John Salerno schreef:
>> pattern = '([a-z][A-Z]{3}[a-z][A-Z]{3}[a-z])+'
>> print re.search(pattern, mess).groups()
>>
>> Anyway, this returns one matching string, but when I put this letter in
>> as the solution to the problem, I get a message saying "yes, but there
>> are more", so assuming this means that there is more than one character
>> with three caps on either side, is my RE written correctly to find them
>> all? I didn't have the parentheses or + sign at first, but I added them
>> to find all the possible matches, but still only one comes up.
>>
>> Thanks.
>
> A quick note: I found nine more matches by using findall() instead of
> search(), but I'm still curious how to write the RE so that it works
> with search, especially since findall wouldn't have returned overlapping
> matches. I guess I didn't write it to properly check multiple times.
It seems to me you should be able to find all matches with search(). Not
with the pattern you mention above: that will only find matches if they
come right after each other, as in
xXXXxXXXxyYYYyYYYyzZZZzZZZz
You'll need something more like
pattern = '([a-z][A-Z]{3}[a-z][A-Z]{3}[a-z]+)+'
so that it will find matches that are further apart from each other.
That said, I think findall() is a better solution for this problem. I
don't think search() will find overlapping matches either, so that's no
reason not to use findall(), and the pattern is simpler with findall();
I solved this challenge with findall() and this regular expression:
pattern = r'[a-z][A-Z]{3}[a-z][A-Z]{3}[a-z]'
--
If I have been able to see further, it was only because I stood
on the shoulders of giants. -- Isaac Newton
Roel Schroeven
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