Python 用正则表达式匹配复杂字符串

Question

I have the following test string:

test_str = `It isn't directed at all,' said the White Rabbit;

My current regular expression uses re.sub to filter out the punctuation so that I can do my own operations.

My current regex is re.sub(r"[^A-Za-z0-9'\\s]", '', test_str)

The output from above is:

['It', "isn't", 'directed', 'at', "all'", 'said', 'the', 'White', 'Rabbit']

The error can be seen at all' when it is suppose to be storing all only.

How do you store words with 's and also ignore ' that comes after a punctuation? In this case, all,' .

Answer 1

Try the following:

import re
test_str = "`It isn't directed at all,' said the White Rabbit;"
a = re.sub(r"[^A-Za-z0-9'\s]", '', test_str)
a = re.sub(r"'[ ]", ' ', a)
print(a)

Answer 2

Try using this regular expression:

print(re.sub('["!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~''](?!\w+)', '', test_str))

Output:

It isn't directed at all said the White Rabbit

Answer 3

Here are other solutions

re.sub("\'[^\w]",' ', test_str)
re.sub("\'[\s]",' ', test_str)
re.sub("\'(?!\w)",'', test_str)
re.sub("\'(?=\s)",'', test_str)