问题
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?
回答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%']
回答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.
回答3:
Use an outer group, with the inner group a non-capturing group:
>>> re.findall("((?:100|[0-9][0-9]|[0-9])%)","89%")
['89%']
来源：https://stackoverflow.com/questions/16045643/python-re-findall-returns-unwanted-result