⍺ ⊆ ⍵ begins a new partition in ⍵ whenever the corresponding positive int in ⍺ is greater than its left-hand neighbour. Zeros in ⍺ omit their corresponding items in ⍵.

⍺ ⊂ ⍵ begins a new partition in ⍵ whenever the corresponding Bool in ⍺ is 1. Zeros in ⍺ join their corresponding item in ⍵ to the current partition, or drops them if there is none (i.e. the leftmost partition has not begun yet.)

@Adám I tried {(2≡/⍵)⊂⍵} for that first CMC you posted but it raises a length error (because (2≡/⍵) generates a list 1 size smaller than the ⍵ length I assume?)

Also tried swapping ≡ for = to no avail.

@J.Salle The first element is always "new", so you need to prepend a 1, so {(1,2≡/⍵)⊂⍵}

@J.Salle Nicely done, btw. Now you can also make it into a train.

@Adám Yeah I usually try to make them work as a Dfn before trying to go for a train. It's still not doing what I need, though. Gotta think some more

@J.Salle Oh, right, you've got it the wrong way: use ≠ or ≢.

Would the 1, count as the X in a Xgh train? Or would that be the 2 and the 1, is the ⍺?

Or is the X the whole 1,2?

@J.Salle Both. 1 is X in 1 , (2≠/⊢) and 2 is X in 2 ≠/ ⊢

@Adám Okay, I'll try working with that

So, my chat wasn't connecting so I went out to get some lunch. I can't seem to get a train to work; I've tried multiple combinations of 1,2≠/ with ∘ and ⊂ and ⊢⊣, all of them raise a RANK ERROR or a SYNTAX ERROR. 1,2≠/ followed by the list works properly, I just can't seem to get it to work with ⊂

@J.Salle As ≠/ has a left argument (namely 2), it needs a right argument too, and since you just want the unmodified argument, that's ⊢.

{(2≠/⍵)⊂⍵}
 (2≠/⊢)⊂⊢

However, there is a solution half the length of that, using ⊆.

@Adám for the record: since I need the 1,, 1,(2≠/⊢)⊂⊢ doesn't work. I had to do ((1,2≠/⊢)⊂⊢) for that to work properly. I'll look into the ⊆ solution

@Adám I don't see how ⊆ would work here. I'd need to get the list to be enclosed with parens to nest it, no?

Idk how I'd make (1 1 1) 4 5 (6 6) (7 7) 3 to pass it as an argument to ⊆, especially not in 3 characters

Actually I just noticed that's not right either since it groups 4 and 5 together facedesk

@J.Salle Uh, the result should be approximately (1 1 1) 4 5 (6 6) (7 7) 3 (except each element should be a vector). ⊆ will do the enclosing for you, you just need to tell it where the partitions begin, i.e. where list changes. So, for strictly positive ascending integers, you can just do ⊆⍨ but how about numbers in general? Actually, now I realise that my solution will only work on ascending numbers (i.e. even non-ints), but only when sorted. Never mind.

@Adám Okay, I managed to get to ⊆⍨ on my own and found that exact problem hahahahahah

@J.Salle Well, by idea was ⍳⍨⊆⊢

I got a result of ⊆⍨(⍋⊃¨⊂) which is technically correct, but it sorts the list, so {(1,2≠/⍵)⊂⍵} would be the best solution?

@J.Salle Yes. And you're getting good at this! They can be shortened to ⊆⍨⍋⊃¨⊂ and ⊢⊂⍨1,2≠/⊢

@Adám I tried ⊆⍨⍋⊃¨⊂ but it raises a Rank Error? ⊢⊂⍨1,2≠/⊢ returns what would be the ⍺ argument for ⊂ (which is 1 0 0 1 1 1 0 1 0 1), not the list.

Weird, if I enclose ⊆⍨⍋⊃¨⊂ in parens, then it works fine.

@J.Salle ⊆⍨⍋⊃¨⊂ is a train; you need to name it or paren it to apply it.

@J.Salle Same goes for ⊢⊂⍨1,2≠/⊢

Oh, duh. I knew I was forgetting something >.>