# Regular expression **URL:** \ **Category:** ruby-talk\ **Created:** [16 January 2008 15:57 UTC](https://rubytalk.org/t/regular-expression/43822 "2008-01-16T15:57:59Z")\ **Posts on this page:** 11\ **Page:** 1
**Author:** ![Johnathan\_Smith](https://avatars.discourse-cdn.com/v4/letter/j/7ea924/32.png) [@Johnathan\_Smith](https://rubytalk.org/u/Johnathan_Smith)\ **Post date:** [16 January 2008 15:57 UTC](https://rubytalk.org/t/regular-expression/43822/1 "2008-01-16T15:57:59Z")
hi there i wanting to write a regular expression which matches words which contain the vowels [aeiou] in alphabetical order im not sure where to start so any help would be greatly appreciated thans > **路路路** > > -- > Posted via [http://www.ruby-forum.com/](http://www.ruby-forum.com/). ---
**Author:** ![Andrew\_Stewart](https://avatars.discourse-cdn.com/v4/letter/a/9de053/32.png) [@Andrew\_Stewart](https://rubytalk.org/u/Andrew_Stewart)\ **Post date:** [16 January 2008 16:52 UTC](https://rubytalk.org/t/regular-expression/43822/2 "2008-01-16T16:52:29Z")
> i wanting to write a regular expression which matches words which > contain the vowels [aeiou] in alphabetical order > > im not sure where to start so any help would be greatly appreciated This will get you started: IRB: \>\> re = /[^aeiou]\*[a][^aeiou]\*[e][^aeiou]\*[i][^aeiou]\*[o][^aeiou]\*[u][^aeiou]\*/ =\> [aeiou][a][aeiou][e][aeiou][i][aeiou][o][aeiou][u][aeiou] \>\> 'abstemious' =~ re =\> 0 \>\> 'facetious' =~ re =\> 0 \>\> 'bstemious' =~ re =\> nil \>\> 'ebstamious' =~ re =\> nil You can see there's a lot of repetition in the regular expression so the next step would probably be to DRY it. I leave that as an exercise to the reader (because I can't work out how to do it 馃檪 You'll may also wish to anchor the regular expression so it doesn't cross words. Regards, Andy Stewart > **路路路** > > On 16 Jan 2008, at 15:57, Johnathan Smith wrote: > > ------- > > > **[AirBlade Software](https://airbladesoftware.com/)** > > > > Andrew Stewart of AirBlade Software. ---
**Author:** ![Rick\_DeNatale1](https://avatars.discourse-cdn.com/v4/letter/r/bbce88/32.png) [@Rick\_DeNatale1](https://rubytalk.org/u/Rick_DeNatale1)\ **Post date:** [16 January 2008 16:53 UTC](https://rubytalk.org/t/regular-expression/43822/3 "2008-01-16T16:53:12Z")
Something like this? irb(main):001:0\> "frabelious" =~ /a.\*?e.\*?i.\*?o.\*?u/ =\> 2 irb(main):002:0\> "frabeelious" =~ /a.\*?e.\*?i.\*?o.\*?u/ =\> 2 irb(main):003:0\> "friaeelaous" =~ /a.\*?e.\*?i.\*?o.\*?u/ =\> nil > **路路路** > > On 1/16/08, Johnathan Smith \ wrote: > > > hi there > > > > i wanting to write a regular expression which matches words which > > contain the vowels [aeiou] in alphabetical order > > > > im not sure where to start so any help would be greatly appreciated > > > > thans > > -- > > Posted via [http://www.ruby-forum.com/\](http://www.ruby-forum.com/%5C). > > -- > Rick DeNatale > > My blog on Ruby > [http://talklikeaduck.denhaven2.com/](http://talklikeaduck.denhaven2.com/) ---
**Author:** ![Eivind\_Eklund](https://avatars.discourse-cdn.com/v4/letter/e/f17d59/32.png) [@Eivind\_Eklund](https://rubytalk.org/u/Eivind_Eklund)\ **Post date:** [17 January 2008 12:55 UTC](https://rubytalk.org/t/regular-expression/43822/4 "2008-01-17T12:55:03Z")
Start with regular expression documentation; I think that's what your teacher would want, instead of just having ruby-talk do your homework for you. [http://ysomeya.hp.infoseek.co.jp/eng-quick\_regex.html](http://ysomeya.hp.infoseek.co.jp/eng-quick_regex.html) is one place to start (very short); [http://evolt.org/article/thelist/20/22700/](http://evolt.org/article/thelist/20/22700/) is another, more extensive tutorial. Neither of these are Ruby specific - this is OK, as what you need is to understand the language regular expressions, not the language Ruby. Eivind. > **路路路** > > On Jan 16, 2008 4:57 PM, Johnathan Smith \ wrote: > > > hi there > > > > i wanting to write a regular expression which matches words which > > contain the vowels [aeiou] in alphabetical order > > > > im not sure where to start so any help would be greatly appreciated ---
**Author:** ![Andrew\_Stewart](https://avatars.discourse-cdn.com/v4/letter/a/9de053/32.png) [@Andrew\_Stewart](https://rubytalk.org/u/Andrew_Stewart)\ **Post date:** [16 January 2008 17:13 UTC](https://rubytalk.org/t/regular-expression/43822/5 "2008-01-16T17:13:32Z")
> \>\> re = /[^aeiou]\*[a][^aeiou]\*[e][^aeiou]\*[i][^aeiou]\*[o][^aeiou]\*[u][^aeiou]\*/ Oops, that should have been: \>\> re = /[^aeiou]\*a[^aeiou]\*e[^aeiou]\*i[^aeiou]\*o[^aeiou]\*u[^aeiou]\*/ Regards, Andy Stewart > **路路路** > > On 16 Jan 2008, at 16:52, Andrew Stewart wrote: > > ------- > > > **[AirBlade Software](https://airbladesoftware.com/)** > > > > Andrew Stewart of AirBlade Software. ---
**Author:** ![Wolfgang\_Nadasi-Donn](https://avatars.discourse-cdn.com/v4/letter/w/9fc348/32.png) [@Wolfgang\_Nadasi-Donn](https://rubytalk.org/u/Wolfgang_Nadasi-Donn)\ **Post date:** [16 January 2008 22:03 UTC](https://rubytalk.org/t/regular-expression/43822/6 "2008-01-16T22:03:16Z")
Rick Denatale wrote: > Something like this? > irb(main):001:0\> "frabelious" =~ /a.\*?e.\*?i.\*?o.\*?u/ > =\> 2 This doesn't work as expected... irb(main):001:0\> "xaxxixxexxixxuxxoxxuxx".match(/a.\*?e.\*?i.\*?o.\*?u/)[0] =\> "axxixxexxixxuxxoxxu" ..., because it allows all wovels in any order in the word, if there is one possible grouping with the wanted ordering - The subpattern "a.\*?e" for example allows any number of "a"s, "i"s, "o"s, and "u"s between the "a" and the "e". Wolfgang N谩dasi-Donner > **路路路** > > -- > Posted via [http://www.ruby-forum.com/\](http://www.ruby-forum.com/%5C). ---
**Author:** ![Wolfgang\_Nadasi-Donn](https://avatars.discourse-cdn.com/v4/letter/w/9fc348/32.png) [@Wolfgang\_Nadasi-Donn](https://rubytalk.org/u/Wolfgang_Nadasi-Donn)\ **Post date:** [16 January 2008 22:19 UTC](https://rubytalk.org/t/regular-expression/43822/7 "2008-01-16T22:19:25Z")
Andrew Stewart wrote: > You can see there's a lot of repetition in the regular expression so > the next step would probably be to DRY it. O.K. - it really increases the readability 馃槈 re = Regexp.compile((v=p='')+"^#{p="[^#{v='aeiou'}]\*"}"+v.split('').      join("[^#{v}]\*")+p+'$') p re # =\> /^[^aeiou]\*a[^aeiou]\*e[^aeiou]\*i[^aeiou]\*o[^aeiou]\*u[^aeiou]\*$/ Wolfgang N谩dasi-Donner > **路路路** > > -- > Posted via [http://www.ruby-forum.com/\](http://www.ruby-forum.com/%5C). ---
**Author:** ![Rick\_DeNatale1](https://avatars.discourse-cdn.com/v4/letter/r/bbce88/32.png) [@Rick\_DeNatale1](https://rubytalk.org/u/Rick_DeNatale1)\ **Post date:** [16 January 2008 22:47 UTC](https://rubytalk.org/t/regular-expression/43822/8 "2008-01-16T22:47:57Z")
Ahh, but "xaxxixxexxixxuxxoxxuxx" DOES contain the vowels in alphabetical order, the original problem statement said nothing about disallowing additional vowels. I really posted my "solution" to drive out what he really wanted. Kind of a test-driven approach. > **路路路** > > On 1/16/08, Wolfgang N谩dasi-Donner \ wrote: > > > Rick Denatale wrote: > > \> Something like this? > > \> irb(main):001:0\> "frabelious" =~ /a.\*?e.\*?i.\*?o.\*?u/ > > \> =\> 2 > > > > This doesn't work as expected... > > > > irb(main):001:0\> "xaxxixxexxixxuxxoxxuxx".match(/a.\*?e.\*?i.\*?o.\*?u/)[0] > > =\> "axxixxexxixxuxxoxxu" > > > > ..., because it allows all wovels in any order in the word, if there is > > one possible grouping with the wanted ordering - The subpattern "a.\*?e" > > for example allows any number of "a"s, "i"s, "o"s, and "u"s between the > > "a" and the "e". > > -- > Rick DeNatale > > My blog on Ruby > [http://talklikeaduck.denhaven2.com/](http://talklikeaduck.denhaven2.com/) ---
**Author:** ![Moises\_Trovo](https://avatars.discourse-cdn.com/v4/letter/m/bc79bd/32.png) [@Moises\_Trovo](https://rubytalk.org/u/Moises_Trovo)\ **Post date:** [16 January 2008 23:15 UTC](https://rubytalk.org/t/regular-expression/43822/9 "2008-01-16T23:15:47Z")
try something like: /a+[^a]\*e+[^ae]\*i+[^aei]\*o+[^aeio]\*u+/ > **路路路** > > On Jan 16, 2008 8:03 PM, Wolfgang N谩dasi-Donner \ wrote: > > > Rick Denatale wrote: > > \> Something like this? > > \> irb(main):001:0\> "frabelious" =~ /a.\*?e.\*?i.\*?o.\*?u/ > > \> =\> 2 > > > > This doesn't work as expected... > > > > irb(main):001:0\> "xaxxixxexxixxuxxoxxuxx".match(/a.\*?e.\*?i.\*?o.\*?u/)[0] > > =\> "axxixxexxixxuxxoxxu" > > > > ..., because it allows all wovels in any order in the word, if there is > > one possible grouping with the wanted ordering - The subpattern "a.\*?e" > > for example allows any number of "a"s, "i"s, "o"s, and "u"s between the > > "a" and the "e". > > > > Wolfgang N谩dasi-Donner > > -- > > > > Posted via [http://www.ruby-forum.com/\](http://www.ruby-forum.com/%5C). ---
**Author:** ![Andrew\_Stewart](https://avatars.discourse-cdn.com/v4/letter/a/9de053/32.png) [@Andrew\_Stewart](https://rubytalk.org/u/Andrew_Stewart)\ **Post date:** [17 January 2008 11:51 UTC](https://rubytalk.org/t/regular-expression/43822/10 "2008-01-17T11:51:37Z")
Yikes! That's clever though my brain copes better with this halfway house:    consonant = /[^aeiou]/    re = /#{consonant}\*a#{consonant}\*e#{consonant}\*i#{consonant}\*o#{consonant}\*u#{consonant}\*/ Or, perhaps, in-between the in-betweens:    vowels = %w( a e i o u )    consonant = '[^aeiou]'    re = /#{vowels.push('').unshift('').join("#{consonant}\*")}/ Regards, Andy Stewart > **路路路** > > On 16 Jan 2008, at 22:19, Wolfgang N谩dasi-Donner wrote: > > > Andrew Stewart wrote: > > > > > You can see there's a lot of repetition in the regular expression so > > > the next step would probably be to DRY it. > > > > O.K. - it really increases the readability 馃槈 > > > > re = Regexp.compile((v=p='')+"^#{p="[^#{v='aeiou'}]\*"}"+v.split(''). > >      join("[^#{v}]\*")+p+'$') > > p re # =\> > > /^[^aeiou]\*a[^aeiou]\*e[^aeiou]\*i[^aeiou]\*o[^aeiou]\*u[^aeiou]\*$/ > > ------- > > > **[AirBlade Software](https://airbladesoftware.com/)** > > > > Andrew Stewart of AirBlade Software. ---
**Author:** ![Wolfgang\_Nadasi-Donn](https://avatars.discourse-cdn.com/v4/letter/w/9fc348/32.png) [@Wolfgang\_Nadasi-Donn](https://rubytalk.org/u/Wolfgang_Nadasi-Donn)\ **Post date:** [17 January 2008 00:43 UTC](https://rubytalk.org/t/regular-expression/43822/11 "2008-01-17T00:43:58Z")
Rick Denatale wrote: > I really posted my "solution" to drive out what he really wanted. > Kind of a test-driven approach. Oh - I see. Let's wait for the more complete specification 馃槈 Wolfgang N谩dasi-Donner > **路路路** > > -- > Posted via [http://www.ruby-forum.com/\](http://www.ruby-forum.com/%5C).