如何转义字符串中的特殊正则表达式字符？

Question

I use re.findall(p, text) to match a pattern generally, but now I came across a question:

I just want p to be matched as a normal string, not regex.

For example: p may contain '+' or '*', I don't want these characters have special meanings as in regex. In another word, I want p to be matched character by character.

In this case p is unknown to me, so I can't add '\' into it to ignore special character.

Answer 1

You can use re.escape :

>>> p = 'foo+*bar'
>>> import re
>>> re.escape(p)
'foo\\+\\*bar'

Or just use string operations to check if p is inside another string:

>>> p in 'blablafoo+*bar123'
True
>>> 'foo+*bar foo+*bar'.count(p)
2

By the way, this is mainly useful if you want to embed p into a proper regex:

>>> re.match(r'\d.*{}.*\d'.format(re.escape(p)), '1 foo+*bar 2')
<_sre.SRE_Match object at 0x7f11e83a31d0>

Answer 2

If you don't need a regex, and just want to test if the pattern is a substring of the string, use:

if pattern in string:

If you want to test at the start or end of the string:

if string.startswith(pattern): # or .endswith(pattern)

See the string methods section of the docs for other string methods.

If you need to know all locations of a substring in a string, use str.find :

offsets = []
offset = string.find(pattern, 0)
while offset != -1:
    offsets.append(offset)
    # start from after the location of the previous match
    offset = string.find(pattern, offset + 1)

Answer 3

You can use .find on strings. This returns the index of the first occurence of the "needle" string (or -1 if it's not found). eg

>>> a = 'test string 1+2*3'
>>> a.find('str')
5
>>> a.find('not there')
-1
>>> a.find('1+2*')
12