zipping lists in R

Question

As a guideline I prefer apply functions on elements of a list using lapply or *ply (from plyr) rather than explicitly iterating through them. However, this works well when I have to process one list at a time. When the function takes multiple arguments, I usually do a cycle.

I was wondering if it's possible to have a cleaner construct, still functional in nature. One possible approach could be to define a function similar to Python, zip(x,y), which takes the input lists, and returns a list, whose i-th element is list(x, y), and then apply the function to this list. But my question is whether I am using the cleanest approach or not. I am not worried about performance optimization, but rather clarity/elegance.

Below is the naive example.

        A <- as.list(0:9)
        B <- as.list(0:9)
        f <- function(x, y) x^2+y

        OUT <- list()
        for (n in 1:10) OUT[[n]] <- f(A[[n]], B[[n]])
        OUT
        [[1]]
        [1] 0

        [[2]]
        [1] 2

        ...

And here is the zipped example (which could be extended to arbitrary arguments):

zip <- function(x, y){
    stopifnot(length(x)==length(y))
    z <- list()
    for (i in seq_along(x)){
        z[[i]] <- list(x[[i]], y[[i]]) 
    }
    z
}
E <- zip(A, B)

lapply(E, function(x) f(x[[1]], x[[2]]))

[[1]]
[1] 0

[[2]]
[1] 2

 ...

score 29 · Accepted Answer · edited Jun 04 '21 at 15:22

29

I think you're looking for mapply:

   ‘mapply’ is a multivariate version of ‘sapply’.  ‘mapply’ applies
     ‘FUN’ to the first elements of each ...  argument, the second
     elements, the third elements, and so on.  Arguments are recycled
     if necessary.

For your example, use mapply(f, A, B)

edited Jun 04 '21 at 15:22

Josiah Yoder

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answered May 26 '11 at 22:25

Ben Bolker

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3

There is also `plyr` family of `m*ply` functions. – Marek May 27 '11 at 08:28
2

And Map which is a thin wrapper around mapply – hadley May 27 '11 at 22:22

score 4 · Answer 2 · edited Jun 04 '21 at 15:24

4

I came across a similar problem today. And after learning the usage of the func mapply, I know how to solve it now.

mapply is so cool!!

Here is an examples:

en = c("cattle", "chicken", "pig")
zh = c("牛",      "鸡",      "猪")

dict <- new.env(hash = TRUE)
Add <- function(key, val) dict[[key]] <- val

mapply(Add, en, zh)
##  cattle chicken     pig 
##    "牛"    "鸡"    "猪"

edited Jun 04 '21 at 15:24

Josiah Yoder

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answered Jan 23 '19 at 10:25

D. Chern Liang

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score 1 · Answer 3 · answered May 26 '11 at 22:31

I think you could do this with what I call an 'implicit loop' (this name does not hit it fully, but whatever), taking into account that you can loop over vectors within *apply:

OUT <- lapply(1:10, function(x) (A[[x]]^2 + B[[x]]))

or

OUT <- lapply(1:10, function(x) f(A[[x]], B[[x]]))

Note that you then could also use vapply (or 'sapply`) for output managing (i.e. if you don't want a list).

(by the way, I am not getting what you want with the zip function, so I am sorry, if I missed your point.)

zipping lists in R

3 Answers3