Python RegEx using re.sub with multiple patterns

Question

I'm trying to use Python RegEx re.sub to remove a colon before the antepenultimate vowel [aeiou] of a word if the antepenultimate vowel (from the end) is preceded by another vowel.

So the colon has to be between the 3rd and 4th vowel counting from the end of the word.

So the 1st example given would break down like this w4:32ny1h.

we:aanyoh > weaanyoh    # w4:32ny1h
hiru:atghigu > hiruatghigu
yo:ubeki > youbeki

Below is the RegEx statement I'm trying to use but I can't get it to work.

word = re.sub(ur"([aeiou]):([aeiou])(([^aeiou])*([aeiou])*([aeiou])([^aeiou])*([aeiou]))$", ur'\1\2\3\4', word)

So you basically search for two vowels connected by a colon ? — ion
– ion, Commented Nov 15, 2015 at 19:47
the colon has to be between the 3rd and 4th vowel counting from the end of the word. so the 1st example given would break down like this w4:32ny1h. — user2743
– user2743, Commented Nov 15, 2015 at 19:49
Nice to specify :) thought of that after decrypting your pattern :) — ion
– ion, Commented Nov 15, 2015 at 19:50
What about wef:fewee ? The colon is before the third-from-last vowel, there is a colon preceeding it, and there are two vowels later on. It satisfies your description, but not your code. — TessellatingHeckler
– TessellatingHeckler, Commented Nov 15, 2015 at 20:20

Jeff Y · Accepted Answer · 2015-11-15 20:17:00Z

1

Don't you just have too many parentheses (and other extra stuff)?:

word = re.sub(ur"([aeiou]):(([aeiou][^aeiou]*){3})$", ur'\1\2', word)

answered Nov 15, 2015 at 20:17

Jeff Y

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2 Comments

ion Over a year ago

This will not make sure the first character after colon is a vowel , see my answer.

dawg Over a year ago

[^aeiou] matches spaces, line feeds, other colons and punctuation.

Tom Zych · Accepted Answer · 2015-11-15 20:23:50Z

1

Not sure if you want to completely ignore consonants; this regex will. Otherwise similar to Jeff's.

import re

tests = [
    'we:aanyoh',
    'hiru:atghigu',
    'yo:ubeki',
    'yo:ubekiki',
    'yo:ubek'
]

for word in tests:
    s = re.sub(r'([^aeiou]*[aeiou][^aeiou]*):((?:[^aeiou]*[aeiou]){3}[^aeiou]*)$', r'\1\2', word)
    print '{} > {}'.format(word, s)

answered Nov 15, 2015 at 20:23

Tom Zych

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1 Comment

dawg Over a year ago

[^aeiou] matches spaces, line feeds, other colons and punctuation.

dawg · Accepted Answer · 2015-11-16 04:20:59Z

You state that you are targeting a word versus a line, so first set anchors to only deal with words:

\b[regex will go here]\b
^                      ^     assert a word boundary

Next, a colon proceeded by and followed by a [aeiou] with two more [aeiou] in the portion following the colon. I assume case independent?

(?i)(\b\w+[aeiou]):((?:[aeiou][^aeiou\s\W]*){3}\b)
                                   ^  match a character that is NOT vowel, space or not a 
                                         ^   \W=[^a-zA-Z0-9_]

Demo

(Note the use of [^aeiou\W] which is consonant letters, numbers and _ but not other characters Demo.)

Python demo:

import re

tests={
    'matches':[
        'we:aanyoh',
        'hiru:atghigu',
        'yo:ubeki'
        ],
    'no match':[
        'wz:ubeki',
        'we:a anyoh',
        'yo:ubek',
        'hiru:atghiguu'
    ]    
}

for k, v in tests.items():
    print k
    for e in v:
        s=re.sub(r'(?i)(\b\w+[aeiou]):((?:[aeiou][^aeiou\s\W]*){3}\b)', r'\1\2', e)
        print '\t{} > {}'.format(e, s)

Prints:

matches
    we:aanyoh > weaanyoh
    hiru:atghigu > hiruatghigu
    yo:ubeki > youbeki
no match
    wz:ubeki > wz:ubeki
    we:a anyoh > we:a anyoh
    yo:ubek > yo:ubek
    hiru:atghiguu > hire:atghiguu

This will only handle words with a single colon. If you want to match words that have multiple colons but have this same pattern, change the LH pattern to have the character class that includes a colon and an anchor that is not \b.

Example: (?i)(^[\w:]+[aeiou]):((?:[aeiou][^aeiou\s\W]*){3}\b)

Alexander · Accepted Answer · 2015-11-15 20:13:34Z

0

It should work with this:

word = re.sub(ur"(?<=[aeiou]):(?=[aeiou]([^aeiou]*[aeiou]){2}[^aeiou]*$)", ur'', word)

see example here: https://regex101.com/r/kA8xH3/2

notice that I only capture the colon and replace it with an empty string rather than capturing groups and concatenate them.

Tt checks for the colon combination, then does a lookahead to check that there are 2 additional vowels (and maybe consonants). It allso allows additional consonants at the end but makes sure that there are no more vowel through the $

edited Nov 15, 2015 at 20:13

answered Nov 15, 2015 at 20:09

Alexander

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4 Comments

Alexander Over a year ago

[^aeiou]*$ the star says that it can be 0 or more, so it can end with 0 consonants which means a vowel

ion Over a year ago

this will fail if there's a space or another character :) Just trying to get even lol

Alexander Over a year ago

@lonut Where? I use [^aeiou] so the second should include space and special characters, shouldn't it? And if mine does, your would as well, right? ;)

ion Over a year ago

:) how about now ? :)

ion · Accepted Answer · 2015-11-15 21:05:55Z

0

This will do it:

    word = re.sub(ur"([aeiou]):([aeiou])([^\Waeiou]*[aeiou][^\Waeiou]*[aeiou][^\Waeiou]*)$", ur'\1\2\3', word)

http://www.phpliveregex.com/p/dCa

edited Nov 15, 2015 at 21:05

answered Nov 15, 2015 at 20:22

ion

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3 Comments

Alexander Over a year ago

that won't work. In your expression there have to be 2 vowels after the colon which means something like the 2nd and 3rd example won't be captured

Alexander Over a year ago

yep i saw that. But you have ([aeiou]):([aeiou])(([aeiou] which says that there have to be 3 vowels in a row, with the 1st and 2nd one having a colon between them

Alexander Over a year ago

yep and what is between :([aeiou]) and (([aeiou][^aeiou]*){2})? there is not another consonant allowed, so something like dsa:adaa wouldn't be allowed

7stud · Accepted Answer · 2015-11-15 21:37:49Z

-1

Roundup(I used a capital vowel to indicate where in the word the substitution should take place). Let me know if you want me to add other test strings.

import re

strings = [
    'wE:aanyoh',
    'hirU:atghigu',
    'yO:ubeki',

    'xE:aaa',
    'xx:aaa',
    'xa:aaaxA:aaa',
    'xa:aaaxA:aaaxx',
    'xa:aaaxA:aaxax',
    'a:aaaxA:aaxax',
    'e:aeixA:aexix',
]


pattern = r"""
    (
        .*
        [aeiou]
    )
    :
    (
        [aeiou]
        .*?
        [aeiou]
        .*?
        [aeiou]
    )
"""

template = "{:>15}: {}"
for string in strings:
    print(
        template.format('original', string)
    )

    print(template.format('Alexander:', 
        re.sub(ur"(?<=[aeiou]):(?=[aeiou]([^aeiou]*[aeiou]){2}[^aeiou]*$)", ur'', string, flags=re.I)
    ))

    print(template.format('lonut:', 
        re.sub(ur"([aeiou]):([aeiou])([^\Waeiou]*[aeiou][^\Waeiou]*[aeiou][^\Waeiou]*)$", ur'\1\2\3', string, flags=re.I)
    ))

    print(template.format('Tom Zych:', 
        re.sub(r'([^aeiou]*[aeiou][^aeiou]*):((?:[^aeiou]*[aeiou]){3}[^aeiou]*)$', r'\1\2', string, flags=re.I)
    ))

    print(template.format('Jeff Y:', 
        re.sub(ur"([aeiou]):(([aeiou][^aeiou]*){3})$", ur'\1\2', string, flags=re.I)
    ))

    print(template.format('7stud:', 
        re.sub(pattern, r'\1\2', string, count=1, flags=re.X|re.I)
    ))

    print("\n")

       original: wE:aanyoh
     Alexander:: wEaanyoh
         lonut:: wEaanyoh
      Tom Zych:: wEaanyoh
        Jeff Y:: wEaanyoh
         7stud:: wEaanyoh


       original: hirU:atghigu
     Alexander:: hirUatghigu
         lonut:: hirUatghigu
      Tom Zych:: hirUatghigu
        Jeff Y:: hirUatghigu
         7stud:: hirUatghigu


       original: yO:ubeki
     Alexander:: yOubeki
         lonut:: yOubeki
      Tom Zych:: yOubeki
        Jeff Y:: yOubeki
         7stud:: yOubeki


       original: xE:aaa
     Alexander:: xEaaa
         lonut:: xEaaa
      Tom Zych:: xEaaa
        Jeff Y:: xEaaa
         7stud:: xEaaa


       original: xx:aaa
     Alexander:: xx:aaa
         lonut:: xx:aaa
      Tom Zych:: xx:aaa
        Jeff Y:: xx:aaa
         7stud:: xx:aaa


       original: xa:aaaxA:aaa
     Alexander:: xa:aaaxAaaa
         lonut:: xa:aaaxAaaa
      Tom Zych:: xa:aaaxAaaa
        Jeff Y:: xa:aaaxAaaa
         7stud:: xa:aaaxAaaa


       original: xa:aaaxA:aaaxx
     Alexander:: xa:aaaxAaaaxx
         lonut:: xa:aaaxAaaaxx
      Tom Zych:: xa:aaaxAaaaxx
        Jeff Y:: xa:aaaxAaaaxx
         7stud:: xa:aaaxAaaaxx


       original: xa:aaaxA:aaxax
     Alexander:: xa:aaaxAaaxax
         lonut:: xa:aaaxAaaxax
      Tom Zych:: xa:aaaxAaaxax
        Jeff Y:: xa:aaaxAaaxax
         7stud:: xa:aaaxAaaxax


       original: a:aaaxA:aaxax
     Alexander:: a:aaaxAaaxax
         lonut:: a:aaaxAaaxax
      Tom Zych:: a:aaaxAaaxax
        Jeff Y:: a:aaaxAaaxax
         7stud:: a:aaaxAaaxax


       original: e:aeixA:aexix
     Alexander:: e:aeixAaexix
         lonut:: e:aeixAaexix
      Tom Zych:: e:aeixAaexix
        Jeff Y:: e:aeixAaexix
         7stud:: e:aeixAaexix

answered Nov 15, 2015 at 21:37

7stud

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8 Comments

TessellatingHeckler Over a year ago

in your pattern - [aeiou].*?[aeiou].*?[aeiou] - vowel followed by anything? Won't that allow hundreds of vowels? Your test cases seem to only test the cases where there are exactly three vowels in the text after the colon.

7stud Over a year ago

@TessellatingHeckler, Yep. :(

Mariano Over a year ago

I don't understand why you're comparing other answers. Would you mind explaining what's the difference?

Mariano Over a year ago

@7stud In this case, all answers yield the same result, that's why I'm asking (there's no "hello" vs. "goodbye" cases here). I'd suggest testing against the strings a:eiouaeiou or a:exxx io where your pattern fails. Also, I don't understand why you left out dawg's answer, that I think should be the accepted answer for this question.

7stud Over a year ago

@Mariano, Suppose that with one sample string one answer produced the result "hello" and another answer produced the result "goodbye" what would you make of that? For example, THeckler suggested a thought experiment, which was like adding an additional test string, and my answer essentially produced 'hello' while all the other answers produced 'goodbye'. If everyone is testing different strings, then you can't know if all of the answers will work on all the strings that were tested.

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