Access and preserve list names in lapply function

Question

I need to access list names inside the lapply function. I've found some threads online where it's said I should iterate through the names of the list to be able to fetch each list element name in my function:

> n = names(mylist)
> mynewlist = lapply(n, function(nameindex, mylist) { return(mylist[[nameindex]]) }, mylist)
> names(mynewlist)
NULL
> names(mynewlist) = n

The problem is that mynewlist loses the original mylist indexes and I have to add that last names() assignment to restore them.

Is there a way to give an explicit index name to each element returned by the lapply function? Or a different way to make sure mynewlist elements have the correct index names set? I feel mynewlist index names could be wrong if lapply does not return the list elements in the same order than mylist.

Answer 1

Have you looked into llply() from the package plyr?

It does exactly what you are asking for. For each element of a list, apply function, keeping results as a list. llply is equivalent to lapply except that it will preserve labels and can display a progress bar. from ?llply

mylist <- list(foo1=1:10,foo2=11:20)
>names(mylist)
[1] "foo1" "foo2"
newlist<- llply(mylist, function(x) mean(x))

>names(newlist)
[1] "foo1" "foo2"

Answer 2

imap() from the purrr package is nice for your problem.

library(purrr)
mylist <- list(foo1=1:10,foo2=11:20)
imap(mylist, function(x, y) mean(x)) ## x is the value, y is the name

or you can use a more compact version of imap:

imap(mylist, ~ mean(.x))

Note that you can use variations of imap_xxx depending on the type of vector you want:

imap_dbl(mylist, ~ mean(.x)) ## will return a named numeric vector. 

Answer 3

Building on joran's answer, and precising it:

The sapply(USE.NAMES=T) wrapper will indeed set as names of the final result the values of the vector you are iterating over (and not its names attribute like lapply), but only if these are characters.

As a result, passing indices will not help. If you want to pass indexes with sapply, you need to resort to some (ugly) casting:

sapply(as.character(c(1,11)), function(i) TEST[[as.numeric(i)]], USE.NAMES = TRUE)

In this case, a cleaner solution is to directly set and use names of your original object. Here is an exhaustive list of solutions:

TEST <- as.list(LETTERS[1:12])

### lapply ##
## Not working because no name attribute
lapply(c(1,11), function(i) TEST[[i]])

## working but cumbersome
index <- c(1,11)
names(index) <- index
lapply(index, function(i) TEST[[i]])

### sapply ##
## Not working because vector elements are not strings
sapply(c(1,11), function(i) TEST[[i]], simplify = F) 

## Working with the casting trick
sapply(as.character(c(1,11)), function(i) TEST[[as.numeric(i)]], simplify = F)

## Cleaner, using names with sapply:
names(TEST) <- LETTERS[26:15] 
sapply(names(TEST)[c(1,11)], function(name) TEST[[name]], simplify = F) 

Answer 4

I believe that lapply by default keeps the names attribute of whatever you are iterating over. When you store the names of myList in n, that vector no longer has any "names". So if you add that back in via,

names(n) <- names(myList)

and the use lapply as before, you should get the desired result.

Edit

My brains a bit foggy this morning. Here's another, perhaps more convenient, option:

sapply(n,FUN = ...,simplify = FALSE,USE.NAMES = TRUE)

I was groping about, confused that lapply didn't have a USE.NAMES argument, and then I actually looked at the code for sapply and realized I was being silly, and this was probably a better way to go.

Answer 5

the setNames function is a useful shortcut here

mylist <- list(a = TRUE, foo = LETTERS[1:3], baz = 1:5)
n <- names(mylist)
mynewlist <- lapply(setNames(n, n), function(nameindex) {mylist[[nameindex]]})

which preserves the names

> mynewlist
$a
[1] TRUE

$foo
[1] "A" "B" "C"

$baz
[1] 1 2 3 4 5