Re: [Bug-apl] IOTA

[Elias Mårtenson]

On 2 March 2016 at 10:28, Christian Robert <address@hidden> wrote:

IOTA ← {∘.,/⍳¨⍵}

yes, it does the trick (you are really good btw)

I don't consider myself good. :-)

Anyway, it's quite simple. Let me break it down:

First of all,  ¨⍳⍵ will simply create the individual iotas:

      ⍳¨2 3 4

┌→──────────────────────┐

│┌→──┐ ┌→────┐ ┌→──────┐│

││1 2│ │1 2 3│ │1 2 3 4││

│└───┘ └─────┘ └───────┘│

└∊──────────────────────┘

The ∘., function will take two lists, and apply the function , (concatenate) on each permutation:

      (⍳2) ∘., (⍳3)

┌→────────────────┐

↓┌→──┐ ┌→──┐ ┌→──┐│

││1 1│ │1 2│ │1 3││

│└───┘ └───┘ └───┘│

│┌→──┐ ┌→──┐ ┌→──┐│

││2 1│ │2 2│ │2 3││

│└───┘ └───┘ └───┘│

└∊────────────────┘

You've probably seen the ∘.× function used to create a multiplication table in a similar manner.

Finally, the / operator acts on a function and a list as if the function was placed between every element in the list. Thus:

      ∘., / (1 2) (1 2 3) (1 2 3 4)

will be equivalent to:

      (1 2) ∘., (1 2 3) ∘., (1 2 3 4)

Regards,

Elias