如何获得列表内的每个列表的均值，从而避免某些值? [英] How to get mean of each list inside the list avoiding certain value?

问题描述

如何在列表内分别计算每个列表的平均值，避免使用特殊值(-999)?

How to calculate mean of values of each list separately inside the list avoiding a special value (-999)?

A = [[4,5,7,8,-999],[3,8,5,7,-999]]
M = [sum(x)/len(x) for x in [y for y in A if y!= -999]
print (M)

有什么主意吗?

为了获得最佳速度:有人可以更正以下代码吗? @ alexanderlukanin13

For best speed: can someone correct the following code? @alexanderlukanin13

M = [np.mean(L [L!=-999])for L in A]

推荐答案

您提到的numpy:

>>> import numpy as np
>>> 
>>> A = [[4,5,7,8,-999],[3,8,5,7,-999]]
>>> M = [np.mean([x for x in L if x > -999]) for L in A]
>>> print M
[6.0, 5.75]

编辑

正如您提到的速度是一项重要要求，您可以执行以下操作:

As you mentioned speed as an important requirement, you can do this:

>>> B = np.array(A)
>>> np.average(B, axis=1, weights=B!=-999)
array([ 6.  ,  5.75])

所有事情都在numpy(即C)空间中发生，这应该会使它变得非常快.

Everything happens in numpy (i.e. C) space, which should get it pretty fast.

解释一下正在发生的事情:

To explain a bit what is happening:

np.mean(A, axis=1)和等效的np.average(A, axis=1)计算数组各列的平均值，即您想要的每一行的均值向量. average 允许使用权重:我们使用B!=-999是一个布尔数组，当用作权重时将其评估为1 s和0 s，即忽略评估为False的值.

np.mean(A, axis=1) and the equivalent np.average(A, axis=1) compute the average over the columns of your array, i.e. the vector of means of each row, which is what you want. average allows the use of weights: we use B!=-999, which is a boolean array, evaluated as 1s and 0s when used as weights, i.e. ignoring the values evaluated as False.

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