ch_27_solutions

Prerequisites:

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27.2.4 Exercises:

1. All 3 parts below:

   even_position <- function(vector) {
     vector[seq(2, length(vector), 2)]
   }

   cut_last_element <- function(vector) {
     vector[-length(vector)]
   }

   even_position_no_missing <- function(vector) {
     vector[(vector %% 2 == 0 & !is.na(vector))]
   }

2. the which function only checks if true and does not coerce the vector to logical. This matters because the base R solution coerces NaN to the logical NA but the which function is able to return NaN.

   vector_x <- c(10, NA, NaN)

   vector_x[-which(vector_x > 0)]
   ## [1]  NA NaN
   vector_x[vector_x <= 0]
   ## [1] NA NA

27.3.4 Exercises:

1. Subsetting with [[ with a positive integer out of bounds returns an error. Similarly, subsetting with [[ on a name that doesn’t exist also returns an error. This is sensible because you are trying to extract a component of a vector which doesn’t exist.

   • Can see in the example below that before drilling down into the component, the single bracket for names and integers out of range returns a sublist without a name, so consequently the double brackets have nothing to extract.

   vector_y <- c(n1 = 'red', n2 = 'blue', n3 = 'green', n4= 'yellow')
   
   str(vector_y[4])
   ##  Named chr "yellow"
   ##  - attr(*, "names")= chr "n4"
   str(vector_y[5])
   ##  Named chr NA
   ##  - attr(*, "names")= chr NA
   
   str(vector_y['n4'])
   ##  Named chr "yellow"
   ##  - attr(*, "names")= chr "n4"
   str(vector_y['n5'])
   ##  Named chr NA
   ##  - attr(*, "names")= chr NA

2. It depends on the type of the “pepper packet”.

   • If the packet is a list, than [ returns a list and [[ returns the component of the list.

   • If the packet is a vector, than [ returns the first element and [[ also returns the element (since there is no outer structure to a vector).

   pepper_1 <- list(
     packet = list(1,2)
   )
   
   pepper_2 <- list(
     packet = c(1,2)
   )
   
   str(pepper_1[[1]][1])
   ## List of 1
   ##  $ : num 1
   str(pepper_1[[1]][[1]])
   ##  num 1
   
   str(pepper_2[[1]][1])
   ##  num 1
   str(pepper_2[[1]][[1]])
   ##  num 1