How to check a list contained by another list without a loop?

Question

As the title mentions,is there any builtins to do this job?I looked for that in dir(list) but got no usable one.Thanks.

Answer 1

Assuming that you want to see if all elements of sublist are also elements of superlist:

all(x in superlist for x in sublist)

Answer 2

Depends on what you mean by "contained". Maybe this:

if set(a) <= set(b):
    print "a is in b"

Answer 3

the solution depends on what values you expect from your lists.

if there is the possiblity of a repetition of a value, and you need to check that there is enough values in the tested container, then here is a time-inefficient solution:

def contained(candidate, container):
    temp = container[:]
    try:
        for v in candidate:
            temp.remove(v)
        return True
    except ValueError:
        return False

test this function with:

>>> a = [1,1,2,3]
>>> b = [1,2,3,4,5]
>>> contained(a,b)
False    
>>> a = [1,2,3]
>>> contained(a,b)
True
>>> a = [1,1,2,4,4]
>>> b = [1,1,2,2,2,3,4,4,5]
>>> contained(a,b)
True

of course this solution can be greatly improved: list.remove() is potentially time consuming and can be avoided using clever sorting and indexing. but i don't see how to avoid a loop here...

(anyway, any other solution will be implemented using sets or list-comprehensions, which are using loops internally...)

Answer 4

You might want to use a set

if set(a).issubset(b):
    print('a is contained in b')