what is `[[` looking for in this sapply example?

Question

A followup to this How to use `[[` and `$` as a function? question: I started playing a bit with the original setup (reduced the size from 10000 to 3 for simplicity)

JSON <- rep(list(x,y),3)
x <- list(a=1, b=1)
y <- list(a=1)
JSON <- rep(list(x,y),3)
sapply(JSON, "[[", "a")
[1] 1 1 1 1 1 1
sapply(JSON,"[[",'b')
[[1]]
[1] 1

[[2]]
NULL

[[3]]
[1] 1

[[4]]
NULL

[[5]]
[1] 1

[[6]]
NULL

sapply(JSON,'[[',1)
[1] 1 1 1 1 1 1
sapply(JSON,'[[',2)
Error in FUN(X[[2L]], ...) : subscript out of bounds

That I think I understand -- searching for "b" is different from demanding the existence of a second element. But then, I created a deeper list:

NOSJ<-rep(list(JSON),3)

sapply(NOSJ,'[[',1)
  [,1] [,2] [,3]
a 1    1    1   
b 1    1    1   
sapply(NOSJ,'[[',2)
$a
[1] 1

$a
[1] 1

$a
[1] 1

And now my head's hurting. Can someone expand on what [[ (or its sapply method) is doing here?

Answer 1

You could think of sapply and lapply as a for-loop that operates on seq_along(NOSJ) as an index vector.

 for( i in seq_along(NOSJ) NOSJ[[i]]  .... then use "[[" with the 3rd argument 

So the first and second results would be:

> NOSJ[[1]][[1]]
$a
[1] 1

$b
[1] 1

> NOSJ[[2]][[1]]
$a
[1] 1

$b
[1] 1

The difference between sapply and lapply is that sapply attempts to use simply2array to return a matrix or array if the dimensions of the returned values are all the same (as they are in this case when using 1, 3 or 5 as the 3rd argument. Quite honestly I do not know why using 2,4,or 6 as the third argument does not return an atomic vector. I thought it should.

Answer 2

sapply(NOSJ,'[[',1) returns the first list element of each of the lists passed to [[ by sapply from NOSJ. Try...

sapply( NOSJ , length )
[1] 6 6 6

Makes sense right? So [[ is operating on the second level lists, the first element of which always contain only a and b so are coercible to a matrix. The second element of those lists of 6, always contain only a.