In the definition for Under {⍺←{⍵ ⋄ ⍺⍺} ⋄ ⍵⍵⍣¯1⊢(⍵⍵ ⍺)⍺⍺(⍵⍵ ⍵)}, what does {⍵ ⋄ ⍺⍺} mean?

@rabbitgrowth It is a monadic operator which, when ⍺ is later mentioned, consumes the ⍵⍵ function without applying it. Now I think of it, I believe it could also have been written ⍺←⍣0

Ah, so the ⋄ ⍺⍺ part doesn't actually do anything and is just there to force the definition into that of an operator?

Correct.

Got it. Thank you!

I had no idea you could write ⍣0 on its own

> Write a function that, given a right argument which is an integer vector and a left argument which is an integer scalar, reorders the right argument so any elements equal to the left argument come first while all other elements keep their order.

I think we can do better than this.

I had {⍵[⍋⍺≠⍵]} as well

Try to do it in terms of sets. We've got one set (3) and another (1 2 3 4 1 3 1 4 5).

@RubenVerg Might run even faster to write (=,≠) as (,∘~⍨=) or ,∘~⍨⍤=

(∩,~)⍨ ?

Yes, perfect, or distribute the ⍨ over the outer tines: ∩⍨,~⍨

Nice that ,∘~⍨⍤=⊢⍤/⊢,⊢ has no parens. Looking forward to ≠~⍛,⍛/⊢,⊢ in 20.0…

That is nice.

Your "nice" was probably on ∩⍨,~⍨ 😃

Yes, that's what I meant :)

@Adám Slightly with a million random elements 1…10:

  3((=,≠)⊢⍤/⊢,⊢)y  → 5.3E¯4 |  0% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕
  3(,∘~⍨⍤=⊢⍤/⊢,⊢)y → 5.2E¯4 | -2% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕

I also wrote down ~⍨,⍨=⊢⍤/⊢ and ~⍨,⍨+.=⍴⊣ but forgot about ∩

Really nice how it preserves the repetitions

Yes. Sometimes people think that's strange, but it is actually more powerful.

OK, I think this is it.

Could always do ∪⍤∩

(looks like a running guy)

How does that work?

@Adám just gets rid of the repetitions

Ah, you mean if one needs a pure set function. Right.

Is it better to do ∩⍥∪ ?

Possibly faster, but probably depends on the data. After all, ∩ might significanty lessens the amound of data to be ∪'ed.

I see, that makes sense

Can't stop seeing ∪⍤∩ and ∩⍥∪ as someone who's running and panting heavily

See you next week for 2020-7: See You in a Bit!

See you!

Sorry, too late. Had no access. See you next week!

@rabbitgrowth while certainly very elegant, I would have never guessed set operations to be about as fast as mine

(that is, until I realized I was basically exactly reimplementing intersection and difference)

also, seems like the version with the double selfies is consistently slower

@Adám j solution: 0<17 b.

(while I think of a good way to do it in dyalog :) )

oh I misread the question, it wants all flags not any flag

doesn't work then

alright, I have ∧/∊⍥(⍸⍤⌽2⊥⍣¯1⊢)