首页 \ 问答 \ python正则表达式不检测方括号(python regex not detecting square brackets)

python正则表达式不检测方括号(python regex not detecting square brackets)

我有一个场景,我想从给定的内容中除去空格以外的所有特殊字符,我正在使用Python,并且我正在使用这个正则表达式

re.sub(r"[^a-zA-z0-9 ]+","",content)

Itt是删除所有特殊字符,但没有删除方括号[ ] ,我只是不知道为什么会发生这种情况?

之后,我只是使用这个正则表达式

content = re.sub(r"[^a-zA-z0-9 ]+|\[|\]","",content)

它在IDLE IDE完美地工作,并删除所有类型的特殊字符,但是当我想替换像维基百科页面这样的大文件时,它现在不会删除方括号]我只是不知道为什么Python做这种奇怪的行为,


I have a scenario where I want to remove all special characters except spaces from given content and I am working with Python and I was using this regex

re.sub(r"[^a-zA-z0-9 ]+","",content)

Itt was removing all special characters but was not removing square brackets [ ] and I just don't know why this happening??

after that I just use this regex

content = re.sub(r"[^a-zA-z0-9 ]+|\[|\]","",content)

It's working flawlessly in IDLE IDE and removing all kind of special characters but when I want to replace large files like Wikipedia's page then its now not removing closing square brackets ] I just dont why Python doing this weird behavior and


原文:https://stackoverflow.com/questions/31225304
更新时间:2023-01-04 12:01

最满意答案

假设您沿着axis (即0或1)工作。 首先查找1的行数/列数:

counts = (arr==1).sum(axis=axis)

您现在可以找到最小计数的索引,从计数列表中选择适当的计数,并将其除以所有其他计数:

counts / counts[counts.argmin()]

请记住最小的数量可以是0。


Assume you work along the axis axis (which is 0 or 1). Start by finding the row/column counts of 1's:

counts = (arr==1).sum(axis=axis)

You can now find the index of the smallest count, select the appropriate count from the list of counts, and divide all other counts by it:

counts / counts[counts.argmin()]

Just remember that the smallest count can be 0.

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