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
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)
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
当Python有一个完美的cromulent sum()函数时,为什么要使用reduce()呢?
print sum(l) / float(len(l))
(float()在Python 2中强制Python执行浮点除法是必需的。)
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)