```
>>> import re
>>> text = '121212 and 121212'
>>> pattern = '(12)+'
>>> print(re.findall(pattern, text))
['12', '12']
>>> 
>>> 
>>> print(re.search(pattern, text))
<re.Match object; span=(0, 6), match='121212'>
>>> 
>>> 
>>> print(re.sub(pattern, '', text))
 and 
# The re.findall have different search result against re.search or re.sub
# re.findall 和 re.search 、 re.sub 的匹配结果不相同
```

With same pattern and string, the re.findall is supposed to have same match with re.search, but it didn't. 

This result is from python3.8.5

It looks correct to me. Of course, the result is different, because they are different functions. re.match() and re.search() return a match object (or None), and re.findall returns a list. What result did you expect?

When groups are present in the regex, findall() returns the subgroups rather than the entire match:

    >>> mo = re.search('(12)+', '121212 and 121212')
    >>> mo[0]                  # Entire match
    '121212'
    >>> mo[1]                  # Group match
    '12'

To get the result you were looking for use a non-capturing expression:

    >>> re.findall('(?:12)+', '121212 and 121212')
    ['121212', '121212']

Also consider using finditer() which gives more fine grained control:

    >>> for mo in re.finditer('(12)+', '121212 and 121212'):
            print(mo.span())
            print(mo[0])
            print(mo[1])
            print()
        
    (0, 6)
    121212
    12

    (11, 17)
    121212
    12

AhAh, got it, I misunderstood the usage, the findall returns tuple of groups the expression set. Thanks @serhiy.storchaka @rhettinger

I was frustrated for hours when I couldn't figure out why this won't match:

>>> re.findall(r'(foo)?bar|cool', 'cool')

Now I know, I have to make this change: (?:foo)
But this isn't obvious.
Should it be mentioned in the docs of re.findall() to use (?:...) for non-capturing groups?