如何在Python中找到列表的平均值?
[1, 2, 3, 4] ⟶ 2.5
如何在Python中找到列表的平均值?
[1, 2, 3, 4] ⟶ 2.5
当前回答
l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
l = map(float,l)
print '%.2f' %(sum(l)/len(l))
其他回答
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
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
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
我想补充另一种方法
import itertools,operator
list(itertools.accumulate(l,operator.add)).pop(-1) / len(l)
xs = [15, 18, 2, 36, 12, 78, 5, 6, 9]
sum(xs) / len(xs)