40 questions to test your skill on R for Data Science

B) csv(‘Dataframe.csv’,header=TRUE)

C) dataframe(‘Dataframe.csv’)

D) csv2(‘Dataframe.csv’,header=FALSE,sep=’,’)

B) Xlsx::read.xlsx(“Dataframe.xlsx”,sheetIndex=3,header=FALSE)

C)XLConnect::readWorksheetFromFile(“Dataframe.xlsx”,sheet=3,header=FALSE)

D)All of the above

B) csv(‘Dataframe.csv’,header=FALSE, sep=’,’,na.strings=c(‘?’))

C) csv2(‘Dataframe.csv’,header=FALSE,sep=’,’,na.strings=c(‘?’,’!’))

D) dataframe(‘Dataframe.csv’)

B) csv2(‘Train.csv’,header=TRUE, row.names=TRUE)

C) dataframe(‘Train.csv’,header=TRUE,sep=’,’)

D) csv(‘Train.csv’,,header=TRUE,sep=’,’)

Solution: (D)

row.names argument in options A and B takes only the vector containing the actual row names or a single number giving the column of the table which contains the row names and not a logical value. Option C doesn’t exist. Therefore, option D is the correct solution.

B) csv2(‘Dataframe.csv’,row.names=1,nrows=2)

C) delim2(‘Dataframe.csv’,header=T,row.names=1,sep=’,’,nrows=2)

D) dataframe(‘Dataframe.csv’,header=TRUE,row.names=1,sep=’,’,skip.last=2)

Solution: (A)

Option B will not be able to read the csv file correctly since the default separator in csv2 function is “;” whereas csv files are of type “,”. Option C has wrong header argument value. Option D doesn’t exist. Therefore, Option  A is the correct answer.

A) merge(dataframe[,1:3],dataframe2)

B) merge(dataframe1,dataframe2)[,1:3]

C) merge(dataframe1,dataframe2,all=TRUE)

D) Both 1 and 2

E) All of the above

Solution: (D)

Option C will result in feature 4 being included in the merged dataframe which is what we do not want. Therefore, Option D is the correct solution.

B) stats(dataframe)

C) summarize(dataframe)

D) summarise(dataframe)

E) None of the above

Solution: (E)

Option A will give only the mean and the median but not the mode. Option B, C and D will also fail to provide the required statistics. Therefore, Option E is the correct solution.

B) apply(is.na(dataframe),2,sum)

C) sapply(dataframe,function(x) sum(is.na(x))

D) table(is.na(dataframe))

Solution: (D)

Option D will give the overall count of the missing values but not column wise. Therefore, Option D is the correct solution.

B) ggplot2::qplot(dataframed$Value,geom=”Histogram”)

C)ggplot2::ggplot(data=dataframed,aes(dataframe$Value))+geom_histogram()

D) All of the above

Solution: (D)

All of the given options will plot a histogram and that can be used to see the skewness of the desired data.

B) dataframe$Parameter_Alpha=0

dataframe$Gende_Beta=0

dataframe$Parameter_Alpha[which(dataframe$Parameter==’Alpha’)]=1

dataframe$Parameter_Beta[which(dataframe$Parameter==’Alpha’)]=0

dataframe$Parameter_Alpha[which(dataframe$Parameter==’Beta’]=0

dataframe$Parameter_Beta[which(dataframe$Parameter==’Beta’]=1

C) contrasts(dataframe$Parameter)

D) Both 1 and 2

Solution: (D)

Option C will encode the Parameter column will 2 levels but will not perform one hot encoding. Therefore, option D is the correct solution.

B) (cov(dataframe$column2,dataframe$column3))/(var(dataframe$column2)*sd(dataframe$column3))

C)

(sum(dataframe$Column2*dataframe$Column3)-                 (sum(dataframe$Column2)*sum(dataframe$Column3)/nrow(dataframe)))/(sqrt((sum(dataframe$Column2*dataframe$Column2)-(sum(dataframe$Column2)^3)/nrow(dataframe))*     (sum(dataframe$Column3*dataframe$Column3)-(sum(dataframe$Column3)^2)/nrow(dataframe))))

D) None of the Above

Solution: (D)

In option A, corr is the wrong function name. Actual function name to calculate correlation is cor. In option B, it is the standard deviation which should be the denominator and not variance. Similarly, the formula in Option C is wrong. Therefore, Option D is the correct solution.

B) subset(dataframe, Parameter==’Alpha’)

C) filter(dataframe,Parameter==’Alpha’)

D) Both 2 and 3

E) All of the above

Solution: (D)

In option A, there should be an equality operator instead of the assignment operator. Therefore, option D is the correct solution.

15) Which of the following function is used to view the dataset in spreadsheet like format?

A) disp()

B) View()

C) seq()

D) All of the Above

Solution : (B)

Option B is the only option that will show the dataset in the spreadsheet format. Therefore, option B is the correct solution.

B) boxplot(A~B,data=data)

C) boxplot(A|B,data=data)

D) None of the above

Solution: (B)

Boxplot function in R requires a formula input to draw different boxplots by levels of a factor variable. Therefore, Option B is the correct solution.

B) par(mfcol=c(4,3))

C) par(mfrow=c(4,3))

D) par(col=c(4,3))

Solution: (B)

mfcol argument will ensure that the plots enter the plotting window column wise. Therefore, Option B is the correct solution.

B) format(df,”%D %A %b %y”)

C) format(df,”%D %a %B %Y”)

D) None of above

Solution: (D)

None of the above options will produce the desired output. Therefore, Option D is the correct solution.

B) colnames(table)[which(colnames==’alpha’)]=’beta’

C) setnames(table,’alpha’,’beta’)

D) All of the above

Solution: (D)

All of the above options are different methods to rename the column names of a dataframe.Therefore, option D is the correct solution.

Solution : (C)

remove and rm , both can be used to clear the workspace. Therefore, option C is the correct solution.

B) filter()

C) arrange()

D) summary()

Solution: (D)

summary is a function in the R base package and not dplyr.

B) table[,3:6]

C) subset(table,select=c(‘Column3’,’Column4’,’Column5’,’Column6’))

D) All of the above

Solution: (D)

Option A, B and C are different column sub setting methods in R. Therefore, option D is the correct solution.

A) dplyr::filter(table,Column1==’Alpha’, Column4<50)

B) dplyr::filter(table,Column1==’Alpha’ & Column4<50)

C) Both of the above

D) None of the above

Solution: (C)

1. filter function in dplyr package uses “,”  and “&” to add the condition. Therefore, Option C is the correct solution.

B) table[order(-Column3,Column2),]

C) Both of the above

D) None of the above

Solution: (C)

Both order and arrange functions can be used to order the columns in R. Therefore, Option C is the correct solution.

A<-paste(“alpha”,”beta”,”gamma”,sep=” ”)

B←paste(“phi”,”theta”,”zeta”,sep=””)

parts←strsplit(c(A,B),split=” ”)

A) alpha

B) beta

C) gamma

D) phi

E) theta

F) zeta

Solution : (B)

c(A.B) would concatenate A=”alpha beta gamma” and B=”phithetazeta” separated by a white space. Upon using strsplit, the two strings will be separated at the white space between A and B into two lists. Parts[[1]][2] tells us to print the second sub element of the first element of the list which is “beta”. Therefore, option B is the correct solution.

grepl(“neeraj”,c(“dheeraj”,”Neeraj”,”neeraj”,”is”,”NEERAJ”))

A) [FALSE TRUE TRUE FALSE TRUE]

B) [FALSE TRUE TRUE FALSE FALSE]

C) [FALSE FALSE TRUE FALSE FALSE]

D) None of the above

Solution: (C)

The above command will go for the exact match of the passed argument and therefore Option C is the correct solution.

A<-c("I can use because thrice in a sentence because because is a special word.")

A) gsub(“because”,”since”,A)

B) sub(“because”,”since”,A

C) regexec(“because”,”since”,A)

D) None of the above

Solution: (A)

sub command will replace only the first occurrence in a string whereas regexec will return a list of positions of the match or -1 if no match occurs. Therefore, Option A is the correct solution.

B) unique(df)

C) dplyr::distinct(df)

D) All of the above

Solution: (D)

All the above methods are different ways of removing the duplicate rows based on both the columns. Therefore, Option D is the correct solution.

B) table::df[,mean(bar),by=foo]

C) dplyr::table%>%group_by(foo)%>%summarize(mean=mean(bar))

D) All of the above

Solution: (D)

All the above methods are used to calculate the grouped statistic of a column. Therefore, Option D is the correct solution.

B) a matrix with 3 columns and 2 rows

C) a data frame with 2 columns and 3 rows

D) a data frame with 3 columns and 2 rows

Solution: (D)

All of the above options define messy data and hence Option D is the correct solution.

Output dataframe

A) tidyr::Gather(maverick, Sex,Count,-Grade)

B) tidyr::spread(maverick, Sex,Count,-Grade

C) tidyr::collect(maverick, Sex,Count,-Grade)

D) None of the above

Solution: (A)

Spread command converts rows into columns whereas there is no collect command in tidyr or base package.

Therefore, Option A is the correct solution.

B) sub(“Delhi”,”Delhi_NCR”,C)

C) Both of the above

D) None of the above

Solution: (C)

Though sub command only replaces the first occurrence of a pattern. In this case, strings have just a single appearance of Delhi. Hence, both gsub and sub command will work in this situation. Therefore, Option C is the correct solution.

Question Context 33

Sometimes creating a feature which represents whether another variable has missing values or not can prove to be very useful for a predictive model.

Below is a dataframe which has missing values in one of its columns.

A)

dataframe$missing<-0

dataframe$Missing[is.na(dataframe$Feature2)]<-1

B)

dataframe$missing<-0

dataframe$Missing[which(is.na(dataframe$Feature2))]<-1

C) Both of the above

D) None of the above

Solution: (C)

Option C is the correct answer.

B) 12

C) 34

D) 80

Solution: (C)

all.x forces the merging to take place on the basis of A and hence will contain the same number of rows as of A. Therefore, Option C is the correct solution.

A) dim()

B) str()

C) View()

D) None of the above

Solution: (C)

View command will print the dataset to the console in a spreadsheet like format but will not help us to view the dimensions. Therefore, option C is the correct solution.

B)FALSE

C) Can’t Say

Solution: (B)

The order of arguments matter in setdiff function. Therefore, option B is the correct solution.

B) ggplot(frame,aes(A,B))+geom_bar(stat=”bin”)

C) ggplot(frame,aes(A,B))+geom_bar()

D) None of the above

Solution: (A)

stat=”identity” will ensure the values in column B become the height of the bar. Therefore, Option A is the correct solution.

B) ggplot(mtcars,aes(factor(cyl),fill=factor(vs)))+geom_bar()

C) All of the above

D) None of the above

Solution: (C)

Both options A and B will create a stacked bar chart guided by the “fill” parameter. Therefore, option C is the correct solution.

B) [1:3x y z]

C) [1x 2y 3z]

D) None of the above

Solution: (C)

B) Construct the plot on the PNG device with png(), then copy it to a PDF with dev.copy2pdf().

C) Open the PostScript device with postscript(), construct the plot, then close the device with dev.off().

D) Open the screen device with quartz(), construct the plot, and then close the device with dev.off().

Solution: (A)

The plots are first created on the screen device and then can be copied easily to a pdf file. Therefore, option A is the correct solution.