Am 16.08.2011 10:00, schrieb Laszlo Nagy:
>
>> Error free? Consider this stated requirement:
>>> l1 = [1,2,2,], l2 = [1,2,3,4,5] -> l1 is not contained in l2
> If you look it the strict way, "containment" relation for lists is meant
> this way:
>
>
> l1 = []
> l2 = [1,l1,2] # l2 CONTAINS l1
>
> But you are right, I was wrong. So let's clarify what the OP wants!
>
> For example:
>
> l1 = [1,2,2,], l2 = [2,1,2,3,4,5]
I dont care about this case, because all list are ordered for me.
I've chosen the following solution
> def _list_contained_in_list(l1,l2):
> d1 = {}
> d2 = {}
> for i in l1:
> if i in d1:
> d1[i] += 1
> else:
> d1[i] = 1
> for i in l2:
> if i in d2:
> d2[i] += 1
> else:
> d2[i] = 1
> if not all([k in d2.keys() for k in d1.keys()]):
> return false
> return all([d1[i] <= d2[i] for i in d1])
greatz Johannes
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