CODEWITHSUNDEEP
pythonlist
LWZ
LWZ

Reputation: 12348

In Python, find out number of differences between two ordered lists

I want to compare two lists of same length

a = [1, 3, 5, 7, 9]
b = [1, 2, 5, 7, 3]

and find out the number of differences n, in this case it'll be n = 2, and also return an error if the length are not equal. What's the pythonic way of doing this?

Upvotes: 12

Views: 9993

Answers (3)

Max Li
Max Li

Reputation: 5219

One-liner solution that also produces an error if the length is not equal:

>>> sum(map(lambda x,y: bool(x-y),a,b))
2

Now try the input of different length:

>>> sum(map(lambda x,y: bool(x-y),[1,2],[1]))
TypeError

How it works: bool(x,y) returns True if elements are different. Then we map this function on 2 lists and get the list [False, True, False, True, False].
If we put into the function map() the lists of different length, we get the TypeError

Finally, the function sum() of this boolean list gives 2.

Upvotes: 1

Gareth Latty
Gareth Latty

Reputation: 89077

The simplest way to do this is to use the sum() built-in and a generator expression:

def differences(a, b):
    if len(a) != len(b):
        raise ValueError("Lists of different length.")
    return sum(i != j for i, j in zip(a, b))

We loop over the lists together using zip() and then compare them. As True == 1 and False == 0, we just sum this to get the number of differences. Another option would be to use the conditional part of the generator expression:

sum(1 for i, j in zip(a, b) if i != j)

I can't really say I feel one is more readable than the other, and doubt there will be a performance difference.

Upvotes: 21

user2079098
user2079098

Reputation: 132

You could use sets. Cast both to a set, then find difference between the two. For example:

>>> a = [1,3,5,7,9]
>>> b = [1,2,5,7,2]
>>> len(set(a) -  set(b))
2

This could be wrapped up in a function to check for length differences first.

Upvotes: -2

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