Like this?
arr=: ;:'a b a c b c a b b b c a b'
(;:'a b c')i.arr
0 1 0 2 1 2 0 1 1 1 2 0 1
((;:'a b c')i.arr){;:'e f g'
┌─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┐
│e│f│e│g│f│g│e│f│f│f│g│e│f│
└─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┘
I hope this helps,
--
Raul
On Sat, May 27, 2017 at 4:43 AM, 'Jon Hough' via Programming
<programm...@jsoftware.com> wrote:
> Let
> a=: 'a'
> b=: 'b'
> c=: 'c'
>
> and let
> arr=: a;b;a;c;b;c;a;b;b;b;c;a;b NB. i.e. all items from a,b,c
>
> I want to relabel the items
> with
> e=: 'e'
> f=: 'f'
> g=: 'g'
> so a -> e, b -> f, c -> g.
> so arr becomes
> e;f;e;g;f;g;e;f;f;f;g;e;f
>
> Is there a succinct (preferably tacit) way to do this?
> I feel like there should be, but can't find a way, so my current
> method is to use a for__i. loop, which feels very clumsy for
> something that seems should have a nice solution.
>
> Thanks,
> Jon
>
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