I am trying to understand the behavior of purrr::map() vs purrr::pmap() for the following use case. I was expecting the same result, but looks like purrr::map() iterates only once through the list.
# using pmap
pmap(.l = list(x = c(1,6)) ,.f = function(x) {substr("costcopull",x,x-1+5)})
# result
[[1]]
[1] "costc"
[[2]]
[1] "opull"
# using map
map(.x = list(x = c(1,6)) ,.f = function(x) {substr("costcopull",x,x-1+5)})
$x
[1] "costc"
I was expecting both the results to be the same, as the function has the singular input "x"
But they aren't.
map applies a function to each element of a vector / list.
In your example, the list you provide as argument has only one element : c(1,6)
As substr isn't a vectorized function, it will only use the first index of the vector : 1
x <- c(1,6)
substr("abcdef",x,1)
[1] "a"
pmap is especially useful for dataframes (which are lists of vectors) to process columns in parallel row by row:
df <- data.frame(x = c(1, 3), y = c(2, 4))
dput(df)
#> structure(list(x = c(1, 3), y = c(2, 4)), class = "data.frame",...)
pmap(df, \(x,y) paste(x,y))
#> [[1]]
#> [1] "1 2"
#>
#> [[2]]
#> [1] "3 4"
When you give a single element list as argument, pmap browses this single column row by row :
pmap(.l = list(x = c(1,3)) ,.f = function(x) {x})
#[[1]]
#[1] 1
#[[2]]
#[1] 3
This is what happens in your example.
To sum up, as pointed out by @Darren Tsai, inputing directly c(1,6) to map seems to be what you're looking for, you don't need to put this vector into a list.
In this case, the parallel column processing capability of pmap doesn't seem to be needed.