Python：定义数字列表的方差
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Python: Variance of a list of defined numbers
提问by GiamPy
I am trying to make a function that prints the variance of a list of defined numbers:
我正在尝试制作一个打印定义数字列表方差的函数：
grades = [100, 100, 90, 40, 80, 100, 85, 70, 90, 65, 90, 85, 50.5]
So far, I have tried proceeding on making these three functions:
到目前为止，我已经尝试继续制作这三个功能：
def grades_sum(my_list):
total = 0
for grade in my_list:
total += grade
return total
def grades_average(my_list):
sum_of_grades = grades_sum(my_list)
average = sum_of_grades / len(my_list)
return average
def grades_variance(my_list, average):
variance = 0
for i in my_list:
variance += (average - my_list[i]) ** 2
return variance / len(my_list)
When I try to execute the code, however, it gives me the following error at the following line:
但是，当我尝试执行代码时，它在以下行出现以下错误：
Line: variance += (average - my_list[i]) ** 2
Error: list index out of range
Apologies if my current Python knowledges are limited, but I am still learning - so please if you wish to help solving this issue try not to suggest extremely-complicated ways on how to solve this, thank you really much.
抱歉，如果我目前的 Python 知识有限，但我仍在学习 - 所以如果你想帮助解决这个问题，请尽量不要提出关于如何解决这个问题的极其复杂的方法，非常感谢你。
采纳答案by Magsol
First I would suggest using Python's built-in summethod to replace your first custom method. grades_averagethen becomes:
首先，我建议使用 Python 的内置sum方法来替换您的第一个自定义方法。grades_average然后变成：
def grades_average(my_list):
sum_of_grades = sum(my_list)
average = sum_of_grades / len(my_list)
return average
Second, I would strongly recommend looking into the NumPy library, as it has these methods built-in. numpy.mean()and numpy.std()would cover both these cases.
其次，我强烈建议查看NumPy 库，因为它内置了这些方法。numpy.mean()并且numpy.std()将涵盖这两种情况。
If you're interested in writing the code for yourself first, that's totally fine too. As for your specific error, I believe @gnibbler above nailed it. If you want to loop using an index, you can restructure the line in grades_varianceto be:
如果您有兴趣先为自己编写代码，那也完全没问题。至于你的具体错误，我相信上面的@gnibbler 已经解决了。如果要使用索引进行循环，可以将 in 行重组grades_variance为：
for i in range(0, len(my_list)):
As Lattywarenoted, looping by index is not particularly "Pythonic"; the way you're currently doing it is generally superior. This is just for your reference.
正如Lattyware 所指出的，按索引循环并不是特别“Pythonic”；你目前的做法通常更胜一筹。这仅供您参考。
回答by John La Rooy
When you say
当你说
for i in my_list:
iisn't the indexof the item. iisthe item
i不是项目的索引。i是项目
for i in my_list:
variance += (average - i) ** 2
回答by Gareth Latty
While gnibbler has solved the problem with your code, you can achieve this much more easily using built-in functionsand a generator expression:
虽然gnibbler 已经用您的代码解决了这个问题，但您可以使用内置函数和生成器表达式更轻松地实现这一点：
average = sum(grades) / len(grades)
varience = sum((average - value) ** 2 for value in grades) / len(grades)
It might look a little scary at first, but if you watch the video I link about list comprehensions and generator expressions - they are actually really simple and useful.
乍一看可能有点吓人，但是如果您观看我链接的有关列表推导式和生成器表达式的视频 - 它们实际上非常简单且有用。
回答by zengr
python 3.4 has a statistics lib which does this.
python 3.4 有一个统计库可以做到这一点。
import statistics
grades = [100, 100, 90, 40, 80, 100, 85, 70, 90, 65, 90, 85, 50.5]
statistics.pvariance(grades)
=> 334.07100591715977
https://docs.python.org/3/library/statistics.html#statistics.pvariance
https://docs.python.org/3/library/statistics.html#statistics.pvariance
回答by Bharatwaja
the below code is used to get the average of values
下面的代码用于获取值的平均值
def grades_average(my_list):
sum_of_grades = sum(my_list)
average = sum(my_list) / len(my_list)
return average
variance formula -> The average of the squared differences from the Mean. This code below is used to get the variance of values
方差公式 -> 均值的平方差的平均值。下面的代码用于获取值的方差
def grades_variance(my_list, average):
variance = 0
for i in my_list:
variance += (average - i) ** 2
return variance / len(my_list)
回答by Don Juan
I suppose you would like the sample variance i.e. the unbiased estimator of the variance. I think this function might do the job. It will print the variance and the mean of a vector n.
我想你想要样本方差，即方差的无偏估计量。我认为这个功能可以完成这项工作。它将打印向量 n 的方差和均值。
n = [5, 3, 1, 2, 4]
def variance1337(n):
var1 = []
mean1 = sum(n)/len(n)
for xs in n:
var1.append((xs - mean1) ** 2)
print(sum(var1)/(len(n) - 1))
print(mean1)
回答by TRINADH NAGUBADI
The below code is used to get the variance I create a custom function
下面的代码用于获取方差我创建了一个自定义函数
def variance(val):
total_sum=sum(val)
average=total_sum/len(val)
a=[]
for i in val:
a.append((i-average)**2)
return sum(a)/len(a)
val=[2.18,2.22,2.24,1.62,1.32,1.85,1.85,2.70,3.60,4.60,1.38,2.34,2.71]
variance(val)
