What is the Pythonic way to find the longest common prefix of a list of lists?

Question

Given: a list of lists, such as [[3,2,1], [3,2,1,4,5], [3,2,1,8,9], [3,2,1,5,7,8,9]]

Todo: Find the longest common prefix of all sublists.

Exists: In another thread "Common elements between two lists not using sets in Python", it is suggested to use "Counter", which is available above python 2.7. However our current project was written in python 2.6, so "Counter" is not used.

I currently code it like this:

l = [[3,2,1], [3,2,1,4,5], [3,2,1,8,9], [3,2,1,5,7,8,9]]
newl = l[0]
if len(l)>1:
    for li in l[1:]:
    newl = [x for x in newl if x in li]

But I find it not very pythonic, is there a better way of coding?

Thanks!

New edit: Sorry to mention: in my case, the shared elements of the lists in 'l' have the same order and alway start from the 0th item. So you wont have cases like [[1,2,5,6],[2,1,7]]

Answer 1

os.path.commonprefix() works well for lists :)

>>> x = [[3,2,1], [3,2,1,4,5], [3,2,1,8,9], [3,2,1,5,7,8,9]]
>>> import os
>>> os.path.commonprefix(x)
[3, 2, 1]