How to detect multiple spaces inside code and not at the beginning of line-CodePudding

I try to refactor python code written by someone else. QA does not want tabs, hence I replaced them by four whitespaces, but there were also a lot of tabs used inside the code, and not only for indentation at beginnings of lines.

Here is a sample piece of code:

    def    post(self,     request,     format=None):
        action     =     request.query_params.get('action',     None)
        res     =     'UNKNOWN ACTION'
        hhtpstatus = status.HTTP_400_BAD_REQUEST

I tried these regex:

[^\S]([ ]{2,})[^\S]

This one selects correctly the multiple spaces inside the code, but select also the ones used for indentation, which I don't want. If I use this one, I'll loose the indentations...

[^\s]([ ]{2,})[^\S]

This one is better, as it only selects the multiple spaces inside the code, but takes the last character of the preceding code, which obviously I don't want neither. If I use this, I ruin all the code by scraping the last character with the multiple space following it...

I work in Atom, and the regex should detect only the multiple whitespaces encompassed by code before and code after, and select only the spaces, so I can replace them by single space

I also tried with [^\W] and [^\w] with less success: small capital still detects indentation spaces, and select all the non words code characters (, : = and so on). Big capital selects only spaces following words, not spaces following other code characters, and selects also the last word character before multiple space...

Is there so a way to use [^\s]([ ]{2,})[^\S], which detects properly only the multiple spaces inside the code, but modify it to contain/select ONLY spaces, and not last preceding code character? So begin selection one character later, if this makes sense...

CodePudding user response：

You can change [^\s] to \S and change [^\S] to \s and assert them instead using lookarounds.

If you match 1 or more spaces, and assert a single space to the right, you will keep 1 space when replacing with an empty string.

(?<=\S)[ ] (?=\s)

Regex demo

If you don't want to match the spaces at the end when there is no whitespace character following, you can match 2 or more spaces and assert a non whitspace char to the right.

Then in the replacement you can use a single space.

(?<=\S)[ ]{2,}(?=\S)

Regex demo