如何用空字符串替换所有方括号及其值？

Question

如果您不知道有多少个方括号，如何用空字符串替换所有方括号及其值？

data = "[a] 1 [b] test [c] other characters"

我希望数据是" 1 test other characters" 。

Answer 1

您可以使用正则表达式在括号部分拆分字符串：

import re

data = "[a] 1 [b] 2 [c]"

parts = re.split(r'\[.*?\]', data)
# ['', ' 1 ', ' 2 ', '']
out = '[' + ' '.join(parts) + ']'
print(out)
# [  1   2  ]

Answer 2

另一个冗长的答案是：

import re
data = "[a] 1 [b] test [c] other characters"

pattern = '\[.*?\]' #any character(s) with [ ]

unwanted = set(re.findall(pattern,data))

results = ' '.join(i for i in data.split() 
                   if i not in unwanted)

print(results)
# 1 test other characters

更新
加法条件
问题 > “[a] 1 [b] test [1] other [c]haracters” 由于.split()这不会被替换

import re

data = "[a] 1 [b] test [1] other [c]haracters"

pattern = '\[.*?\]' #any character(s) with [ ]

print(re.sub(pattern,'',data))
# " 1  test  other  haracters"

删除多余的空格

_ = re.sub(pattern,'',data).split()
print(' '.join(i for i in _ if i))

# "1 test other haracters"

Answer 3

尝试正则表达式替代。

import re

data = "[a] 1 [b] 2 [c]"
new = re.sub("([[a-z]*])", "", data)