将Matlab（bsxfun，rdivide）行转换为Python [英] Translating a line of Matlab (bsxfun, rdivide) to Python

问题描述

我正在将Matlab函数转换为Python。不幸的是，我不是Matlab专家，所以我很难理解一些内容，例如。 G。这个：

I am translating a Matlab function to Python. Unfortunately I am not a Matlab expert and it is hard for me to understand some lines, e. g. this one:

a = [[0, 1]; [2, 3]]
bsxfun(@rdivide, sqrt(a), a)

我还不太了解，但是我认为这行会

I did not really understand it yet, but I think this line does

r / a

对于sqrt（a）的每一行r（或者是每一列？）和r / sqrt（a）通常可以转换为numpy as

for each row r of sqrt(a) (or is it each column?) and r / sqrt(a) can usually be translated to numpy as

numpy.linalg.solve(sqrt(a).T, r.T).T

问题是：Matlab说结果是

The problem with this is: Matlab says the result is

       NaN   1.00000
   0.70711   0.57735

numpy表示这是

[ 1.  0.]
[ 0.55051026  1.41421356]

是由

for i in range(2): print linalg.solve(sqrt(a).T, a[i, :].T).T

错误在哪里？矩阵sqrt（a）和a只是示例。您可以将它们替换为任何其他矩阵。我只是想了解bsxfun对rdivide的作用。

Where is the error? The matrices sqrt(a) and a are just examples. You can replace them by any other matrix. I am just trying to understand what bsxfun does with rdivide.

推荐答案

>>> import numpy as np
>>> a = np.array([[0,1],[2,3]])
>>> a
array([[0, 1],
       [2, 3]])
>>> b = np.sqrt(a)
>>> b/a
Warning: invalid value encountered in divide
array([[        nan,  1.        ],
       [ 0.70710678,  0.57735027]])
>>>

因为您需要按元素划分，而不是矩阵乘以逆， numpy.linalg 不是你想要什么。

Since you need an element-wise division, not matrix multiplication by the inverse, numpy.linalg is not what you want.

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