Not sure how elegant this is but it seems to work:
({:"1 A) =/ {."1 B
0 1 0 0 1
1 0 0 1 0
0 0 1 0 0
(([:$])#:[:I.[:,]) ({:"1 A) =/ {."1 B
0 1
0 4
1 0
1 3
2 2
<"1 |:(([:$])#:[:I.[:,]) ({:"1 A) =/ {."1 B
+---------+---------+
|0 0 1 1 2|1 4 0 3 2|
+---------+---------+
ixs=. <"1 |:(([:$])#:[:I.[:,]) ({:"1 A) =/ {."1 B
,.&>/0 1}."1&.>ixs{&.>A;<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 Mon, Oct 15, 2012 at 12:50 PM, Aai <agroeneveld...@gmail.com> wrote:
> 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
>>
>> ----------------------------------------------------------------------
>> For information about J forums see http://www.jsoftware.com/forums.htm
>
>
> --
> Met vriendelijke groet,
> @@i = Arie Groeneveld
>
>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
--
Devon McCormick, CFA
^me^ at acm.
org is my
preferred e-mail
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
