I did not write stringreplace, so am not prepared to discuss its
implementation tradeoffs.
But, yes, this works:
r=: , LF,.~": _7 ]\ 0 >. i.&.(+&5) 32
r
0 0 0 0 0 0 1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31 0 0 0 0 0
r rplc ' 0';' '
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31
Thanks,
--
Raul
On Thu, Sep 28, 2017 at 9:21 AM, Rudolf Sykora <[email protected]> wrote:
> On 28 September 2017 at 14:50, Raul Miller <[email protected]> wrote:
>> Another way to work around this problem would be to use stringreplace"1
>>
>> r=: _7 ]\ 0 >. i.&.(+&5) 32
>> (":r) rplc"1 ' 0';' '
>
> Ok. Thanks. So is there a good reason for txt=. ,y in stringreplace?
>
> Or, can the whole of r be represented as just a (single) list of characters,
> like
>
> a =. 0 : 0
> 1 0 3
> 0 4 5
> )
>
> a
> 1 0 3
> 0 4 5
>
> $a
> 14
>
> a rplc ' 0';' '
> 1 3
> 4 5
>
> i.e., here the rplc can be used directly...
>
> Thanks
> Ruda
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