I'm working with some filters in OpenCV and don't know how to multiply a number (1/5) in this example
CvMat* kernel=0;
IplImage* dst = cvCreateImage(cvGetSize( entrada ), IPL_DEPTH_8U, 3);
kernel = cvCreateMat(3, 3,CV_32FC1);
cvSet2D( kernel, 0, 0, cvRealScalar(1));
cvSet2D( kernel, 0, 1, cvRealScalar(1));
cvSet2D( kernel, 0, 2, cvRealScalar(1));
cvSet2D( kernel, 1, 0, cvRealScalar(1));
cvSet2D( kernel, 1, 1, cvRealScalar(2));
cvSet2D( kernel, 1, 2, cvRealScalar(1));
cvSet2D( kernel, 2, 0, cvRealScalar(1));
cvSet2D( kernel, 2, 1, cvRealScalar(1));
cvSet2D( kernel, 2, 2, cvRealScalar(1));
// Matriz utilizada para el filtrado paso alto
// 1 1 1
// 1 2 1
// 1 1 1
cvFilter2D(entrada, dst, kernel, cvPoint(-1,-1));
What about cvScale(src, dst, scale) with scale being the number, whatever matrix you want to multiply it with. If you want to multiply it with the kernel, what about just initializing the kernel with the multiplied values?
Related
Suppose I have an image with mask, valid pixels are masked as 1 and others 0, how to perform bilinear interpolation to fill all the invalid pixels?
for example, image:
1, 0, 0, 4
mask:
1, 0, 0, 1
interpolation result should be:
1, 2, 3, 4
The valid pixels are not regularly arranged, a more complicated sample, image:
4, 0, 6, 0,
0, 8, 5, 0
5, 3, 0, 0
mask:
1, 0, 1, 0,
0, 1, 1, 0
1, 1, 0, 0
interpolate with scipy.interpolate.interp2d and the result has many holes and noise
I know that to implement the following
I would use this code:
Mat o_k;
Mat Lapl;
double lambda;
Laplacian(o_k, Lapl, o_k.depth(), 1, 1, 0, BORDER_REFLECT);
Lapl = 1.0 - 2.0*lambda*Lapl;
However, I am trying to implement in OpenCV the following equation:
I know the div, or divergence, term would be like this, right?
int ksize = parser.get<int>("ksize");
int scale = parser.get<int>("scale");
int delta = parser.get<int>("delta");
Sobel(res, sobelx, CV_64F, 1, 0, ksize, scale, delta, BORDER_DEFAULT);
Sobel(res, sobely, CV_64F, 0, 1, ksize, scale, delta, BORDER_DEFAULT);
div = sobelx + sobely;
Where res is the result of the term in parenthesis. But how I get the term in parenthesis?
Or am I doing this wrong? Would div above actually be equal to the gradient of res? If so, then how do I get the divergence?
EDIT:
According to this link, the magnitude can also be computed as mag = abs(x) + abs(y): https://docs.opencv.org/2.4/doc/tutorials/imgproc/imgtrans/sobel_derivatives/sobel_derivatives.html#sobel-derivatives
And since the div of a gradient is the Laplacian, would the below code be equivalent to the 2nd equation?
Sobel(res, sobelx, CV_64F, 1, 0, ksize, scale, delta, BORDER_DEFAULT);
Sobel(res, sobely, CV_64F, 0, 1, ksize, scale, delta, BORDER_DEFAULT);
convertScaleAbs( sobelx, abs_grad_x );
convertScaleAbs( sobely, abs_grad_y );
/// Total Gradient (approximate)
Mat mag;
addWeighted( abs_grad_x, 1, abs_grad_y, 1, 0, mag);
Laplacian(o_k, Lapl, o_k.depth(), 1, 1, 0, BORDER_REFLECT);
Mat top;
top = lambda * Lapl;
Mat result;
divide(top, mag, result);
I have some BGR image:
cv::Mat image;
I want to subtract from all the pixels in the image the vector:
[10, 103, 196]
Meaning that the blue channel for all the pixels will be reduced by 10, the green by 103 and the red by 196.
Is there a standard way to do that, or should I run for loops over all the channels and all the pixels?
suppose we have image that all channels filled with zero and for instance it's dimension is 2x3
cv::Mat image = cv::Mat::zeros(2,3,CV_32SC3)
output will be:
[0, 0, 0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0, 0]
then if we want to add or subtract a singleton variable, then we can use cv::Scalar
1- suppose we want to add 3 in blue channel:
image = image + Scalar(3,0,0); // the result will be same as image=image+3;
with above code our matrix is now:
[3, 0, 0, 3, 0, 0, 3, 0, 0;
3, 0, 0, 3, 0, 0, 3, 0, 0]
2- if you want to add to another channel you can use second or third argument(or forth) of cv::Scalar like below
image = image +Scalar(3,2,-3);
output will be
[3, 2, -3, 3, 2, -3, 3, 2, -3;
3, 2, -3, 3, 2, -3, 3, 2, -3]
Using cv::subtract
cv::Mat image = cv::Mat::zeros(2,3,CV_32SC3);
subtract(image,Scalar(2,3,1),image);
output
[-2, -3, -1, -2, -3, -1, -2, -3, -1;
-2, -3, -1, -2, -3, -1, -2, -3, -1]
For example, I have a 2by3 matrix [1,0,5;1,0,-5], and a Mat trans_mat( 2, 3, CV_32FC1).
How can I assign those values to the trans_mat matrix?
Mat trans_mat( 2, 3, CV_32FC1);
trans_mat = (Mat_<float>(2, 3) << 1, 0, 5, 1, 0, -5);
Given a 3 x 3 rotation matrix,R, and a 3 x 1 translation matrix,T, I am wondering how to multiply the T and R matrices to an image?
Lets say the Iplimage img is 640 x 480.
What I want to do is R*(T*img).
I was thinking of using cvGemm, but that didn't work.
The function you are searching for is probably warpPerspective() : this is a use case...
// Projection 2D -> 3D matrix
Mat A1 = (Mat_<double>(4,3) <<
1, 0, -w/2,
0, 1, -h/2,
0, 0, 0,
0, 0, 1);
// Rotation matrices around the X axis
Mat R = (Mat_<double>(4, 4) <<
1, 0, 0, 0,
0, cos(alpha), -sin(alpha), 0,
0, sin(alpha), cos(alpha), 0,
0, 0, 0, 1);
// Translation matrix on the Z axis
Mat T = (Mat_<double>(4, 4) <<
1, 0, 0, 0,
0, 1, 0, 0,
0, 0, 1, dist,
0, 0, 0, 1);
// Camera Intrisecs matrix 3D -> 2D
Mat A2 = (Mat_<double>(3,4) <<
f, 0, w/2, 0,
0, f, h/2, 0,
0, 0, 1, 0);
Mat transfo = A2 * (T * (R * A1));
Mat source;
Mat destination;
warpPerspective(source, destination, transfo, source.size(), INTER_CUBIC | WARP_INVERSE_MAP);
I hope it could help you,
Julien
PS : I gave the example with a projection from 2D to 3D but you can use directly transfo = T* R;