From: Ross Bamford <rossrt@...>
Date: 2006-03-01T21:36:59+09:00
Subject: Re: Regex replacement problem---replace every n-th

On Wed, 2006-03-01 at 21:03 +0900, junk5@microserf.org.uk wrote:
> Hi all
> 
> I have the string '1 20 3 400 5 60 7 800 9 0 ' and need to replace
> every n-th space in the string with another character. So if n=2 and
> the replacement character is '\n', then the above would become
> 
> '1 20\n3 400\n5 60\n7 800\n9 0\n'.
> 
> I'm sure it must be possible (easy, even) to do this with a regex
> substitution, but unfortunately I'm no regex ninja.

There's probably a better way to do this, but here's a 'metaregex' idea:

	str =  "1 20 3 400 5 60 7 800 9 0 "
	# => "1 20 3 400 5 60 7 800 9 0 "

	n = 2
	# => 2

	r = Regexp.new('(' + ('\d+\s' * (n - 1)) + '\d+)(\s)')
	# => /(\d+\s\d+)(\s)/

	str.gsub(r) { $1 + rep }
	# => "1 20\n3 400\n5 60\n7 800\n9 0\n"

	n = 3
	# => 3

	r = Regexp.new('(' + ('\d+\s' * (n - 1)) + '\d+)(\s)')
	# => /(\d+\s\d+\s\d+)(\s)/

	str.gsub(r) { $1 + rep }
	# => "1 20 3\n400 5 60\n7 800 9\n0 "

-- 
Ross Bamford - rosco@roscopeco.REMOVE.co.uk