Python的正則表達式來刪除所有方括號和它們的內容

Question

我想用這個正則表達式來從字符串中刪除所有方括號（和它們中的所有）的實例。例如，這個工程時，有字符串中只有一對括號組成：

import re 
pattern = r'\[[^()]*\]' 
s = """Issachar is a rawboned[a] donkey lying down among the sheep pens.""" 
t = re.sub(pattern, '', s) 
print t 

我得到的是正確的：

>>>Issachar is a rawboned donkey lying down among the sheep pens. 

但是，如果我的字符串包含一個以上的組方括號，它不起作用。例如：

s = """Issachar is a rawboned[a] donkey lying down among the sheep pens.[b]""" 

我得到：

>>>Issachar is a rawboned 

我需要的正則表達式不管方括號中有多少字符串中工作。正確的答案應該是：

>>>Issachar is a rawboned donkey lying down among the sheep pens. 

我已經研究並嘗試了許多排列無濟於事。

Answer 1

默認情況下*（或+）貪婪地匹配，所以在這個問題給出的模式將匹配高達最後]。

>>> re.findall(r'\[[^()]*\]', "Issachar is a rawboned[a] donkey lying down among the sheep pens.[b]") 
['[a] donkey lying down among the sheep pens.[b]'] 

通過重複操作符（*）後追加?，你可以把它非貪婪的方式匹配。

>>> import re 
>>> pattern = r'\[.*?\]' 
>>> s = """Issachar is a rawboned[a] donkey lying down among the sheep pens.[b]""" 
>>> re.sub(pattern, '', s) 
'Issachar is a rawboned donkey lying down among the sheep pens.' 

Answer 2

嘗試：

import re 
pattern = r'\[[^\]]*\]' 
s = """Issachar is a rawboned[a] donkey lying down among the sheep pens.[b]""" 
t = re.sub(pattern, '', s) 
print t 

輸出：

Issachar is a rawboned donkey lying down among the sheep pens.