R Project - Identifying individuals most likely to click an ad
Geoffrey Chege
2022-03-28
1. Introduction
1.1 Defining the question
- Determine which individuals are most likely to click on an ad using supervised learning prediction models.
1.2 The Context
- A Kenyan entrepreneur has created an online cryptography course and would want to advertise it on her blog.
- She currently targets audiences originating from various countries.
- In the past, she ran ads to advertise a related course on the same blog and collected data in the process.
- She would now like to employ my services as a Data Science Consultant to help her identify which individuals are most likely to click on her ads.
1.3 Metric for success
- Accuracy score of 85% and above.
1.4 Experimental Design Taken
- Installing packages and loading libraries needed
- Loading the data
- Data Cleaning
- Exploratory Data Analysis:
- Univariate Analysis
- Bivariate Analysis
- Modelling
- Predictions and evaluation of the model
- Conclusion
1.5 Appropriateness of the available data
- The columns in the dataset include:
- Daily_Time_Spent_on_Site
- Age
- Area_Income
- Daily_Internet_Usage
- Ad_Topic_Line
- City
- Male
- Country
- Timestamp
- Clicked_on_Ad
2. Installing and loading Necessary Packages
3. Loading the Data
ad <- read.csv("C:/Users/user/Downloads/advertising.csv")
head(ad)## Daily.Time.Spent.on.Site Age Area.Income Daily.Internet.Usage
## 1 68.95 35 61833.90 256.09
## 2 80.23 31 68441.85 193.77
## 3 69.47 26 59785.94 236.50
## 4 74.15 29 54806.18 245.89
## 5 68.37 35 73889.99 225.58
## 6 59.99 23 59761.56 226.74
## Ad.Topic.Line City Male Country
## 1 Cloned 5thgeneration orchestration Wrightburgh 0 Tunisia
## 2 Monitored national standardization West Jodi 1 Nauru
## 3 Organic bottom-line service-desk Davidton 0 San Marino
## 4 Triple-buffered reciprocal time-frame West Terrifurt 1 Italy
## 5 Robust logistical utilization South Manuel 0 Iceland
## 6 Sharable client-driven software Jamieberg 1 Norway
## Timestamp Clicked.on.Ad
## 1 2016-03-27 00:53:11 0
## 2 2016-04-04 01:39:02 0
## 3 2016-03-13 20:35:42 0
## 4 2016-01-10 02:31:19 0
## 5 2016-06-03 03:36:18 0
## 6 2016-05-19 14:30:17 04. Data Cleaning
4.1 Checking the attribute types
## Daily.Time.Spent.on.Site Age Area.Income
## "numeric" "integer" "numeric"
## Daily.Internet.Usage Ad.Topic.Line City
## "numeric" "character" "character"
## Male Country Timestamp
## "integer" "character" "character"
## Clicked.on.Ad
## "integer"4.2 converting time variable from character to date and time (POSIXct) format
ad$Timestamp <- as.POSIXct(ad$Timestamp, "%Y-%m-%d %H:%M:%S",tz = "GMT")4.3 Checking for duplicates
duplicates <- ad[duplicated(ad),]
duplicates## [1] Daily.Time.Spent.on.Site Age Area.Income
## [4] Daily.Internet.Usage Ad.Topic.Line City
## [7] Male Country Timestamp
## [10] Clicked.on.Ad
## <0 rows> (or 0-length row.names)There are no duplicates in the dataset
4.4 checking for null values
colSums(is.na(ad))## Daily.Time.Spent.on.Site Age Area.Income
## 0 0 0
## Daily.Internet.Usage Ad.Topic.Line City
## 0 0 0
## Male Country Timestamp
## 0 0 0
## Clicked.on.Ad
## 0There are no null values in the dataset
4.5 checking column names
names(ad)## [1] "Daily.Time.Spent.on.Site" "Age"
## [3] "Area.Income" "Daily.Internet.Usage"
## [5] "Ad.Topic.Line" "City"
## [7] "Male" "Country"
## [9] "Timestamp" "Clicked.on.Ad"Replacing the periods “.” with underscores “_”
names(ad) <- gsub("[.]", "_", names(ad))names(ad)## [1] "Daily_Time_Spent_on_Site" "Age"
## [3] "Area_Income" "Daily_Internet_Usage"
## [5] "Ad_Topic_Line" "City"
## [7] "Male" "Country"
## [9] "Timestamp" "Clicked_on_Ad"4.6 Outliers
I will use boxplots to check for outliers.
Boxplot for “Area_Income”
Boxplot for “Age”
Boxplot for “Daily_Time_Spent_on_Site”
Boxplot for “Daily_Internet_Usage”
5. Exploratory Data Analysis
5.1 Univariate Analysis
Summary statistics of the dataset
summary(ad)## Daily_Time_Spent_on_Site Age Area_Income Daily_Internet_Usage
## Min. :32.60 Min. :19.00 Min. :13996 Min. :104.8
## 1st Qu.:51.36 1st Qu.:29.00 1st Qu.:47032 1st Qu.:138.8
## Median :68.22 Median :35.00 Median :57012 Median :183.1
## Mean :65.00 Mean :36.01 Mean :55000 Mean :180.0
## 3rd Qu.:78.55 3rd Qu.:42.00 3rd Qu.:65471 3rd Qu.:218.8
## Max. :91.43 Max. :61.00 Max. :79485 Max. :270.0
## Ad_Topic_Line City Male Country
## Length:1000 Length:1000 Min. :0.000 Length:1000
## Class :character Class :character 1st Qu.:0.000 Class :character
## Mode :character Mode :character Median :0.000 Mode :character
## Mean :0.481
## 3rd Qu.:1.000
## Max. :1.000
## Timestamp Clicked_on_Ad
## Min. :2016-01-01 02:52:10 Min. :0.0
## 1st Qu.:2016-02-18 02:55:42 1st Qu.:0.0
## Median :2016-04-07 17:27:29 Median :0.5
## Mean :2016-04-10 10:34:06 Mean :0.5
## 3rd Qu.:2016-05-31 03:18:14 3rd Qu.:1.0
## Max. :2016-07-24 00:22:16 Max. :1.0From the summary statistics, the following can be concluded about these columns:
Daily_Time_Spent_on_Site:
mean: 65
median: 68.22
Age:
mean: 36.01
median: 35
Area Income:
mean: 55,000
median: 57,012
Daily_Internet_Usage:
mean: 180
median: 183.1
Using describe() to get range, skewness, kurtosis and standard deviation among others:
describe(ad)## Warning in FUN(newX[, i], ...): no non-missing arguments to min; returning Inf## Warning in FUN(newX[, i], ...): no non-missing arguments to max; returning -Inf## vars n mean sd median trimmed mad
## Daily_Time_Spent_on_Site 1 1000 65.00 15.85 68.22 65.74 17.92
## Age 2 1000 36.01 8.79 35.00 35.51 8.90
## Area_Income 3 1000 55000.00 13414.63 57012.30 56038.94 13316.62
## Daily_Internet_Usage 4 1000 180.00 43.90 183.13 179.99 58.61
## Ad_Topic_Line* 5 1000 500.50 288.82 500.50 500.50 370.65
## City* 6 1000 487.32 279.31 485.50 487.51 356.57
## Male 7 1000 0.48 0.50 0.00 0.48 0.00
## Country* 8 1000 116.41 69.94 114.50 115.82 89.70
## Timestamp 9 1000 NaN NA NA NaN NA
## Clicked_on_Ad 10 1000 0.50 0.50 0.50 0.50 0.74
## min max range skew kurtosis se
## Daily_Time_Spent_on_Site 32.60 91.43 58.83 -0.37 -1.10 0.50
## Age 19.00 61.00 42.00 0.48 -0.41 0.28
## Area_Income 13996.50 79484.80 65488.30 -0.65 -0.11 424.21
## Daily_Internet_Usage 104.78 269.96 165.18 -0.03 -1.28 1.39
## Ad_Topic_Line* 1.00 1000.00 999.00 0.00 -1.20 9.13
## City* 1.00 969.00 968.00 0.00 -1.19 8.83
## Male 0.00 1.00 1.00 0.08 -2.00 0.02
## Country* 1.00 237.00 236.00 0.08 -1.23 2.21
## Timestamp Inf -Inf -Inf NA NA NA
## Clicked_on_Ad 0.00 1.00 1.00 0.00 -2.00 0.02Mode
A function to determine the mode:
mode <- function(v){
uniq <- unique(v)
uniq[which.max(tabulate(match(v,uniq)))]
}The most recurrent Ad Topic Line:
## [1] "Cloned 5thgeneration orchestration"The most recurrent City:
## [1] "Lisamouth"The most recurrent Country:
## [1] "Czech Republic"Checking the modal age using a barplot:
From the plot, the modal age is 31.
Checking the distribution in terms of gender where 1 is Male and 0 is Female:
## gender
## 0 1
## 519 481
From this, there are More women than men, making female the modal gender.
5.2 Bivariate Analysis
Scatterplots
# scatterplot
plot((ad$Daily_Time_Spent_on_Site), (ad$Age),
main = "A scatterplot of Time Spent on site against age",
xlab = 'Time spent',
ylab = 'Age')
# scatterplot of Time on site vs income
plot((ad$Daily_Time_Spent_on_Site), (ad$Area_Income),
main = "A scatterplot of Time Spent on site against income",
xlab = 'Time Spent on Site',
ylab = 'Income')
# scatterplot of Time on site vs Internet usage
plot((ad$Daily_Time_Spent_on_Site), (ad$Daily_Internet_Usage),
main = "A scatterplot of Time Spent on site against Daily Internet Usage",
xlab = 'Time Spent on Site',
ylab = 'Daily Internet Usage')
Heatmap
# Heat map
# Checking the relationship between the variables
# Using Numeric variables only
numeric_tbl <- ad %>%
select_if(is.numeric) %>%
select(Daily_Time_Spent_on_Site, Age, Area_Income,Daily_Internet_Usage)
# Calculate the correlations
corr <- cor(numeric_tbl, use = "complete.obs")
ggcorrplot(round(corr, 2),
type = "full", lab = T)
#### Those who clicked on ads:
Analysis of people who click on the ads:
# Analysis of people who click on the ads
ad_click <- ad[which(ad$Clicked_on_Ad == 1),]Most popular age group of people clicking on ads:
# Most popular age group of people clicking on ads
hist((ad_click$Age),
main = "Histogram of Age of those who click ads",
xlab = 'Age',
ylab = 'Frequency',
col = "blue")
40 - 45 year olds click on the most ads
Plotting to visualize the gender distribution:
gender2 <- (ad_click$Male)
gender2.frequency <- table(gender2)
gender2.frequency## gender2
## 0 1
## 269 231# plotting to visualize the gender distribution
barplot(gender2.frequency,
main="A bar chart showing Gender of those who clicked",
xlab="Gender(0 = Female, 1 = Male)",
ylab = "Frequency",
col=c("darkblue","red"),
)
Females clicked more ads than males.
Scatterplots of those who clicked:
# scatterplot
plot((ad_click$Daily_Time_Spent_on_Site), (ad_click$Age),
main = "A scatterplot of Time Spent on site and clicked ad against age",
xlab = 'Time spent',
ylab = 'Age')
# scatterplot of Time on site vs income
plot((ad_click$Daily_Time_Spent_on_Site), (ad_click$Area_Income),
main = "A scatterplot of Time Spent on site and ad clicked against income",
xlab = 'Time Spent on Site',
ylab = 'Income')
# scatterplot of Time on site vs Internet usage
plot((ad_click$Daily_Time_Spent_on_Site), (ad_click$Daily_Internet_Usage),
main = "A scatterplot of Time Spent on site and ad clicked against Daily Internet Usage",
xlab = 'Time Spent on Site',
ylab = 'Daily Internet Usage')
# Heat map
# Checking the relationship between the variables
# Using Numeric variables only
numeric_tbl <- ad_click %>%
select_if(is.numeric) %>%
select(Daily_Time_Spent_on_Site, Age, Area_Income,Daily_Internet_Usage)
# Calculate the correlations
corr <- cor(numeric_tbl, use = "complete.obs")
ggcorrplot(round(corr, 2),
type = "full", lab = T)
The country with the most ad clicks:
mode(ad_click$Country)## [1] "Australia"The income that clicks most:
mode(ad_click$Area_Income)## [1] 24593.33Ad title that garners most clicks:
## [1] "Reactive local challenge"6. Modelling
# Heat map
# Checking the relationship between the variables
# Using Numeric variables only
numeric_tbl2 <- ad %>%
select_if(is.numeric) %>%
select(Daily_Time_Spent_on_Site, Age, Area_Income,Daily_Internet_Usage, Clicked_on_Ad)
# Calculate the correlations
corr <- cor(numeric_tbl2, use = "complete.obs")
ggcorrplot(round(corr, 2),
type = "full", lab = T)
Clicked_on_Ad has high correlation. It is the target variable so I will not include it in the correlation.
# Heat map
# Checking the relationship between the variables
# Using Numeric variables only
numeric_tbl3 <- ad %>%
select_if(is.numeric) %>%
select(Daily_Time_Spent_on_Site, Age, Area_Income,Daily_Internet_Usage)
# Calculate the correlations
corr <- cor(numeric_tbl3, use = "complete.obs")
ggcorrplot(round(corr, 2),
type = "full", lab = T)
There are no highly correlated numeric independent variables, so I will use them all in analysis.
Decision Tree Classifier
# Specifying target and predictor variables
m <- rpart(Clicked_on_Ad ~ . ,
data = numeric_tbl2,
method = "class")# Plotting model
rpart.plot(m)
# Making predictions
# Printing the confusion matrix
p <- predict(m, numeric_tbl2, type ="class")
table(p, numeric_tbl2$Clicked_on_Ad)##
## p 0 1
## 0 485 28
## 1 15 472# Printing the Accuracy
(mean(numeric_tbl2$Clicked_on_Ad == p))*100## [1] 95.7- The model has an accuracy of 95.7%
- This is a good model for making predictions
7. Conclusion
Decision Tree gives an accuracy of 95.7%