From: Robert Klemme <shortcutter@...>
Date: 2008-11-22T19:50:54+09:00
Subject: Re: search reg-exp for exact match

On 20.11.2008 14:17, Jesús Gabriel y Galán wrote:
> On Thu, Nov 20, 2008 at 1:49 PM, John Butler <johnnybutler7@gmail.com> wrote:
>> Hi,
>>
>> I have a regular expression
>> /\b2003|\2004|\2005|\2006|\2007|\2008|\2009\b/
>> and i want to check if various years are present.
>>
>> "2003" =~  /\b2003|\2004|\2005|\2006|\2007|\2008|\2009\b/
>> returns 0 as expected
>>
>> "2010" =~  /\b2003|\2004|\2005|\2006|\2007|\2008|\2009\b/
>> returns nil as expected
>>
>> But i want only exact matches so when i search for "2003 - 2008" i want
>> nil returned as there is no exact match for that particular string.  I
>> thought the \b would give me this but it doesnt.
>>
>> "2003 - 2008" =~  /\b2003|\2004|\2005|\2006|\2007|\2008|\2009\b/
>> returns 0 i want nil returned.
>>
>> Anyone help?
> 
> To do exactly what you are asking for: you can anchor the regexp
> to the beggining or end of the string:
> 
> irb(main):013:0> re = /\A(2003|2004|2005|2006|2007|2008|2009)\Z/
> => /\A(2003|2004|2005|2006|2007|2008|2009)\Z/
> irb(main):014:0> "2003" =~ re
> => 0
> irb(main):015:0> "2003 - 2008" =~ re
> => nil

I'd rather use /\A200[3-9]\z/.

> In this case you don't need the \b anymore. BTW, you had typos there
> because you had \2 instead of \b.
> Anyway, if you want exact matches of strings you don't need regexps:
> 
> irb(main):018:0> years = (2003..2009).map {|x| x.to_s}
> => ["2003", "2004", "2005", "2006", "2007", "2008", "2009"]
> irb(main):020:0> years.include? "2003"
> => true
> irb(main):021:0> years.include? "2003 - 2008"
> => false

Or

irb(main):001:0> s="2005"
=> "2005"
irb(main):002:0> (2003..2009) === s[/\A\d{4}\z/].to_i
=> true
irb(main):003:0> s="2010"
=> "2010"
irb(main):004:0> (2003..2009) === s[/\A\d{4}\z/].to_i
=> false
irb(main):006:0> (2003..2009).include? s[/\A\d{4}\z/].to_i
=> false

Note, this works because 0 (= nil.to_i) is not part of the range!

> If you have many numbers and many lookups, a Set should be better,
> performance-wise.

You can even use a bit set:

irb(main):007:0> t = (2003..2009).inject(0) {|mask,y| mask | 1 << y}
=> 
116650078639864259662055853239652489576667478532211432368528502061497852157464823887836603809757037023714110007321126217782227286423686421672874625786531963635756068971637276480699799614611885589371789821904502024698121311064730577770474098457113815634439476503092997189887743679313284635928742849521858004245675611528209841692017556564840683843349732924435866760173931843810360262352061792429448169450281904579322760817054128336138506834410834183565543664844525391283837108127106791786643268532096672079466512393065631776802367002142967381057920196424747178242497261636008255151052901022379808767413846016
irb(main):008:0> t[s.to_i]
=> 0
irb(main):009:0> s="2005"
=> "2005"
irb(main):010:0> t[s.to_i]
=> 1
irb(main):011:0>

There are many ways... :-)

> Now, if we are talking about ranges of years we can do even better:

... or use the range test (as above) directly.

Kind regards

	robert