On 2014-03-19 23:16:32, Elias Mårtenson wrote:
> I see. So the way I see it, I should actually use 2 OP/X instead? I seems to
> actually do what I expected.
It if does what you need then sure, but it's a very different thing
than scan:
,\'abcd'
a ab abc abcd
2,/'abcd'
ab bc cd
-
I see. So the way I see it, I should actually use 2 *OP*/X instead? I seems
to actually do what I expected.
Regards,
Elias
On 19 March 2014 22:25, Kacper Gutowski wrote:
> On 2014-03-19 21:22:04, Elias Mårtenson wrote:
> > This can't possibly be correct?
>
> But it is!
> Let me just quote rele
I've been struggling with this one for days now, and I really need some
advice.
Assume I have a binary sequence. Say, something like this:
0 0 1 1 0 0 1 1 1 1 1 0 0 0 1 0 1 0 1 1 0 1 1 1
Here, we can see a few sequences of consecutive ones. I want to zero out
every second in each sequence. I.e,
The following code returns the expected results:
* {⍺+⍵}\⍳10*
1 3 6 10 15 21 28 36 45 55
However, if I also print all the calls, I get a lot more evaluations of the
function than expected:
*{(⍺+⍵)⊣⎕←('[',(⍕⍺),',',(⍕⍵),']')}\⍳10*
[1,2]
[2,3]
[1,5]
[3,4]
[2,7]
[1,9]
[4,5]
[3,9]
[2,12]
