[英]Regular expression substitution in Python
I have a string 我有一个字符串
line = "haha (as jfeoiwf) avsrv arv (as qwefo) afneoifew"
From this I want to remove all instances of "(as...)"
using some regular expression. 从这里我想删除使用一些正则表达式的
"(as...)"
所有实例。 I want the output to look like 我希望输出看起来像
line = "haha avsrv arv afneoifew"
I tried: 我试过了:
line = re.sub(r'\(+as .*\)','',line)
But this yields: 但这会产生:
line = "haha afneoifew"
To get non-greedy behaviour , you have to use *?
要获得非贪婪的行为 ,你必须使用
*?
instead of *
, ie re.sub(r'\\(+as .*?\\) ','',line)
. 而不是
*
,即re.sub(r'\\(+as .*?\\) ','',line)
。 To get the desired string, you also have to add a space, ie re.sub(r'\\(+as .*?\\) ','',line)
. 要获得所需的字符串,还必须添加一个空格,即
re.sub(r'\\(+as .*?\\) ','',line)
。
The problem is that your regexp matches this whole group : (as jfeoiwf) avsrv arv (as qwefo)
, hence your result. 问题是你的正则表达式匹配整个组:(
(as jfeoiwf) avsrv arv (as qwefo)
,因此你的结果。
You can use : 您可以使用 :
>>> import re
>>> line = "haha (as jfeoiwf) avsrv arv (as qwefo) afneoifew"
>>> line = re.sub(r'\(+as [a-zA-Z]*\)','',line)
>>> line
'haha avsrv arv afneoifew'
Hope it'll be helpful. 希望它会有所帮助。
You were very close. 你非常接近。 You need to use lazy quantifier '?'
你需要使用懒惰的量词'?' after .*.
之后。*。 In default it will try to capture biggest group it possibly can.
默认情况下,它会尝试捕获它可能的最大组。 With lazy quantifier it'll actually try to match smallest possible groups.
使用惰性量词,它实际上会尝试匹配最小的可能组。
line = re.sub(r'\(+as .*?\) ','',line)
尝试:
re.sub(u".\(as \w+\).", ' ',line)
尝试:
re.sub(r'\(as[^\)]*\)', '', line)
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