非零值的整数平均值 [英] Numpy mean of nonzero values

问题描述

我有一个大小为N * M的矩阵，我想找到每一行的平均值.值是从1到5，并且没有任何值的条目设置为0.但是，当我想使用以下方法查找均值时，它给了我错误的均值，因为它还计算了具有值的条目0.

I have a matrix of size N*M and I want to find the mean value for each row. The values are from 1 to 5 and entries that do not have any value are set to 0. However, when I want to find the mean using the following method, it gives me the wrong mean as it also counts the entries that have value of 0.

matrix_row_mean= matrix.mean(axis=1)

如何获取非零值的均值?

How can I get the mean of only nonzero values?

推荐答案

获取每一行中非零的计数，并将其用于平均每一行的总和.因此，实现看起来像这样-

Get the count of non-zeros in each row and use that for averaging the summation along each row. Thus, the implementation would look something like this -

np.true_divide(matrix.sum(1),(matrix!=0).sum(1))

如果您使用的是较旧版本的NumPy，则可以使用计数的浮点转换来替换np.true_divide，就像这样-

If you are on an older version of NumPy, you can use float conversion of the count to replace np.true_divide, like so -

matrix.sum(1)/(matrix!=0).sum(1).astype(float)

样品运行-

In [160]: matrix
Out[160]: 
array([[0, 0, 1, 0, 2],
       [1, 0, 0, 2, 0],
       [0, 1, 1, 0, 0],
       [0, 2, 2, 2, 2]])

In [161]: np.true_divide(matrix.sum(1),(matrix!=0).sum(1))
Out[161]: array([ 1.5,  1.5,  1. ,  2. ])

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解决问题的另一种方法是用NaNs替换零，然后使用np.nanmean，这将忽略那些NaNs而实际上是那些原始的zeros，就像-

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Another way to solve the problem would be to replace zeros with NaNs and then use np.nanmean, which would ignore those NaNs and in effect those original zeros, like so -

np.nanmean(np.where(matrix!=0,matrix,np.nan),1)

从性能的角度来看，我建议使用第一种方法.

From performance point of view, I would recommend the first approach.

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