Comparing lists in Python

Question

Say I have list1 = [1,2,3,4] and list2 = [5,6,7,8]. How would I compare the first element, 1, in list1 with the first el

Answer 1

The default comparison operators compare lists in lexicographical order. So you can say things like:

>>> [1, 2, 3, 4] < [5, 6, 7, 8]
True

If instead you want to compute the elementwise comparison, you can use map and cmp (or any other operator:

>>> map(cmp, [1, 2, 3, 4], [5, 6, 7, 8])
[-1, -1, -1, -1]

Answer 2

If your result is going to be a new list, then you can use a list comprehension:

new_list = [ some_function(i, j) for i, j in zip(list1, list2) ]

Here's a real example of the above code:

>>> list1 = [1, 2, 3, 4]
>>> list2 = [1, 3, 4, 4]
>>> like_nums = [ i == j for i, j in zip(list1, list2) ]
>>> print like_nums
[True, False, False, True]

This will make a list of bools that show whether items of the same index in two lists are equal to each other.

Furthermore, if you use the zip function, there is a way to unzip the result when you are done operating on it. Here's how:

>>> list1 = [1, 2, 3, 4]
>>> list2 = [1, 3, 4, 4]
>>> new_list = zip(list1, list2)         # zip
>>> print new_list
[(1, 1), (2, 3), (3, 4), (4, 4)]
>>> newlist1, newlist2 = zip(*new_list)  # unzip
>>> print list(newlist1)
[1, 2, 3, 4]
>>> print list(newlist2)
[1, 3, 4, 5]

This could be useful if you need to modify the original lists, while also comparing the elements of the same index in some way.

Answer 3

You can traverse both lists simultaneously using zip:

for (x, y) in zip(list1, list2): do_something

The 'zip' function gives you [(1,5), (2,6), (3,7), (4,8)], so in loop iteration N you get the Nth element of each list.