import re
s = 'a12a34a56a78 ‘
reg1 = r'a(\d+)a(\d+)a(\d+)a(\d+)'
reg2 = r'(?:a(\d+)){4}' #我的错误示范
This topic created in 3375 days ago, the information mentioned may be changed or developed.
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shoalyAug 7, 2017
1
真要是这种有规律的东西 , 用循环加字符串切割吧
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sxmAug 7, 2017 via Android
2
(a(\d+))+
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L2AKnG8GXx60bc6PAug 7, 2017 via iPhone
3
you need regex
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L2AKnG8GXx60bc6PAug 7, 2017 via iPhone
4
i mean
import regex as re |
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momocraftAug 7, 2017
5
ruby 中這個操作叫 scan. 你可以看看 py 有沒有類似的
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inflationaaronAug 8, 2017
8
没办法，与其用 regex 不如` split('a')`
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code42Aug 20, 2017
9
re.findall(r'a(\d+)', 'a12a34a56a78')
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