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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