What is `` [[`looking for in this sample example?

Question

Then follow How to use `[` and `$` as a function? question: I started playing a little with the initial setting (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 

I think I understand - the search for "b" is different from the requirement for the second element to exist. 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 hurts. Can someone expand on what [[ (or his sapply method) does here?

Answer 1

You can think of sapply and lapply as a loop that works in 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 will be as follows:

 > 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 tries to use simply2array to return the matrix or array if the sizes of the returned values ​​are the same (as in this case when using 1 , 3 or 5 as the third argument. Honestly, I don’t know why using 2,4 or 6, since the third argument does not return an atomic vector, I thought it should be.

Answer 2

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

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

Does it make sense right? Thus, [[ works on second-level lists, the first element of which always contains only a and b , so they can be copied to the matrix. The second element of these lists from 6 always contains only a .