Elijah's reply offers lots of ideas, with thoughts about filling histogram
results.
However, this might also be of interest:
xs NB. a small example
1 2 4 6 7
4 6 3 8 5
expand NB. I think this is included in J at startup
#^:_1
expxs =: {{ix,:y expand~x e.~ix=. >:i. {:x}}/
expxs xs
1 2 3 4 5 6 7
4 6 0 3 0 8 5
This follows your apparent requirement to use origin 1. It would need refining
if you
wished the expanded x to start with {.x rather than 1 .
Cheers,
Mike
Sent from my iPad
> On 24 Jul 2023, at 03:03, Fr. Daniel Gregoire <daniel.l.grego...@gmail.com>
> wrote:
>
> Hi!
>
> (Apologies if this is a double-post, I believe I sent it before I was
> officially on the mailing list, and I don't see it in the archives).
>
> Given:
>
> |:xs
>
> 1 23
> 2 4
> 3 5
> 4 10
> 5 397
> 6 3
> 7 190
> 8 44
> 9 4
> 10 5
> 11 13
> 12 1011
> 13 10
> 14 1119
> 15 72
> 16 1
> 17 1
> 19 6
> 21 3
> 22 2
> 23 1
> 26 2
> 28 2
> 29 2
> 30 1
> 31 2
>
>
> I'd like $xs to be 2 31 but you can see that 18, 20, 24, 25, and 27
> are "missing".
>
>
> I'd like to "fill the holes" so the first row of xs is all the
> integers 1 through 31 sequentially, and to put corresponding zeros in
> the second row where I've filled the holes in the first.
>
>
> My thoughts have centered around using (1+i.31),.31#0 to have a 2 by
> 31 array with all zeros in the second row, and then shifting my
> original to find the non-sequential parts with something like:
>
>
> (1|.{.xs) - {.xs
>
>
> But then I'm struggling to reason about a non-loopy way to put it all
> together. Any help is greatly appreciated.
>
>
> Kind regards,
>
> Daniel
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