Is there a better way to compare two lists in Python, stopping at the shorter list?

Question

I'm trying to compare two lists with short-circuiting logic such that if one list is shorter than the other stop the comparison and return True. I'd like to know if what I have is sufficiently Pythonic or if there is a better way to do it.

def compareLists(list1,list2):
    # Comparison invalid if either list is empty
    if not list1 or not list2:
        return False

    equalList = True    # initialize as true
    for (l1,l2) in zip(list1,list2):
        if l1 != l2:
            equalList = False
            break
    return equalList

ipdb> list1 = [1,2,3]
ipdb> list2 = [1,2,3,4]    
ipdb> compareLists(list1,list2)
True

Answer 1

A more pythonic way would be to use all :

def compare_lists(list1, list2):
    if not list1 or not list2:
        return False
    else:
        return all(x1 == x2 for x1, x2 in zip(list1, list2))

Also note that the naming convention in Python is not camel case, but snake case.

Answer 2

You don't need a loop. You can use the min function to get the length of the shorter list (L), then use the == operator to compare the first L elements of each list, like so:

def compareLists(list1, list2):
    L=min(len(list1), len(list2))
    return list1[:L]==list2[:L]

The following returns True :

lista=[1,3,4,6,7,8,4,6]
listb=[1,3,4,6,7,8,4,6,3,7,5,2,4]
print (compareLists(lista, listb))

Whereas, the following returns False :

lista=[1,3,4,6,7,-8,4,6]
listb=[1,3,4,6,7,8,4,6,3,7,5,2,4]
print (compareLists(lista, listb))

Answer 3

This is a better way

Change from def to Function

function compareLists(list1,list2): # Comparison invalid if either list is empty if not list1 or not list2: return False

equalList = True    # initialize as true
for (l1,l2) in zip(list1,list2):
    if l1 != l2:
        equalList = False
        break
  return equalList
  list1 = [1,2,3]list2 = [1,2,3,4]     compareLists(list1,list2)

this is correct way

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