python's re: return True if string contains regex pattern

Question

import re
word = 'fubar'
regexp = re.compile(r'ba[rzd]')
if regexp.search(word):
  print 'matched'

The best one by far is

bool(re.search('ba[rzd]', 'foobarrrr'))

Returns True

Match objects are always true, and None is returned if there is no match. Just test for trueness.

Code:

>>> st = 'bar'
>>> m = re.match(r"ba[r|z|d]",st)
>>> if m:
...     m.group(0)
...
'bar'

Output = bar

If you want search functionality

>>> st = "bar"
>>> m = re.search(r"ba[r|z|d]",st)
>>> if m is not None:
...     m.group(0)
...
'bar'

and if regexp not found than

>>> st = "hello"
>>> m = re.search(r"ba[r|z|d]",st)
>>> if m:
...     m.group(0)
... else:
...   print "no match"
...
no match

As @bukzor mentioned if st = foo bar than match will not work. So, its more appropriate to use re.search.

Here's a function that does what you want:

import re

def is_match(regex, text):
    pattern = re.compile(regex)
    return pattern.search(text) is not None

The regular expression search method returns an object on success and None if the pattern is not found in the string. With that in mind, we return True as long as the search gives us something back.

Examples:

>>> is_match('ba[rzd]', 'foobar')
True
>>> is_match('ba[zrd]', 'foobaz')
True
>>> is_match('ba[zrd]', 'foobad')
True
>>> is_match('ba[zrd]', 'foobam')
False