Data Science Quiz For Practice

Welcome to your Data Science Quiz

How would you access the ‘StreamingTV’ column from the ‘customer_churn’ data.frame?

customer_churn@StreamingTV

customer_churn#StreamingTV

customer_churn$StreamingTV

customer_churn&StreamingTV

How would you create a list consisting of these elements: 100, ‘sparta’, TRUE

List(100,’sparta’,TRUE)

list(100,’sparta’,TRUE)

list(c(100,’sparta’,TRUE))

list(list(100,’sparta’,TRUE))

How would you get the last 100 records from the ‘customer_churn’ dataframe

last(100)

tail(100)

last(customer_churn,100)

tail(customer_churn,100)

How would you give a discount of 33% to the 5th cell of ‘MonthlyCharges’ column?

customer_churn$MonthlyCharges[5]*(.33)

customer_churn$MonthlyCharges[5]*(33)

customer_churn$MonthlyCharges[5]*(.67)

customer_churn$MonthlyCharges[5]*(67)

Which of these is the correct code to get the count of number of customers whose ‘MonthlyCharges’ is greater than 100

count=0 for(iin1:nrow(customer_churn)){ if(customer_churn$MonthlyCharges[i]>100){ count=count+1 } } count

count=0 i=1 while(i100){ count=count+1 } } count

count=1 for(i in 1:nrow(customer_churn)){ if(customer_churn$MonthlyCharges[i]>100){ count=count+50 } } count

count=0 if(customer_churn$MonthlyCharges[i]>100){ count=count+1

How would you extract only the female customers from the ‘customer_churn’ data.frame?

sqldf (select gender='Female' from customer_churn)

sqldf(select from customer_churn where gender=='Female')

sqldf(select * from customer_churn where gender=='Female')

sqldf(select * from customer_churn where gender in 'Female')

How would you extract a random sample of 33 records from the ‘customer_churn’ dataframe?

sample(customer_churn,33)

sample_frac(customer_churn,.33)

sample_frac(customer_churn,33)

sample_n(customer_churn,33)

How would you select the 3rd, 4th& 5th columns from the ‘customer_churn’dataframe?

select(customer_churn,(3,4,5))

select(customer_churn,3,4,5)

select(customer_churn,list(3,4,5))

select(3:5,customer_churn)

How would you get a summarized result for the mean of ‘MonthlyCharges’ grouped w.r.t ‘PaymentMethod’?

summarise(group_by(customer_churn,PaymentMethod),mean_MC=mean(MonthlyCharges))

group_by(mean_MC=mean(MonthlyCharges),summarise(customer_churn,PaymentMethod))

group_by(summarise(customer_churn,PaymentMethod),mean_MC=mean(MonthlyCharges))

summarise(group_by(PaymentMethod),mean_MC=mean(customer_churn,MonthlyCharges))

How would you extract those customers who have subscribed to both ‘StreamingTV’ & ‘StreamingMovies’?

filter(customer_churn, StreamingTV=='Yes' &StreamingMovies=='Yes')

filter(customer_churn, StreamingTV=='Yes' &&StreamingMovies=='Yes')

filter(customer_churn, StreamingTV=='Yes' andStreamingMovies=='Yes')

filter(customer_churn, StreamingTV=='Yes' AND StreamingMovies=='Yes')

To which of these geometries can you add the facet_grid()?

geom_bar()

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