Python regex删除[Number]的所有出现

Question

我正在寻找python中的regex语句，它将从字符串中删除所有出现的[1]或[17]或基本上[*]。发生情况如下，其中*等于某个数：

[*]
[ * ]
[ *]
[* ]

目前我有：

re.sub(r'\[*\]', '', origional_string)

引发invalid expression

输入字符串示例：

makeup of organisms.[10] In 1997, while working at the University of Tennessee, Pigliucci received the Theodosius Dobzhansky Prize,[11] 

预期产量：

makeup of organisms. In 1997, while working at the University of Tennessee, Pigliucci received the Theodosius Dobzhansky Prize,

Answer 1

 >>> import re
 >>> re.sub(r'\[\s*\d+\s*\]', '', original_string)

Answer 2

我想这样应该行得通：

import re

origional_string = "makeup of organisms.[10] In 1997, while working at the University of Tennessee, Pigliucci received the Theodosius Dobzhansky Prize,[11]"

result = re.sub(r'\[ *[0-9]+ *\]', '', origional_string)

print(result)

[0-9]+匹配一个或多个数字，而 *匹配空格（如果有）。你知道吗

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Answer 3

\d表示正则表达式中的数字。我还在这里的数字周围添加了选项\s。你知道吗

re.sub(r'\[\s*\d+\s*\]', '', origional_string)