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  1. import re
  2. s_l = ['康熙十年','咸丰三年','民国二十二年', 'abc']
  3. rx = re.compile(r'(?!\u5E74)[\u2E80-\u2E99\u2E9B-\u2EF3\u2F00-\u2FD5\u3005\u3007\u3021-\u3029\u3038-\u303B\u3400-\u4DBF\u4E00-\u9FFF\uF900-\uFA6D\uFA70-\uFAD9\U00016FE2\U00016FE3\U00016FF0\U00016FF1\U00020000-\U0002A6DF\U0002A700-\U0002B738\U0002B740-\U0002B81D\U0002B820-\U0002CEA1\U0002CEB0-\U0002EBE0\U0002F800-\U0002FA1D\U00030000-\U0003134A]')
  4. print( [len(rx.findall(s)) for s in s_l] )
  5. # => [3, 3, 5, 0]
Success #stdin #stdout 0.03s 9628KB
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Standard input is empty
stdout
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[3, 3, 5, 0]
https://ideone.com/hNhktm
language:
Python 3 (python  3.12)
created:
3 years ago
visibility:
secret

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