python - Find all possible sublists of a list

Question

Let's say I have the following list

[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]

I want to find all possible sublists of a certain lenght where they don't contain one certain number and without losing the order of the numbers.

For example all possible sublists with length 6 without the 12 are:

[1,2,3,4,5,6]
[2,3,4,5,6,7]
[3,4,5,6,7,8]
[4,5,6,7,8,9]
[5,6,7,8,9,10]
[6,7,8,9,10,11]
[13,14,15,16,17,18]

The problem is that I want to do it in a very big list and I want the most quick way.

Update with my method:

oldlist = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]
newlist = []
length = 6
exclude = 12
for i in oldlist:
   if length+i>len(oldlist):
       break
   else:
       mylist.append(oldlist[i:(i+length)]
for i in newlist:
    if exclude in i:
       newlist.remove(i)

I know it's not the best method, that's why I need a better one.

Answer 1

Use itertools.combinations:

import itertools
mylist = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]
def contains_sublist(lst, sublst):
    n = len(sublst)
    return any((sublst == lst[i:i+n]) for i in xrange(len(lst)-n+1))
print [i for i in itertools.combinations(mylist,6) if 12 not in i and contains_sublist(mylist, list(i))]

Prints:

[(1, 2, 3, 4, 5, 6), (2, 3, 4, 5, 6, 7), (3, 4, 5, 6, 7, 8), (4, 5, 6, 7, 8, 9), (5, 6, 7, 8, 9, 10), (6, 7, 8, 9, 10, 11), (13, 14, 15, 16, 17, 18)]