Key (/.) is cool! For the boolean case, plus-insert-prefix (+/\) is your friend
b =: 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0
+/\ b
0 0 1 1 2 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5
(+/\b) </. az
+--+--+-----+--------+------+---+
|ab|cd|efghi|jklmnopq|rstuvw|xyz|
+--+--+-----+--------+------+---+
Thanks for making me read that page.[1]
... peter
[1] http://www.jsoftware.com/docs/help701/dictionary/d421.htm
Christoph von Basum wrote:
Number 1 the brute force way: (i#i.#i)</.az
On Tue, Oct 9, 2012 at 4:37 PM, Roger Hui <[email protected]> wrote:
I mean to say <;.2 instead of <;._2 .
On Tue, Oct 9, 2012 at 7:36 AM, Roger Hui <[email protected]>
wrote:
1. (;i{.&.>1) <;.1 az (various ways of generating left argument; also
use
<;._2 depending on left argument).
2. b <;._2 az
On Tue, Oct 9, 2012 at 7:32 AM, Ian Clark <[email protected]> wrote:
1. Suppose:
az=: 'abcdefghijklmnopqrstuvwxyz'
i=: 3 2 5 8 5 3 NB. -(for example)
How best can I partition az into parts pp where (n{pp) has length (n{i)
?
viz.
pp -: 'abc';'de';'fghij';'klmnopqr';'stuvw';'xyz'
2. Suppose instead of (i) I have a boolean (b) marking where to cut az
...
b =: 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0
(c/f dyadic {enclose} in APL+)
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