减去矩阵行方向在OpenCV中

Question

是否有任何内置的方法在OpenCV中（C++ API）至减法两个矩阵逐行。

我有以下矩阵：

Mat A(10,2, CV_64F); 
Mat B(1,2, CV_64F); 
Mat C(10,2, CV_64F); 

C = A - B; 

// B is a 1 x 2 matrix, A is a 10 x 2 matrix, and C is a 10 x 2 matrix. 

所述的每一行必须被选自B用C

Answer 1

我解决了它下面的方式减去存储为行：

Mat A(10,2, CV_64F); 
Mat B(1,2, CV_64F); 
Mat C(0,2, CV_64F); 
Mat D(0,2, CV_64F); 

for(int i=0;i<A.rows;i++) 
{ 
C=(A.row(i)-B.row(0)); 
D.push_back(C.row(0)); 
} 

cout<<"A\n"<<A<<endl; 
cout<<"B\n"<<B<<endl; 
cout<<"C\n"<<C<<endl; 
cout<<"D\n"<<D<<endl; 

Answer 2

根据OpenCV文档，可以使用方法subtract（）。在你的情况下，它是

subtract(A.row(0), B.row(0), tempMat); 

然后将tempMat推到C或CopyTo。

下面是引用： OpenCV Array Operation

Answer 3

使用cv::repeat() http://docs.opencv.org/2.4/modules/core/doc/operations_on_arrays.html#repeat

cv::Mat A = cv::Mat::ones(10, 2, CV_64F) * 5.0; 
cv::Mat B = cv::Mat::ones(1, 2, CV_64F) * 3.0; 
cv::Mat BRepeat = cv::repeat(B, A.rows, 1); 
cv::Mat C = A - BRepeat; 

cout << endl; 
cout << "A" << endl << A << endl << endl; 
cout << "B" << endl << B << endl << endl; 
cout << "BRepeat" << endl << BRepeat << endl << endl; 
cout << "C" << endl << C << endl << endl;