在Python中如何找到列表的中值?列表可以是任意大小的，并且数字不保证是任何特定的顺序。
如果列表包含偶数个元素，则函数应返回中间两个元素的平均值。
以下是一些例子(为了便于展示，进行了排序):
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
当前回答
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
其他回答
(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
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])
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
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
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
