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

[Geoff Canyon]

All of those return a single value; I wanted to convey the concept of
returning multiple values. To me listOffset implies it does the same thing
as itemOffset, since items come in a list. How about:

offsets -- not my favorite because it's almost indistinguishable from offset
offsetsOf -- seems a tad clumsy

On Fri, Nov 2, 2018 at 7:41 AM Bob Sneidar via use-livecode <
use-livecode at lists.runrev.com> wrote:

> 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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