From: Ryan Pavlik <rpav@...>
Date: 2003-06-25T17:49:28+09:00
Subject: Re: another problem with regular expression

On Wed, 25 Jun 2003 15:52:07 +0900
"Pe�a, Botp" <botp@delmonte-phil.com> wrote:

<snip>
> > The second way is probably more efficient, since it can 
> > optimize the match and do it in one pass.
> 
> not sure if that would scale up to a 10000 entry array though..
<snip>

This is a good point... it wouldn't.  OTOH, it depends on the
problem; the first way would likely be poor for a few words on
many megs of text; the second would be worse in the way you
mentioned.

The first is probably the best for the general case, since even
though it would be poor for huge amounts of text, it would
eventually get done.  Of course, you could opt for a hybrid
solution (experiment with COUNT for best results):

    COUNT = 20
    i = 0; loop {
        re = array[i...i+COUNT].join('|')
        s.gsub(re, '*')

        i += COUNT
        break if(i >= array.size)
    }

;-)

-- 
Ryan Pavlik <rpav@users.sf.net>

"I sure as hell didn't learn anything, if thats what you mean." - 8BT