop.table(table(wbcd$diagnosis)) * 100, digits = 1)##
## Benign Malignant
## 62.7 37.3normalize <- function(x) {
return ((x - min(x)) / (max(x) - min(x)))
}
wbcd_n <- as.data.frame(lapply(wbcd[2:31], normalize))
summary(wbcd_n)## radius_mean texture_mean perimeter_mean area_mean
## Min. :0.0000 Min. :0.0000 Min. :0.0000 Min. :0.0000
## 1st Qu.:0.2233 1st Qu.:0.2185 1st Qu.:0.2168 1st Qu.:0.1174
## Median :0.3024 Median :0.3088 Median :0.2933 Median :0.1729
## Mean :0.3382 Mean :0.3240 Mean :0.3329 Mean :0.2169
## 3rd Qu.:0.4164 3rd Qu.:0.4089 3rd Qu.:0.4168 3rd Qu.:0.2711
## Max. :1.0000 Max. :1.0000 Max. :1.0000 Max. :1.0000
## smoothness_mean compactness_mean concavity_mean points_mean
## Min. :0.0000 Min. :0.0000 Min. :0.00000 Min. :0.0000
## 1st Qu.:0.3046 1st Qu.:0.1397 1st Qu.:0.06926 1st Qu.:0.1009
## Median :0.3904 Median :0.2247 Median :0.14419 Median :0.1665
## Mean :0.3948 Mean :0.2606 Mean :0.20806 Mean :0.2431
## 3rd Qu.:0.4755 3rd Qu.:0.3405 3rd Qu.:0.30623 3rd Qu.:0.3678
## Max. :1.0000 Max. :1.0000 Max. :1.00000 Max. :1.0000
## symmetry_mean dimension_mean radius_se texture_se
## Min. :0.0000 Min. :0.0000 Min. :0.00000 Min. :0.0000
## 1st Qu.:0.2823 1st Qu.:0.1630 1st Qu.:0.04378 1st Qu.:0.1047
## Median :0.3697 Median :0.2439 Median :0.07702 Median :0.1653
## Mean :0.3796 Mean :0.2704 Mean :0.10635 Mean :0.1893
## 3rd Qu.:0.4530 3rd Qu.:0.3404 3rd Qu.:0.13304 3rd Qu.:0.2462
## Max. :1.0000 Max. :1.0000 Max. :1.00000 Max. :1.0000
## perimeter_se area_se smoothness_se compactness_se
## Min. :0.00000 Min. :0.00000 Min. :0.0000 Min. :0.00000
## 1st Qu.:0.04000 1st Qu.:0.02064 1st Qu.:0.1175 1st Qu.:0.08132
## Median :0.07209 Median :0.03311 Median :0.1586 Median :0.13667
## Mean :0.09938 Mean :0.06264 Mean :0.1811 Mean :0.17444
## 3rd Qu.:0.12251 3rd Qu.:0.07170 3rd Qu.:0.2187 3rd Qu.:0.22680
## Max. :1.00000 Max. :1.00000 Max. :1.0000 Max. :1.00000
## concavity_se points_se symmetry_se dimension_se
## Min. :0.00000 Min. :0.0000 Min. :0.0000 Min. :0.00000
## 1st Qu.:0.03811 1st Qu.:0.1447 1st Qu.:0.1024 1st Qu.:0.04675
## Median :0.06538 Median :0.2070 Median :0.1526 Median :0.07919
## Mean :0.08054 Mean :0.2235 Mean :0.1781 Mean :0.10019
## 3rd Qu.:0.10619 3rd Qu.:0.2787 3rd Qu.:0.2195 3rd Qu.:0.12656
## Max. :1.00000 Max. :1.0000 Max. :1.0000 Max. :1.00000
## radius_worst texture_worst perimeter_worst area_worst
## Min. :0.0000 Min. :0.0000 Min. :0.0000 Min. :0.00000
## 1st Qu.:0.1807 1st Qu.:0.2415 1st Qu.:0.1678 1st Qu.:0.08113
## Median :0.2504 Median :0.3569 Median :0.2353 Median :0.12321
## Mean :0.2967 Mean :0.3640 Mean :0.2831 Mean :0.17091
## 3rd Qu.:0.3863 3rd Qu.:0.4717 3rd Qu.:0.3735 3rd Qu.:0.22090
## Max. :1.0000 Max. :1.0000 Max. :1.0000 Max. :1.00000
## smoothness_worst compactness_worst concavity_worst points_worst
## Min. :0.0000 Min. :0.0000 Min. :0.00000 Min. :0.0000
## 1st Qu.:0.3000 1st Qu.:0.1163 1st Qu.:0.09145 1st Qu.:0.2231
## Median :0.3971 Median :0.1791 Median :0.18107 Median :0.3434
## Mean :0.4041 Mean :0.2202 Mean :0.21740 Mean :0.3938
## 3rd Qu.:0.4942 3rd Qu.:0.3025 3rd Qu.:0.30583 3rd Qu.:0.5546
## Max. :1.0000 Max. :1.0000 Max. :1.00000 Max. :1.0000
## symmetry_worst dimension_worst
## Min. :0.0000 Min. :0.0000
## 1st Qu.:0.1851 1st Qu.:0.1077
## Median :0.2478 Median :0.1640
## Mean :0.2633 Mean :0.1896
## 3rd Qu.:0.3182 3rd Qu.:0.2429
## Max. :1.0000 Max. :1.0000#split training dataset then train the model
wbcd_train <- wbcd_n[1:469, ]
wbcd_test <- wbcd_n[470:569, ]
wbcd_train_labels <- wbcd[1:469, 1]
wbcd_test_labels <- wbcd[470:569, 1]
if (!require("class")) {
install.packages("class")
library(class)
}## Loading required package: classwbcd_test_pred <- knn(train = wbcd_train, test = wbcd_test, cl=wbcd_train_labels, k = 21)#Evaluation
(wbcd_tbl <- table(wbcd_test_pred, wbcd_test_labels))## wbcd_test_labels
## wbcd_test_pred Benign Malignant
## Benign 61 2
## Malignant 0 37(Accuracy <- (wbcd_tbl[1]+wbcd_tbl[4])/sum(wbcd_tbl)*100)## [1] 98News Popularity problem
Q4- Now let’s get back to our problem of news popularity and see if we can apply KNN (K- nearest neighbors) to improve the accuracy of the model. Use the same strategy of training and testing that we did on first 2 labs, and don’t forget that whenever it is required you should use: set.seed(12345).
It appears that everything is straightforward. The only unknown value is k and it is your job to find the best value of k. In your report you should build this model, test it on your test data and write about the model evaluation and why you think the k that yoou have chosen is the best value.
#News Popularity problem
news <- read.csv("~/lab3/OnlineNewsPopularity_for_R.csv")
newsShort <- data.frame(news$n_tokens_title, news$n_tokens_content, news$n_unique_tokens, news$n_non_stop_words, news$num_hrefs, news$num_imgs, news$num_videos, news$average_token_length, news$num_keywords, news$kw_max_max, news$global_sentiment_polarity, news$avg_positive_polarity, news$title_subjectivity, news$title_sentiment_polarity, news$abs_title_subjectivity, news$abs_title_sentiment_polarity, news$shares)
colnames(newsShort) <- c("n_tokens_title", "n_tokens_content", "n_unique_tokens", "n_non_stop_words", "num_hrefs", "num_imgs", "num_videos", "average_token_length", "num_keywords", "kw_max_max", "global_sentiment_polarity", "avg_positive_polarity", "title_subjectivity", "title_sentiment_polarity", "abs_title_subjectivity", "abs_title_sentiment_polarity", "shares")
newsShort$popular = rep('na', nrow(newsShort))
for(i in 1:39644) {
if(newsShort$shares[i] >= 1400) {
newsShort$popular[i] = "yes"}
else {newsShort$popular[i] = "no"}
}
newsShort$shares = newsShort$popular
#for(i in 1:39644) {
# if(newsShort$shares[i] >= 1400) {
# newsShort$shares[i] = "yes"}
# else {newsShort$shares[i] = "no"}
#cat("i=,",i," shares=",newsShort$shares[i],"\n")
#}
newsShort$shares <- as.factor(newsShort$shares)
newsShort <- newsShort[-18]
news_n <- as.data.frame(lapply(newsShort[1:16], normalize))
#news_rand <- news_n[order(runif(10000)), ]
#set.seed(12345)
#Split the data into training and test datasets
news_train <- news_n[1:9000, ]
news_test <- news_n[9001:10000, ]
news_train_labels <- newsShort[1:9000, 17]
news_test_labels <- newsShort[9001:10000, 17]