Python Numpy 中的均方误差？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/16774849/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Mean Squared Error in Numpy?
提问by TheMeaningfulEngineer
Is there a method in numpy for calculating the Mean Squared Error between two matrices?
numpy 中是否有计算两个矩阵之间的均方误差的方法？
I've tried searching but found none. Is it under a different name?
我试过搜索，但没有找到。它是在不同的名称下吗？
If there isn't, how do you overcome this? Do you write it yourself or use a different lib?
如果没有，你如何克服这个？您是自己编写还是使用不同的库？
采纳答案by Saullo G. P. Castro
You can use:
您可以使用：
mse = ((A - B)**2).mean(axis=ax)
Or
或者
mse = (np.square(A - B)).mean(axis=ax)
- with
ax=0the average is performed along the row, for each column, returning an array - with
ax=1the average is performed along the column, for each row, returning an array - with
ax=Nonethe average is performed element-wise along the array, returning a scalar value
- 与
ax=0平均值沿着行进行的，对于每一列，返回一个数组 - 与
ax=1平均值沿着列进行的，对于每一行，返回一个数组 - 与
ax=None平均值沿着阵列进行逐元素，返回一个标量值
回答by Charity Leschinski
This isn't part of numpy, but it will work with numpy.ndarrayobjects. A numpy.matrixcan be converted to a numpy.ndarrayand a numpy.ndarraycan be converted to a numpy.matrix.
这不是 的一部分numpy，但它适用于numpy.ndarray对象。Anumpy.matrix可以转换为 a numpy.ndarray，anumpy.ndarray可以转换为 a numpy.matrix。
from sklearn.metrics import mean_squared_error
mse = mean_squared_error(A, B)
See Scikit Learn mean_squared_errorfor documentation on how to control axis.
有关如何控制轴的文档，请参阅Scikit Learn mean_squared_error。
回答by nickandross
Another alternative to the accepted answer that avoids any issues with matrix multiplication:
已接受答案的另一种替代方法，可避免矩阵乘法的任何问题：
def MSE(Y, YH):
return np.square(Y - YH).mean()
From the documents for np.square: "Return the element-wise square of the input."
来自np.square的文档：“返回输入的元素平方。”
回答by Mark Swardstrom
Even more numpy
更麻木
np.square(np.subtract(A, B)).mean()
