Using nested lists as input to non-vectorized a function

Question

I need to use nested elements in a list for a query. For example:

sha_list <- list(c("ed76504696469470dcbf", "164f798524fd6904236a"),
c("669dfeccad88cd4348f7", "af70a76691aacf05c1bb"))

This is a list which has the following structure:

[[1]]
[1] "ed76504696469470dcbf" "164f798524fd6904236a"

[[2]]
[1] "669dfeccad88cd4348f7" "af70a76691aacf05c1bb"

Here's what I've tried so far:

library("devtools")
lapply(sha_list, source_gist)

this gives:

Error: length(id) == 1 is not TRUE

The expected output would be a list with two elements in each of the two entries of the list:

[[1]]
[1] gist1
[2] gist2

[[2]]
[1] gist3
[2] gist4

What I need to do here is to iterate across this list to execute a function (which is non-vectorized, so it can only take one of the elements as an argument at a time). I was hoping that this would work, since I want to recursively use each item in each list:

rapply(sha_list, source_gist, how = "list")

This doesn't work either.

unlist() will not work, because I need to maintain the structure of the list (i.e. the output needs to be structured in the same way, as a list).

Or lapply(sha_list, function(x) lapply(x, source_gist)) to maintain the structure of the list — maccruiskeen
– maccruiskeen, Commented Jul 24, 2015 at 20:07
@choff - The structure of the list will remain unchanged in my call, so long as you do not re-assign unlist(sha_list) to sha_list — Rich Scriven
– Rich Scriven, Commented Jul 24, 2015 at 20:09
@RichardScriven But it will change the length of the list to 4, no? — maccruiskeen
– maccruiskeen, Commented Jul 24, 2015 at 20:11
@histelheim, what output are you expecting? You are sourcing gists here, which may become complicated in the manor you are describing. I would suggest one gist in each element of a four element list — Rich Scriven
– Rich Scriven, Commented Jul 24, 2015 at 20:14

maccruiskeen · Accepted Answer · 2015-07-25 13:27:39Z

2

You can do:

library("devtools")
sha_list <- list(c("ed76504696469470dcbf", "164f798524fd6904236a"),
                 c("669dfeccad88cd4348f7", "af70a76691aacf05c1bb"))
my.gists <- lapply(sha_list, function(x) lapply(x, source_gist))

Then, for example, the gist associated with sha_list[[1]][2] can be accessed with my.gists[[1]][[2]].

answered Jul 25, 2015 at 13:27

maccruiskeen

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