匹配包含所有元音的单词的正则表达式是什么？

Question

I am learning regex in python but can't seem to get the hang of it. I am trying the filter out all the words containing all the vowels in english and this is my regex:

r'\b(\S*[aeiou]){5}\b'

seems like it is too vague since any vowel(even repeated ones) can appear at any place and any number is times so this is throwing words like 'actionable', 'unfortunate' which do have count of vowels as 5 but not all the vowels. I looked around the internet and found this regex:

r'[^aeiou]*a[^aeiou]*e[^aeiou]*i[^aeiou]*o[^aeiou]*u[^aeiou]*

But as it appears, its only for the sequential appearance of the vowels, pretty limited task than the one I am trying to accomplish. Can someone 'think out loud' while crafting the regex for the problem that I have?

Answer 1

If you plan to match words as chunks of text only consisting of English letters you may use a regex like

\b(?=\w*?a)(?=\w*?e)(?=\w*?i)(?=\w*?o)(?=\w*?u)[a-zA-Z]+\b

See the regex demo

To support languages other than English, you may replace [a-zA-Z]+ with [^\\W\\d_]+ .

If a "word" you want to match is a chunk of non-whitespace chars you may use

(?<!\S)(?=\S*?a)(?=\S*?e)(?=\S*?i)(?=\S*?o)(?=\S*?u)\S+

See this regex demo .

Define these patterns in Python using raw string literals, eg:

rx_AllVowelWords = r'\b(?=\w*?a)(?=\w*?e)(?=\w*?i)(?=\w*?o)(?=\w*?u)[a-zA-Z]+\b'

Details

• \\b(?=\\w*?a)(?=\\w*?e)(?=\\w*?i)(?=\\w*?o)(?=\\w*?u)[a-zA-Z]+\\b :
  • \\b - a word boundary, here, a starting word boundary
  • (?=\\w*?a)(?=\\w*?e)(?=\\w*?i)(?=\\w*?o)(?=\\w*?u) - a sequence of positive lookaheads that are triggered right after the word boundary position is detected, and require the presence of a , e , i , o and u after any 0+ word chars (letters, digits, underscores - you may replace \\w*? with [^\\W\\d_]*? to only check letters)
  • [a-zA-Z]+ - 1 or more ASCII letters (replace with [^\\W\\d_]+ to match all letters)
  • \\b - a word boundary, here, a trailing word boundary

The second pattern details:

• (?<!\\S)(?=\\S*?a)(?=\\S*?e)(?=\\S*?i)(?=\\S*?o)(?=\\S*?u)\\S+ :
  • (?<!\\S) - a position at the start of the string or after a whitespace
  • (?=\\S*?a)(?=\\S*?e)(?=\\S*?i)(?=\\S*?o)(?=\\S*?u) - all English vowels must be present - in any order - after any 0+ chars other than whitespace
  • \\S+ - 1+ non-whitespace chars.

Answer 2

I can't think of an easy way to find "words with all vowels" with a single regexp, but it can easily be done by anding-together regex matches to a, e, i, o, and u separately. For example, something like the following Python script should determine whether a given English word has all vowels (in any order, any multiplicity) or not:

#! /usr/bin/python3
# all-vowels.py
import sys
import re
if len(sys.argv) != 2: sys.exit()
word=sys.argv[1]
if re.search(r'a', word) and re.search(r'e', word) and re.search(r'i', word) and re.search(r'o', word) and re.search(r'u', word):
   print("Word has all vowels!")
else:
   print("Word does NOT have all vowels.")