How to find all occurrences of a substring?

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

import re[m.start() for m in re.finditer('test', 'test test test test')]#[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]

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

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

>>> help(str.find)Help on method_descriptor:find(...)    S.find(sub [,start [,end]]) -> int

Thus, we can build it ourselves:

def find_all(a_str, sub):    start = 0    while True:        start = a_str.find(sub, start)        if start == -1: return        yield start        start += len(sub) # use start += 1 to find overlapping matcheslist(find_all('spam spam spam spam', 'spam')) # [0, 5, 10, 15]

No temporary strings or regexes required.

Here's a (very inefficient) way to get all (i.e. even overlapping) matches:

>>> string = "test test test test">>> [i for i in range(len(string)) if string.startswith('test', i)][0, 5, 10, 15]