If elegant == short-tacit then Aai's way seems to be the best path to take.
Otherwise, `catalogue` presents some nice options, like:
(=/S:0 { ({:"1 A);({."1 B)) # (;S:1 { <"1&.>A;}."1 B)
On Mon, Oct 15, 2012 at 8:21 PM, Raul Miller <rauldmil...@gmail.com> wrote:
> Ok, thank you! I am going to blame my confusion about {: vs. {. on my
> browser font, where both : and . get only a single pixel per dot on my
> laptop's screen. I can see the difference if I focus on it, but I
> should have copied those phrases rather than retyping them.
>
> That said... note that the asymmetric treatment of A and B ({: vs. {.
> and the dropping of one column of B and the reordering of the
> representation of B in the result) all require code to express.
>
> Also, the most concise expression would be O(A */@,&$ B) and and it's
> tempting to try to make this more efficient, by keying on the join
> column. But that leaves me with something like this:I
>
> join1=:4 :0
> jA=. {:"1 A=.x
> jB=. {."1 B=.y
> u=. jA ([ -. -.)&~. jB
> uA=. (jA e. u) # A
> tuB=. (jB e. u) # B
> uB=. (/: u&i.@:({."1)) tuB
> juA=. {:"1 uA
> juB=. {."1 uB
> bA=. juA </. uA
> bB=. juB </. uB
> ; bA ,/@:(,"1/)each bB
> )
>
> Something like Aai's proposal (;A ((,}.)"1 <@#~ (=&{: |.)"1)"1 _ B) is
> going to be much more elegant, in my opinion...
>
> --
> Raul
>
> On Mon, Oct 15, 2012 at 12:33 PM, R.E. Boss <r.e.b...@planet.nl> wrote:
> > NB. ({:"1 A) equals {."1 B
> > so not the leading column of A
> >
> > Perhaps I should have added that ({:"1 A) =&~. {."1 B
> > or all not matching rows are deleted.
> >
> >
> > R.E. Boss
> >
> >
> >> -----Oorspronkelijk bericht-----
> >> Van: programming-boun...@forums.jsoftware.com [mailto:
> programming-boun...@forums.jsoftware.com] Namens Raul Miller
> >> Verzonden: maandag 15 oktober 2012 18:24
> >> Aan: programm...@jsoftware.com
> >> Onderwerp: Re: [Jprogramming] stitching matrices
> >>
> >> Given your description, I would have expected a result like this:
> >>
> >> 0 1 2 0 1 1 2 0
> >> 0 1 2 0 1 1 2 0
> >> 2 0 1 2 0 0 1 2
> >> 1 2 0 1 2 2 0 1
> >> 1 2 0 1 2 2 0 1
> >>
> >> I do not understand the result you proposed.
> >>
> >> Also, you have not specified what happens for rows where the leading
> >> column of A has a value which does not appear in the leading column of
> >> B. I can imagine several treatments for this case:
> >>
> >> 1) error
> >> 2) join to a row of fills
> >> 3) eliminate the row from the result
> >>
> >> Likewise for the reverse case, where the leading column of B has a
> >> value which does not appear in the leading column of A. (And, the
> >> asymmetric treatment of A and B, where one of the columns of B does
> >> not contribute to the result, makes me wonder if the treatment for
> >> {."1 A not in {."1 B might be different from from the treatment of
> >> {."1 B not in {."1 A.)
> >>
> >> So... I am confused. Any clarification or explanation you can provide
> >> could help.
> >>
> >> Thanks,
> >>
> >> --
> >> Raul
> >>
> >> On Mon, Oct 15, 2012 at 10:59 AM, R.E. Boss <r.e.b...@planet.nl> wrote:
> >> > Given
> >> >
> >> > ['A B' =. 3|L:0<@i."(1)3 5,: 5 4
> >> > +---------+-------+
> >> > |0 1 2 0 1|0 1 2 0|
> >> > |2 0 1 2 0|1 2 0 1|
> >> > |1 2 0 1 2|2 0 1 2|
> >> > | |0 1 2 0|
> >> > | |1 2 0 1|
> >> > +---------+-------+
> >> >
> >> > I want to stitch every row from A with all rows from B where ({:"1 A)
> equals {."1 B and one of these columns is deleted.
> >> > This will give me
> >> >
> >> > A foo B
> >> > 0 1 2 0 1 2 0 1
> >> > 0 1 2 0 1 2 0 1
> >> > 2 0 1 2 0 1 2 0
> >> > 2 0 1 2 0 1 2 0
> >> > 1 2 0 1 2 0 1 2
> >> >
> >> > Any elegant solutions?
> >> >
> >> >
> >> > R.E. Boss
> >> >
> >> > ----------------------------------------------------------------------
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> http://www.jsoftware.com/forums.htm
> >> ----------------------------------------------------------------------
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> >
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