[2, 3, 1, 0] indicates that the smallest element is at index 2, the next smallest at index 3, then index 1, then index 0.

There are a number of ways to get the result you are looking for:

import numpy as np
import scipy.stats as stats

def using_indexed_assignment(x):
    "https://stackoverflow.com/a/5284703/190597 (Sven Marnach)"
    result = np.empty(len(x), dtype=int)
    temp = x.argsort()
    result[temp] = np.arange(len(x))
    return result

def using_rankdata(x):
    return stats.rankdata(x)-1

def using_argsort_twice(x):
    "https://stackoverflow.com/a/6266510/190597 (k.rooijers)"
    return np.argsort(np.argsort(x))

def using_digitize(x):
    unique_vals, index = np.unique(x, return_inverse=True)
    return np.digitize(x, bins=unique_vals) - 1

For example,

In [72]: x = np.array([1.48,1.41,0.0,0.1])

In [73]: using_indexed_assignment(x)
Out[73]: array([3, 2, 0, 1])

This checks that they all produce the same result:

x = np.random.random(10**5)
expected = using_indexed_assignment(x)
for func in (using_argsort_twice, using_digitize, using_rankdata):
    assert np.allclose(expected, func(x))

These IPython %timeit benchmarks suggests for large arrays using_indexed_assignment is the fastest:

In [50]: x = np.random.random(10**5)
In [66]: %timeit using_indexed_assignment(x)
100 loops, best of 3: 9.32 ms per loop

In [70]: %timeit using_rankdata(x)
100 loops, best of 3: 10.6 ms per loop

In [56]: %timeit using_argsort_twice(x)
100 loops, best of 3: 16.2 ms per loop

In [59]: %timeit using_digitize(x)
10 loops, best of 3: 27 ms per loop

For small arrays, using_argsort_twice may be faster:

In [78]: x = np.random.random(10**2)

In [81]: %timeit using_argsort_twice(x)
100000 loops, best of 3: 3.45 µs per loop

In [79]: %timeit using_indexed_assignment(x)
100000 loops, best of 3: 4.78 µs per loop

In [80]: %timeit using_rankdata(x)
100000 loops, best of 3: 19 µs per loop

In [82]: %timeit using_digitize(x)
10000 loops, best of 3: 26.2 µs per loop

Note also that stats.rankdata gives you more control over how to handle elements of equal value.

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.