Python Regex:匹配不在单词前面或后面有数字的字符串

这是一个可以使用的正则表达式:

(?:^\s+|[^\w\s]+\s*|\b[^\d\s]+\s+)(April)\b(?!\s*\w*\d)

设置了case-independent标志。在捕获组1中捕获目标单词。

Demo

Python的regex引擎执行以下操作:

(?:           # begin non-cap grp
  ^           # match beginning of line
  \s*         # match 0+ whitespace characters
  |           # or
  [^\w\s]+    # match 1+ chars other than word chars and whitespace
  \s*         # match 0+ whitespace chars
  |           # or
  \b          # match word break
  [^\d\s]+    # match 1+ chars other than digits and whitespace
  \s+         # match 1+ whitespace chars
)             # end non-cap grp  
(April)       # match 'April' in capture group
\b            # match word break
(?!           # begin negative lookahead
  \s*         # match 0+ whitespace chars         
  \w*         # match 0+ word chars
  \d          # match a digit
)             # end negative lookahead

我所采取的方法是指明可能发生的事 "April" 为什么不跟着它走呢。我不能说明什么不能先于 “四月” 因为这需要一个否定的lookbehind,这是Python的regex引擎不支持的。

我认为

• 在字符串的开头,后面可能有空格;
• 前接不含数字的单词,后面可能有空格。

我也这么认为 后跟一个分词符,该分词不能后跟包含数字的单词,前面可能有空格。

这是可以做到的 Pypi regex library 它支持可变长度的lookbehind。

import regex

str = 'Today is 4th April. Her name is April. Tomorrow is April 5th.'
res = regex.sub(r'(?<!\d[a-z]* )April(?! [a-z]*\d)', 'PERSON', str)
print(res)

输出:

Today is 4th April. Her name is PERSON. Tomorrow is April 5th.

说明:

(?<!\d[a-z]* )      # negative lookbehind, make sure we haven't a digit followed by 0 or more letters and a space before
April               # literally
(?! [a-z]*\d)       # negative lookahead, make sure we haven't a space, 0 or more letters and a digit after

更新 re

import re

str = 'Today is 4th April. Her name is April. Tomorrow is April 5th.'
res = re.sub(r'(\b[a-z]+ )April(?! [a-z]*\d)', '\g<1>PERSON', str)
print(res)