Thanks Elijah, Mike, and Pascal!
Elijah: I'd forgotten the idiom of putting one's default value at the end
of an array and using it as the "not found" value, so that's good to have
reinforced, thank you. Your insight about the data's origins and your last
solution hit the nail on the head, and that's the approach I went with,
since it comes across less as "clean up the mess" and more as "clearly
establish your domain."
Mike: Yep, expand is defined as that in my uncustomized J installation:
JVERSION
Engine: j903/j64/darwin
Release-b: commercial/2022-01-28T04:07:40
Library: 9.03.08
Qt IDE: 1.9.5/5.15.2(5.15.2)
Platform: Darwin 64
Installer: J903 install
InstallPath: /users/dlg/j903
Contact: www.jsoftware.com
-Daniel
On Sun, Jul 23, 2023 at 10:36 PM Elijah Stone <elro...@elronnd.net> wrote:
> Starting with
>
> 'k v'=. xs
> n=. >./k
> k2=. 1+i.n
>
> For the second row, the first approach that comes to mind is to treat it
> as an
> association:
>
> v2=. (k i. k2) { v , 0
>
> Then recombine as k2 ,: v2. An alternative, slightly less nice, is to add
> the
> requisite number of zeroes onto the end and then sort:
>
> missing=. k2 -. k
> v2=. (k , missing) /: (v , (#missing)#0)
>
> Or recombine before sorting:
>
> xs2=. /:~&.|: (k , missing) ,: (v , (#missing)#0)
>
> Your idea to combine with an existing bed of zeroes is workable too; don't
> use
> take and drop, though, but amend:
>
> v2=. v (k-1)} n#0
>
> (The k-1 being necessary only because your indices start at 1.)
>
> Or, use a complex left argument to # to pad the result with zeroes in the
> appropriate places:
>
> v2=. (1 j. _1 + 2 -~/\ 0,k) # v
>
> Though this has one pitfall: it doesn't work if there is a gap at the
> _beginning_ of k (e.g., the first element is 2 rather than 1), because the
> padding is added after the replicated element.
>
> How did you end up with this array in the first place? It looks a bit
> like a
> histogram showing how frequently some event happened on each day of the
> month.
> Assuming so, there is a trick to avoid the problem in the first place.
> Consider the following example histogram (only going up to 7 for brevity,
> and
> starting from zero rather than one out of sensibility):
>
> a=. 0 0 0 1 1 3 6 6 6 6
> #/.~ a
> 3 2 1 4
>
> Here 2, 4, and 5 didn't show up in the histogram, even though we would
> like to
> know that they appeared zero times. The solution is to add one of each
> element we are interested in, and then subtract it after:
>
> _1 + #/.~ (i.7) , a
> 3 2 0 1 0 0 4
>
> Prepending i.7 rather than appending it also ensures that the result
> histogram
> comes out in order, even if a happened not to be.
>
> -E
>
> Sun, 23 Jul 2023, Fr. Daniel Gregoire 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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> ----------------------------------------------------------------------
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