I derived a slower and more space-consuming function than Raul's,
using cut ( ;. ) , so won't bore you with it....
However, in seeking a solution, I wondered how you wanted to treat
the case where the first item is zero, or the same as the left argument.
Raul's f : (I'm forcing extra line feeds) ...
1 f 1 0 2 3 4 5 0 0 0 4 3 2 4
1 0 2 3 4 5 0 0 0 4 3 2 4
... is a sensible, do-nothing, answer, but is it what you wanted?
Should it be
0 0 2 3 4 5 0 0 0 4 3 2 4
or
_ 0 2 3 4 5 0 0 0 4 3 2 4
or something else?
Mike
Matthew Brand wrote:
Thanks Raul.
On Wed, Oct 29, 2008 at 3:21 PM, Raul Miller <[EMAIL PROTECTED]> wrote:
On Wed, Oct 29, 2008 at 10:12 AM, Matthew Brand
<[EMAIL PROTECTED]> wrote:
0 f 1 0 2 3 4 5 0 0 0 4 3 2 4
1 1 2 3 4 5 5 5 5 4 3 2 4
f=: ] {~ ~: >./\@:* [EMAIL PROTECTED]@]
--
Raul
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