library(magrittr)
library(dplyr)
library(tidyr)
library(ggplot2)
library(knitr)
library(kableExtra)
library(reshape2)setwd(“C:_homeworks_2023”)
## customerID gender SeniorCitizen Partner Dependents tenure PhoneService
## 1 7590-VHVEG Female 0 Yes No 1 No
## 2 5575-GNVDE Male 0 No No 34 Yes
## 3 3668-QPYBK Male 0 No No 2 Yes
## 4 7795-CFOCW Male 0 No No 45 No
## 5 9237-HQITU Female 0 No No 2 Yes
## 6 9305-CDSKC Female 0 No No 8 Yes
## MultipleLines InternetService OnlineSecurity OnlineBackup DeviceProtection
## 1 No phone service DSL No Yes No
## 2 No DSL Yes No Yes
## 3 No DSL Yes Yes No
## 4 No phone service DSL Yes No Yes
## 5 No Fiber optic No No No
## 6 Yes Fiber optic No No Yes
## TechSupport StreamingTV StreamingMovies Contract PaperlessBilling
## 1 No No No Month-to-month Yes
## 2 No No No One year No
## 3 No No No Month-to-month Yes
## 4 Yes No No One year No
## 5 No No No Month-to-month Yes
## 6 No Yes Yes Month-to-month Yes
## PaymentMethod MonthlyCharges TotalCharges Churn
## 1 Electronic check 29.85 29.85 No
## 2 Mailed check 56.95 1889.50 No
## 3 Mailed check 53.85 108.15 Yes
## 4 Bank transfer (automatic) 42.30 1840.75 No
## 5 Electronic check 70.70 151.65 Yes
## 6 Electronic check 99.65 820.50 Yeskbl(df[1:5,]) %>%
kable_paper("hover", full_width = F)| customerID | gender | SeniorCitizen | Partner | Dependents | tenure | PhoneService | MultipleLines | InternetService | OnlineSecurity | OnlineBackup | DeviceProtection | TechSupport | StreamingTV | StreamingMovies | Contract | PaperlessBilling | PaymentMethod | MonthlyCharges | TotalCharges | Churn |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 7590-VHVEG | Female | 0 | Yes | No | 1 | No | No phone service | DSL | No | Yes | No | No | No | No | Month-to-month | Yes | Electronic check | 29.85 | 29.85 | No |
| 5575-GNVDE | Male | 0 | No | No | 34 | Yes | No | DSL | Yes | No | Yes | No | No | No | One year | No | Mailed check | 56.95 | 1889.50 | No |
| 3668-QPYBK | Male | 0 | No | No | 2 | Yes | No | DSL | Yes | Yes | No | No | No | No | Month-to-month | Yes | Mailed check | 53.85 | 108.15 | Yes |
| 7795-CFOCW | Male | 0 | No | No | 45 | No | No phone service | DSL | Yes | No | Yes | Yes | No | No | One year | No | Bank transfer (automatic) | 42.30 | 1840.75 | No |
| 9237-HQITU | Female | 0 | No | No | 2 | Yes | No | Fiber optic | No | No | No | No | No | No | Month-to-month | Yes | Electronic check | 70.70 | 151.65 | Yes |
str(df)## 'data.frame': 7043 obs. of 21 variables:
## $ customerID : chr "7590-VHVEG" "5575-GNVDE" "3668-QPYBK" "7795-CFOCW" ...
## $ gender : chr "Female" "Male" "Male" "Male" ...
## $ SeniorCitizen : int 0 0 0 0 0 0 0 0 0 0 ...
## $ Partner : chr "Yes" "No" "No" "No" ...
## $ Dependents : chr "No" "No" "No" "No" ...
## $ tenure : int 1 34 2 45 2 8 22 10 28 62 ...
## $ PhoneService : chr "No" "Yes" "Yes" "No" ...
## $ MultipleLines : chr "No phone service" "No" "No" "No phone service" ...
## $ InternetService : chr "DSL" "DSL" "DSL" "DSL" ...
## $ OnlineSecurity : chr "No" "Yes" "Yes" "Yes" ...
## $ OnlineBackup : chr "Yes" "No" "Yes" "No" ...
## $ DeviceProtection: chr "No" "Yes" "No" "Yes" ...
## $ TechSupport : chr "No" "No" "No" "Yes" ...
## $ StreamingTV : chr "No" "No" "No" "No" ...
## $ StreamingMovies : chr "No" "No" "No" "No" ...
## $ Contract : chr "Month-to-month" "One year" "Month-to-month" "One year" ...
## $ PaperlessBilling: chr "Yes" "No" "Yes" "No" ...
## $ PaymentMethod : chr "Electronic check" "Mailed check" "Mailed check" "Bank transfer (automatic)" ...
## $ MonthlyCharges : num 29.9 57 53.9 42.3 70.7 ...
## $ TotalCharges : num 29.9 1889.5 108.2 1840.8 151.7 ...
## $ Churn : chr "No" "No" "Yes" "No" ...summary(df)## customerID gender SeniorCitizen Partner
## Length:7043 Length:7043 Min. :0.0000 Length:7043
## Class :character Class :character 1st Qu.:0.0000 Class :character
## Mode :character Mode :character Median :0.0000 Mode :character
## Mean :0.1621
## 3rd Qu.:0.0000
## Max. :1.0000
##
## Dependents tenure PhoneService MultipleLines
## Length:7043 Min. : 0.00 Length:7043 Length:7043
## Class :character 1st Qu.: 9.00 Class :character Class :character
## Mode :character Median :29.00 Mode :character Mode :character
## Mean :32.37
## 3rd Qu.:55.00
## Max. :72.00
##
## InternetService OnlineSecurity OnlineBackup DeviceProtection
## Length:7043 Length:7043 Length:7043 Length:7043
## Class :character Class :character Class :character Class :character
## Mode :character Mode :character Mode :character Mode :character
##
##
##
##
## TechSupport StreamingTV StreamingMovies Contract
## Length:7043 Length:7043 Length:7043 Length:7043
## Class :character Class :character Class :character Class :character
## Mode :character Mode :character Mode :character Mode :character
##
##
##
##
## PaperlessBilling PaymentMethod MonthlyCharges TotalCharges
## Length:7043 Length:7043 Min. : 18.25 Min. : 18.8
## Class :character Class :character 1st Qu.: 35.50 1st Qu.: 401.4
## Mode :character Mode :character Median : 70.35 Median :1397.5
## Mean : 64.76 Mean :2283.3
## 3rd Qu.: 89.85 3rd Qu.:3794.7
## Max. :118.75 Max. :8684.8
## NA's :11
## Churn
## Length:7043
## Class :character
## Mode :character
##
##
##
## The dataset contains the data of the customers in telecommunication industry. Therefore the unit of observation is a customer. The dataset includes 21 variables and data of 7043 customers.
The variable description is as below.
variables=read.csv("variables.csv")
variables## customerID
## 1 gender
## 2 SeniorCitizen
## 3 Partner
## 4 Dependents
## 5 tenure
## 6 PhoneService
## 7 MultipleLines
## 8 InternetService
## 9 OnlineSecurity
## 10 OnlineBackup
## 11 DeviceProtection
## 12 TechSupport
## 13 StreamingTV
## 14 StreamingMovies
## 15 Contract
## 16 PaperlessBilling
## 17 PaymentMethod
## 18 MonthlyCharges
## 19 TotalCharges
## 20 Churn
## ID.of.the.Customer
## 1 Whether the customer is a male or a female
## 2 Whether the customer is a senior citizen or not (1, 0)
## 3 Whether the customer has a partner or not (Yes, No)
## 4 Whether the customer has dependents or not (Yes, No)
## 5 Number of months the customer has stayed with the company
## 6 Whether the customer has a phone service or not (Yes, No)
## 7 Whether the customer has multiple lines or not (Yes, No, No phone service)
## 8 Customer\x92s internet service provider (DSL, Fiber optic, No)
## 9 Whether the customer has online security or not (Yes, No, No internet service)
## 10 Whether the customer has online backup or not (Yes, No, No internet service)
## 11 Whether the customer has device protection or not (Yes, No, No internet service)
## 12 Whether the customer has tech support or not (Yes, No, No internet service)
## 13 Whether the customer has streaming TV or not (Yes, No, No internet service)
## 14 Whether the customer has streaming movies or not (Yes, No, No internet service)
## 15 The contract term of the customer (Month-to-month, One year, Two year)
## 16 Whether the customer has paperless billing or not (Yes, No)
## 17 The customer\x92s payment method (Electronic check, Mailed check, Bank transfer (automatic), Credit card (automatic))
## 18 The amount charged to the customer monthly ($)
## 19 The total amount charged to the customer ($)
## 20 Whether the customer churned or not (Yes or No)The main goal of this study is to build a prediction model for predicting the customers who are likely to churn from the network. The second goal of the study is to conduct a descriptive analaysis to identyfy trends, factors associate with the customer prediction.
#Checking the existence of null values
sapply(df, function(x) sum(is.na(x)))## customerID gender SeniorCitizen Partner
## 0 0 0 0
## Dependents tenure PhoneService MultipleLines
## 0 0 0 0
## InternetService OnlineSecurity OnlineBackup DeviceProtection
## 0 0 0 0
## TechSupport StreamingTV StreamingMovies Contract
## 0 0 0 0
## PaperlessBilling PaymentMethod MonthlyCharges TotalCharges
## 0 0 0 11
## Churn
## 0Only TotalCharges variable contains 11 missing values. Since the sample size is large enough, the 11 observations will be removed from the datasets.
df = na.omit(df)
str(df)## 'data.frame': 7032 obs. of 21 variables:
## $ customerID : chr "7590-VHVEG" "5575-GNVDE" "3668-QPYBK" "7795-CFOCW" ...
## $ gender : chr "Female" "Male" "Male" "Male" ...
## $ SeniorCitizen : int 0 0 0 0 0 0 0 0 0 0 ...
## $ Partner : chr "Yes" "No" "No" "No" ...
## $ Dependents : chr "No" "No" "No" "No" ...
## $ tenure : int 1 34 2 45 2 8 22 10 28 62 ...
## $ PhoneService : chr "No" "Yes" "Yes" "No" ...
## $ MultipleLines : chr "No phone service" "No" "No" "No phone service" ...
## $ InternetService : chr "DSL" "DSL" "DSL" "DSL" ...
## $ OnlineSecurity : chr "No" "Yes" "Yes" "Yes" ...
## $ OnlineBackup : chr "Yes" "No" "Yes" "No" ...
## $ DeviceProtection: chr "No" "Yes" "No" "Yes" ...
## $ TechSupport : chr "No" "No" "No" "Yes" ...
## $ StreamingTV : chr "No" "No" "No" "No" ...
## $ StreamingMovies : chr "No" "No" "No" "No" ...
## $ Contract : chr "Month-to-month" "One year" "Month-to-month" "One year" ...
## $ PaperlessBilling: chr "Yes" "No" "Yes" "No" ...
## $ PaymentMethod : chr "Electronic check" "Mailed check" "Mailed check" "Bank transfer (automatic)" ...
## $ MonthlyCharges : num 29.9 57 53.9 42.3 70.7 ...
## $ TotalCharges : num 29.9 1889.5 108.2 1840.8 151.7 ...
## $ Churn : chr "No" "No" "Yes" "No" ...
## - attr(*, "na.action")= 'omit' Named int [1:11] 489 754 937 1083 1341 3332 3827 4381 5219 6671 ...
## ..- attr(*, "names")= chr [1:11] "489" "754" "937" "1083" ...sapply(df, function(x) sum(is.na(x)))## customerID gender SeniorCitizen Partner
## 0 0 0 0
## Dependents tenure PhoneService MultipleLines
## 0 0 0 0
## InternetService OnlineSecurity OnlineBackup DeviceProtection
## 0 0 0 0
## TechSupport StreamingTV StreamingMovies Contract
## 0 0 0 0
## PaperlessBilling PaymentMethod MonthlyCharges TotalCharges
## 0 0 0 0
## Churn
## 0Renaming the Senior Citizen variable
levels(df$SeniorCitizen) = list("No" = 0, "Yes" = 1)
df$SeniorCitizen = as.factor(df$SeniorCitizen)All the categorical variables in the dataset are in ‘char’ format. Therefore they will be converted to factors before moving to the descriptive analysis.
df$gender = as.factor(df$gender)
df$Partner = as.factor(df$Partner)
df$Dependents = as.factor(df$Dependents)
df$PhoneService = as.factor(df$PhoneService)
df$MultipleLines= as.factor(df$MultipleLines)
df$InternetService= as.factor(df$InternetService)
df$OnlineSecurity= as.factor(df$OnlineSecurity)
df$OnlineBackup= as.factor(df$OnlineBackup)
df$DeviceProtection= as.factor(df$DeviceProtection)
df$TechSupport= as.factor(df$TechSupport)
df$StreamingTV= as.factor(df$StreamingTV)
df$StreamingMovies= as.factor(df$StreamingMovies)
df$Contract= as.factor(df$Contract)
df$PaperlessBilling= as.factor(df$PaperlessBilling)
df$PaymentMethod= as.factor(df$PaymentMethod)
df$Churn= as.factor(df$Churn)
str(df)## 'data.frame': 7032 obs. of 21 variables:
## $ customerID : chr "7590-VHVEG" "5575-GNVDE" "3668-QPYBK" "7795-CFOCW" ...
## $ gender : Factor w/ 2 levels "Female","Male": 1 2 2 2 1 1 2 1 1 2 ...
## $ SeniorCitizen : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
## $ Partner : Factor w/ 2 levels "No","Yes": 2 1 1 1 1 1 1 1 2 1 ...
## $ Dependents : Factor w/ 2 levels "No","Yes": 1 1 1 1 1 1 2 1 1 2 ...
## $ tenure : int 1 34 2 45 2 8 22 10 28 62 ...
## $ PhoneService : Factor w/ 2 levels "No","Yes": 1 2 2 1 2 2 2 1 2 2 ...
## $ MultipleLines : Factor w/ 3 levels "No","No phone service",..: 2 1 1 2 1 3 3 2 3 1 ...
## $ InternetService : Factor w/ 3 levels "DSL","Fiber optic",..: 1 1 1 1 2 2 2 1 2 1 ...
## $ OnlineSecurity : Factor w/ 3 levels "No","No internet service",..: 1 3 3 3 1 1 1 3 1 3 ...
## $ OnlineBackup : Factor w/ 3 levels "No","No internet service",..: 3 1 3 1 1 1 3 1 1 3 ...
## $ DeviceProtection: Factor w/ 3 levels "No","No internet service",..: 1 3 1 3 1 3 1 1 3 1 ...
## $ TechSupport : Factor w/ 3 levels "No","No internet service",..: 1 1 1 3 1 1 1 1 3 1 ...
## $ StreamingTV : Factor w/ 3 levels "No","No internet service",..: 1 1 1 1 1 3 3 1 3 1 ...
## $ StreamingMovies : Factor w/ 3 levels "No","No internet service",..: 1 1 1 1 1 3 1 1 3 1 ...
## $ Contract : Factor w/ 3 levels "Month-to-month",..: 1 2 1 2 1 1 1 1 1 2 ...
## $ PaperlessBilling: Factor w/ 2 levels "No","Yes": 2 1 2 1 2 2 2 1 2 1 ...
## $ PaymentMethod : Factor w/ 4 levels "Bank transfer (automatic)",..: 3 4 4 1 3 3 2 4 3 1 ...
## $ MonthlyCharges : num 29.9 57 53.9 42.3 70.7 ...
## $ TotalCharges : num 29.9 1889.5 108.2 1840.8 151.7 ...
## $ Churn : Factor w/ 2 levels "No","Yes": 1 1 2 1 2 2 1 1 2 1 ...
## - attr(*, "na.action")= 'omit' Named int [1:11] 489 754 937 1083 1341 3332 3827 4381 5219 6671 ...
## ..- attr(*, "names")= chr [1:11] "489" "754" "937" "1083" ...require('gridExtra')
plot1 <- ggplot(df, aes(Churn)) +
geom_bar() +
labs(title = "Customer Churn Status",
x = "Whether the customer churned or not (Yes or No",
y = "Count")
#Gender
plot2 <- ggplot(df, aes(gender)) +
geom_bar() +
labs(title = "Customer Gender",
x = "Whether the customer Male/Female",
y = "Count")
#Senior Citizen
plot3 <- ggplot(df, aes(SeniorCitizen)) +
geom_bar() +
labs(title = "Senior Citizen Customers",
x = "Whether the customer Senior Citizen/not",
y = "Count")
#Dependents
plot4 <- ggplot(df, aes(Dependents)) +
geom_bar() +
labs(title = "Dependents",
x = "Whether the customer have dependent/not",
y = "Count")
#Partner
plot5 <- ggplot(df, aes(Partner)) +
geom_bar() +
labs(title = "Partners",
x = "Whether the customer having a partner/not",
y = "Count")
grid.arrange(plot1, plot2, plot3, ncol=3)
Churn: The response variable of the analysis is churn status of the customer, from the sample of 7032 of the customers about 1850 of the customers have churned from the network, which makes it about 25% of the customers have churned. The churn percentage is considerable high; therefore, the objective of the study is also clear, that the factors associated with the customer churn should be identified to prevent the customer churn.
Gender: The percentages of males and females are approximately same according to the bar chart. Therefore, it can be observed that both males and females use this network equally.
Senior Citizen Customer: From the bar chart it is observed that about 1100 of the customers out of 7032, are senior citizens in this sample. That is about 15% from the sample.
require('gridExtra')
grid.arrange(plot4, plot5, ncol=2)
About 2100 (30%) of the sample have dependents while 3400 (50%) customers have partners. Here a great opportunity can be identified that the network can be promoted to the dependents and partners via the customer.
for all 6 variables have three categories ‘Yes’, ‘No’, ‘No internet connection’. Therefore, the customers who do not have internet connection will not be considered and the percentage of ‘Yes’ are obtained for each variable.
Internet_Addons_df = data.frame('Online Security' = df$OnlineSecurity,
'Online Backup' = df$OnlineBackup,
'Device Protection' = df$DeviceProtection,
'Tech Support' = df$TechSupport,
'Streaming TV'= df$StreamingTV,
'Streaming Movies' = df$StreamingMovies)
perc = function(x){
l = length(x)
v = length(grep("Yes", x))
v/l*100
}
sumer = Internet_Addons_df %>%
filter(Online.Security != "No internet service")%>%
summarise_all(funs(perc))%>%
gather('metric','percent')
ggplot(sumer, aes(reorder(metric,percent), percent))+
geom_bar( stat = "identity") +
coord_flip()+
labs(title = "Popularity of Internet Add-ons",
subtitle = "Only considering customers who have internet",
x = "",
y = "Percentage of Internet Subscribers with Service")
plot6 <- ggplot(df, aes(MonthlyCharges)) +
geom_histogram(bins = 20) +
labs(title = "Histogram of MonthlyCharges",
x = "MonthlyCharges",
y = "Count")
plot7 <- ggplot(df, aes(TotalCharges)) +
geom_histogram(bins = 20) +
labs(title = "Histogram of TotalCharges",
x = "TotalCharges",
y = "Count")
grid.arrange(plot6, plot7, ncol=2)
# Tenure
ggplot(data = df, aes(x = tenure)) +
geom_bar() +
labs(title = "Distribution of tenure",
x = "Tenure (No of months)",
y = "No of customers")
Mode of the customers’ monthly charge is below 25. Both the monthly charges and the total charges histograms are skewed to the right and do not follow normal distributions.
plot8 <- ggplot(df, aes(Partner, fill = Churn)) +
geom_bar() +
labs(title = "Partner Vs Churn",
x = "Does the Customer have a Partner(Yes or No)?",
y = "Count")
plot9 <- ggplot(df, aes(Dependents, fill = Churn)) +
geom_bar(position = 'fill') +
labs(title = "Dependents Status Vs Churn",
x = "Does the Customer have Dependents (Yes or No)?",
y = "Count")
plot10 <-ggplot(df, aes(gender, fill = Churn)) +
geom_bar() +
labs(title = "Gender Vs Churn",
x = "Gender",
y = "Count")
plot8
The number of churned customers are high in the customer segment who do not have partners.
plot9
The number of churned customers are high in the customer segment who do not have dependents.
plot10
Churn rates are pretty much similar to both gernders.
#grid.arrange(plot8, plot9, plot10, ncol=3)
plot11 <- ggplot(df, aes(PaperlessBilling, fill = Churn)) +
geom_bar() +
labs(title = "Paperless Billing Status",
x = "Does the Customer Use Paperless Billing?",
y = "Count")
plot12 <- ggplot(df, aes(PaymentMethod, fill = Churn)) +
geom_bar() +
labs(title = "Payment Method",
x = "What Payment Method does the Customer Use?",
y = "Count") +
theme(axis.text.x = element_text(angle = 90, vjust = 0.5, hjust=1))
plot11
In overall, More than 4000 customers have subscribed for paperless billing services and also the higher count of churned customers are also identified in this category.
plot12
#grid.arrange(plot11, plot12, ncol=2)Most people tend to use electronic check method as their payment method. The percentages are approximately similar for other three categories. The highest churn count is also observed in the customers who use electronic checking.
ggplot(df, aes(SeniorCitizen, fill = PaymentMethod)) +
geom_bar(position = 'fill') +
labs(title = "Payment Method of Seniors Citizen",
x = "Is the Customer a Senior Citizen?",
y = "Fraction")
#Monthly Charge vs Churn
plot13 <- ggplot(data = df, aes(MonthlyCharges, color = Churn))+
geom_freqpoly(binwidth = 5, size = 1)+
labs(title = "Churn Count vs MonthlyCharges",
x = "Monthly Charges",
y = "Churn Count")
#Total Charge vs Churn
plot14 <-ggplot(data = df, aes(TotalCharges, color = Churn))+
geom_freqpoly(binwidth = 200, size = 1)+
labs(title = "Churn Count vs TotalCharges",
x = "Total Charges",
y = "Churn Count")
#tenure vs churn
plot15 <-ggplot(data = df, aes(tenure, colour = Churn))+
geom_freqpoly(binwidth = 5, size = 1)+
labs(title = "Churn Count vs tenure",
x = "tenure",
y = "Churn Count")
plot13
From the monthly charges univariate histogram also, it was observed that the highest frequency of customers belonged to below 25 category. The same pattern is here also observed for the customers who are staying on the network. But the pattern is different for the customers who have churned from the network. More customers who have churned, have used between monthly charges between 75-100.
plot14
The total charge two-line charts follow the same pattern for both churned and current customers. Although the count of churned customers is low in the dataset, still the count who use below about 250, are same for both categories.
plot15
#grid.arrange(plot13, plot14, plot15, ncol=3)
ggplot(df, aes(x=gender, y=TotalCharges)) +
geom_boxplot() +
labs(title = "Gender vs Total Charges",
x = "Gender",
y = "Total Charge($)")
The distributions of the total charges are similar for both males and females. The median total charge is about 1350 while 25% of the customers use more than total charge of 3750 for both categories.
ggplot(df, aes(x=gender, y=MonthlyCharges)) + geom_boxplot() +
labs(title = "Gender vs Monthly Charges",
x = "Gender",
y = "Monthly Charge($)")
The distributions of the monthly charges are similar for both males and females. The median monthly charge is about 71 while 25% of the customers use more than total charge about 90 in both categories
#scatter plots
ggplot(df, aes(x=MonthlyCharges, y=TotalCharges)) +
geom_point(size=1)+
labs(title = "MonthlyCharges vs TotalCharges",
x = "MonthlyCharges",
y = "TotalCharges")
As the monthly charge increase, the total charge also increases which is a fact. A positive moderate strong relationship can be identified in the two factors considered.
ggplot(df, aes(x=tenure, y=TotalCharges)) +
geom_point(size=1)+
labs(title = "TotalCharges vs tenure",
x = "tenure",
y = "TotalCharges")
As the tenure increase, the total charge also increases according to the scatterplot. A positive moderate strong relationship can be identified in the two factors considered.
ggplot(df, aes(x=tenure, y=MonthlyCharges)) + geom_point(size=1)+
labs(title = "MonthlyCharges vs tenure",
x = "tenure",
y = "MonthlyCharges")
There is no relationship can be identified between the monthly charge and the tenure from the scatterplot.
ggplot(df, aes(x = MonthlyCharges, y = TotalCharges, color = Churn)) +
geom_point()
The same positive moderate strong relationship can be identified for both churned and current customers.
ggplot(data=df, mapping = aes(x = MonthlyCharges, y = TotalCharges)) +
geom_point() +
facet_wrap(~Churn) +
theme_bw()
The same positive moderate strong relationship can be identified for both churned and current customers.
data <- cor(df[sapply(df,is.numeric)])
data1 <- melt(data)
ggplot(data1, aes(x = Var1,
y = Var2,
fill = value))+geom_tile()+
labs(title = "Heatmp",
x = "",
y = "")