Grabbing multiple patterns in a string using regex

You could try something like: re.findall("\d+", your_string).

You cannot do this with a single regex (unless you were using .NET), because each capturing group will only ever return one result even if it is repeated (the last one in the case of Python).

Since variable length lookbehinds are also not possible (in which case you could do (?<=inputs.*=.*)\d+), you will have to separate this into two steps:

match = re.match(r'\s*inputs\s*=\s*(\d+(?:\s*\d+)+)', string)
integers = re.split(r'\s+',match.group(1))

So now you capture the entire list of integers (and the spaces between them), and then you split that capture at the spaces.

The second step could also be done using findall:

integers = re.findall(r'\d+',match.group(1))

The results are identical.

You can embed your regular expression:

import re
s = 'inputs       =    12 1  345 543 2'
print re.findall(r'(\d+)', re.match(r'inputs\s*=\s*([\s\d]+)', s).group(1))
>>> 
['12', '1', '345', '543', '2']

Or do it in layers:

import re

def get_inputs(s, regex=r'inputs\s*=\s*([\s\d]+)'):
    match = re.match(regex, s)
    if not match:
        return False # or raise an exception - whatever you want
    else:
        return re.findall(r'(\d+)', match.group(1))

s = 'inputs       =    12 1  345 543 2'
print get_inputs(s)
>>> 
['12', '1', '345', '543', '2']

You should look at this answer: https://stackoverflow.com/a/4651893/1129561

In short:

In Python, this isn’t possible with a single regular expression: each capture of a group overrides the last capture of that same group (in .NET, this would actually be possible since the engine distinguishes between captures and groups).