问题
Is there any way to directly replace all groups using regex syntax?
The normal way:
re.match(r"(?:aaa)(_bbb)", string1).group(1)
But I want to achieve something like this:
re.match(r"(\d.*?)\s(\d.*?)", "(CALL_GROUP_1) (CALL_GROUP_2)")
I want to build the new string instantaneously from the groups the Regex just captured.
回答1:
Have a look at re.sub:
result = re.sub(r"(\d.*?)\s(\d.*?)", r"\1 \2", string1)
This is Python's regex substitution (replace) function. The replacement string can be filled with so-called backreferences (backslash, group number) which are replaced with what was matched by the groups. Groups are counted the same as by the group(...) function, i.e. starting from 1, from left to right, by opening parentheses.
回答2:
The accepted answer is perfect. I would add that group reference is probably better achieved by using this syntax:
r"\g<1> \g<2>"
for the replacement string. This way, you work around syntax limitations where a group may be followed by a digit. Again, this is all present in the doc, nothing new, just sometimes difficult to spot at first sight.
来源：https://stackoverflow.com/questions/14007545/python-regex-instantly-replace-groups