Yes, you're right in a way about @: , but I'd skipped defining a verb
to do the job.
What I had in mind, but didn't explicitly show, was something like
remz =: ($$rplc & (' 0';' ')) @ ": NB. @ also works ok!
so allowing
remz a
etc.
And yes, the $$ restores the shape of the ravelled output of rplc.
There are lots of other ways too, as shown by the correspondence
you've generated!
Mike
On 28/09/2017 12:42, Rudolf Sykora wrote:
On 27 September 2017 at 19:13, 'Mike Day' via Programming
<[email protected]> wrote:
($$rplc & (' 0';' ')) @: ": a
Yes. (I guess the @: is unnecessary here, right?)
I actually had tried something similar first:
(' 0';' ') stringreplace ":r
which, however, yielded
1 2 3 4 5 6 7 8 9 10 11 12 13 14 1516 17 18 19 20 21 2223 24 25 26 27 28 2930 31
I didn't (and still don't) understand why I got a single line, and
decided to ask for help then...
I see (if I understand) you put the shape back in with $$...
so in my case it would be
($$ (' 0';' ')& stringreplace) ":r
Thanks
Ruda
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