R Language quiz questions

R Language interview questions

  • 1.

    What will be the output of the following code ?

    > x <- list(a = 1:4, b = rnorm(10), c = rnorm(20, 1), d = rnorm(100, 5))
    > lapply(x, mean)

     

    1. $a
      [1] 2.5
      $b
      [1] 1.248845
      $c
      [1] 1.9935285
      Loop Functions 90
      $d
      [1] 5.051388
    2. $a
      [1] 2.5
      $b
      [1] 0.248845
      $c
      [1] 0.9935285
      Loop Functions 90
      $d
      [1] 5.051388
    3. $a
      [1] 3.5
      $b
      [1] 0.248845
      $c
      [1] 0.9935285
      Loop Functions 90
      $d
      [1] 5.051388
    4. None of the mentioned

    Answer
  • 2.

    What will be the output of following code ?

    > x <- list(a = 1:5, b = rnorm(10))
    > lapply(x, mean)

     

    1. $a
      [1] 3
      $b
      [1] 0.1322028
    2. $a
      [1] 4
      $b
      [1] 0.1322028
    3. $a
      [1] 5
      $b
      [1] 0.1322028
    4. Error

    Answer
  • 3.

    Body of lapply function is :

    A)

    function (X, FUN, ...)
    {
        FUN <- match.fun(FUN)
        if (!is.vector(X) || is.object(X))
           X <- as.list(X)
        .Internal(lapply(X, FUN))
    }

    B)

    function (X, FUN, ...)
    {
        FUN <- match.fun(FUN)
        if (!is.vector(X) | is.object(X))
          X <- as.list(X)
        .Internal(lapply(X, FUN))
    }

    C)

    function (X, FUN, ...)
    {
        FUN <- match.fun(FUN)
        if (is.vector(X) || is.object(X))
          X <- as.list(X)
        .Internal(lapply(X, FUN))
    }

    D) None of the mentioned

    1. A

    2. B

    3. C

    4. D

    Answer
  • 4.

    lappy functions takes _________ arguments in R language.

    1. two

    2. three

    3. four

    4. All of the mentioned

    Answer
  • 5.

    _______ is used to apply a function over subsets of a vector.

    1. apply()

    2. lapply()

    3. tapply()

    4. mapply()

    Answer
  • 6.

    ________ applies a function over the margins of an array.

    1. apply()

    2. lapply()

    3. tapply()

    4. mapply()

    Answer
  • 7.

    Point out the correct statement :

    1. lapply() takes elements of the list and passes them as the first argument of the function you are applying

    2. You can use lapply() to evaluate a function multiple times each with a different argument

    3. Functions that you pass to lapply() may have other arguments

    4. None of the mentioned

    Answer
  • 8.

    Which of the following is multivariate version of lapply ?

    1. apply()

    2. lapply()

    3. sapply()

    4. mapply()

    Answer
  • 9.

    __________ function is same as lapply in R

    1. apply()

    2. lapply()

    3. sapply()

    4. mapply()

    Answer
  • 10.

    Point out the wrong statement :

    1. Multi-line expressions with curly braces are just not that easy to sort through when working on the command line

    2. lappy() loops over a list, iterating over each element in that list

    3. lapply() does not always returns a list

    4. All of the mentioned

    Answer
  • 11.

    ________ loop over a list and evaluate a function on each element

    1. apply()

    2. lapply()

    3. sapply()

    4. mapply()

    Answer
  • 12.

    What will be the output of following code ?

    > g <- function(x) {
    +               a <- 3
    +               x+a+y
    +             ## 'y' is a free variable
    + }
    > y <- 3
    > g(2)

     

    1. 9

    2. 42

    3. 8

    4. Error

    Answer
  • 13.

    The _________ function is used to plot negative likelihood.

    1. plot()

    2. graph()

    3. graph.plot()

    4. None of the mentioned

    Answer
  • 14.

    What will be the output of the following code ?

    > nLL <- make.NegLogLik(normals, c(1, FALSE))
    > optimize(nLL, c(1e-6, 10))$minimum

     

    1. 1.217775

    2. 1.800596

    3. 3.73424

    4. empty

    Answer
  • 15.

    What will be the output of the following code ?

    > nLL <- make.NegLogLik(normals, c(FALSE, 2))
    > optimize(nLL, c(-1, 3))$minimum

     

    1. 1.217775

    2. 2.217775

    3. 3

    4. empty

    Answer
  • 16.

    What will be the output of the following code ?

    function(p) {
              params[!fixed] <- p
              mu <- params[1]
              sigma <- params[2]
    ## Calculate the Normal density
              a <- -0.5*length(data)*log(2*pi*sigma^2)
              b <- -0.5*sum((data-mu)^2) / (sigma^2)
              -(a + b)
    }
    > ls(environment(nLL))

     

    1. “data” “fixed” “param”

    2. “data” “variable” “params”

    3. “data” “fixed” “params”

    4. All of the mentioned

    Answer
  • 17.

    Point out the correct statement :

    1. An environment is a collection of (symbol, value) pairs, i.e. x is a symbol and 3.14 might be its value

    2. If the value of a symbol is not found in the environment in which a function was defined, then the search is continued in the child environment

    3. After the top-level environment, the search continues down the search list until we hit the parent environment

    4. None of the mentioned

    Answer
  • 18.

    _________ require you to pass a function whose argument is a vector of parameters

    1. optimize()

    2. optimise()

    3. opt()

    4. All of the mentioned

    Answer
  • 19.

    Which of the following language supports lexical scoping ?

    1. Perl

    2. Python

    3. Common Lisp

    4. All of the mentioned

    Answer
  • 20.

    Point out the wrong statement :

    1. Dynamic scoping turns out to be particularly useful for simplifying statistical computations

    2. Lexical scoping turns out to be particularly useful for simplifying statistical computations

    3. The scoping rules of a language determine how values are assigned to free variables

    4. All of the mentioned

    Answer

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