字符串模式正则表达式python

Question

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 .

Answer 1

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

Answer 2

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])

Debuggex Demo

This one only matches abababab9 .

Edit 1:

Here is an explanation of the grouping concept of regular expressions: groups and capturing

Answer 3

在内部使用?:删除第二个组捕获(ab) ：

testing= re.findall(r'((?:ab)*[0-9])',temp);