摘要: import re ret = re.findall(r'((?<=\d)\d{4})+\b','12345678901 38473456 2305578 349579174 1786986745 18360869123') print(ret) ['8901', '3456', '5578', '9174', '6745', '9123'] #running result 阅读全文
posted @ 2018-08-02 14:26 pie_thn 阅读(1768) 评论(0) 推荐(0) 编辑