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
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