Python正则匹配含特殊字符单词的问题及修正方法问询

Hey there! Let's sort out that regex problem you're facing. Your current pattern \w{1,15} only matches alphanumeric characters and underscores—so it's no wonder it's missing words like single-word and b**m, and correctly ignoring the !? junk (which is what we want, but we need to capture the right words too).

The Solution Regex #

The pattern you need is r'[\w*-]+'. Let's break down what each part does:

• [\w*-]: This character set matches any alphanumeric character (\w covers letters, numbers, and underscores), hyphens (-), and asterisks (*). Placing the hyphen at the end of the set keeps it from being misinterpreted as a range marker (like a-z).
• +: This quantifier tells the regex to match one or more of the characters in the set, so it grabs entire words instead of single characters.

Updated Code #

Here's your revised code that will produce the exact output you want:

import re
str1 = "These should be counted as a single-word, b**m !?"
match_pattern = re.findall(r'[\w*-]+', str1)
print(match_pattern)

When you run this, you'll get the desired result:

['These', 'should', 'be', 'counted', 'as', 'a', 'single-word', 'b**m']

If you ever need to include other special characters in your "words" (like @ or #), just add them to the character set—for example, r'[\w*-@#]+' would include those symbols too.

内容的提问来源于stack exchange，提问作者James Rudolf