Python – Extract list substring from list with regex

I am trying to extract a list from a string, for example, the list always behind id=000-

l = "machine status: id=000-[{'89c57212e14694d3488e3924c46219dc_2': True}, {'45b5eab7caf0773f07a5f3a1b9f0c545_2': True}]"

My desired output should be list of dict

"[{'89c57212e14694d3488e3924c46219dc_2': True}, {'45b5eab7caf0773f07a5f3a1b9f0c545_2': True}]"

However with my regex

re.findall(r'(id=000-\[.*\])', l)

I got output below, which included the id=000 prefix

["id=000-[{'89c57212e14694d3488e3924c46219dc_2': True}, {'45b5eab7caf0773f07a5f3a1b9f0c545_2': True}]"]

>Solution :

Simply use a "positive lookbehind", which is with (?<=pattern):

import re

l = "machine status: id=000-[{'89c57212e14694d3488e3924c46219dc_2': True}, {'45b5eab7caf0773f07a5f3a1b9f0c545_2': True}]"

for result in re.findall(r'(?<=id=000-)\[.*\]', l):
    print(result)

Output:

[{'89c57212e14694d3488e3924c46219dc_2': True}, {'45b5eab7caf0773f07a5f3a1b9f0c545_2': True}]