Sort a list of lists with a custom compare function

Question

I know there are several questions named like this, but they don't seem to work for me.

I have a list of lists, 50 times 5 elements. I want to sort this list by applying a custom compare function to each element. This function calculates the fitness of the list by which the elements shall be sorted. I created two functions, compare and fitness:

def compare(item1, item2):
    return (fitness(item1) < fitness(item2))

and

def fitness(item):
    return item[0]+item[1]+item[2]+item[3]+item[4]

Then I tried to call them by:

sorted(mylist, cmp=compare)

or

sorted(mylist, key=fitness)

or

sorted(mylist, cmp=compare, key=fitness)

or

sorted(mylist, cmp=lambda x,y: compare(x,y))

Also I tried list.sort() with the same parameters. But in any case the functions don't get a list as an argument but a None. I have no idea why that is, coming from mostly C++ this contradicts any idea of a callback function for me. How can I sort this lists with a custom function?

Edit I found my mistake. In the chain that creates the original list one function didn't return anything but the return value was used. Sorry for the bother

Answer 1

Also, your compare function is incorrect. It needs to return -1, 0, or 1, not a boolean as you have it. The correct compare function would be:

def compare(item1, item2):
    if fitness(item1) < fitness(item2):
        return -1
    elif fitness(item1) > fitness(item2):
        return 1
    else:
        return 0

# Calling
list.sort(key=compare)

Answer 2

Since the OP was asking for using a custom compare function (and this is what led me to this question as well), I want to give a solid answer here:

Generally, you want to use the built-in sorted() function which takes a custom comparator as its parameter. We need to pay attention to the fact that in Python 3 the parameter name and semantics have changed.

How the custom comparator works

When providing a custom comparator, it should generally return an integer/float value that follows the following pattern (as with most other programming languages and frameworks):

• return a negative value (< 0) when the left item should be sorted before the right item
• return a positive value (> 0) when the left item should be sorted after the right item
• return 0 when both the left and the right item have the same weight and should be ordered "equally" without precedence

In the particular case of the OP's question, the following custom compare function can be used:

def compare(item1, item2):
    return fitness(item1) - fitness(item2)

Using the minus operation is a nifty trick because it yields to positive values when the weight of left item1 is bigger than the weight of the right item2. Hence item1 will be sorted after item2.

If you want to reverse the sort order, simply reverse the subtraction: return fitness(item2) - fitness(item1)

Calling sorted() in Python 2

sorted(mylist, cmp=compare)

or:

sorted(mylist, cmp=lambda item1, item2: fitness(item1) - fitness(item2))

Calling sorted() in Python 3

from functools import cmp_to_key
sorted(mylist, key=cmp_to_key(compare))

or:

from functools import cmp_to_key
sorted(mylist, key=cmp_to_key(lambda item1, item2: fitness(item1) - fitness(item2)))

Answer 3

You need to slightly modify your compare function and use functools.cmp_to_key to pass it to sorted. Example code:

import functools

lst = [list(range(i, i+5)) for i in range(5, 1, -1)]

def fitness(item):
    return item[0]+item[1]+item[2]+item[3]+item[4]
def compare(item1, item2):
    return fitness(item1) - fitness(item2)

sorted(lst, key=functools.cmp_to_key(compare))

Output:

[[2, 3, 4, 5, 6], [3, 4, 5, 6, 7], [4, 5, 6, 7, 8], [5, 6, 7, 8, 9]]

Works :)