Finding median of list in Python

2019-01-03 05:01发布

How do you find the median of a list in Python? The list can be of any size and the numbers are not guaranteed to be in any particular order.

If the list contains an even number of elements, the function should return the average of the middle two.

Here are some examples (sorted for display purposes):

median([1]) == 1
median([1, 1]) == 1
median([1, 1, 2, 4]) == 1.5
median([0, 2, 5, 6, 8, 9, 9]) == 6
median([0, 0, 0, 0, 4, 4, 6, 8]) == 2

18条回答
Explosion°爆炸
2楼-- · 2019-01-03 05:53
import numpy as np
def median(x):

    if len(x)%2==0:
        x = sorted(x)
        num = round(len(x)/2)
        num2 = num-1
        middlenum = (x[num]+x[num2])/2
    else:
        x = sorted(x)
        listlength = len(x) 
        num = round(listlength / 2)
        middlenum = x[num]
    return middlenum
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Juvenile、少年°
3楼-- · 2019-01-03 05:53

I had some problems with lists of float values. I ended up using a code snippet from the python3 statistics.median and is working perfect with float values without imports. source

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
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Lonely孤独者°
4楼-- · 2019-01-03 05:54

Of course you can use build in functions, but if you would like to create your own you can do something like this. The trick here is to use ~ operator that flip positive number to negative. For instance ~2 -> -3 and using negative in for list in Python will count items from the end. So if you have mid == 2 then it will take third element from beginning and third item from the end.

def median(data):
    data.sort()
    mid = len(data) // 2
    return (data[mid] + data[~mid]) / 2
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劳资没心,怎么记你
5楼-- · 2019-01-03 05:55

You can try the quickselect algorithm if faster average-case running times are needed. Quickselect has average (and best) case performance O(n), although it can end up O(n²) on a bad day.

Here's an implementation with a randomly chosen pivot:

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)

You can trivially turn this into a method to find medians:

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

This is very unoptimised, but it's not likely that even an optimised version will outperform Tim Sort (CPython's built-in sort) because that's really fast. I've tried before and I lost.

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我只想做你的唯一
6楼-- · 2019-01-03 05:56

Here's the tedious way to find median without using the median function:

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
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仙女界的扛把子
7楼-- · 2019-01-03 05:58

Here what I came up with during this exercise in Codecademy:

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])
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