From: Alan Burch <orotone@...>
Date: 2006-03-12T00:44:17+09:00
Subject: Re: Help me understand why the Ruby block is slower than wit

Benjohn Barnes wrote:
> On 11 Mar 2006, at 02:13, George Ogata wrote:
> 
>> Another speedup:  replace:
>>
>>   w.split(//).uniq.size == 11
>>
>> with:
>>
>>   w !~ /(.).*\1/
> 
> !? :) How on earth does that work? Every time I think I've sort of
> got the hang of regexp, they spring something new on me.
> 
> I was also going to ask why everyone was doing "split( // )" instead
> of "split( '' )"?
> 
> - oooh, coffee's ready...
> 
> Cheers,
> 	Benjohn

Benjohn:
A . matches any char except the \n, putting it in (), makes it save in 
\1 each time it matches, the .* matches zero or more chars that's not a 
\n, so if we try to match a string such as "profligate\n" the regex 
would first look for (p).*\p, with the second p being anywhere in the 
string then (r).*(r), etc.  A string with a repeating set of letters 
"wristwatch\n" matches (w)rist\w and returns a match.  I highly 
recommend O'reilly's "Mastering Regular Expressions", I've only read the 
first edition, but it's an eye-opener (or maybe the opposite if you try 
to read it in bed :))

A note for those following along in the DOS world, the dos string ends 
\r\n and won't return the expected result as a matching DOS string will 
need to be 12 long.  This sacrifices portability for speed (I didn't 
want to use chop after each gets).

As to split, I just used what I'm used to from perl.  It's an empty 
pattern and makes sense to me that way.
Alan


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