I am a novice in regular expressions. I have written the following regex to find abababab9 in the given string. The regular expression returns two results, however I was expecting one result.
testing= re.findall(r'((ab)*[0-9])',temp);
**Output**: [('abababab9', 'ab')]
According to my understanding, it should have returned only abababab9, why has it returned ab alone.
You didnt' read the findall documentation:
Return a list of all non-overlapping matches in the string.
If one or more capturing groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group.
Empty matches are included in the result.
And if you take a look at the re module capturing groups are subpatterns enclosed in parenthesis like (ab).
If you want to only get the complete match you can use one of the following solutions:
re.findall(r'(?:ab)*[0-9]', temp) # use non-capturing groups
[groups[0] for groups in re.findall(r'(ab)*[0-9]', temp)] # take the first group
[match.group() for match in re.finditer(r'(ab)*[0-9]', temp)] # use finditer
You have configured by (...) two matching groups the first group is ((ab)*[0-9]) and the second group is (ab). Therefore you get these two results. To get only the first group you could make the second a non-capturing group. This is done by ?:. So this result is not delivered.
((?:ab)*[0-9])
This one only matches abababab9.
Edit 1:
Here is an explanation of the grouping concept of regular expressions: groups and capturing
capturing group?
Remove the second group capturing (ab) using ?: inside:
testing= re.findall(r'((?:ab)*[0-9])',temp);
() is captured inside a group from the regex matching. In your regex it had two ().?:. Please check this python doc.
abababab9"? (there's one morebain your code)