How can I aggregate elements of list of lists to get one vector?

Question

I have list of lists (same lengths: n). How can I concatenate them, to get one vector of the same input length (n) ? For example I have:

[[1]] 
[1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
[[2]] 
[1] "a" "a" NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
[[3]]
[1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA "d" "e" NA NA NA

I want to get:

"a" "a" NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA "d" "e" NA NA NA

I have tried:

Reduce('union',lapply(l,function(x){x$AB}))

But I got:

"a" NA "d" "e"

I want to get a vector with same length as the input l[[1]]$AB. Any ideas?

Could you share you input data with `dput(x$AB)` (if `x$AB` is the name of your data object) ? That would help and clarify if your object is really a nested list. — cbo, Mar 12 '20 at 16:37

iod · Answer 1 · 2020-03-12T16:59:30.260

I believe you're looking for dplyr's coalesce:

library(dplyr)
mylist<-lapply(mylist,as,"character")
do.call(coalesce,mylist)
[1] "a" "a" NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  "d" "e" NA  NA  NA

NOTE: Using your exact example, the first element on the list is automatically interpreted as logical class because it's all NAs. I added row 2 in the code above to get around that by coercing all rows in the list into a character class.

akrun · Answer 2 · 2020-03-12T16:42:04.520

1

We can use pmax

do.call(pmax, c(l, na.rm = TRUE))
#[1] "a" "a" NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  "d" "e" NA  NA  NA

data

l <- list(c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA), c("a", "a", NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA), c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, "d", "e", NA, NA, NA))

edited Mar 12 '20 at 16:42

answered Mar 12 '20 at 16:24

akrun

874,273
37
540
662

Your solution `do.call(pmax, c(list_obj, na.rm = TRUE))` posted [here](https://stackoverflow.com/a/29634492/8245406) works. I am closing as duplicate. – Rui Barradas Mar 12 '20 at 16:40
I have upvoted the one in the link, believing that it's the right place to do it. But I can reopen/upvote this one/close :(. – Rui Barradas Mar 12 '20 at 17:41
It is a dupe but the question is not very clear especially wiith the `x$AB` – akrun Mar 12 '20 at 17:42

jay.sf · Answer 3 · 2020-03-12T16:39:07.083

You could coerce list to a data frame and use ifelse case handling like this:

apply(as.data.frame(l), 1, function(x) ifelse(all(is.na(x)), NA, x[!is.na(x)]))
# [1] "a" "a" NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA  NA 
# [17] NA  NA  "d" "e" NA  NA  NA

Data

l <- list(c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA), c("a", "a", NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA), c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, "d", "e", NA, NA, NA))

score 0 · Accepted Answer · answered Mar 12 '20 at 16:44

You could also perform row operation with conversion to a data.frame :

list_obj <- list(rep(NA, 20), c("a", "a", rep(NA, 18)), c(rep(NA,16), "d", "e", NA, NA))
df_l <- as.data.frame(list_obj)
apply(df_l, 1, FUN = function(x) paste(ifelse(is.na(x), "", x), collapse = ""))
#>  [1] "a" "a" ""  ""  ""  ""  ""  ""  ""  ""  ""  ""  ""  ""  ""  ""  "d" "e" "" 
#> [20] ""

How can I aggregate elements of list of lists to get one vector?

4 Answers4

data