RegEx获取除字母，空格，'和-以外的所有内容

Question

Is there a more simplified regular expression to match anything that is not a letter, hypen, space, or apostrophe?

This is the regex I was using...

[^\w\s'-]|\d|_|\xa0

It's working, I was just curious if there was a more simplified expression

Answer 1

[^a-zA-Z-' ]

匹配除字母Az，连字符，空格和撇号以外的所有字符

Answer 2

\\w already includes \\d and _ . So the simplest regex will be:

[^\w\s\-']

Answer 3

The following pattern...

[^a-z- ']

...is simpler and should do what you want with case-insensitivity set:

import re
p = re.compile(ur'[^a-z- \']', re.IGNORECASE)
test_str1 = u"9"
test_str2 = u"["
test_str3 = u"_"

re.search(p, test_str1)
re.search(p, test_str2)
re.search(p, test_str3)

Mirroring Maroun Maroun's comment, \\w matches _ ; it also matches 0-9 : so saying "not az or AZ or 0-9 or _ " with [^\\w ... ] ...then saying " 0-9 or _ " with |\\d|_ is a bit confusing and needlessly complicating.

Same with \\s , as it matches more than a space (specifically a carriage return, new line, tab, or form feed), which does not jive with wanting to match "anything that is not... a space ...": given your description then, use a literal over the \\s character class.