How to use re to find consecutive, repeated chars

I want to 开发者_如何学Pythonfind all consecutive, repeated character blocks in a string. For example, consider the following:

s = r'http://www.google.com/search=ooo-jjj'

What I want to find this: www, ooo and jjj.

I tried to do it like this:

m = re.search(r'(\w)\1\1', s)

But it doesn't seem to work as I expect. Any ideas?

Also, how can I do it in Bash?

((\w)\2{2,}) matches 3 or more consecutive characters:

In [71]: import re
In [72]: s = r'http://www.google.com/search=ooo-jjjj'
In [73]: re.findall(r'((\w)\2{2,})', s)
Out[73]: [('www', 'w'), ('ooo', 'o'), ('jjjj', 'j')]

In [78]: [match[0] for match in re.findall(r'((\w)\2{2,})', s)]
Out[78]: ['www', 'ooo', 'jjjj']

(\w) matches any alphanumeric character.

((\w)\2) matches any alphanumeric character followed by the same character, since \2 matches the contents of group number 2. Since I nested the parentheses, group number 2 refers to the character matched by \w.

Then putting it all together, ((\w)\2{2,}) matches any alphanumeric character, followed by the same character repeated 2 or more additional times.

In total, that means the regex require the character to be repeated 3 or more times.

The following code should solve your problem:

s="abc def aaa bbb ccc def hhh"

for match in re.finditer(r"(\w)\1\1", s):
    print s[match.start():match.end()]

It works almost right, just replace search with finditer. It returns an iterator, not a match but...:

m = [(x.start(),x.end()) for x in re.finditer(r'(\w)\1\1', s)]