Re: inner sublist positions ?

[Steven Bethard]

Bernard A. wrote:
hello, 

while trying to play with generator, i was looking for an idea to get
the position of a inner list inside another one, here is my first idea
:
- first find position of first inner element, 
- and then see if the slice starting from here is equal to the inner
->

def subPositions(alist, innerlist):
    if innerlist == []:
                return
        first, start = innerlist[0], 0
        while 1:
                try:
                        p = alist[start:].index(first)
                except ValueError:
                        break # or should i better use return ?
                start = start + p               
                if alist[start: start + len(innerlist)] == innerlist:
                        yield start, start + len(innerlist)
                start += 1

list(subPositions(range(5) + range(5), [2,3]))
[(2, 4), (7, 9)]
maybe have you some better / faster ideas / implementations ? or even
a more pythonic to help me learning that
I don't know if this is any faster, but you could try:
py> def indexes(lst, sublist):
...     sublen = len(sublist)
...     for i in range(len(lst)):
...         if lst[i:i+sublen] == sublist:
...             yield i, i+sublen
...
py> list(indexes(range(5) + range(5), [2, 3]))
[(2, 4), (7, 9)]
Use the timeit module to see if it's any faster.  Also, in your version 
you should probably use
    p = alist.index(first, start)
instead of
    p = alist[start:].index(first)
so you don't make so many new lists.

STeVe
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