testing if a list contains a sublist

[Peter Otten]

Johannes wrote:

> Am 16.08.2011 09:44, schrieb Peter Otten:
>> Johannes wrote:
>> 
>>> 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.
>> 
>> Would [1, 2, 1] be contained in [1, 1, 2]? You have received an answer
>> that assumes yes and another that assume no.
>> 
> 
> yes it should be contained in the other one. I just work with sorted
> lists, so i did not think about this case.
> 
> geatz Johannes

For "short" lists l1:

>>> pairs = [
... ([1,2], [1,2,3,4,5]),
... ([1,2,2,], [1,2,3,4,5]),
... ([1,2,3], [1,3,5,7])]
>>> for l1, l2 in pairs:
...     contained = all(l1.count(item) <= l2.count(item) for item in 
set(l1))
...     print l1, ["not in", "in"][contained], l2
...
[1, 2] in [1, 2, 3, 4, 5]
[1, 2, 2] not in [1, 2, 3, 4, 5]
[1, 2, 3] not in [1, 3, 5, 7]