如何在Python中找到列表的平均值?

[1, 2, 3, 4]  ⟶  2.5

当前回答

我尝试使用上面的选项,但没有工作。 试试这个:

from statistics import mean

n = [11, 13, 15, 17, 19]

print(n)
print(mean(n))

使用过python 3.5

其他回答

使用numpy.mean:

xs = [15, 18, 2, 36, 12, 78, 5, 6, 9]

import numpy as np
print(np.mean(xs))
numbers = [0,1,2,3]

numbers[0] = input("Please enter a number")

numbers[1] = input("Please enter a second number")

numbers[2] = input("Please enter a third number")

numbers[3] = input("Please enter a fourth number")

print (numbers)

print ("Finding the Avarage")

avarage = int(numbers[0]) + int(numbers[1]) + int(numbers[2]) + int(numbers [3]) / 4

print (avarage)

你可以为平均值,使用率做一个函数:

average(21,343,2983) # You can pass as many arguments as you want.

代码如下:

def average(*args):
    total = 0
    for num in args:
        total+=num
    return total/len(args)

*args允许任意数量的答案。

Sum (l) / float(len(l))是正确答案,但为了完整起见,你可以用一个reduce来计算平均值:

>>> reduce(lambda x, y: x + y / float(len(l)), l, 0)
20.111111111111114

注意,这可能会导致轻微的舍入误差:

>>> sum(l) / float(len(l))
20.111111111111111

假设

x = [
    [-5.01,-5.43,1.08,0.86,-2.67,4.94,-2.51,-2.25,5.56,1.03],
    [-8.12,-3.48,-5.52,-3.78,0.63,3.29,2.09,-2.13,2.86,-3.33],
    [-3.68,-3.54,1.66,-4.11,7.39,2.08,-2.59,-6.94,-2.26,4.33]
]

你可以注意到x的维数是3*10如果你需要得到每一行的平均值,你可以输入这个

theMean = np.mean(x1,axis=1)

不要忘记将numpy导入为np