>>> a = ["[Test](Link)", "[Test2](link2)", "[test3](link3)"]
>>> b1,b2,b3,b4,b5,b6 = (y.strip('[]()') for x in a for y in x.split(']'))
>>> print (b1,b2,b3,b4,b5,b6)
Test Link Test2 link2 test3 link3
import re
a = ["[Test](Link)", "[Test2](link2)", "[test3](link3)"]
for s in a:
m = re.match('(\[.*\])(\(.*\))$', s)
print(m.group(1))
print(m.group(2))
( # Matching group 1
\[.*?\] # Matches non-greedily in between brackets
| # OR
\(.*?\) # Matches non-greedily between parenthesis
) # End of matching group
结果:
在列表中使用
join
之后,可以使用re.findall
:正则表达式解释:
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