Finding The Index Of Sorted Elements In Python Array

I have seen answers to the question: Is it possible to arrange a numpy array (or python list) by using the indexes of the elements in decreasing order? (eg. Finding the Index of N

Solution 1:

EDIT when I first read the question, I thought you were looking for numpy.argsort.

Having read it again, I realized I misread.

scipy.stats.rankdata is what you're looking for (offset by 1, and reversed)

(scipy.stats.rankdata([4, 1, 0, 8, 5, 2])-1)[::-1]
=> array([ 2.,  4.,  5.,  0.,  1.,  3.])

(Original wrong answer, referring to argsort:)

from numpy import array, argsort
L = array([4, 1, 0, 8, 5, 2])
argsort(L)
=> array([2, 1, 5, 0, 4, 3])

Solution 2:

You can simply use your technique twice to get the indices in the sorted list:

A=[4, 1, 0, 8, 5, 2]
B=sorted(range(len(A)),key=lambda x:A[x],reverse=True)
C=sorted(range(len(A)),key=lambda x:B[x])
print C

prints

[2, 4, 5, 0, 1, 3]

Explanation

The idea is that the first iteration produces a list:

B = [3, 4, 0, 5, 1, 2]

giving the locations in the original list of the sorted sequence.

In other words, A[3]=8 is the largest element in the original list, A[4]=5 is the next largest, etc.

The second stage then sorts these indices in B back into the order 0,1,2,3,4,5 and produces C which contains the index in the list B.

It may help to think of B as sorting the list into descending order, and C as reversing the sort back into the original unsorted order, while keeping track of the indices in the sorted list.

Solution 3:

are you using numpy? if so, use numpy.argsort

import numpy as np
fakedata = np.array([40,20,60,50,10,20,80,30,50,40])
sorted_index = np.argsort(fakedata)
print(sorted_index)
print(fakedata[sorted_index])
[4157093826] # sorted index
[10202030404050506080] # values selected by the sorted index