用2个方阵模拟matlab的mrdivide [英] Simulating matlab's mrdivide with 2 square matrices

问题描述

我有2个19x19方阵（a& b），我想使用斜线（mrdivide）运算符执行除法，使

  c = a / b 
  

我试图在OpenCV中实现。我发现有几个人建议使用 cv :: solve ，但到目前为止我一直无法找到任何给我一个结果任何接近matlab。

有没有人知道我如何用opencv实现mrdivide？

我试过下面的代码： p>

  cv :: Mat mldivide（const cv :: Mat& A，const cv :: Mat& B）
 {
 // return b * A.inv（）; 
 cv :: Mat a; 
 cv :: Mat b; 
 A.convertTo（a，CV_64FC1）; 
 B.convertTo（b，CV_64FC1）; 
 
 cv :: Mat ret; 
 cv :: solve（a，b，ret，cv :: DECOMP_NORMAL）; 
 
 cv :: Mat ret2; 
 ret.convertTo（ret2，A.type（））; 
 return ret2; 
} 
  

然后我执行mrdivide如下：

  cv :: Mat mrdivide（const cv :: Mat& A，const cv :: Mat& B）
 {
 return mldivide （At（），Bt（））.t（）; 
} 
  

（ 编辑 ：

这给我一个错误的答案，即没有什么像matlab。根据我的意见，我也尝试了

  cv :: Mat mrdivide（const cv :: Mat& A，const cv :: Mat& ; B）
 {
 return A * B.inv（）; 
} 
  

这给出了与上面相同的答案，但也是错误的。

解决方案

在MATLAB中，使用 a / b 是等价的，因此可以使用rel =nofollow> mrdivide 到 a * b ^ { - 1} 其中 b ^ { - 1} c> b 。因此，你可以做的是先反转矩阵 b ，然后预乘这个 a 。

一种方法是使用 > cv :: invert ，然后用 a 。这可以通过以下函数定义来完成（从你上面的代码中借用）：

  cv :: Mat mrdivide const cv :: Mat& A，const cv :: Mat& B）
 {
 cv :: Mat bInvert; 
 cv :: invert（B，bInvert）; 
 return A * bInvert; 
} 
  

另一种方法是使用 inv（） 方法 cv :: Mat 接口，只需使用并乘以矩阵本身：

  cv :: Mat mrdivide（const cv :: Mat& A，const cv :: Mat& B）
 {
 return A * B.inv（）; 
} 
  



我不知道哪一个更快，所以你可能要做一些测试，但两种方法都应该工作。然而，为了提供对可能的时序的一点洞察，有三种方式来反转OpenCV中的矩阵。您只需重写第三个参数 cv :: invert 或指定 cv :: Mat.inv（）以方便这一点。

这个StackOverflow文章使用以下三种方法来处理相对较大矩阵大小的矩阵的定时：与矩阵相反的最快方法

I have 2 19x19 square matrices (a & b) and I am trying to use the slash (mrdivide) operator to perform a division such that

c = a / b

I am trying to implement this in OpenCV. I've found a few people suggesting using cv::solve but so far I've been unable to find anything that gives me a result anything close to matlab.

Does anyone have any idea how I could implement mrdivide with opencv?

I've tried the following code:

cv::Mat mldivide(const cv::Mat& A, const cv::Mat& B ) 
{
    //return  b * A.inv();
    cv::Mat a;
    cv::Mat b;
    A.convertTo( a, CV_64FC1 );
    B.convertTo( b, CV_64FC1 );

    cv::Mat ret;
    cv::solve( a, b, ret, cv::DECOMP_NORMAL );

    cv::Mat ret2;
    ret.convertTo( ret2, A.type() );
    return ret2;
}

I've then implemented mrdivide as follows:

cv::Mat mrdivide(const cv::Mat& A, const cv::Mat& B ) 
{
   return mldivide( A.t(), B.t() ).t();
}

(Edit: As per the answers this does actually give me the correct answer when I used it correctly!)

This gives me a wrong answer, ie nothing like matlab. As per the comments I also tried

cv::Mat mrdivide(const cv::Mat& A, const cv::Mat& B ) 
{
    return A * B.inv();
}

This give the same answer as above but is also wrong.

解决方案

In MATLAB, using mrdivide on two matrices of compatible dimension such that a / b is equivalent to a * b^{-1} where b^{-1} is the inverse of b. As such, what you could do is perhaps invert the matrix b first, then pre-multiply this a.

One method is to use cv::invert on the matrix b then pre-multiplying with a. This could be done with the following function definition (borrowing from your code in your post above):

cv::Mat mrdivide(const cv::Mat& A, const cv::Mat& B) 
{
    cv::Mat bInvert;
    cv::invert(B, bInvert);
    return A * bInvert;
}

Another way is to use the inv() method that's builtin to the cv::Mat interface and just use that and multiply the matrices themselves:

cv::Mat mrdivide(const cv::Mat& A, const cv::Mat& B) 
{
    return A * B.inv();
}

I'm not sure which one is faster, so you may have to do some tests, but either of the two methods should work. However, to provide a bit of insight in terms of possible timings, there are three ways to invert a matrix in OpenCV. You simply override the third parameter to cv::invert or specify the method in cv::Mat.inv() to facilitate this.

This StackOverflow post goes through the timings of inverting a matrix for a relatively large matrix size using the three methods: Fastest method in inverse of matrix

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