Is there a way to get the index of the median in python in one command?

a quick approximation:

numpy.argsort(data)[len(data)//2]

In general, this is an ill-posed question because an array does not necessarily contain its own median for numpy's definition of the median. For example:

>>> np.median([1, 2])
1.5

But when the length of the array is odd, the median will generally be in the array, so asking for its index does make sense:

>>> np.median([1, 2, 3])
2

For odd-length arrays, an efficient way to determine the index of the median value is by using the np.argpartition function. For example:

import numpy as np

def argmedian(x):
  return np.argpartition(x, len(x) // 2)[len(x) // 2]

# Works for odd-length arrays, where the median is in the array:
x = np.random.rand(101)

print("median in array:", np.median(x) in x)
# median in array: True

print(x[argmedian(x)], np.median(x))
# 0.5819150016674371 0.5819150016674371

# Doesn't work for even-length arrays, where the median is not in the array:
x = np.random.rand(100)

print("median in array:", np.median(x) in x)
# median in array: False

print(x[argmedian(x)], np.median(x))
# 0.6116799104572843 0.6047559243909065

This is quite a bit faster than the accepted sort-based solution as the size of the array grows:

x = np.random.rand(1000)
%timeit np.argsort(x)[len(x)//2]
# 10000 loops, best of 3: 25.4 µs per loop
%timeit np.argpartition(x, len(x) // 2)[len(x) // 2]
# 100000 loops, best of 3: 6.03 µs per loop

You can keep the indices with the elements (zip) and sort and return the element on the middle or two elements on the middle, however sorting will be O(n.logn). The following method is O(n) in terms of time complexity.

import numpy as np

def arg_median(a):
    if len(a) % 2 == 1:
        return np.where(a == np.median(a))[0][0]
    else:
        l,r = len(a) // 2 - 1, len(a) // 2
        left = np.partition(a, l)[l]
        right = np.partition(a, r)[r]
        return [np.where(a == left)[0][0], np.where(a == right)[0][0]]

print(arg_median(np.array([ 3,  9,  5,  1, 15])))
# 1 3 5 9 15, median=5, index=2
print(arg_median(np.array([ 3,  9,  5,  1, 15, 12])))
# 1 3 5 9 12 15, median=5,9, index=2,1

Output:

2
[2, 1]

The idea is if there is only one median (array has a odd length), then it returns the index of the median. If we need to average to elements (array has even length) then it returns the indices of these two elements in an list.

It seems old question, but i found a nice way to make it so:

import random
import numpy as np
#some random list with 20 elements
a = [random.random() for i in range(20)]
#find the median index of a
medIdx = a.index(np.percentile(a,50,interpolation='nearest'))

The neat trick here is the percentile builtin option for nearest interpolation, which return a "real" median value from the list, so it is safe to search for it afterwards.