def avg(l):
    """uses floating-point division."""
    return sum(l) / float(len(l))

例子:

l1 = [3,5,14,2,5,36,4,3]
l2 = [0,0,0]

print(avg(l1)) # 9.0
print(avg(l2)) # 0.0

def list_mean(nums):
    sumof = 0
    num_of = len(nums)
    mean = 0
    for i in nums:
        sumof += i
    mean = sumof / num_of
    return float(mean)

NumPy有一个NumPy。这是算术平均数。用法如下:

>>> import numpy
>>> a = [1, 2, 4]
>>> numpy.mean(a)
2.3333333333333335

其他人已经发布了非常好的答案，但有些人可能仍然在寻找一个经典的方法来查找Mean(avg)，所以我在这里发布这个(在Python 3.6中测试的代码):

def meanmanual(listt):

mean = 0
lsum = 0
lenoflist = len(listt)

for i in listt:
    lsum += i

mean = lsum / lenoflist
return float(mean)

a = [1, 2, 3, 4, 5, 6]
meanmanual(a)

Answer: 3.5

def avg(l):
    """uses floating-point division."""
    return sum(l) / float(len(l))

例子:

l1 = [3,5,14,2,5,36,4,3]
l2 = [0,0,0]

print(avg(l1)) # 9.0
print(avg(l2)) # 0.0

你问题的正确答案是用统计学。但是为了好玩，这里有一个不使用len()函数的mean版本，所以它(像statistics.mean一样)可以用于不支持len()的生成器:

from functools import reduce
from operator import truediv
def ave(seq):
    return truediv(*reduce(lambda a, b: (a[0] + b[1], b[0]), 
                           enumerate(seq, start=1), 
                           (0, 0)))