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

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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
  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.

import re

regexes = [

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

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

import re

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

mystring = 'hi'

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

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):

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.*",

reg_lst = []
for raw_regex in raw_lst:

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

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.

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