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01:00 - 19:0020:00 - 22:00

1:16 AM
@Adám APLcart has a typo on the top: C, D: chara_ra_cter array
4 hours later…
5:00 AM
Any idea to improve this?
5 hours later…
10:12 AM
@Bubbler Why is there an issue with length-1 arrays?
10:25 AM
⋄⎕←(⍴-0(⊥⍨=)⊢⌹∘⍉∘↑1+⍴⍴¨(⊢,1+⌽)∘⍳¨∘⍳∘⍴)1⋄⎕←(⍴-0(⊥⍨=)⊢⌹∘⍉∘↑1+⍴⍴¨(⊢,1+⌽)∘⍳¨∘⍳∘⍴)1 1
Ack should have split them up
Using ≢ in place of every monadic ⍴ will fix that
Ahh right
10:27 AM
Ah and I see that the other way you've gotten around that in the test cases is by using 1⍴1. Pardon me, I should have read the test cases more carefully
2 hours later…
12:44 PM
No problem, your question made me recall the "tally" function :)
12:55 PM
it probably doesn't work on (m)any other cases, might get in an infinite loop and is pretty damn slow, but i'm happy
The opposite of pointle^W pointfree.io?
@Ven well i do have a trainifier too, but it's way more simple
3 hours later…
4:04 PM
is there a way to see a train tree in the interpreter like we can see here?
@rcabaco ]box on -trains=tree
thank you!
this works as i expect, for each of the arguments it gives me a boolean vector where the data is not '='
now i want to partition each argument with this result
this was obviously my first approach, which does not work
⊢ ┌─┼─────┐
  ⍨ ~ ┌───┼─┐
┌─┘   '=' ¨ ⊢
⊆       ┌─┘
(⊢⊆⍨(~'='⍷¨⊢)) should work
i still get a domain error
but looking at the tree i cannot spot the problem
4:15 PM
Well, I'm not sure what '='⍷¨⊢ is supposed to do. What sort of argument do you have? Maybe you want (⊢⊆⍨(~'='⍷⊢))¨
i am missing an each for the partition enclosed
or i can try that
you are right, one each is enough
thank you
It might make more sense to remove the like ((~'='⍷⊢)⊆⊢)¨
(⊢⊆⍨∘~'='⍷⊢)¨ also works
yep :)
4:56 PM
@Bubbler ≢{⊃⌽⍋0≠⍵⌹⍉1+↑(⍺⍴⍳,⊢-⍳)¨⍳⍺}⊢ with ⎕io←0
1 hour later…
6:24 PM
how can i break a vector in groups of two?
i have been on this for an hour and didn't think of reshape
01:00 - 19:0020:00 - 22:00

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