Test whether list A is contained in list B

Use any with list slicing:

def contained_in(lst, sub):
    n = len(sub)
    return any(sub == lst[i:i+n] for i in range(len(lst)-n+1))

Or, use join to join both lists to strings and use in operator:

def contained_in(lst, sub):
    return ','.join(map(str, sub)) in ','.join(map(str, lst))

Usage:

>>> contained_in([1, 2, 3, 4, 5], [2, 3, 4])
True
>>> contained_in([1, 2, 2, 4, 5], [2, 3, 4])
False

many people have posted their answers. but I want to post my efforts anyway ;) this is my code:

def containedin(a,b):
    for j in range(len(b)-len(a)+1):
        if a==b[j:j+len(a)]:
            return True
    return False

print(containedin([2, 3, 4],[1, 2, 3, 4, 5]))
print(containedin([2, 3, 4],[1, 1, 2, 2, 3, 3, 4, 4, 5, 5]))
print(containedin([2, 3, 4],[5, 4, 3, 2, 1]))
print(containedin([2, 2, 2],[1, 2, 3, 4, 5]))
print(containedin([2, 2, 2],[1, 1, 1, 2, 2, 2, 3, 3, 3]))

this is the output: True False False False True

Assuming a always shorter than b what you can do is as follows.

 any(a == b[i:i+len(a)] for i in range(len(b)-len(a)+1))