Description:

The mean (or average) is the most popular measure of central tendency; however it does not behave very well when the data is skewed (i.e. wages distribution). In such cases, it's better to use the median.

Your task for this kata is to find the median of an array consisting of n elements.

You can assume that all inputs are arrays of numbers in integer format. For the empty array your code should return `NaN` (false in JavaScript/`NULL` in PHP/`nil` in Ruby).

Examples:

Input `[1, 2, 3, 4]` --> Median `2.5`

Input `[3, 4, 1, 2, 5]` --> Median `3`

Source: codewars | Difficulty: EASY

Starter code: def median(arr):
	

Solutions:
from statistics import median


def median(arr):
	arr.sort()
	return (arr[(len(arr) - 1) // 2] + arr[len(arr) // 2]) / 2


from numpy import median


import numpy as np

def median(arr):
	return np.median(arr) if len(arr) > 0 else None


from random import choice

def median(lst):
	return quickselect(lst, len(lst) // 2, len(lst) % 2 == 0)

def quickselect(lst, nth, even):
	pivot = choice(lst)
	smaller = [n for n in lst if n < pivot]
	n_equal = sum((n == pivot for n in lst))
	bigger = [n for n in lst if n > pivot]
	if even:
		if len(smaller) > nth:
			return quickselect(smaller, nth, True)
		elif len(smaller) == nth:
			return (pivot + quickselect(smaller, nth - 1, False)) / 2
		elif len(smaller) + n_equal > nth:
			return pivot
		elif len(smaller) + n_equal == nth:
			return (pivot + quickselect(bigger, nth - len(smaller) - n_equal, False)) / 2
		else:
			return quickselect(bigger, nth - len(smaller) - n_equal, True)
	elif len(smaller) > nth:
		return quickselect(smaller, nth, False)
	elif len(smaller) + n_equal > nth:
		return pivot
	else:
		return quickselect(bigger, nth - len(smaller) - n_equal, False)


from numpy import median as median_

def median(arr):
	return median_(arr)


import numpy

def median(A):
	return None if len(A) < 1 else numpy.median(A)


import numpy as np

def median(arr):
	return np.median(arr)