Pythonic way to compare two lists and print out the differences

Question

I have two lists which are guaranteed to be the same length. I want to compare the corresponding values in the list (except the first item) and print out the ones which dont match. The way I am doing it is like this

i = len(list1)
if i == 1:
    print 'Nothing to compare'
else:
    for i in range(i):
        if not (i == 0):
            if list1[i] != list2[i]:
                print list1[i]
                print list2[i]

Is there a better way to do this? (Python 2.x)

Answer 1

list1=[1,2,3,4]
list2=[1,5,3,4]
print [(i,j) for i,j in zip(list1,list2) if i!=j]

Output:

[(2, 5)]

Edit: Easily extended to skip n first items (same output):

list1=[1,2,3,4]
list2=[2,5,3,4]
print [(i,j) for i,j in zip(list1,list2)[1:] if i!=j]

Answer 2

Nobody's mentioned filter:

a = [1, 2, 3]
b = [42, 3, 4]

aToCompare = a[1:]
bToCompare = b[1:]

c = filter( lambda x: (not(x in aToCompare)), bToCompare)
print c