How to add a column to lists within a list without losing their names?

Question

I did several attempts to add a specific column to data frames resp. lists within a list, but all *apply() attempts failed to preserve the names of the data frames.

For example for list l,

l <- list(alpha=data.frame(1:3), bravo=data.frame(4:6), charly=data.frame(7:9))

> l
$`alpha`
  X1.3
1    1
2    2
3    3

$bravo
  X4.6
1    4
2    5
3    6

$charly
  X7.9
1    7
2    8
3    9

I want the initial letters of the lists' names as a second id column. I tried these attempts who give me basically what I want:

lapply(seq_along(l), function(x) cbind(l[[x]], id=substr(names(l)[x], 1, 1)))
# or
lapply(seq_along(l), function(x) data.frame(l[[x]], id=substr(names(l)[x], 1, 1)))
# [[1]]
# X1.3 id
# 1    1  a
# 2    2  a
# 3    3  a
# 
# [[2]]
# X4.6 id
# 1    4  b
# 2    5  b
# 3    6  b
# 
# [[3]]
# X7.9 id
# 1    7  c
# 2    8  c
# 3    9  c

but the inner lists have lost their names. Option USE.NAMES=TRUE from lapply() documentation didn't work.

I also tried these two attempts, but they failed even worse.

lapply(seq_along(l), function(x) mapply(cbind, l[[x]], id=substr(names(l)[x], 1, 1), 
                                        SIMPLIFY=FALSE))
rapply(l, function(x) cbind(x, id=substr(names(l)[x], 1, 1)), how="list")

I know I could do this like so:

l1 <- lapply(seq_along(l), function(x) cbind(l[[x]], id=substr(names(l)[x], 1, 1)))
names(l1) <- names(l)

or do a for loop:

for(i in seq_along(l)) {
  l[[i]] <- data.frame(l[[i]], id=substr(names(l)[i], 1, 1))
}

but I'd like to know whether an *apply() solution could be improved to bring the expected output, which would be:

$`alpha`
  X1.3 id
1    1  a
2    2  a
3    3  a

$bravo
  X4.6 id
1    4  b
2    5  b
3    6  b

$charly
  X7.9 id
1    7  c
2    8  c
3    9  c

Answer 1

Try Map

Map(`[<-`, l, "id", value = substr(names(l), 1, 1))
#$alpha
#  X1.3 id
#1    1  a
#2    2  a
#3    3  a

#$bravo
#  X4.6 id
#1    4  b
#2    5  b
#3    6  b

#$charly
#  X7.9 id
#1    7  c
#2    8  c
#3    9  c

The first argument is a function. Map then applies the function "to the first elements of each ... argument, the second elements, the third elements, and so on.", see ?mapply.