Sorry, by far from elegant:
;A <@((,"1}.)"1#~(=&{: |.)"1)"1 _ 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
On 15-10-12 16:59, R.E. Boss 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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Met vriendelijke groet,
@@i = Arie Groeneveld
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