Most of the time I'm too slow to come up with a complete and
readysolution, so I showed this:
;A <@((,"1}.)"1#~(=&{: |.)"1)"1 _ B
If I had given myself some more time I would have come up with the more
complete and obvious solution R.E. Boss came up with:
A ;@:(<@((,"1}.)"1#~(=&{: |.)"1)"1 _) B
In my opinion not very important, but after I posted my solution I
modified my phrase to what Raul Miller showed by 'moving' the < to the
other possible position ('connected' with #). I practice this because
I'm curious to this kind of variations of a solution.
Finally answering your question: No I don't like your alternative more
because I don't like the use of caps. But ok, that's a kind of nonsense
preference (as is my drive to use as much as possible the natural rank
of primitives/verbs). It happens several times I'm not smart enough to
avoid the use caps. Also the alternative could cause the use of lots of
parenthesis. Not elegant according to R.E.Boss's definition with which I
do agree.
On 17-10-12 08:14, Linda Alvord wrote:
Would you like this more? I just modified your sentence a little.
j=: 13 :' ;x <@((D=:,"1}.)"1#~(C=:=&{:|.)"1)"1 _ y'
j
4 : ' ;x <@((D=:,"1}.)"1#~(C=:=&{:|.)"1)"1 _ y'
f=: 13 :';x ([:<(,"1}.)"1#~(=&{:|.)"1)"1 _ y'
f
[: ; ([: < (,"1 }.)"1 #~ (=&{: |.)"1)"1 _
A j 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
A f 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
Linda
-----Original Message-----
From: programming-boun...@forums.jsoftware.com
[mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Aai
Sent: Monday, October 15, 2012 12:50 PM
To: programm...@jsoftware.com
Subject: Re: [Jprogramming] stitching matrices
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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