I am wondering how to actually sample the data I am passing to the shader file. I am using two methods, is it the same for both? Are there any resources online for me to actually read up on this sort of thing?
Compiling at 5.0 but the version number does not matter so much.
I have two methods to pass the data.
The first;
for( UINT row = 0; row < textureDesc.Height; row++ )
{
UINT rowStart = row * mappedResource.RowPitch;
for( UINT col = 0; col < textureDesc.Width; col++ )
{
//width * number of channels (r,g,b,a)
UINT colStart = col * 4;
pTexels[rowStart + colStart + 0] = 10.0f; // Red
pTexels[rowStart + colStart + 1] = 10.0f; // Green
pTexels[rowStart + colStart + 2] = 255.0f; // Blue
pTexels[rowStart + colStart + 3] = 255.0f; // Alpha
}
}
The second;
float elements[416][416];
int elementsCount = 416*416;
for( int i = 0; i < 416; i++ )
{
for( int k = 0; k < 416; k++ )
{
elements[i][k] = 0;
}
}
memcpy(mappedResource.pData, elements, sizeof(float) * elementsCount);
Seems that I missed an important part of all of this.
When creating a texture, in the texture description, the Format is the type that will be returned when the object is sampled. Many thanks to Drop for the help.
Related
I tried to code histogram equalization operation in order to enhance contrast of the images, but my code didn't work. When I displayed image's original histogram and histogram after processed by my code I saw that output histogram only have a value at 0 and there aren't values for other pixel intensities. I don't know why. Here is my code:
private: System::Void histogramEqualizationToolStripMenuItem_Click(System::Object^ sender, System::EventArgs^ e) {
Raw_Intensity = ConvertBMPToIntensity(Buffer, Width, Height);
int histogram[256] = { 0 };
int equalizedHistogram[256] = { 0 };
int runningSum = 0;
int numberOfPixels = Width * Height;
for (int row = 0; row < Height; row++)
{
for (int column = 0; column < Width; column++)
{
histogram[Raw_Intensity[row * Width + column]]++;
}
}
for (int i = 0; i < 256; i++)
{
runningSum += histogram[i];
int index = round(((runningSum / numberOfPixels) * 255));
equalizedHistogram[index] += histogram[i];
}
}
I think the casting issue is occuring.
change this line :
int index = round(((runningSum / numberOfPixels) * 255));
to
int index = round(((runningSum*1.0 / numberOfPixels) * 255));
How to efficiency linearized Mat (symmetric matrix) to one row by right triangle.
For example, when I have:
0aabbb
b0aaaa
ba0bba
bac0aa
aaaa0c
abcab0
and then from that I get:
aabbbaaaabbaaac
Something like this:
...
template<class T>
Mat SSMJ::triangleLinearized(Mat mat){
int c = mat.cols;
Mat row = Mat(1, ((c*c)-c)/2, mat.type());
int i = 0;
for(int y = 1; y < mat.rows; y++)
for(int x = y; x < mat.cols; x++) {
row.at<T>(i)=mat.at<T>(y, x);
i++;
}
return row;
}
...
Since data in your mat is just a 1d array stored in row.data you can do whatever you want with it. I don't think you will find anything more special (w/o using vectorized methods) than just copying from this array.
int rows = 6;
char data[] = { 0,1,2,3,4,5,
0,1,2,3,4,5,
0,1,2,3,4,5,
0,1,2,3,4,5,
0,1,2,3,4,5};
char result[100];
int offset = 0;
for (int i = 0; i < 5; offset += 5-i, i++) {
memcpy(&result[offset] , &data[rows * i + i + 1], 5 - i);
}
Or with opencv Mat it would be
int rows = mat.cols;
char result[100]; // you can calculate how much data u need
int offset = 0;
for (int i = 0; i < 5; offset += 5-i, i++) {
memcpy(&result[offset] , &mat.data[rows * i + i + 1], 5 - i);
}
Mat resultMat(1, offset, result);
I want process image so each pixel value will be mean of its value and 4 neighbours.
Created two different functions:
Mat meanImage(cv::Mat& inputImage)
{
Mat output;
Mat kernel(3,3,CV_32F,0.0);
kernel.at<float>(0,1) = 0.2;
kernel.at<float>(1,0) = 0.2;
kernel.at<float>(1,1) = 0.2;
kernel.at<float>(1,2) = 0.2;
kernel.at<float>(2,1) = 0.2;
filter2D(inputImage,output,-1,kernel);
return output;
}
and:
Mat meanImage2(Mat& inputImage)
{
Mat temp;
Mat output(inputImage.rows,inputImage.cols,inputImage.type());
copyMakeBorder(inputImage,temp,1,1,1,1,BORDER_REPLICATE);
CV_Assert(output.isContinuous());
CV_Assert(temp.isContinuous());
const int len = output.rows * output.cols * output.channels();
const int rowLenTemp = temp.cols * temp.channels();
const int twoRowLenTemp = 2 * rowLenTemp;
const int rowLen = output.cols * output.channels();
uchar* outPtr = output.ptr<uchar>(0);
uchar* tempPtr = temp.ptr<uchar>(0);
for(int i = 0; i < len; ++i)
{
const int a = 6 * (i / rowLen) + 3;
outPtr[i] = (tempPtr[i+rowLenTemp+a] + tempPtr[i+a] +
tempPtr[i+rowLenTemp+a+3] + tempPtr[i+rowLenTemp+a-3] +
tempPtr[i+twoRowLenTemp+a]) / 5;
}
return output;
}
I've assumed that the result will be the same. So I've compared images:
Mat diff;
compare(meanImg1,meanImg2,diff,CMP_NE);
printf("Difference: %d\n",countNonZero(diff));
imshow("diff",diff);
And get a lot off differences. What is the difference between this functions?
Edit:
Difference for lena image taken from Lena
Beware that when you do the sum of pixels, you add unsigned chars and you may overflow.
Test your code by casting these pixels values to int.
outPtr[i] = ((int)tempPtr[i+rowLenTemp+a] + (int)tempPtr[i+a] +
(int)tempPtr[i+rowLenTemp+a+3] + (int)tempPtr[i+rowLenTemp+a-3] +
(int)tempPtr[i+twoRowLenTemp+a]) / 5;
Edit: I'd rather code this like (assuming image type is uchar and it has 3 channels)
for (int r = 0; r < output.rows; r++)
{
uchar* previousRow = temp.ptr<uchar>(r) + 3;
uchar* currentRow = temp.ptr<uchar>(r+1) + 3;
uchar* nextRow = temp.ptr<uchar>(r+2) + 3;
uchar* outRow = output.ptr<uchar>(r);
for (int c = 0; c < 3*output.cols; c++)
{
int value = (int)previousRow[c] +
(int)currentRow[c-3] + (int)currentRow [c] + (int)currentRow[c+3] +
(int)nextRow [c];
outRow[c] = value / 5;
}
}
I'm writing a gaussian filter, and my goal is to match the gaussian blur filter in photoshop as closely as possible. This is my first image processing endeavor. Some problems/questions I have are...
Further blurring an image with my filter darkens it, while photoshop’s seems to lighten it.
The deviation value (“sigma,” in my code) I’m using is r/3, which results in the gaussian curve having approached about 0.0001 within the matrix...is there a better way to determine this value?
How does photoshop (or most people) handle image borders for this type of blur?
int matrixDimension = (radius*2)+1;
float sigma = radius/3;
float twoSigmaSquared = 2*pow(sigma, 2);
float oneOverSquareRootOfTwoPiSigmaSquared = 1/(sqrt(M_PI*twoSigmaSquared));
float kernel[matrixDimension];
int index = 0;
for (int offset = -radius; offset <= radius; offset++) {
float xSquared = pow(offset, 2);
float exponent = -(xSquared/twoSigmaSquared);
float eToThePower = pow(M_E, exponent);
float multFactor = oneOverSquareRootOfTwoPiSigmaSquared*eToThePower;
kernel[index] = multFactor;
index++;
}
//Normalize the kernel such that all its values will add to 1
float sum = 0;
for (int i = 0; i < matrixDimension; i++) {
sum += kernel[i];
}
for (int i = 0; i < matrixDimension; i++) {
kernel[i] = kernel[i]/sum;
}
//Blur horizontally
for (int row = 0; row < imageHeight; row++) {
for (int column = 0; column < imageWidth; column++) {
int currentPixel = (row*imageWidth)+column;
int sum1 = 0;
int sum2 = 0;
int sum3 = 0;
int sum4 = 0;
int index = 0;
for (int offset = -radius; offset <= radius; offset++) {
if (!(column+offset < 0) && !(column+offset > imageWidth-1)) {
int firstByteOfPixelWereLookingAtInSrcData = (currentPixel+offset)*4;
int in1 = srcData[firstByteOfPixelWereLookingAtInSrcData];
int in2 = srcData[firstByteOfPixelWereLookingAtInSrcData+1];
int in3 = srcData[firstByteOfPixelWereLookingAtInSrcData+2];
int in4 = srcData[firstByteOfPixelWereLookingAtInSrcData+3];
sum1 += (int)(in1 * kernel[index]);
sum2 += (int)(in2 * kernel[index]);
sum3 += (int)(in3 * kernel[index]);
sum4 += (int)(in4 * kernel[index]);
}
index++;
}
int currentPixelInData = currentPixel*4;
destData[currentPixelInData] = sum1;
destData[currentPixelInData+1] = sum2;
destData[currentPixelInData+2] = sum3;
destData[currentPixelInData+3] = sum4;
}
}
//Blur vertically
for (int row = 0; row < imageHeight; row++) {
for (int column = 0; column < imageWidth; column++) {
int currentPixel = (row*imageWidth)+column;
int sum1 = 0;
int sum2 = 0;
int sum3 = 0;
int sum4 = 0;
int index = 0;
for (int offset = -radius; offset <= radius; offset++) {
if (!(row+offset < 0) && !(row+offset > imageHeight-1)) {
int firstByteOfPixelWereLookingAtInSrcData = (currentPixel+(offset*imageWidth))*4;
int in1 = destData[firstByteOfPixelWereLookingAtInSrcData];
int in2 = destData[firstByteOfPixelWereLookingAtInSrcData+1];
int in3 = destData[firstByteOfPixelWereLookingAtInSrcData+2];
int in4 = destData[firstByteOfPixelWereLookingAtInSrcData+3];
sum1 += (int)(in1 * kernel[index]);
sum2 += (int)(in2 * kernel[index]);
sum3 += (int)(in3 * kernel[index]);
sum4 += (int)(in4 * kernel[index]);
}
index++;
}
int currentPixelInData = currentPixel*4;
finalData[currentPixelInData] = sum1;
finalData[currentPixelInData+1] = sum2;
finalData[currentPixelInData+2] = sum3;
finalData[currentPixelInData+3] = sum4;
}
}
To reverse engineer a filter, you need to find its impulse response. On a background of a very dark value, say 32, place a nearly white pixel, say 223. You don't want to use 0 and 255 because some filters will try to create values beyond the starting values. Run the filter on this image, and take the output values and stretch them from 0.0 to 1.0: (value-32)/(223-32). Now you have the exact weights needed to emulate the filter.
There are lots of ways to treat the image edges. I would suggest taking the filter weights and summing them, then dividing the result by that sum; if you're trying to go beyond the edge, use 0.0 for both the pixel value and the filter weight on that pixel.
Boundary conditions sometimes depend on exactly what you're doing and what kind of data you're working with, but I think for general purpose image manipulation the best thing to do is to extend the values at the borders beyond the edges of the image. Not literally of course, but if the filter tries to read a pixel that's outside the image borders, you substitute the value of the nearest pixel on the edge of the image. Which is really the same as just clamping the row to be between 0 and height, and the column to be between 0 and width.
I'm using the Hough transform in OpenCV to detect lines. However, I know in advance that I only need lines within a very limited range of angles (about 10 degrees or so). I'm doing this in a very performance sensitive setting, so I'd like to avoid the extra work spent detecting lines at other angles, lines I know in advance I don't care about.
I could extract the Hough source from OpenCV and just hack it to take min_rho and max_rho parameters, but I'd like a less fragile approach (have to manually update my code w/ each OpenCV update, etc.).
What's the best approach here?
Well, i've modified the icvHoughlines function to go for a certain range of angles. I'm sure there's cleaner ways that plays with memory allocation as well, but I got a speed gain going from 100ms to 33ms for a range of angle going from 180deg to 60deg, so i'm happy with that.
Note that this code also outputs the accumulator value. Also, I only output 1 line because that fit my purposes but there was no gain really there.
static void
icvHoughLinesStandard2( const CvMat* img, float rho, float theta,
int threshold, CvSeq *lines, int linesMax )
{
cv::AutoBuffer<int> _accum, _sort_buf;
cv::AutoBuffer<float> _tabSin, _tabCos;
const uchar* image;
int step, width, height;
int numangle, numrho;
int total = 0;
float ang;
int r, n;
int i, j;
float irho = 1 / rho;
double scale;
CV_Assert( CV_IS_MAT(img) && CV_MAT_TYPE(img->type) == CV_8UC1 );
image = img->data.ptr;
step = img->step;
width = img->cols;
height = img->rows;
numangle = cvRound(CV_PI / theta);
numrho = cvRound(((width + height) * 2 + 1) / rho);
_accum.allocate((numangle+2) * (numrho+2));
_sort_buf.allocate(numangle * numrho);
_tabSin.allocate(numangle);
_tabCos.allocate(numangle);
int *accum = _accum, *sort_buf = _sort_buf;
float *tabSin = _tabSin, *tabCos = _tabCos;
memset( accum, 0, sizeof(accum[0]) * (numangle+2) * (numrho+2) );
// find n and ang limits (in our case we want 60 to 120
float limit_min = 60.0/180.0*PI;
float limit_max = 120.0/180.0*PI;
//num_steps = (limit_max - limit_min)/theta;
int start_n = floor(limit_min/theta);
int stop_n = floor(limit_max/theta);
for( ang = limit_min, n = start_n; n < stop_n; ang += theta, n++ )
{
tabSin[n] = (float)(sin(ang) * irho);
tabCos[n] = (float)(cos(ang) * irho);
}
// stage 1. fill accumulator
for( i = 0; i < height; i++ )
for( j = 0; j < width; j++ )
{
if( image[i * step + j] != 0 )
//
for( n = start_n; n < stop_n; n++ )
{
r = cvRound( j * tabCos[n] + i * tabSin[n] );
r += (numrho - 1) / 2;
accum[(n+1) * (numrho+2) + r+1]++;
}
}
int max_accum = 0;
int max_ind = 0;
for( r = 0; r < numrho; r++ )
{
for( n = start_n; n < stop_n; n++ )
{
int base = (n+1) * (numrho+2) + r+1;
if (accum[base] > max_accum)
{
max_accum = accum[base];
max_ind = base;
}
}
}
CvLinePolar2 line;
scale = 1./(numrho+2);
int idx = max_ind;
n = cvFloor(idx*scale) - 1;
r = idx - (n+1)*(numrho+2) - 1;
line.rho = (r - (numrho - 1)*0.5f) * rho;
line.angle = n * theta;
line.votes = accum[idx];
cvSeqPush( lines, &line );
}
If you use the Probabilistic Hough transform then the output is in the form of a cvPoint each for lines[0] and lines[1] parameters. We can get x and y co-ordinated for each of the two points by pt1.x, pt1.y and pt2.x and pt2.y.
Then use the simple formula for finding slope of a line - (y2-y1)/(x2-x1). Taking arctan (tan inverse) of that will yield that angle in radians. Then simply filter out desired angles from the values for each hough line obtained.
I think it's more natural to use standart HoughLines(...) function, which gives collection of lines directly in rho and theta terms and select nessessary angle range from it, rather than recalculate angle from segment end points.