⍝ Given a vector 1 2 3 4 , try to generate vector of vector [[1 2 3 4], [2, 3, 4], [3, 4], [4] []]

I have the function, but I could not figure out how to apply the function on each element of a number such 0 1 2 3 4

you may be interested in ,\

@elliptic00 ⎕←{(⌽0,-⍵)↑¨⊂⍵}⍳4 ⋄ (⊂↑¨⍨∘⌽0,-)⍳4

@Adám
┌───────┬─────┬───┬─┬┐
│1 2 3 4│2 3 4│3 4│4││
└───────┴─────┴───┴─┴┘
┌───────┬─────┬───┬─┬┐
│1 2 3 4│2 3 4│3 4│4││
└───────┴─────┴───┴─┴┘

⎕IO←0⋄(⍳∘≢↓¨⊂∘⊢)1 2 3 4

⎕← ⎕IO←0⋄(⍳∘(1+≢)↓¨⊂∘⊢)1 2 3 4

@doug
0
┌───────┬─────┬───┬─┬┐
│1 2 3 4│2 3 4│3 4│4││
└───────┴─────┴───┴─┴┘

@rak1507 D'oh! I missed this. Sorry if I spoiled the hint.

⎕←⌽⌽¨,\(⊂⍬),⌽⍳4

@Adám
┌───────┬─────┬───┬─┬┐
│1 2 3 4│2 3 4│3 4│4││
└───────┴─────┴───┴─┴┘

⎕←⌽⌽¨1↓¨,\0,⌽⍳4

@Adám
┌───────┬─────┬───┬─┬┐
│1 2 3 4│2 3 4│3 4│4││
└───────┴─────┴───┴─┴┘

⎕←⍬,⍨⌽⌽¨,\⌽1 2 3 4

@PyGamer0
┌───────┬─────┬───┬─┐
│1 2 3 4│2 3 4│3 4│4│
└───────┴─────┴───┴─┘

Nice one. Technically we've been making the last box empty.

That's what the ⊂⍬ is doing in Adam's version above.

You could do ⍬,⍥⊆⍨

{(⌽(-⍳≢⍵)↑¨⊂⍵),⊂⍬} - a solution without ,\

{⌽¨⌽,\⌽⍵,⊂⍬} also seems pretty cool and short

even cooler when you realise that you can do stuff like ⋄⌽¨⌽,\⌽1 2 3 4 ⍬

@KamilaSzewczyk
┌───────┬─────┬───┬─┬┐
│1 2 3 4│2 3 4│3 4│4││
└───────┴─────┴───┴─┴┘

Is there a way to limit the output of long results in the repl?

@pamphlet Yes, have a look at ]rows

@Adám Looks perfect. Thanks!

k ← {x ← {⍺} ⋄ ⍵ + x}

APL does not support inner function?

What are you aiming to do?

APL does support nested functions: ⎕ ← {x ← {⍵} ⋄ - x 1 }0

Each dfn will create it's own scope however

If you wanted to use ⍺ from the top-level function, you'd have to assign it to a variable.

@awagga, Ok, my mistaken,