How to find the offset of the last instance of a repeating character in a string? (Geoff Canyon)

[Bob Sneidar]

It probably should be named listOffset, like itemOffset or lineOffset. 

Bob S


> On Nov 1, 2018, at 17:04 , Geoff Canyon via use-livecode <use-livecode at lists.runrev.com> wrote:
> 
> Nice! I *just* finished creating a github repository for it, and adding
> support for multi-char search strings, much as you did. I was coming to the
> list to post the update when I saw your post.
> 
> Here's the GitHub link: https://github.com/gcanyon/offsetlist
> 
> Here's my updated version:
> 
> function offsetList D,S,pCase
>   -- returns a comma-delimited list of the offsets of D in S
>   set the caseSensitive to pCase is true
>   set the itemDel to D
>   put length(D) into dLength
>   put 1 - dLength into C
>   repeat for each item i in S
>      add length(i) + dLength to C
>      put C,"" after R
>   end repeat
>   set the itemDel to comma
>   if char -dLength to -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
> 
> On Thu, Nov 1, 2018 at 8:28 AM Niggemann, Bernd via use-livecode <
> use-livecode at lists.runrev.com> wrote:
> 
>> 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 at 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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