Question

1. The following R code sorts each vector in a list (lst) and stores the new list in lst1. The code uses a for loop.

   lst <- list(c1=sample(1:10,10,replace=TRUE),c2=sample(1:20,10,replace=TRUE),c3=sample(1:10,8,replace=TRUE))
   lst1 <- list()
   for(i in 1:length(lst))
   {
   lst1[[i]] <- sort(lst[[i]])
   }
   lst1

   Which of the following options does the same thing as the given R code above ?

   		lst <- list(c1=sample(1:10,10,replace=TRUE),c2=sample(1:20,10,replace=TRUE),c3=sample(1:10,8,replace=TRUE)) library(purrr) lst1 <- map(lst,sort) lst1
   		lst <- list(c1=sample(1:10,10,replace=TRUE),c2=sample(1:20,10,replace=TRUE),c3=sample(1:10,8,replace=TRUE)) library(purrr) lst1 <- map_dbl(lst,sort) lst1
   		None of the given options
   		lst <- list(c1=sample(1:10,10,replace=TRUE),c2=sample(1:20,10,replace=TRUE),c3=sample(1:10,8,replace=TRUE)) library(purrr) lst1 <- map_int(lst,sort) lst1

5 points   

QUESTION 2

1. Match the R code on the left (with loop) to its equivalent one (without loop) on the right.

   -       A.       B.       C.    V <- c(1,NA,0,4,-1,2) y <- c() for(i in 1:3) { y <- c(y,V[i]) }       -       A.       B.       C.    V <- c(1,NA,0,4,-1,2) y <- c() for(i in 1:4) { y <- c(y,V[i]) }       -       A.       B.       C.    df <- data.frame(col1 = sample(1:3,3,replace=FALSE),                  col2 = sample(1:3,3,replace=FALSE),                  col3 = sample(1:3,3,replace=FALSE)) y <- c() for(i in 1:ncol(df)) { y <- c(y,sum(df[,i])) }

   A. V <- c(1,NA,0,4,-1,2) y <- V[1:3] B. df <- data.frame(col1 = sample(1:3,3,replace=FALSE),                  col2 = sample(1:3,3,replace=FALSE),                  col3 = sample(1:3,3,replace=FALSE)) y <- colSums(df) C. V <- c(1,NA,0,4,-1,2) y <- V[1:4]

15 points   

QUESTION 3

1. Match the R code with the value of x

   (Here: try to work out the solution with hand and double check by running the code. Don't just run the code and get the answer without understanding the logic)

   -       A.       B.       C.       D.    V <- c(1,NA,0,4,-1,2) for(i in 1:3) { y <- V[i]   } x <- y       -       A.       B.       C.       D.    V <- c(1,NA,0,4,-1,2) for(i in 1:4) { y <- 1 } x <- y x       -       A.       B.       C.       D.    df <- data.frame(col1 = sample(1:3,3,replace=FALSE),                  col2 = sample(1:3,3,replace=FALSE),                  col3 = sample(1:3,3,replace=FALSE)) for(i in 1:ncol(df)) { y <- sum(df[,i]) } x <- y x       -       A.       B.       C.       D.    library(purrr) df <- data.frame(col1 = sample(1:4,4,replace=FALSE),                  col2 = sample(1:4,4,replace=FALSE),                  col3 = sample(1:4,4,replace=FALSE)) y <- map_dbl(df,max) x <- min(y) x

   A. 6 B. 0 C. 1 D. 4

20 points   

QUESTION 4

1. The following R code extracts the unique items in each vector in a list (lst) and stores the new list in lst1. The code uses a for loop.

   lst <- list(c1=sample(1:10,10,replace=TRUE),c2=sample(1:20,10,replace=TRUE),c3=sample(1:10,8,replace=TRUE))
   lst1 <- list()
   for(i in 1:length(lst))
   {
   lst1[[i]] <- unique(lst[[i]])
   }
   lst1

   Which of the following options does the same thing as the given R code above ?

   		lst <- list(c1=sample(1:10,10,replace=TRUE),c2=sample(1:20,10,replace=TRUE),c3=sample(1:10,8,replace=TRUE)) library(purrr) lst1 <- map_int(lst,unique) lst1
   		lst <- list(c1=sample(1:10,10,replace=TRUE),c2=sample(1:20,10,replace=TRUE),c3=sample(1:10,8,replace=TRUE)) library(purrr) lst1 <- map_dbl(lst,unique) lst1
   		None of the given options
   		lst <- list(c1=sample(1:10,10,replace=TRUE),c2=sample(1:20,10,replace=TRUE),c3=sample(1:10,8,replace=TRUE)) library(purrr) lst1 <- map(lst,unique) lst1

5 points   

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Answer 1

1.

answer:
lst <- list(c1=sample(1:10,10,replace=TRUE),c2=sample(1:20,10,replace=TRUE),c3=sample(1:10,8,replace=TRUE))
library(purrr)
lst1 <- map(lst,sort)
lst1

2. ANSWER:

V <- c(1,NA,0,4,-1,2)

y <- c()
for(i in 1:3)   
{
y <- c(y,V[i])
}   

  

C.

expalantion:

3.Answer :   B  

C

   A

D

20.

ANswer: D
lst <- list(c1=sample(1:10,10,replace=TRUE),c2=sample(1:20,10,replace=TRUE),c3=sample(1:10,8,replace=TRUE))
library(purrr)
lst1 <- map(lst,unique)
lst1