Hi Geoff,
thank you for this beautiful script.
I modified it a bit to accept multi-character search string and also for case
sensitivity.
It definitely is a lot faster for unicode text than anything I have seen.
-----------------------------
function offsetList D,S, pCase
-- returns a comma-delimited list of the offsets of D in S
-- pCase is a boolean for caseSensitive
set the caseSensitive to pCase
set the itemDel to D
put the length of D into tDelimLength
repeat for each item i in S
add length(i) + tDelimLength to C
put C - (tDelimLength - 1),"" after R
end repeat
set the itemDel to comma
if char -1 of S is D then return char 1 to -2 of R
put length(C) + 1 into lenC
put length(R) into lenR
if lenC = lenR then return 0
return char 1 to lenR - lenC - 1 of R
end offsetList
------------------------------
Kind regards
Bernd
>
> Date: Thu, 1 Nov 2018 00:15:37 -0700
> From: Geoff Canyon
> To: How to use LiveCode <use-livecode@lists.runrev.com>
> Subject: Re: How to find the offset of the last instance of a
> repeating character in a string?
>
> I was curious if using the itemDelimiter might work for this, so I wrote
> the below code out of curiosity; but in my quick testing with single-byte
> characters it was only about 30% faster than the above methods, so I didn't
> bother to post it.
>
> But Ben Rubinstein just posted about a terrible slow-down doing pretty much
> this same thing for text with unicode characters. So I ran a simple test
> with 8000 character long strings that start with a single unicode
> character, this is about 15x faster than offset() with skip. For
> 100,000-character lines it's about 300x faster, so it seems to be immune to
> the line-painter issues skip is subject to. So for what it's worth:
>
> function offsetList D,S
> -- returns a comma-delimited list of the offsets of D in S
> set the itemDel to D
> repeat for each item i in S
> add length(i) + 1 to C
> put C,"" after R
> end repeat
> set the itemDel to comma
> if char -1 of S is D then return char 1 to -2 of R
> put length(C) + 1 into lenC
> put length(R) into lenR
> if lenC = lenR then return 0
> return char 1 to lenR - lenC - 1 of R
> end offsetList
>
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