Python - How to perform indirect sorting with a user-defined function?

Question

numpy.argsort returns a sorted list to perform an indirect sorting, but it doesn't seem to accept a user-defined function to compare two elements.

I wonder how one can get the sorted list based on a comparison with a user-defined function.

In my case, I have a table of results:

a = [[1,2,3,5,6,7,8],[4,6,2,5,6,3,4],...]

And my function to compare is:

def argMedian(A):
    s   = np.array([sum(A[:i+1]) for i in range(Nmentions)])
    mid = float(s[Nmentions-1])/2
    return np.argwhere(mid < s)[0][0]

def tieBreaking(A, B):
Ac = A
Bc = B
medA = argMedian(Ac)
medB = argMedian(Bc)
while medA == medB:
    Ac[medA] -= 1
    Bc[medB] -= 1
    medA = argMedian(Ac)
    medB = argMedian(Bc)
return 1 if medA > medB else -1

This is an implementation of the majority judgment . A list contains the number of votes for each grade. In case of the median between two lists is equal, a median vote is deleted for both lists and the comparison of the median is again tested. Here, I consider index 0 as the best grade and index 7 as the worst.

I need to perform an indirect sorting, because I want to know the ranking.

Answer 1

If you are using Python >= 3.4, you can use statistics.median_low() .

from random import randrange
from statistics import median_low

a = [[randrange(8) for _ in range(7)] for _ in range(10)]

print("unsorted")
for item in a:
    print(item)

a.sort(key=median_low)

print("\nsorted")
for item in a:
    print(item)

output:

unsorted
[4, 2, 2, 1, 4, 7, 4]
[2, 2, 2, 7, 5, 5, 6]
[3, 0, 0, 5, 5, 3, 1]
[4, 7, 6, 2, 6, 7, 3]
[4, 7, 7, 1, 2, 7, 7]
[6, 2, 6, 5, 6, 7, 2]
[7, 1, 6, 0, 0, 7, 1]
[0, 5, 1, 2, 1, 7, 7]
[2, 7, 6, 7, 5, 4, 7]
[6, 5, 2, 3, 5, 0, 3]

sorted
[7, 1, 6, 0, 0, 7, 1]
[0, 5, 1, 2, 1, 7, 7]
[3, 0, 0, 5, 5, 3, 1]
[6, 5, 2, 3, 5, 0, 3]
[4, 2, 2, 1, 4, 7, 4]
[2, 2, 2, 7, 5, 5, 6]
[4, 7, 6, 2, 6, 7, 3]
[6, 2, 6, 5, 6, 7, 2]
[2, 7, 6, 7, 5, 4, 7]
[4, 7, 7, 1, 2, 7, 7]

EDIT full alternate solution:

Consider sorting a list as follows: 1. Find the median_low of the list and move it to the front of the list 2. Find the median_low of list[1:] and move it to the second spot 3. Find the median_low of list[2:] ...

You can sort the original lists by median values, or create keys in which the elements are sorted by median values.

def def keyfunc(x):
    t = x[:]
    return [t.pop(t.index(median_low(t))) for _ in range(len(t))]

a = [
    [4, 2, 2, 1, 4, 7, 4],
    [2, 2, 2, 7, 5, 5, 6],
    [3, 0, 0, 5, 5, 3, 1],
    [7, 1, 6, 0, 0, 6, 1],    # tie for first four rounds, but then wins
    [4, 7, 6, 2, 6, 7, 3],
    [6, 2, 6, 5, 6, 7, 2],
    [7, 1, 6, 0, 0, 7, 1],    # tie for first four rounds
    [0, 5, 1, 2, 1, 7, 7],
    [2, 7, 6, 7, 5, 4, 7],
    [6, 5, 2, 3, 5, 0, 3]
]

a.sort(key=keyfunc)

print("\nsorted")
for item in a:
    print(item)

output:

sorted
[7, 1, 6, 0, 0, 6, 1]
[7, 1, 6, 0, 0, 7, 1]
[0, 5, 1, 2, 1, 7, 7]
[3, 0, 0, 5, 5, 3, 1]
[6, 5, 2, 3, 5, 0, 3]
[4, 2, 2, 1, 4, 7, 4]
[2, 2, 2, 7, 5, 5, 6]
[4, 7, 6, 2, 6, 7, 3]
[6, 2, 6, 5, 6, 7, 2]
[2, 7, 6, 7, 5, 4, 7]

Answer 2

You don't need to use numpy to sort the elements of your list. You can the sort() fuctions for lists or sorted(). Define your custom comparator there if needed.

Answer 3

For computing an indirect sorting with basic python, few tricks exist. They were compared in Equivalent of Numpy.argsort() in basic python? .

Taking the fastest solution, one can add a comparison function.

In Python 2.7:

sorted(range(len(results)), cmp=tieBreaking, key=results.__getitem__)

In Python 3, one should take care that cmp has disappeared, but the function cmp_to_key can replace it.