```
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.

# Best How To :

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 `1`

s.

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
```