Why `lapply` returns result of assignment automatically? - r

Question

q <- lapply(1:3, function(x) x ** 2)
## returns nothing, because it is an assignment
# however, how you explain this?:
> lapply(list(1:3, 4:6, 7:9, 10:11), function(v) q <- lapply(v, function(x) x ** 2))
[[1]]
[[1]][[1]]
[1] 1
[[1]][[2]]
[1] 4
[[1]][[3]]
[1] 9
[[2]]
[[2]][[1]]
[1] 16
[[2]][[2]]
[1] 25
[[2]][[3]]
[1] 36
[[3]]
[[3]][[1]]
[1] 49
[[3]][[2]]
[1] 64
[[3]][[3]]
[1] 81
[[4]]
[[4]][[1]]
[1] 100
[[4]][[2]]
[1] 121
# while this gives the same but is logical (q is stated as return value).
> lapply(list(1:3, 4:6, 7:9, 10:11), function(v) {q <- lapply(v, function(x) x ** 2);q})
[[1]]
[[1]][[1]]
[1] 1
[[1]][[2]]
[1] 4
[[1]][[3]]
[1] 9
[[2]]
[[2]][[1]]
[1] 16
[[2]][[2]]
[1] 25
[[2]][[3]]
[1] 36
[[3]]
[[3]][[1]]
[1] 49
[[3]][[2]]
[1] 64
[[3]][[3]]
[1] 81
[[4]]
[[4]][[1]]
[1] 100
[[4]][[2]]
[1] 121
why in the second expression, although the inner lapply is just assigned to q but q not called at end of function, the value of the assignment
is returned to the outer lapply and thus collected?
Please, anybody has an explanation for this phenomenon?
It also works with =
lapply(list(1:3, 4:6, 7:9, 10:11), function(v) q = lapply(c(v), function(x) x ** 2))
[[1]]
[[1]][[1]]
[1] 1
[[1]][[2]]
[1] 4
[[1]][[3]]
[1] 9
[[2]]
[[2]][[1]]
[1] 16
[[2]][[2]]
[1] 25
[[2]][[3]]
[1] 36
[[3]]
[[3]][[1]]
[1] 49
[[3]][[2]]
[1] 64
[[3]][[3]]
[1] 81
[[4]]
[[4]][[1]]
[1] 100
[[4]][[2]]
[1] 121

Answer 1

The answer lies in the return value of an assignment operation. The assignment operator <- not only writes a value to a variable in the calling environment, it actually invisibly returns the assigned value itself to the caller.
Remember all operations in R are actually functions. When you do
x <- 3
You are actually doing
`<-`(x, 3)
Which not only creates the symbol "x" in the calling environment and assigns the value 3 to that symbol, but invisibly returns the value 3 to the caller. To see this, consider:
y <- 2
y
#> [1] 2
y <- `<-`(x, 3)
y
#> [1] 3
Or equivalently,
y <- (x <- 4)
y
#> [1] 4
And in fact, because of R's order of evaluation, we can even do:
y <- x <- 5
y
#> [1] 5
Which is a neat way of setting multiple variables to the same value on the same line.
Now consider the lambda function you use inside your lapply:
function(v) q <- lapply(v, function(x) x ** 2)
Look what happens when we consider this function as a stand-alone:
func <- function(v) q <- lapply(v, function(x) x ** 2)
func(1:3)
As predicted, nothing happens. But what happens when we do:
a <- func(1:3)
If func(1:3) doesn't return anything, then presumably a should be empty now.
But it isn't...
a
#> [[1]]
#> [1] 1
#>
#> [[2]]
#> [1] 4
#>
#> [[3]]
#> [1] 9
Because the value of assignation was returned to the caller invisibly, we were able to assign it to a value in the calling scope. Therefore, doing
lapply(list(1:3, 4:6, 7:9, 10:11), function(v) q <- lapply(v, function(x) x ** 2))
assigns the value of your inner function applied to all the list elements to a new list. This list is not returned invisibly, but just returned as normal.
So this is expected behaviour.