我在“中位数的中位数”算法的Python实现中发布了我的解决方案，这比使用sort()稍微快一点。我的解决方案每列使用15个数字，速度~5N比每列使用5个数字的速度~10N快。最佳速度是~4N，但我可能是错的。

根据Tom在评论中的要求，我在这里添加了我的代码，以供参考。我认为速度的关键部分是每列使用15个数字，而不是5个。

#!/bin/pypy
#
# TH @stackoverflow, 2016-01-20, linear time "median of medians" algorithm
#
import sys, random


items_per_column = 15


def find_i_th_smallest( A, i ):
    t = len(A)
    if(t <= items_per_column):
        # if A is a small list with less than items_per_column items, then:
        #
        # 1. do sort on A
        # 2. find i-th smallest item of A
        #
        return sorted(A)[i]
    else:
        # 1. partition A into columns of k items each. k is odd, say 5.
        # 2. find the median of every column
        # 3. put all medians in a new list, say, B
        #
        B = [ find_i_th_smallest(k, (len(k) - 1)/2) for k in [A[j:(j + items_per_column)] for j in range(0,len(A),items_per_column)]]

        # 4. find M, the median of B
        #
        M = find_i_th_smallest(B, (len(B) - 1)/2)


        # 5. split A into 3 parts by M, { < M }, { == M }, and { > M }
        # 6. find which above set has A's i-th smallest, recursively.
        #
        P1 = [ j for j in A if j < M ]
        if(i < len(P1)):
            return find_i_th_smallest( P1, i)
        P3 = [ j for j in A if j > M ]
        L3 = len(P3)
        if(i < (t - L3)):
            return M
        return find_i_th_smallest( P3, i - (t - L3))


# How many numbers should be randomly generated for testing?
#
number_of_numbers = int(sys.argv[1])


# create a list of random positive integers
#
L = [ random.randint(0, number_of_numbers) for i in range(0, number_of_numbers) ]


# Show the original list
#
# print L


# This is for validation
#
# print sorted(L)[int((len(L) - 1)/2)]


# This is the result of the "median of medians" function.
# Its result should be the same as the above.
#
print find_i_th_smallest( L, (len(L) - 1) / 2)

中值函数

def median(midlist):
    midlist.sort()
    lens = len(midlist)
    if lens % 2 != 0: 
        midl = (lens / 2)
        res = midlist[midl]
    else:
        odd = (lens / 2) -1
        ev = (lens / 2) 
        res = float(midlist[odd] + midlist[ev]) / float(2)
    return res

def midme(list1):

    list1.sort()
    if len(list1)%2>0:
            x = list1[int((len(list1)/2))]
    else:
            x = ((list1[int((len(list1)/2))-1])+(list1[int(((len(list1)/2)))]))/2
    return x


midme([4,5,1,7,2])

只要两行就够了。

def get_median(arr):
    '''
    Calculate the median of a sequence.
    :param arr: list
    :return: int or float
    '''
    arr = sorted(arr)
    return arr[len(arr)//2] if len(arr) % 2 else (arr[len(arr)//2] + arr[len(arr)//2-1])/2

当然，你可以使用内置函数，但如果你想创建自己的函数，你可以这样做。这里的技巧是使用~运算符将正数转换为负数。例如~2 -> -3和在Python中使用负in For list将从末尾开始计数。如果你有mid == 2，那么它会从开始取第三个元素，从结束取第三个元素。

def median(data):
    data.sort()
    mid = len(data) // 2
    return (data[mid] + data[~mid]) / 2

函数值:

def median(d):
    d=np.sort(d)
    n2=int(len(d)/2)
    r=n2%2
    if (r==0):
        med=d[n2] 
    else:
        med=(d[n2] + d[n2+1]) / 2
    return med