Python: Elegant way to check if at least one regex in list matches a string

Question

I have a list of regexes in python, and a string. Is there an elegant way to check if the at least one regex in the list matches the string? By elegant, I mean something better than simply looping through all of the regexes and checking them against the string and stopping if a match is found.

Basically, I had this code:

list = ['something','another','thing','hello']
string = 'hi'
if string in list:
  pass # do something
else:
  pass # do something else

Now I would like to have some regular expressions in the list, rather than just strings, and I am wondering if there is an elegant solution to check for a match to replace if string in list:.

Thanks in advance.

python has nothing that does emacs lisp's regexp-opt https://www.gnu.org/software/emacs/manual/html_node/elisp/Regexp-Functions.html — ggg, Mar 14 '14 at 09:10
Then the question should not be for "elegant", but for "less than O(n)". — Martin Thoma, Dec 15 '18 at 08:45
a) You mean 'performant' not 'elegant' b) You want to know which individual regex matched (but that doesn't mean you need to iterate over each one individually, which will not be scaleable). — smci, Apr 28 '21 at 03:01

score 109 · Accepted Answer · answered Jun 14 '10 at 20:48

109

import re

regexes = [
    "foo.*",
    "bar.*",
    "qu*x"
    ]

# Make a regex that matches if any of our regexes match.
combined = "(" + ")|(".join(regexes) + ")"

if re.match(combined, mystring):
    print "Some regex matched!"

answered Jun 14 '10 at 20:48

Ned Batchelder

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If you don't need to know which one matched, it's better to bracket them with `(?:regex)` instead of `(regex)` – John Machin Jun 15 '10 at 00:22
4

This method doesn't work if there are more than 100 regexes in the array (Python 2.6). Try nosklo's answer below. – Amjith Aug 18 '11 at 15:13
7

`regexes = '(?:%s)' % '|'.join(regexes)` – alxndr Feb 01 '13 at 21:08
Any idea why "quax" does not match? – radar Mar 12 '19 at 08:52
4

Because "qu*x" means, q, zero or more u's, and an x. – Ned Batchelder Mar 12 '19 at 14:25
1

in the above answer how can i come to know which regex matched? – Prafful Garg Jul 14 '20 at 10:35

nosklo · Answer 2 · 2010-06-14T23:45:08.847

103

import re

regexes = [
    # your regexes here
    re.compile('hi'),
#    re.compile(...),
#    re.compile(...),
#    re.compile(...),
]

mystring = 'hi'

if any(regex.match(mystring) for regex in regexes):
    print 'Some regex matched!'

edited Jun 14 '10 at 23:45

answered Jun 14 '10 at 20:39

nosklo

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If working in python 2.4, you won't have `any` - see http://stackoverflow.com/questions/3785433/python-backports-for-some-methods – Sam Heuck Sep 12 '13 at 19:42
3

How is this *"something better than simply looping through all of the regexes and checking them against the string and stopping if a match is found"*? I guess the combination of Ned's and this answer could be a winner though... – johndodo Jan 21 '14 at 15:26

score 11 · Answer 3 · answered Dec 28 '18 at 05:07

Here's what I went for based on the other answers:

raw_list = ["some_regex","some_regex","some_regex","some_regex"]
reg_list = map(re.compile, raw_list)

mystring = "some_string"

if any(regex.match(mystring) for regex in reg_list):
    print("matched")

score 7 · Answer 4 · answered Oct 30 '17 at 14:40

A mix of both Ned's and Nosklo's answers. Works guaranteed for any length of list... hope you enjoy

import re   
raw_lst = ["foo.*",
          "bar.*",
          "(Spam.{0,3}){1,3}"]

reg_lst = []
for raw_regex in raw_lst:
    reg_lst.append(re.compile(raw_regex))

mystring = "Spam, Spam, Spam!"
if any(compiled_reg.match(mystring) for compiled_reg in reg_lst):
    print("something matched")

score 2 · Answer 5 · answered Dec 15 '18 at 01:50

2

If you loop over the strings, the time complexity would be O(n). A better approach would be combine those regexes as a regex-trie.

answered Dec 15 '18 at 01:50

ospider

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Python: Elegant way to check if at least one regex in list matches a string

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