Get multiple search using regex in python in singe line

.* at the beginning of the regexp causes the first match to include the entire beginning of the input string, up to the first[. And.*` at the end of the regexp causes the first match to include the rest of the input string.

So both of these prevent the regexp from matching multiple times. You shouldn't use them when you're using re.findall().

Then you need to use non-greedy quantifiers, so that .* won't match across multiple sets of brackets. Or you could use \[[^]]*\] instead of .*, so it won't match the close bracket.

And there's no need for the non-capturing group around the parts you want to capture.

Just use:

re.findall(r'(\[.*?\]) (\d+)', stat_log, re.DOTALL)

DEMO

In your example string, the first non capturing capturing group (?:.*) will match until the end of the string. Then it will backtrack and capture the last [50 +5] in group 2 and the 1 in group 3. For the .* there are no more characters to match.

Instead of .* which is greedy you could use a negated character class matching not an opening or a closing bracket:

(\[[^][]+\])\s+(\d+)

Explanation

• ( First capturing group
• \[[^][]+\] Negated character class to match [, then not ] or [ and match ]
• )
• \s+ match 1+ times a whitespace char (or use only a space)
• (\d+) Capture in group 2 matching 1+ times a digit

regex demo | Python demo

For example:

import re
stat_log = '[0 +5] 23 for bucket [5 +5] 1 for bucket [25 +5] 22 for bucket [50 +5] 1'
stats_iter = re.findall('(\[[^][]+\])\s+(\d+)', stat_log,re.DOTALL)
print(stats_iter)