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