CODEWITHSUNDEEP
r
smykes
smykes

Reputation: 312

Which with vectors sometimes works - R

Why does this example work:

which(letters %in% c('j', 'e', 'f', 'f', 'r', 'e', 'y'))

However; this one does not?

name <- c(strsplit("jeffrey", ""))
which(letters %in% name)

Isn't this the exact same thing since in both instance the second argument in the which function is a vector?

Upvotes: 2

Views: 38

Answers (2)

akrun
akrun

Reputation: 887691

The issue is that the strsplit is still a list of length 1 with the first element a vector

strsplit("jeffrey", "")
#[[1]]
#[1] "j" "e" "f" "f" "r" "e" "y"

Wrapping with c is not going to change the scenario as by default recursive = FALSE.

c(strsplit("jeffrey", ""))
#[[1]]
#[1] "j" "e" "f" "f" "r" "e" "y"

Changing the recursive = TRUE will make it possible to convert the list to a vector

c(strsplit("jeffrey", ""), recursive = TRUE)
#[1] "j" "e" "f" "f" "r" "e" "y"

If we use unlist (as in @JohnyCrunch's solution), it unlists the list because by default recursive = TRUE and convert to vector. In our case, another approach would be to extract the list element with [[ (as it is only a list of length 1.

name <- strsplit("jeffrey", "")[[1]]
which(letters %in% name)
#[1]  5  6 10 18 25

Upvotes: 4

LocoGris
LocoGris

Reputation: 4480

If you run class(name) you will note that it is a list. Use unlist()to solve the problem:

name <- unlist(strsplit("jeffrey", ""))
which(letters %in% name)

Best!

Upvotes: 3

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