python regex查找至少包含一个字母的字母数字字符串

您可以在python中将这个基于lookahead的regex用于 findall

(?i)>((?=\d*[a-z])[a-z\d]{4})<

RegEx Demo

代码:

>>> regex = re.compile(r">((?=\d*[a-z])[a-z\d]{4})<", re.I)
>>> s = ">ABCD<>1234<>ABC1<>ABC2"
>>> print (regex.findall(s))
['ABCD', 'ABC1']

正则表达式详细信息:

• re.I :启用忽略大小写修饰符
• > :匹配文字字符 >
• ( :启动捕获组
  • (?=\d*[a-z]) :Lookahead断言0或更多数字后至少有一个字母
  • [a-z\d]{4}
• ) :结束捕获组
• < :匹配文字字符 <

import re
sentence = ">ABCD<>1234<>ABC1<>ABC2"
pattern = "\>((?=[a-zA-Z])(.){4})\<"
m = [m[0] for m in re.findall(pattern, sentence)]
#outputs ['ABCD', 'ABC1']