What's the proper way to write comparator as key in Python 3 for sorting?

Question

I'm not sure how to write a comparator in Python 3 as the cmp parameter is removed. Considering the following code in Python 3, how do I rewrite the comparator using only key?

import functools

def my_cmp(x, y):
    return x*5-y*2

l = [50, 2, 1, 9]
print(sorted(l, key=functools.cmp_to_key(my_cmp)))

thanks.

Answer 1

This "comparison" function that you came up with is inconsistent: it should provide a definite (deterministic) order, meaning, if you change the order of the elements in the list and run sorted - you should get the same result!

In your case, the order of the elements effects the sorting:

import functools

def my_cmp(x, y):
    return x*5-y*2


l = [50, 2, 1, 9]
print(sorted(l, key=functools.cmp_to_key(my_cmp))) # [2, 1, 9, 50]

l = [50, 1, 2, 9]
print(sorted(l, key=functools.cmp_to_key(my_cmp))) # [1, 2, 9, 50]

which means that your "comparison" function is inconsistent. First provide good ordering function, then it should not be very difficult to convert it to a key function.

---

Regards the question that you raised in the comments, key accepts a function that takes only a single argument - and returns a "measurement" of "how big is it". The easiest example would be to compare numbers, in that case your key function can simply be: lambda x: x. For any number the lambda expression will returns itself and the comparison is now trivial!

Modifying your example:

def my_key(x):
    return x    

l = [50, 2, 1, 9]
print(sorted(l, key=my_key)) # [1, 2, 9, 50]

A shorter version of the above would be:

l = [50, 2, 1, 9]
print(sorted(l, key=lambda x: x)) # [1, 2, 9, 50]