> In the following example, collection 0 is 10 30 20 20 and has ranks 3 0
> 1 1, while collection 1 is 200 100 300 and has ranks 1 2 0. The problem
> is the extra 0 appended to make them the same size, [...]
The usual way to avoid those extra fills is to box the interim values
and to flatten afterwards. This is what your left tine of your
rankbykey2 is doing, but with some unnecessarily convoluted
"pre-boxing - pre-de-boxing - post-re-boxing" sequence.
> rankbykey2=: ( ;@((i.~\:~)&.>@</.) /: ;@([</....@#@]) )
Developing it directly from your innitial code
> 0 1 1 1 0 0 0 (i.~\:~)/. 10 200 100 300 30 20 20
> 3 0 1 1
> 1 2 0 0
in the naive way I come to:
0 1 1 1 0 0 0 <@(i.~\:~)/. 10 200 100 300 30 20 20
+-------+-----+
|3 0 1 1|1 2 0|
+-------+-----+
0 1 1 1 0 0 0 ;@(<@(i.~\:~)/.) 10 200 100 300 30 20 20
3 0 1 1 1 2 0
> [...] and the rearranging
> that happens. I want the output to be 3 1 2 0 0 1 1, so that it
> corresponds to the original right-side input.
For that step, I can offer a slight variation, too. Your right tine in
> rankbykey2=: ( ;@((i.~\:~)&.>@</.) /: ;@([</....@#@]) )
reshuffles the index vector 0 1 2 3 ... N according to the key sets:
0 1 1 1 0 0 0 ;@([ </. i...@#@]) 10 200 100 300 30 20 20
0 4 5 6 1 2 3
With these values, there is no difference between simple indexing and
re-grading your rank values:
0 1 1 1 0 0 0 (;@(<@(i.~\:~)/.) /: ;@([</....@#@])) 10 200 100 300 30 20 20
3 1 2 0 0 1 1
0 1 1 1 0 0 0 (;@(<@(i.~\:~)/.) {~ ;@([</....@#@])) 10 200 100 300 30 20 20
3 1 2 0 0 1 1
My idea was to re-grade the rank values a bit more directly on the key:
0 1 1 1 0 0 0 (;@(<@(i.~\:~)/.) /: [) 10 200 100 300 30 20 20
3 1 2 0 0 1 1
works already nicely, but just because the (two) distinct key groups
happen to be introduced in ascending order. (Perhaps your keys
*ARE* always of this nature?) In the general case, self-indexing will
help:
i.~ 8 4 4 4 8 8 8
0 1 1 1 0 0 0
8 4 4 4 8 8 8 (;@(<@(i.~\:~)/.) /: i...@[) 10 200 100 300 30 20 20
3 1 2 0 0 1 1
All in all, this is not very different from what you did. Slighltly
less boxing/unboxing. Can you measure any difference?
Martin Neitzel
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