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?
Answers:
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%']
>>> 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.
Use an outer group, with the inner group a non-capturing group:
>>> re.findall("((?:100|[0-9][0-9]|[0-9])%)","89%")
['89%']
re.findall("/d+/%","89%")
d+ gets a number, regardless of how long.
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?
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%']
>>> 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.
Use an outer group, with the inner group a non-capturing group:
>>> re.findall("((?:100|[0-9][0-9]|[0-9])%)","89%")
['89%']
re.findall("/d+/%","89%")
d+ gets a number, regardless of how long.