Equivalent of Numpy.argsort() in basic python?

is there a builtin function of Python that does on python.array what argsort() does on a numpy.array?

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Answers:

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Method 1

There is no built-in function, but it’s easy to assemble one out of the terrific tools Python makes available:

def argsort(seq):
    # http://stackoverflow.com/questions/3071415/efficient-method-to-calculate-the-rank-vector-of-a-list-in-python
    return sorted(range(len(seq)), key=seq.__getitem__)

x = [5,2,1,10]

print(argsort(x))
# [2, 1, 0, 3]

It works on Python array.arrays the same way:

import array
x = array.array('d', [5, 2, 1, 10])
print(argsort(x))
# [2, 1, 0, 3]

Method 2

I timed the suggestions above and here are my results.

import timeit
import random
import numpy as np

def f(seq):
    # http://stackoverflow.com/questions/3382352/equivalent-of-numpy-argsort-in-basic-python/3383106#3383106
    #non-lambda version by Tony Veijalainen
    return [i for (v, i) in sorted((v, i) for (i, v) in enumerate(seq))]

def g(seq):
    # http://stackoverflow.com/questions/3382352/equivalent-of-numpy-argsort-in-basic-python/3383106#3383106
    #lambda version by Tony Veijalainen
    return [x for x,y in sorted(enumerate(seq), key = lambda x: x[1])]


def h(seq):
    #http://stackoverflow.com/questions/3382352/equivalent-of-numpy-argsort-in-basic-python/3382369#3382369
    #by unutbu
    return sorted(range(len(seq)), key=seq.__getitem__)


seq = list(range(10000))
random.shuffle(seq)

n_trials = 100
for cmd in [
        'f(seq)', 'g(seq)', 'h(seq)', 'np.argsort(seq)',
        'np.argsort(seq).tolist()'
        ]:
    t = timeit.Timer(cmd, globals={**globals(), **locals()})
    print('time for {:d}x {:}: {:.6f}'.format(n_trials, cmd, t.timeit(n_trials)))

output

time for 100x f(seq): 0.323915
time for 100x g(seq): 0.235183
time for 100x h(seq): 0.132787
time for 100x np.argsort(seq): 0.091086
time for 100x np.argsort(seq).tolist(): 0.104226

A problem size dependent analysis is given here.

Method 3

My alternative with enumerate:

def argsort(seq):
    return [x for x,y in sorted(enumerate(seq), key = lambda x: x[1])]

seq=[5,2,1,10]
print(argsort(seq))
# Output:
# [2, 1, 0, 3]

Better though to use answer from https://stackoverflow.com/users/9990/marcelo-cantos answer to thread python sort without lambda expressions

[i for (v, i) in sorted((v, i) for (i, v) in enumerate(seq))]

Method 4

Found this question, but needed argsort for a list of objects based on an object property.

Extending unutbu’s answer, this would be:

sorted(range(len(seq)), key = lambda x: seq[x].sort_property)

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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