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Why, oh, why, little regexp?

Daniel Waite

10/30/2007 11:14:00 PM

'cost * tax'.match(/([a-z]+)*/).to_a
=> ["cost", "cost"]

Why?

I'm reading it as... Take one or more characters between a and z, store
them into a back reference, then repeat the previous match zero or more
times.

Now, that regexp doesn't do what I want it to do, but what it IS doing
doesn't make sense to me.

What I'd like is to grab all the "words" in the string. So in the above
example I'd like two matches, cost and tax.

Any ideas?

PS: match(...).captures always, always returns an empty array...
--
Posted via http://www.ruby-....
14 Answers

Joel VanderWerf

10/30/2007 11:24:00 PM

0

Daniel Waite wrote:
> 'cost * tax'.match(/([a-z]+)*/).to_a
> => ["cost", "cost"]
>
> Why?
>
> I'm reading it as... Take one or more characters between a and z, store
> them into a back reference, then repeat the previous match zero or more
> times.
>
> Now, that regexp doesn't do what I want it to do, but what it IS doing
> doesn't make sense to me.
>
> What I'd like is to grab all the "words" in the string. So in the above
> example I'd like two matches, cost and tax.
>
> Any ideas?

'cost * tax'.scan(/\w+/)
=> ["cost", "tax"]

> PS: match(...).captures always, always returns an empty array...

How are you using it?

"foo".match(/(foo)/).captures
=> ["foo"]
'cost * tax'.match(/([a-z]+)*/).captures
=> ["cost"]

--
vjoel : Joel VanderWerf : path berkeley edu : 510 665 3407

Daniel Waite

10/30/2007 11:33:00 PM

0

Joel VanderWerf wrote:
>> What I'd like is to grab all the "words" in the string. So in the above
>> example I'd like two matches, cost and tax.
>>
>> Any ideas?
>
> 'cost * tax'.scan(/\w+/)
> => ["cost", "tax"]

How do you people do that? The last time I had a regexp question someone
came down from the clouds and handed me something about that short. Why
do I think it's more difficult than it is?

After making the example a little more complex I had to change it
every-so-slightly...

'cost * tax + 0.075'.scan(/[a-z]+/)
=> ["cost", "tax"]

But it's effectively the same. Thank you Joel, you rock!

Is there a book you recommend to learn more about regular expressions?
How did YOU learn them?

>> PS: match(...).captures always, always returns an empty array...
> How are you using it?
>
> "foo".match(/(foo)/).captures
> => ["foo"]
> 'cost * tax'.match(/([a-z]+)*/).captures
> => ["cost"]

LOL I'm an idiot -- *captures* -- back references, right. Gotcha...
--
Posted via http://www.ruby-....

Stanislav Sedov

10/30/2007 11:36:00 PM

0

On Wed, Oct 31, 2007 at 08:14:01AM +0900 Daniel Waite mentioned:
> 'cost * tax'.match(/([a-z]+)*/).to_a
> => ["cost", "cost"]
>
> Why?
>

Well, the regexp always matches the longest possible string.
What did you wrote is effectively equialent to ([a-z]*).
The single regexp can't match multiple strings, it always matches
one. It can't match the space after the 'cost' either, since this
symbol wasn't included to your regexp.

In case, if you want to match two words, you should write e.g.
([[:alpha:]]+)[[:space:]]+([[:alpha:]]+)
This regexp will match two words separated by a space.
Regexp can't match an undefined number of words, you should know
in advance which number of words you want to match.

For more infor on regexps see e.g. re_format(7).

--
Stanislav Sedov
ST4096-RIPE

Daniel Waite

10/30/2007 11:52:00 PM

0

Stanislav Sedov wrote:
> On Wed, Oct 31, 2007 at 08:14:01AM +0900 Daniel Waite mentioned:
>> 'cost * tax'.match(/([a-z]+)*/).to_a
>> => ["cost", "cost"]
>>
>> Why?
>>
>
> Well, the regexp always matches the longest possible string.
> What did you wrote is effectively equialent to ([a-z]*).
> The single regexp can't match multiple strings, it always matches
> one. It can't match the space after the 'cost' either, since this
> symbol wasn't included to your regexp.
>
> In case, if you want to match two words, you should write e.g.
> ([[:alpha:]]+)[[:space:]]+([[:alpha:]]+)
> This regexp will match two words separated by a space.
> Regexp can't match an undefined number of words, you should know
> in advance which number of words you want to match.
>
> For more infor on regexps see e.g. re_format(7).

Hmm... if what you say is true, why does the second poster's solution
capture multiple words? Wait, I know why. String#scan is different than
string#match. Interesting...

So how does that work if I wanted to match ALL occurrences of \w+
WITHOUT scan?
--
Posted via http://www.ruby-....

Jim Clark

10/31/2007 12:13:00 AM

0

Daniel Waite wrote:
> Is there a book you recommend to learn more about regular expressions?
> How did YOU learn them?
>
"Mastering Regular Expressions" by Jeffrey Friedl. I haven't seen the
third edition to see if there is any Ruby specific examples but even
with all the Perl examples in the first edition, I still use it as a
reference because of the similarities between Perl and Ruby's regular
expressions.

-Jim

7stud --

10/31/2007 1:24:00 AM

0

Daniel Waite wrote:
> What I'd like is to grab all the "words" in the string.
> So how does that work if I wanted to match ALL occurrences
> of \w+ WITHOUT scan?

Your using the wrong method. match() only returns the first match:

pattern = /x.x/
str = "xax hello xbx"

puts pattern1.match(str)

--output:--
xax

>
> So how does that work if I wanted to match ALL occurrences
> of \w+ WITHOUT scan?
>

str = " cost * tax"
words = str.split("*").map {|elmt| elmt.strip()}
p words

--output:--
["cost", "tax"]



str = " cost * tax = 123"
words = []

str.split().map do |word|
good_word = true

word.each_byte do |code|
if code < ?a or code > ?z
good_word = false
break
end
end

if good_word
words << word
end
end

p words

--output:--
["cost", "tax"]

--
Posted via http://www.ruby-....

Daniel Waite

10/31/2007 3:31:00 AM

0

7stud -- wrote:
> str = " cost * tax = 123"
> words = []
>
> str.split().map do |word|
> good_word = true
>
> word.each_byte do |code|
> if code < ?a or code > ?z
> good_word = false
> break
> end
> end
>
> if good_word
> words << word
> end
> end
>
> p words
>
> --output:--
> ["cost", "tax"]

That's clever use of ?a, which I recognize but have never seen anyone
use before. Thanks for the example!

Jim Clark wrote:
> "Mastering Regular Expressions" by Jeffrey Friedl. I haven't seen the
> third edition to see if there is any Ruby specific examples but even
> with all the Perl examples in the first edition, I still use it as a
> reference because of the similarities between Perl and Ruby's regular
> expressions.

I shall check that out Jim, thanks much.
--
Posted via http://www.ruby-....

Phrogz

10/31/2007 3:59:00 AM

0

On Oct 30, 9:30 pm, Daniel Waite <rabbitb...@gmail.com> wrote:
> That's clever use of ?a, which I recognize but have never seen anyone
> use before. Thanks for the example!

My current favorite use for the ?x syntax is converting single-
character strings representing digits into their integer form:

# Jenny jenny, who can I turn to?
irb(main):006:0> "8675309".each_byte{ |x| p x - ?0 }
8
6
7
5
3
0
9

Brian Adkins

10/31/2007 4:28:00 AM

0

On Oct 30, 11:58 pm, Phrogz <phr...@mac.com> wrote:
> On Oct 30, 9:30 pm, Daniel Waite <rabbitb...@gmail.com> wrote:
>
> > That's clever use of ?a, which I recognize but have never seen anyone
> > use before. Thanks for the example!
>
> My current favorite use for the ?x syntax is converting single-
> character strings representing digits into their integer form:

Yeah, so you can squeeze Ruby code into small places :)

1.upto(?d){|i|i%3<1&&x=:Fizz;puts i%5<1?"#{x}Buzz":x||i}

Rick DeNatale

10/31/2007 11:17:00 AM

0

On 10/31/07, Brian Adkins <lojicdotcom@gmail.com> wrote:
> On Oct 30, 11:58 pm, Phrogz <phr...@mac.com> wrote:
> > On Oct 30, 9:30 pm, Daniel Waite <rabbitb...@gmail.com> wrote:
> >
> > > That's clever use of ?a, which I recognize but have never seen anyone
> > > use before. Thanks for the example!
> >
> > My current favorite use for the ?x syntax is converting single-
> > character strings representing digits into their integer form:
>
> Yeah, so you can squeeze Ruby code into small places :)
>
> 1.upto(?d){|i|i%3<1&&x=:Fizz;puts i%5<1?"#{x}Buzz":x||i}

Except under the upcoming revision (1.9) of the (Ruby) Rules of Golf,
the R(uby)&A(ncient) has outlawed that usage, and instituted the
penalty that ?d will no longer be 100, but "d".

--
Rick DeNatale

My blog on Ruby
http://talklikeaduck.denh...