f=:13 : '1 (x)}y$0'
3 5 7 f 10$0
0 0 0 1 0 1 0 1 0 0
(correct)
but
f2=:1([)}0$~]
3 5 7 f2 10
0 1 0 0 0 0 0 0 0 0
I can see that there are some challenges from the boxed notation returned
for f2... but wondered if there was a way to achieve this tacitly. I'm
guessing, but f looks like it is less efficient because it resolves to 4:1
(x)}y$0. i.e. interpret a string on-the-fly.
thanks,
-Steven
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