# testing if a list contains a sublist

alex23 wuwei23 at gmail.com
Tue Aug 16 09:14:08 CEST 2011

```On Aug 16, 4:51 pm, Laszlo Nagy <gand... at shopzeus.com> wrote:
> >> hi list,
> >> what is the best way to check if a given list (lets call it l1) is
> >> totally contained in a second list (l2)?
>
> >> for example:
> >> l1 = [1,2], l2 = [1,2,3,4,5] ->  l1 is contained in l2
> >> l1 = [1,2,2,], l2 = [1,2,3,4,5] ->  l1 is not contained in l2
> >> l1 = [1,2,3], l2 = [1,3,5,7] ->  l1 is not contained in l2
>
> >> my problem is the second example, which makes it impossible to work with
> >> sets insteads of lists. But something like set.issubset for lists would
> >> be nice.
>
> >> greatz Johannes
>
> Fastest, error-free and simplest solution is to use sets:
>
>  >>> l1 = [1,2]
>  >>> l2 = [1,2,3,4,5]
>  >>> set(l1)-set(l2)
> set([])
>  >>> set(l2)-set(l1)
> set([3, 4, 5])

Error free? Consider this stated requirement:
> l1 = [1,2,2,], l2 = [1,2,3,4,5] -> l1 is not contained in l2

>>> l1 = [1,2,2]
>>> l2 = [1,2,3,4,5]
>>> set(l1)-set(l2)
set()
>>> set(l2)-set(l1)
{3, 4, 5}

So set([1,2]) == set([1,2,2]), despite [1,2,2] not being a sublist of
the original.

It also completely ignores list order, which would make [9,8,7] a
sublist of [5,6,7,8,9].

```