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
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
其他回答
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
为了使用reduce来获取运行平均值,您需要跟踪到目前为止所看到的元素总数。因为它不是列表中的一个普通元素,所以还必须向reduce传递一个要折叠成的额外参数。
>>> l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
>>> running_average = reduce(lambda aggr, elem: (aggr[0] + elem, aggr[1]+1), l, (0.0,0))
>>> running_average[0]
(181.0, 9)
>>> running_average[0]/running_average[1]
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
而不是强制转换为float,你可以将0.0加到和:
def avg(l):
return sum(l, 0.0) / len(l)
或者使用熊猫系列。意思是方法:
pd.Series(sequence).mean()
演示:
>>> import pandas as pd
>>> l = [15, 18, 2, 36, 12, 78, 5, 6, 9]
>>> pd.Series(l).mean()
20.11111111111111
>>>
从文档中可以看出:
系列。意思是(axis= no, skipna= no, level= no, numic_only = no, kwargs
这里是这个的文档:
https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.mean.html
整个文档:
https://pandas.pydata.org/pandas-docs/stable/10min.html