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

在列表中求平均值 通过使用以下PYTHON代码:

l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
print(sum(l)//len(l))

试着简单一点。

对于Python 3.8+，使用统计信息。浮点数稳定性的平均值。(快)。

对于Python 3.4+，使用统计信息。平均数值稳定性与浮子。(慢)。

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

import statistics
statistics.mean(xs)  # = 20.11111111111111

对于较旧版本的Python 3，请使用

sum(xs) / len(xs)

对于Python 2，将len转换为浮点数以获得浮点除法:

sum(xs) / float(len(xs))

xs = [15, 18, 2, 36, 12, 78, 5, 6, 9]
sum(xs) / len(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)

结合上面的几个答案，我提出了以下与reduce一起工作的方法，并且不假设你在reduce函数中有L可用:

from operator import truediv

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

def sum_and_count(x, y):
    try:
        return (x[0] + y, x[1] + 1)
    except TypeError:
        return (x + y, 2)

truediv(*reduce(sum_and_count, L))

# prints 
20.11111111111111