python re, find expression containing an optional group
python, regex
Solution
The simplest way is to make the first `*` non-greedy:
>>> import re
>>> string = "(src://path/to/foldernames canhave spaces/file.xzy)"
>>> string2 = \
... '(src://path/to/foldernames canhave spaces/file.xzy "optional string")'
>>> re.findall(r'\((.*?)( ".*")?\)', string2)
[('src://path/to/foldernames canhave spaces/file.xzy', ' "optional string"')]
>>> re.findall(r'\((.*?)( ".*")?\)', string)
[('src://path/to/foldernames canhave spaces/file.xzy', '')]
Problem
I have a regular expression that can have either from: ``` (src://path/to/foldernames canhave spaces/file.xzy) (src://path/to/foldernames canhave spaces/file.xzy "optional string") ``` These expressions occur within a much longer string (they are not individual strings). I am having trouble matching both expressions when using `re.search` or `re.findall` (as there may be multiple expression in the string). It's straightforward enough to match either individually but how can I go about matching either case so that two groups are returned, the first with `src://path/...` and the second with the `optional string` if it exists or `None` if not? I am thinking that I need to somehow specify OR groups---for instance, consider: The pattern `\((.*)( ".*")\)` matches the second instance but not the first because it does not contain `"..."`. ``` r = re.search(r'\((.*)( ".*")\)', '(src://path/to/foldernames canhave spaces/file.xzy)' r.groups() # Nothing found AttributeError: 'NoneType' object has no attribute 'groups' ``` While `\((.*)( ".*")?\)` matches the first group but does not individually identify the `"optional string"` as a group in the second instance. ``` r = re.search(r'\((.*)( ".*")?\)', '(src://path/to/foldernames canhave spaces/file.xzy "optional string")') r.groups() ('src://path/to/foldernames canhave spaces/file.xzy "optional string"', None) ``` Any thoughts, ye' masters of expressions (of the regular variety)?