Numpy 的均方差？

有没有一种用 numpy 表示的方法来计算两个矩阵之间的均方差？

我找过了，没找到，是用别的名字吗？

如果没有，你如何克服这个问题? 你是自己写的还是用另一个图书馆？

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最佳答案

You can use:

mse = ((A - B)**2).mean(axis=ax)

mse = (np.square(A - B)).mean(axis=ax)

with ax=0 the average is performed along the row, for each column, returning an array
with ax=1 the average is performed along the column, for each row, returning an array
with omitting the ax parameter (or setting it to ax=None) the average is performed element-wise along the array, returning a scalar value

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This isn't part of numpy, but it will work with numpy.ndarray objects. A numpy.matrix can be converted to a numpy.ndarray and a numpy.ndarray can be converted to a numpy.matrix.

from sklearn.metrics import mean_squared_error
mse = mean_squared_error(A, B)

See Scikit Learn mean_squared_error for documentation on how to control axis.

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Another alternative to the accepted answer that avoids any issues with matrix multiplication:

 def MSE(Y, YH):
return np.square(Y - YH).mean()

From the documents for np.square:

Return the element-wise square of the input.

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Even more numpy

np.square(np.subtract(A, B)).mean()

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Just for kicks

mse = (np.linalg.norm(A-B)**2)/len(A)

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The standard numpy methods for calculation mean squared error (variance) and its square root (standard deviation) are numpy.var() and numpy.std(), see here and here. They apply to matrices and have the same syntax as numpy.mean().

I suppose that the question and the preceding answers might have been posted before these functions became available.