Equivalent of Numpy.argsort() in basic python? [duplicate]
is there a builtin function of Python that does on python.array
what argsort()
does on a numpy.array
?
Solution 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.array
s the same way:
import array
x = array.array('d', [5, 2, 1, 10])
print(argsort(x))
# [2, 1, 0, 3]
Solution 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.
Solution 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))]
Solution 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)