From: Robert Dober <robert.dober@...>
Date: 2009-07-22T17:55:40+09:00
Subject: Re: String spliting and inclusion

On 7/22/09, Robert Dober <robert.dober@gmail.com> wrote:
> On 7/22/09, David A. Black <dblack@rubypal.com> wrote:
>
> > Hi --
>  >
>  >
>  >  On Wed, 22 Jul 2009, Stuart Clarke wrote:
>  >
>  >
>  > > Hi all,
>  > >
>  > > I am having trouble working out some logic for my problem. I basically
>  > > have a long string (320 characters) and I want to split into smaller
>  > > strings no longer than 50 characters in length. At present I have the
>  > > following regex:
>  > >
>  > > data = "big long string"
>  > >
>  > > puts data.scan(/{50}/)
>  > >
>  > > This nicely breaks up the string however there are a few problems with
>  > > it, including:
>  > >
>  > > It only outputs 50 character chunks, therefore when it gets to the end
>  > > and only 20 characters remain it misses them off the output (it outputs
>  > > 6 50 characters strings and ignores the remaining 20)
>  > >
>  > > This regex also splits up words, which is something I don't want. I want
>  > > a script to count to 50 and when it gets there, go backwards to find
>  > > some white space and split it at that point, therefore not breaking up a
>  > > word. As a result a number of sub strings of various sizes will be
>  > > created all less than 50 chars.
>  > >
>  > > I hope this makes sense, to summarise I want to break up a string into a
>  > > max of 50 characters without breaking up words.
>  > >
>  >
>  >  Try this. I don't guarantee robustness.
>  >
>  >   str.scan(/\b.{0,50}(?:$|\b)/m)
>
> Hmm my \b at the end of my solution might have been a problem in some
>  edge cases, however I would suggest the usage of \z instead of $ and
>  the m switch. I fail to see why you put a \b at the beginning David,
>  would you mind to explain?
>
>  In Ruby 1.9 (or Oniguruma that is) the negative lookahead assertion
>  might lead to the most elegant solution:
>
>    /.{,50}(?!\B)/
Nahh that leaves us with spaces at the beginning of the line, of
course we could do
   scan(...).map( &:lstrip ) but that hurts my regex pride ;)

This seems to work (but does not really):

s = "Some words are made of letters! Some are not!"
puts s.scan( /.{,10}\p{Graph}(?:\P{Graph}|\z)/ )

Replace the puts with p and you will see trailing whitespace now :(.

This is a little bastard of a problem indeed. Simplest I could come up
with so far:

  s = "Some words are made of letters! Some are not!"
  p s.scan( /.{,10}\p{Graph}(?:\P{Graph}|\z)/ ).map( &:strip )

HTH
Robert


>
>  BTW it seems that {n,m} does not have a "non greedy" and "possessive"
>  variant, or did I miss it?
Yes I did, they are there {n,m}? and {n,m}+, sorry.
>
>  Cheers
>  Robert
>
>


-- 
Toutes les grandes personnes ont d’abord été des enfants, mais peu
d’entre elles s’en souviennent.

All adults have been children first, but not many remember.

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