python compare items in 2 list of different length - order is important

Question

list_1 = ['a', 'a', 'a', 'b']
list_2 = ['a', 'b', 'b', 'b', 'c']

so in the list above, only items in index 0 is the same while index 1 to 4 in both list are different. also, list_2 has an extra item 'c'. I want to count the number of times the index in both list are different, In this case I should get 3.

I tried doing this:

x = 0
for i in max(len(list_1),len(list_2)):
    if list_1[i]==list_2[i]:
        continue
    else:
        x+=1

I am getting an error.

Answer 1

Use the zip() function to pair up the lists, counting all the differences, then add the difference in length.

zip() will only iterate over the items that can be paired up, but there is little point in iterating over the remainder; you know those are all to be counted as different:

differences = sum(a != b for a, b in zip(list_1, list_2))
differences += abs(len(list_1) - len(list_2))

The sum() sums up True and False values; this works because Python's boolean type is a subclass of int and False equals 0, True equals 1. Thus, for each differing pair of elements, the True values produced by the != tests add up as 1s.

Demo:

>>> list_1 = ['a', 'a', 'a', 'b']
>>> list_2 = ['a', 'b', 'b', 'b', 'c']
>>> sum(a != b for a, b in zip(list_1, list_2))
2
>>> abs(len(list_1) - len(list_2))
1
>>> difference = sum(a != b for a, b in zip(list_1, list_2))
>>> difference += abs(len(list_1) - len(list_2))
>>> difference
3

Answer 2

You can try with this :

list1 = [1,2,3,5,7,8,23,24,25,32]
list2 = [5,3,4,21,201,51,4,5,9,12,32,23]

list3 = []

for i in range(len(list2)):
    if list2[i] not in list1:
        pass
    else :
        list3.append(list2[i])
print list3
print len(list3)