Python 像往常一样在空格处拆分字符串，但保留某些包含空格的子字符串？

Question

Consider I have the following string:

>>> aa="63452 [  0] AAA BB CCC"

If I do the usual .split() which splits at whitespaces, I get this:

>>> aa.split()
['63452', '[', '0]', 'AAA', 'BB', 'CCC']

What I'd want to obtain instead, is this list: ['63452', '[ 0]', 'AAA', 'BB', 'CCC']

Basically, the second part is a string that matches the format: opening square bracket + none or more of whitespace characters + none or more digits + closing square bracket - which I can match with this regex:

>>> import re
>>> re.findall(r'\[\s*\d*\]', aa)
['[  0]']

In essence, I'd first want to identify the "square bracket" item, then split as .split() usually does it, while keeping the "square bracket" item.

So, what would be the most straightforward way, to obtain the desired list from the given string?

Answer 1

You can use an alternation pattern that matches a bracketed string or non-space characters:

re.findall(r'\[.*?]|\S+', aa)