regex - Python - re.findall returns unwanted result

Question

re.findall("(100|[0-9][0-9]|[0-9])%", "89%")

This returns only result [89] and I need to return the whole 89%. Any ideas how to do it please?

Answer 1

The trivial solution:

>>> re.findall("(100%|[0-9][0-9]%|[0-9]%)","89%")
['89%']

More beautiful solution:

>>> re.findall("(100%|[0-9]{1,2}%)","89%")
['89%']

The prettiest solution:

>>> re.findall("(?:100|[0-9]{1,2})%","89%")
['89%']

Answer 2

>>> re.findall("(?:100|[0-9][0-9]|[0-9])%", "89%")
['89%']

When there are capture groups findall returns only the captured parts. Use ?: to prevent the parentheses from being a capture group.

Answer 3

Use an outer group, with the inner group a non-capturing group:

>>> re.findall("((?:100|[0-9][0-9]|[0-9])%)","89%")
['89%']