sorted()函数对此非常有用。使用排序函数 要对列表排序，只需返回中间值(或两个中间值的平均值) 如果列表包含偶数个元素，则为。

def median(lst):
    sortedLst = sorted(lst)
    lstLen = len(lst)
    index = (lstLen - 1) // 2
   
    if (lstLen % 2):
        return sortedLst[index]
    else:
        return (sortedLst[index] + sortedLst[index + 1])/2.0

你可以使用这个列表。排序以避免创建已排序的新列表，并对列表进行排序。

此外，你不应该使用list作为变量名，因为它会掩盖python自己的列表。

def median(l):
    half = len(l) // 2
    l.sort()
    if not len(l) % 2:
        return (l[half - 1] + l[half]) / 2.0
    return l[half]

(适用于python-2.x):

def median(lst):
    n = len(lst)
    s = sorted(lst)
    return (s[n//2-1]/2.0+s[n//2]/2.0, s[n//2])[n % 2] if n else None

>>> median([-5, -5, -3, -4, 0, -1])
-3.5

numpy.median ():

>>> from numpy import median
>>> median([1, -4, -1, -1, 1, -3])
-1.0

python 3。X，使用statistics.median:

>>> from statistics import median
>>> median([5, 2, 3, 8, 9, -2])
4.0

Python 3.4有statistics.median:

返回数值数据的中位数(中间值)。 当数据点数为奇数时，返回中间的数据点。 当数据点数为偶数时，通过取两个中间值的平均值来插值中位数: >>>中位数([1,3,5]) 3. >>>中位数([1,3,5,7]) 4.0

用法:

import statistics

items = [6, 1, 8, 2, 3]

statistics.median(items)
#>>> 3

它对类型也非常小心:

statistics.median(map(float, items))
#>>> 3.0

from decimal import Decimal
statistics.median(map(Decimal, items))
#>>> Decimal('3')

如果需要更快的平均情况运行时间，可以尝试快速选择算法。Quickselect具有平均(和最佳)情况性能O(n)，尽管在糟糕的一天它可能会以O(n²)结束。

下面是一个随机选择枢轴的实现:

import random

def select_nth(n, items):
    pivot = random.choice(items)

    lesser = [item for item in items if item < pivot]
    if len(lesser) > n:
        return select_nth(n, lesser)
    n -= len(lesser)

    numequal = items.count(pivot)
    if numequal > n:
        return pivot
    n -= numequal

    greater = [item for item in items if item > pivot]
    return select_nth(n, greater)

你可以简单地把它变成一个方法来寻找中位数:

def median(items):
    if len(items) % 2:
        return select_nth(len(items)//2, items)

    else:
        left  = select_nth((len(items)-1) // 2, items)
        right = select_nth((len(items)+1) // 2, items)

        return (left + right) / 2

这是非常未优化的，但即使是一个优化的版本也不太可能超过Tim Sort (CPython的内置排序)，因为它真的很快。我以前试过，但失败了。

我为一组数字定义了一个中值函数

def median(numbers):
    return (sorted(numbers)[int(round((len(numbers) - 1) / 2.0))] + sorted(numbers)[int(round((len(numbers) - 1) // 2.0))]) / 2.0

这里有一个更干净的解决方案:

def median(lst):
    quotient, remainder = divmod(len(lst), 2)
    if remainder:
        return sorted(lst)[quotient]
    return sum(sorted(lst)[quotient - 1:quotient + 1]) / 2.

注:答案更改为在评论中加入建议。

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

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)

def median(array):
    """Calculate median of the given list.
    """
    # TODO: use statistics.median in Python 3
    array = sorted(array)
    half, odd = divmod(len(array), 2)
    if odd:
        return array[half]
    return (array[half - 1] + array[half]) / 2.0

以下是我在Codecademy的练习中得出的结论:

def median(data):
    new_list = sorted(data)
    if len(new_list)%2 > 0:
        return new_list[len(new_list)/2]
    elif len(new_list)%2 == 0:
        return (new_list[(len(new_list)/2)] + new_list[(len(new_list)/2)-1]) /2.0

print median([1,2,3,4,5,9])

下面是不使用中值函数就能找到中值的乏味方法:

def median(*arg):
    order(arg)
    numArg = len(arg)
    half = int(numArg/2)
    if numArg/2 ==half:
        print((arg[half-1]+arg[half])/2)
    else:
        print(int(arg[half]))

def order(tup):
    ordered = [tup[i] for i in range(len(tup))]
    test(ordered)
    while(test(ordered)):
        test(ordered)
    print(ordered)


def test(ordered):
    whileloop = 0 
    for i in range(len(ordered)-1):
        print(i)
        if (ordered[i]>ordered[i+1]):
            print(str(ordered[i]) + ' is greater than ' + str(ordered[i+1]))
            original = ordered[i+1]
            ordered[i+1]=ordered[i]
            ordered[i]=original
            whileloop = 1 #run the loop again if you had to switch values
    return whileloop

我在浮点值列表中遇到了一些问题。我最终使用了来自python3统计数据的代码片段。中位数和工作完美的浮动值没有导入。源

def calculateMedian(list):
    data = sorted(list)
    n = len(data)
    if n == 0:
        return None
    if n % 2 == 1:
        return data[n // 2]
    else:
        i = n // 2
        return (data[i - 1] + data[i]) / 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

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])

def median(array):
    if len(array) < 1:
        return(None)
    if len(array) % 2 == 0:
        median = (array[len(array)//2-1: len(array)//2+1])
        return sum(median) / len(median)
    else:
        return(array[len(array)//2])

def median(x):
    x = sorted(x)
    listlength = len(x) 
    num = listlength//2
    if listlength%2==0:
        middlenum = (x[num]+x[num-1])/2
    else:
        middlenum = x[num]
    return middlenum

这很简单;

def median(alist):
    #to find median you will have to sort the list first
    sList = sorted(alist)
    first = 0
    last = len(sList)-1
    midpoint = (first + last)//2
    return midpoint

你可以这样使用返回值median = median(anyList)

import numpy as np
def get_median(xs):
        mid = len(xs) // 2  # Take the mid of the list
        if len(xs) % 2 == 1: # check if the len of list is odd
            return sorted(xs)[mid] #if true then mid will be median after sorting
        else:
            #return 0.5 * sum(sorted(xs)[mid - 1:mid + 1])
            return 0.5 * np.sum(sorted(xs)[mid - 1:mid + 1]) #if false take the avg of mid
print(get_median([7, 7, 3, 1, 4, 5]))
print(get_median([1,2,3, 4,5]))

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

如果您需要关于列表分布的额外信息，百分位数方法可能会很有用。中位数对应于列表的第50个百分位数:

import numpy as np
a = np.array([1,2,3,4,5,6,7,8,9])
median_value = np.percentile(a, 50) # return 50th percentile
print median_value 

更普遍的中位数(和百分位数)方法是:

def get_percentile(data, percentile):
    # Get the number of observations
    cnt=len(data)
    # Sort the list
    data=sorted(data)
    # Determine the split point
    i=(cnt-1)*percentile
    # Find the `floor` of the split point
    diff=i-int(i)
    # Return the weighted average of the value above and below the split point
    return data[int(i)]*(1-diff)+data[int(i)+1]*(diff)

# Data
data=[1,2,3,4,5]
# For the median
print(get_percentile(data=data, percentile=.50))
# > 3
print(get_percentile(data=data, percentile=.75))
# > 4

# Note the weighted average difference when an int is not returned by the percentile
print(get_percentile(data=data, percentile=.51))
# > 3.04

一个返回给定列表中值的简单函数:

def median(lst):
    lst = sorted(lst)  # Sort the list first
    if len(lst) % 2 == 0:  # Checking if the length is even
        # Applying formula which is sum of middle two divided by 2
        return (lst[len(lst) // 2] + lst[(len(lst) - 1) // 2]) / 2
    else:
        # If length is odd then get middle value
        return lst[len(lst) // 2]

一些关于中值函数的例子:

>>> median([9, 12, 20, 21, 34, 80])  # Even
20.5
>>> median([9, 12, 80, 21, 34])  # Odd
21

如果你想使用库，你可以简单地做:

>>> import statistics
>>> statistics.median([9, 12, 20, 21, 34, 80])  # Even
20.5
>>> statistics.median([9, 12, 80, 21, 34])  # Odd
21

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

实现它:

def median(numbers):
    """
    Calculate median of a list numbers.
    :param numbers: the numbers to be calculated.
    :return: median value of numbers.

    >>> median([1, 3, 3, 6, 7, 8, 9])
    6
    >>> median([1, 2, 3, 4, 5, 6, 8, 9])
    4.5
    >>> import statistics
    >>> import random
    >>> numbers = random.sample(range(-50, 50), k=100)
    >>> statistics.median(numbers) == median(numbers)
    True
    """
    numbers = sorted(numbers)
    mid_index = len(numbers) // 2
    return (
        (numbers[mid_index] + numbers[mid_index - 1]) / 2 if mid_index % 2 == 0
        else numbers[mid_index]
    )


if __name__ == "__main__":
    from doctest import testmod

    testmod()

来源

我所做的是:

def median(a):
    a = sorted(a)
    if len(a) / 2 != int:
        return a[len(a) / 2]
    else:
        return (a[len(a) / 2] + a[(len(a) / 2) - 1]) / 2

解释:基本上，如果列表中的项目数量是奇数，则返回中间的数字，否则，如果你是偶数列表的一半，python会自动舍入较大的数字，这样我们就知道在它之前的数字会少一个(因为我们对它进行了排序)，我们可以将默认的较大数字和小于它的数字相加，然后除以2得到中位数。

简单地说，创建一个中值函数，参数为数字列表，并调用该函数。

def median(l):
    l = sorted(l)
    lent = len(l)
    if (lent % 2) == 0:
        m = int(lent / 2)
        result = l[m]
    else:
        m = int(float(lent / 2) - 0.5)
        result = l[m]
    return result

试试这个

import math
def find_median(arr):
    if len(arr)%2==1:
        med=math.ceil(len(arr)/2)-1
        return arr[med]
    else:
        return -1
print(find_median([1,2,3,4,5,6,7,8]))