I want to do an apply / map function-like but using one element of the list as a operand of the calling function.
I give you better an example, let's say I want to subtract to every element of a list, the first element of the list, i.e.: x[i] = x[i] - x[1].
If I understood correctly, when calling the function with apply, the function gets the current element of the list, x[i], only as an argument, but I would need the list itself to access to the first element (x[1]) and subtract it to the current element (x[i]).
I would expect a way to provide to the applying function an extra argument or access to the list itself.
lapply can take multiple arguments.
Consider list lst,
> lst
$X1
[1] 1 2 3
$X2
[1] 4 5 6
$X3
[1] 7 8 9
$X4
[1] 10 11 12
and first as hot element.
> h <- 1
Then we provide lst without that element as first argument to iterate over, and lst[[h]] as second argument after the function. In contrast to Map the latter will always passed as a whole.
> lapply(lst[-h], \(x, y) x - y, l[[h]])
$X2
[1] 3 3 3
$X3
[1] 6 6 6
$X4
[1] 9 9 9
If you aim to replace all but the hot element, do
> lst[-h] <- lapply(lst[-h], \(x, d) x - d[[h]], lst)
> lst
$X1
[1] 1 2 3
$X2
[1] 3 3 3
$X3
[1] 6 6 6
$X4
[1] 9 9 9
Data:
> dput(lst)
list(X1 = 1:3, X2 = 4:6, X3 = 7:9, X4 = 10:12)