How to sort a NumPy array by frequency?

Question

I am attempting to sort a NumPy array by frequency of elements. So for example, if there's an array [3,4,5,1,2,4,1,1,2,4], the output would be another NumPy sorted from most common to least common elements (no duplicates). So the solution would be [4,1,2,3,5]. If two elements have the same number of occurrences, the element that appears first is placed first in the output. I have tried doing this, but I can't seem to get a functional answer. Here is my code so far:

temp1 = problems[j]
indexes = np.unique(temp1, return_index = True)[1]
temp2 = temp1[np.sort(indexes)]
temp3 = np.unique(temp1, return_counts = True)[1]
temp4 = np.argsort(temp3)[::-1] + 1

where problems[j] is a NumPy array like [3,4,5,1,2,4,1,1,2,4]. temp4 returns [4,1,2,5,3] so far but it is not correct because it can't handle when two elements have the same number of occurrences.

Answer 1

You can use argsort on the frequency of each element to find the sorted positions and apply the indexes to the unique element array

unique_elements, frequency = np.unique(array, return_counts=True)
sorted_indexes = np.argsort(frequency)[::-1]
sorted_by_freq = unique_elements[sorted_indexes]

Answer 2

A non-NumPy solution, which does still work with NumPy arrays, is to use an OrderedCounter followed by sorted with a custom function:

from collections import OrderedDict, Counter

class OrderedCounter(Counter, OrderedDict):
    pass

L = [3,4,5,1,2,4,1,1,2,4]

c = OrderedCounter(L)
keys = list(c)

res = sorted(c, key=lambda x: (-c[x], keys.index(x)))

print(res)

[4, 1, 2, 3, 5]

Answer 3

You can count up the number of each element in the array, and then use it as a key to the build-in sorted function

def sortbyfreq(arr):
    s = set(arr)
    keys = {n: (-arr.count(n), arr.index(n)) for n in s}
    return sorted(list(s), key=lambda n: keys[n])

Answer 4

Use zip and itemgetter should help

from operator import itemgetter
import numpy as np
temp1 = problems[j]
temp, idx, cnt = np.unique(temp1, return_index = True, return_counts=True)
cnt = 1 / cnt
k = sorted(zip(temp, cnt, idx), key=itemgetter(1, 2))
print(next(zip(*k)))

Answer 5

If the values are integer and small, or you only care about bins of size 1:

def sort_by_frequency(arr):
    return np.flip(np.argsort(np.bincount(arr))[-(np.unique(arr).size):])

v = [1,1,1,1,1,2,2,9,3,3,3,3,7,8,8]
sort_by_frequency(v)

this should yield

array([1, 3, 8, 2, 9, 7]