[ Python 常見問題 ] How to find all occurrences of a substring?

Source From Here
Question
Python has str.find() and str.rfind() to get the index of a substring in a string.

I'm wondering whether there is something like str.find_all() which can return all found indexes (not only the first from the beginning or the first from the end). For example:

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1. string = "test test test test"  
2.   
3. print string.find('test') # 0  
4. print string.rfind('test') # 15  
5.   
6. #this is the goal  
7. print string.find_all('test') # [0,5,10,15]  

How-To
There is no simple built-in string function that does what you're looking for, but you could use the more powerful regular expression module re:

>>> import re
>>> ts = 'test test test test'
>>> [m.start() for m in re.finditer('test', ts)]
[0, 5, 10, 15]

If you want to find overlapping matches, lookahead will do that:

>>> [m.start() for m in re.finditer('(?=tt)', 'ttt')]
[0, 1]
>>> [m.start() for m in re.finditer('tt', 'ttt')]
[0]

If you want a reverse find-all without overlaps, you can combine positive and negative lookahead into an expression like this:

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1. search = 'tt'  
2. [m.start() for m in re.finditer('(?=%s)(?!.{1,%d}%s)' % (search, len(search)-1, search), 'ttt')]  
3. #[1]  

re.finditer returns a generator, so you could change the [] in the above to () to get a generator instead of a list which will be more efficient if you're only iterating through the results once.