How to get indices of a sorted array in Python

I have a numerical list:

myList = [1, 2, 3, 100, 5]

Now if I sort this list to obtain [1, 2, 3, 5, 100]. What I want is the indices of the elements from the original list in the sorted order i.e. [0, 1, 2, 4, 3] --- ala MATLAB's sort function that returns both values and indices.

Matthew Lewis

If you are using numpy, you have the argsort() function available:

>>> import numpy
>>> numpy.argsort(myList)
array([0, 1, 2, 4, 3])

http://docs.scipy.org/doc/numpy/reference/generated/numpy.argsort.html

This returns the arguments that would sort the array or list.

Roman Bodnarchuk

Something like next:

>>> myList = [1, 2, 3, 100, 5]
>>> [i[0] for i in sorted(enumerate(myList), key=lambda x:x[1])]
[0, 1, 2, 4, 3]

enumerate(myList) gives you a list containing tuples of (index, value):

[(0, 1), (1, 2), (2, 3), (3, 100), (4, 5)]

You sort the list by passing it to sorted and specifying a function to extract the sort key (the second element of each tuple; that's what the lambda is for. Finally, the original index of each sorted element is extracted using the [i[0] for i in ...] list comprehension.

robert king

myList = [1, 2, 3, 100, 5]    
sorted(range(len(myList)),key=myList.__getitem__)

[0, 1, 2, 4, 3]

The answers with enumerate are nice, but I personally don't like the lambda used to sort by the value. The following just reverses the index and the value, and sorts that. So it'll first sort by value, then by index.

sorted((e,i) for i,e in enumerate(myList))

Updated answer with enumerate and itemgetter:

sorted(enumerate(a), key=lambda x: x[1])
# [(0, 1), (1, 2), (2, 3), (4, 5), (3, 100)]

Zip the lists together: The first element in the tuple will the index, the second is the value (then sort it using the second value of the tuple x[1], x is the tuple)

Or using itemgetter from the operatormodule`:

from operator import itemgetter
sorted(enumerate(a), key=itemgetter(1))

If you do not want to use numpy,

sorted(range(len(seq)), key=seq.__getitem__)

is fastest, as demonstrated here.

The other answers are WRONG.

Running argsort once is not the solution. For example, the following code:

import numpy as np
x = [3,1,2]
np.argsort(x)

yields array([1, 2, 0], dtype=int64) which is not what we want.

The answer should be to run argsort twice:

import numpy as np
x = [3,1,2]
np.argsort(np.argsort(x))

gives array([2, 0, 1], dtype=int64) as expected.

I did a quick performance check on these with perfplot (a project of mine) and found that it's hard to recommend anything else but numpy (note the log scale):

Code to reproduce the plot:

import perfplot
import numpy


def sorted_enumerate(seq):
    return [i for (v, i) in sorted((v, i) for (i, v) in enumerate(seq))]


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


def sorted_range(seq):
    return sorted(range(len(seq)), key=seq.__getitem__)


def numpy_argsort(x):
    return numpy.argsort(x)


perfplot.save(
    "argsort.png",
    setup=lambda n: numpy.random.rand(n),
    kernels=[sorted_enumerate, sorted_enumerate_key, sorted_range, numpy_argsort],
    n_range=[2 ** k for k in range(15)],
    xlabel="len(x)",
    logx=True,
    logy=True,
)

Import numpy as np

FOR INDEX

S=[11,2,44,55,66,0,10,3,33]

r=np.argsort(S)

[output]=array([5, 1, 7, 6, 0, 8, 2, 3, 4])

argsort Returns the indices of S in sorted order

FOR VALUE

np.sort(S)

[output]=array([ 0,  2,  3, 10, 11, 33, 44, 55, 66])

We will create another array of indexes from 0 to n-1 Then zip this to the original array and then sort it on the basis of the original values

ar = [1,2,3,4,5]
new_ar = list(zip(ar,[i for i in range(len(ar))]))
new_ar.sort()

来源：https://stackoverflow.com/questions/6422700/how-to-get-indices-of-a-sorted-array-in-python

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