My understanding is that iasc and rank works internally in below order:
iasc: step by step procedure till rank
original: 2 7 3 2 5 / 0->2, 1->7, 2->3, 3->2, 4->5 //Index item mapping
asc original returns 2 2 3 5 7 / 0->2, 1->2, 2->3, 3->5, 4->7 // Index item mapping
iasc original returns 0 3 2 4 1 / 0->0, 1->3, 2->2, 3->4, 5->1 // Index item mapping /// iasc using asc internally
asc iasc original returns 0 1 2 3 4 / 0->0, 1->1, 2->2, 3->3, 4->4 // Index item mapping
iasc iasc original returns 0 4 2 1 3 // hence it is equal to rank original
rank:
original: 2 7 3 2 5 / 0->2, 1->7, 2->3, 3->2, 4->5 //Index item mapping
asc original returns 2 2 3 5 7 / 0->2, 1->2, 2->3, 3->5, 4->7 // Index item mapping
rank original -> 0 4 2 1 3 / rank of orignal items of list in sorted list /// rank using asc internally
But when I see the code of iasc and rank, both of them uses rank internally. Can you please tell me how iasc and rank works internally(is my understanding not correct)?
You are correct in saying iasc iasc is equivalent to using rank. If you look inside the function iasc and rank you can see that they are very similar:
q)rank
k){$[0h>@x;'`rank;<<x]}
q)iasc
k){$[0h>@x;'`rank;<x]}
In this case the k code < will do the same job as iasc, so rank is essentially using iasc iasc internally, since << is the k code for iasc iasc.
When you say that they both use rank internally, you may be referring to '`rank, which is used to throw an error if anything other than a list is used, since the function uses an if-else to determine if the input is a list.