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can anyone tell me how to generate a 2d gaussian filter kernel using the gaussian filter equation? how does the x and y value vary?
ref: http://en.wikipedia.org/wiki/Gaussian_function
To generate the kernel is quite simple. If your problem is in applying the kernel, you need to update the question.
The kernel is simply a square matrix of values, generally an odd number size so that there's a clearly defined center. To fill it, the x and y values go from -(n-1)/2 to (n-1)/2 where n is the size of the matrix.
double half_n = (n - 1) / 2.0;
for (i = 0; i < n; ++i)
{
double x = i - half_n;
for (j = 0; j < n; ++j)
{
double y = j - half_n;
kernel[i][j] = // use formula with x and y here
}
}
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I have have two distance matrices d_X: n x n and d_Y: m x m.
set.seed(1)
n <- 2
m <- 3
d_X <- as.matrix(dist(runif(n)))
d_Y <- as.matrix(dist(runif(m)))
From matrices d_X and d_Y matrix G: nm x nm is formed:
G <- matrix(nrow = n*m,ncol = n*m)
for(i in 1:n) {
for (j in 1:m) {
for(ii in 1:n) {
for(jj in 1:m) {
G[(i-1)*m+j,(ii-1)*m+jj] = abs(d_X[i, ii] - d_Y[j, jj])
}
}
}
}
There is also matrix U: nm*1:
U <- runif(m*n)
My goal is to calculate G%*%U. Now, when n and m are 200, we need 6GB to allocate G. Since G is symmetric we could save half the space needed by restoring it properly.
In practice n and m sizes are up to 5000 which makes allocating G impossible. Since I only need the value of G%*%U, it would be sufficient to calculate it piece by piece. I'm struggling to find an effective way to do it.
*Time also matters
Since I have to run these calculations thousands of times, it is also important, that computing G%*%U takes reasonable time. I have used following function to speed up computing G in cases where n and m are less than a hundred:
Rcpp::cppFunction('NumericMatrix G_mat(NumericMatrix d_X, NumericMatrix d_Y) {
NumericMatrix G(d_X.nrow()*d_Y.nrow(),d_X.nrow()*d_Y.nrow());
for (int i = 0; i <d_X.nrow(); i++) {
for (int j = 0; j < d_Y.nrow(); j++) {
for (int ii = 0; ii < d_X.nrow(); ii++) {
for (int jj = 0; jj < d_Y.nrow(); jj++) {
G(i*d_Y.nrow()+j,ii*d_Y.nrow()+jj) = fabs(d_X(i, ii) - d_Y(j, jj));
};
};
};
};
return(G);
}
')
So I guess this workaround should be also implemented in C++ to get best results (speed wise)? How to do it?
Maybe this
A <- numeric(m*n)
for(i in 1:n) {
for (j in 1:n) {
A[((i-1)*m+1):(i*m)]= A[((i-1)*m+1):(i*m)] + abs(d_Y-d_X[i,j])%*%U[((j-1)*m+1):(j*m)]
}
}
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I have a 2D point A inside the [0,1]² square.
The square is divided into 9 subsquares (of equal dimensions)
http://www.noelshack.com/2015-23-1433689273-capture.png
I want to know which subsquare the point A belongs to.
I can do a if elseif else on the first coordinate, then inside each branch, another if else if else on the second coordinate.
There is a lot of code repeating (the check on the second coordinate)
Is there a better way ?
The trouble is, it is not clear what your 2D point is, what you want the sub-square value as, etc. So a definitive answer is difficult. But anyway, I will make some assumptions and see if I am right about what you are asking...
Assuming you had a point A with a coordinate such as:
float point[2] = {0.1224, 0.4553}
Then to work out where it is inside the square you can do some simple maths:
float x = point[0] * 3;
float y = point[1] * 3; //Multiply both by 3
int xIdx = floor(x);
int yIdx = floor(y); //Floor the result - this gives a number 0 to 2
int cell = yIdx * 3 + xIdx + 1; // Calculate the cell index (based on your diagram)
Now you could generalise this for any point - for example the point might be (1.23343, 2.6768) - by simply removing the integer part from the point - leaving a number between 0 and 1. The integer part would be the super cell, and the fractional part would be converted into the sub cell as above.
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Can anyone tell me how to find the centroid of an object in point cloud?.
I haven't tried any code yet because I have no slight idea as to how to go about it.
If you have the point locations, you should be able to just average the x and y positions.
int x = 0; y = 0;
for ( i = 0; i < num_pts; i++ )
{
x += pt[i].x;
y += pt[i].y;
}
centroid.x = x / num_pts;
centroid.y = y / num_pts;
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How can you define a function to calculate the value of a definite integral in C++? For example to solve the integral of the function x^2 * cos(x)?
Interestingly enough, I ran across this article a little while ago explaining one method for calculating numerical integrals using function pointers.
https://helloacm.com/c-function-to-compute-numerical-integral-using-function-pointers/
For something like x^2 * cos(x):
You would need an overloaded integral function:
double integral(double(*f)(double x), double(*g)(double x, double y), double a, double b, int n)
{
double step = (b - a)/n; // width of rectangle
double area = 0.0;
double y = 0; // height of rectangle
for(int i = 0; i < n; ++i)
{
y = f(a + (i + 0.5) * step) * g(a + (i + 0.5) * step, y);
area += y * step // find the area of the rectangle and add it to the previous area. Effectively summing up the area under the curve.
}
return area;
}
To call:
int main()
{
int x = 3;
int low_end = 0;
int high_end = 2 * M_PI;
int steps = 100;
cout << integral(std::powf, std::cosf, low_end, high_end, steps);
return 0;
}
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I'm having the following code and the number of my channels are 3
IplImage* img_crop_mat = cvLoadImage("....", 1);
...
int b = 0;
uchar* rgb = (uchar*) img_crop_mat->imageData;
I would like to have R, G and B matrices in a loop, skimming the entire image:
for (int y = b; y < height - b; y++)
{
???
for (int x = b; x < width - b; x++)
{
????
}
}
The previous forums regarding my question deal with CvMat but not with pointers as my code.
What are the indexes that I must take into account?
You can use the following macro to access an arbitrary pixel of a 3-channel, 8U-image:
CV_IMAGE_ELEM(myImage, unsigned char, y, x*3 + ChannelOfInterest)
This is an lvalue so you can take and use its value, or you can change the pixel's value.
By default,
ChannelOfInterest = 0, blue
ChannelOfInterest = 1, green
ChannelOfInterest = 2, red
The actual data structure is pretty straightforward, look up the definition of CV_IMAGE_ELEM.