How to match any string from a list of strings in

2020-05-20 04:46发布

Lets say I have a list of strings,

string_lst = ['fun', 'dum', 'sun', 'gum']

I want to make a regular expression, where at a point in it, I can match any of the strings i have in that list, within a group, such as this:

import re
template = re.compile(r".*(elem for elem in string_lst).*")
template.match("I love to have fun.")

What would be the correct way to do this? Or would one have to make multiple regular expressions and match them all separately to the string?

5条回答
做个烂人
2楼-- · 2020-05-20 05:27

Except for the regular expression, you can use list comprehension, hope it's not off the topic.

import re
def match(input_string, string_list):
    words = re.findall(r'\w+', input_string)
    return [word for word in words if word in string_list]

>>> string_lst = ['fun', 'dum', 'sun', 'gum']
>>> match("I love to have fun.", string_lst)
['fun']
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smile是对你的礼貌
3楼-- · 2020-05-20 05:30
string_lst = ['fun', 'dum', 'sun', 'gum']
x="I love to have fun."

print re.findall(r"(?=("+'|'.join(string_lst)+r"))",x)

You cannot use match as it will match from start.Use findall instead.

Output:['fun']

using search you will get only the first match.So use findall instead.

Also use lookahead if you have overlapping matches not starting at the same point.

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Viruses.
4楼-- · 2020-05-20 05:38

In line with @vks reply - I feel this actually does the comeplete task..

finds = re.findall(r"(?=(\b" + '\\b|\\b'.join(string_lst) + r"\b))", x)

Adding word boundary completes the task!

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狗以群分
5楼-- · 2020-05-20 05:39

regex module has named lists (sets actually):

#!/usr/bin/env python
import regex as re # $ pip install regex

p = re.compile(r"\L<words>", words=['fun', 'dum', 'sun', 'gum'])
if p.search("I love to have fun."):
    print('matched')

Here words is just a name, you can use anything you like instead.
.search() methods is used instead of .* before/after the named list.

To emulate named lists using stdlib's re module:

#!/usr/bin/env python
import re

words = ['fun', 'dum', 'sun', 'gum']
longest_first = sorted(words, key=len, reverse=True)
p = re.compile(r'(?:{})'.format('|'.join(map(re.escape, longest_first))))
if p.search("I love to have fun."):
    print('matched')

re.escape() is used to escape regex meta-characters such as .*? inside individual words (to match the words literally).
sorted() emulates regex behavior and it puts the longest words first among the alternatives, compare:

>>> import re
>>> re.findall("(funny|fun)", "it is funny")
['funny']
>>> re.findall("(fun|funny)", "it is funny")
['fun']
>>> import regex
>>> regex.findall(r"\L<words>", "it is funny", words=['fun', 'funny'])
['funny']
>>> regex.findall(r"\L<words>", "it is funny", words=['funny', 'fun'])
['funny']
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冷血范
6楼-- · 2020-05-20 05:43

You should make sure to escape the strings correctly before combining into a regex

>>> import re
>>> string_lst = ['fun', 'dum', 'sun', 'gum']
>>> x = "I love to have fun."
>>> regex = re.compile("(?=(" + "|".join(map(re.escape, string_lst)) + "))")
>>> re.findall(regex, x)
['fun']
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