I want to pixelate an image stored in a 1d array, although i am not sure how to do it, this is what i have comeup with so far...
the value of pixelation is currently 3 for testing purposes.
currently it just creates a section of randomly coloured pixels along the left third of the image, if i increase the value of pixelation the amount of random coloured pixels decreases and vice versa, so what am i doing wrong?
I have also already implemented the rotation, reading of the image and saving of a new image this is just a separate function which i need assistance with.
picture pixelate( const std::string& file_name, picture& tempImage, int& pixelation /* TODO: OTHER PARAMETERS HERE */)
{
picture pixelated = tempImage;
RGB tempPixel;
tempPixel.r = 0;
tempPixel.g = 0;
tempPixel.b = 0;
int counter = 0;
int numtimesrun = 0;
for (int x = 1; x<tempImage.width; x+=pixelation)
{
for (int y = 1; y<tempImage.height; y+=pixelation)
{
//RGB tempcol;
//tempcol for pixelate
for (int i = 1; i<pixelation; i++)
{
for (int j = 1; j<pixelation; j++)
{
tempPixel.r +=tempImage.pixel[counter+pixelation*numtimesrun].colour.r;
tempPixel.g +=tempImage.pixel[counter+pixelation*numtimesrun].colour.g;
tempPixel.b +=tempImage.pixel[counter+pixelation*numtimesrun].colour.b;
counter++;
//read colour
}
}
for (int k = 1; k<pixelation; k++)
{
for (int l = 1; l<pixelation; l++)
{
pixelated.pixel[numtimesrun].colour.r = tempPixel.r/pixelation;
pixelated.pixel[numtimesrun].colour.g = tempPixel.g/pixelation;
pixelated.pixel[numtimesrun].colour.b = tempPixel.b/pixelation;
//set colour
}
}
counter = 0;
numtimesrun++;
}
cout << x << endl;
}
cout << "Image successfully pixelated." << endl;
return pixelated;
}
I'm not too sure what you really want to do with your code, but I can see a few problems.
For one, you use for() loops with variables starting at 1. That's certainly wrong. Arrays in C/C++ start at 0.
The other main problem I can see is the pixelation parameter. You use it to increase x and y without knowing (at least in that function) whether it is a multiple of width and height. If not, you will definitively be missing pixels on the right edge and at the bottom (which edges will depend on the orientation, of course). Again, it very much depends on what you're trying to achieve.
Also the i and j loops start at the position defined by counter and numtimesrun which means that the last line you want to hit is not tempImage.width or tempImage.height. With that you are rather likely to have many overflows. Actually that would also explain the problems you see on the edges. (see update below)
Another potential problem, cannot tell for sure without seeing the structure declaration, but this sum using tempPixel.c += <value> may overflow. If the RGB components are defined as unsigned char (rather common) then you will definitively get overflows. So your average sum is broken if that's the fact. If that structure uses floats, then you're good.
Note also that your average is wrong. You are adding source data for pixelation x pixalation and your average is calculated as sum / pixelation. So you get a total which is pixalation times larger. You probably wanted sum / (pixelation * pixelation).
Your first loop with i and j computes a sum. The math is most certainly wrong. The counter + pixelation * numtimesrun expression will start reading at the second line, it seems. However, you are reading i * j values. That being said, it may be what you are trying to do (i.e. a moving average) in which case it could be optimized but I'll leave that out for now.
Update
If I understand what you are doing, a representation would be something like a filter. There is a picture of a 3x3:
.+. *
+*+ =>
.+.
What is on the left is what you are reading. This means the source needs to be at least 3x3. What I show on the right is the result. As we can see, the result needs to be 1x1. From what I see in your code you do not take that in account at all. (the varied characters represent varied weights, in your case all weights are 1.0).
You have two ways to handle that problem:
The resulting image has a size of width - pixelation * 2 + 1 by height - pixelation * 2 + 1; in this case you keep one result and do not care about the edges...
You rewrite the code to handle edges. This means you use less source data to compute the resulting edges. Another way is to compute the edge cases and save that in several output pixels (i.e. duplicate the pixels on the edges).
Update 2
Hmmm... looking at your code again, it seems that you compute the average of the 3x3 and save it in the 3x3:
.+. ***
+*+ => ***
.+. ***
Then the problem is different. The numtimesrun is wrong. In your k and l loops you save the pixels pixelation * pixelation in the SAME pixel and that advanced by one each time... so you are doing what I shown in my first update, but it looks like you were trying to do what is shown in my 2nd update.
The numtimesrun could be increased by pixelation each time:
numtimesrun += pixelation;
However, that's not enough to fix your k and l loops. There you probably need to calculate the correct destination. Maybe something like this (also requires a reset of the counter before the loop):
counter = 0;
... for loops ...
pixelated.pixel[counter+pixelation*numtimesrun].colour.r = ...;
... (take care of g and b)
++counter;
Yet again, I cannot tell for sure what you are trying to do, so I do not know why you'd want to copy the same pixel pixelation x pixelation times. But that explains why you get data only at the left (or top) of the image (very much depends on the orientation, one side for sure. And if that's 1/3rd then pixelation is probably 3.)
WARNING: if you implement the save properly, you'll experience crashes if you do not take care of the overflows mentioned earlier.
Update 3
As explained by Mark in the comment below, you have an array representing a 2d image. In that case, your counter variable is completely wrong since this is 100% linear whereas the 2d image is not. The 2nd line is width further away. At this point, you read the first 3 pixels at the top-left, then the next 3 pixels on the same, and finally the next 3 pixels still on the same line. Of course, it could be that your image is thus defined and these pixels are really one after another, although it is not very likely...
Mark's answer is concise and gives you the information necessary to access the correct pixels. However, you will still be hit by the overflow and possibly the fact that the width and height parameters are not a multiple of pixelation...
I don't do a lot of C++, but here's a pixelate function I wrote for Processing. It takes an argument of the width/height of the pixels you want to create.
void pixelateImage(int pxSize) {
// use ratio of height/width...
float ratio;
if (width < height) {
ratio = height/width;
}
else {
ratio = width/height;
}
// ... to set pixel height
int pxH = int(pxSize * ratio);
noStroke();
for (int x=0; x<width; x+=pxSize) {
for (int y=0; y<height; y+=pxH) {
fill(p.get(x, y));
rect(x, y, pxSize, pxH);
}
}
}
Without the built-in rect() function you'd have to write pixel-by-pixel using another two for loops:
for (int px=0; px<pxSize; px++) {
for (int py=0; py<pxH; py++) {
pixelated.pixel[py * tempImage.width + px].colour.r = tempPixel.r;
pixelated.pixel[py * tempImage.width + px].colour.g = tempPixel.g;
pixelated.pixel[py * tempImage.width + px].colour.b = tempPixel.b;
}
}
Generally when accessing an image stored in a 1D buffer, each row of the image will be stored as consecutive pixels and the next row will follow immediately after. The way to address into such a buffer is:
image[y*width+x]
For your purposes you want both inner loops to generate coordinates that go from the top and left of the pixelation square to the bottom right.
I using the code of bytefish in order to calculate Local Binary Patterns (LBP) spatial uniform histograms for an image. I am using the the spatial_histogram function which calculates the histogram of local patches of image. Every calculated patch has size 256 so the final Mat hist file size is 1x(n*256). What I am trying to understand is how can I convert that histogram implementation to uniform histogram implementation. Implemented histogram code is the following:
void lbp::histogram_(const Mat& src, Mat& hist, int numPatterns) {
hist = Mat::zeros(1, numPatterns, CV_32SC1);
for(int i = 0; i < src.rows; i++) {
for(int j = 0; j < src.cols; j++) {
int bin = src.at<_Tp>(i,j);
hist.at<int>(0,bin) += 1;
}
}
Uniform process is based on the following paper ( for local binary patterns) here.
A local binary pattern is called uniform if the binary pattern contains at most two bitwise transitions from 0 to 1 or vice versa when the bit pattern is considered circular.
[edit2] color reduction
That is easy it is just recoloring by the table uniform[256] has nothing to do with uniform histograms !!!
create translation(recoloring) table for each possible color
for 8-bit gray-scale it is 256 colors for example:
BYTE table[256] = {
0,1,2,3,4,58,5,6,7,58,58,58,8,58,9,10,11,58,58,58,58,58,58,58,12,58,58,58,13,58,
14,15,16,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,17,58,58,58,58,58,58,58,18,
58,58,58,19,58,20,21,22,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,
58,58,58,58,58,58,58,58,58,58,58,58,23,58,58,58,58,58,58,58,58,58,58,58,58,58,
58,58,24,58,58,58,58,58,58,58,25,58,58,58,26,58,27,28,29,30,58,31,58,58,58,32,58,
58,58,58,58,58,58,33,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,34,58,58,58,58,
58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,
58,35,36,37,58,38,58,58,58,39,58,58,58,58,58,58,58,40,58,58,58,58,58,58,58,58,58,
58,58,58,58,58,58,41,42,43,58,44,58,58,58,45,58,58,58,58,58,58,58,46,47,48,58,49,
58,58,58,50,51,52,58,53,54,55,56,57 };
You can also compute it programaticaly like table[i]=(58*i)/255; for linear distribution but I suggest it is more like this recolor based on histogram example:
//hist[256] - already computed classic histogram
//table[59] - wanted recolor table
void compute_table(int *table,int *hist)
{
int i,c,threshold=1;
for (c=-1,i=0;i<256;i++)
if (hist[i]>threshold) { c++; table[i]=c; }
else table[i]=58;
}
set the threshold by area size or color count or whatever ...
recolor color
color_59=table[color_256]; either recolor source image or just change color value before used in histogram computation
That is all.
[edit1] LBP
I do not think is a good idea to compute histogram for LBP at all
I would compute min and max color for sub image region you work with
then convert colors to binary
if (color>=(max+min)/2) color=1; else color=0;
now shift+or them to form the LBP vector
4x4 LBP example:
LBP =color[0][0];
LBP<<=1; LBP|=color[0][1];
LBP<<=1; LBP|=color[0][2];
...
LBP<<=1; LBP|=color[3][3];
you can do the step #3 directly in step #2
[original answer] - now obsolete
Histogram is the probability/occurrence/count of distinct color(shade)
uniform histogram means that each colors have almost the same count/area in the whole image
by bins I assume you mean distinct colors not the sub-images
To combine sub-histogram
just add them together or use single hist array init it once and then just sum to it as you have something like:
??? hist=Mat::zeros(1, numPatterns, CV_32SC1);
void lbp::histogram_(const Mat& src, Mat& hist, int numPatterns, bool init) {
if (init) hist = Mat::zeros(1, numPatterns, CV_32SC1);
for(int i = 0; i < src.rows; i++) {
for(int j = 0; j < src.cols; j++) {
int bin = src.at<_Tp>(i,j);
hist.at<int>(0,bin) += 1;
}
}
set init to true for the first patch call and false for all the rest. numPatterns is the max used color+1 or max possible colors count (not the distinct colors count)
If you want save only used colors
then you need to remember also the color. int hist[][2],hists=0; or use some dynamic list template for that the hist computation will change (will be much slower).
take color
test if it is in hist[i][0]==color
if yes increment its counter hist[i][1]++;
if not add new color hist[hists][0]=color; hist[hists][1]=1; hists++;
this will save space only if used colors are less then half of the possible ones. To improve performance you can compute hist normally and recompute to this list after that in the same manner (instead of the increment part of coarse)
Lets start with some code:
QByteArray OpenGLWidget::modifyImage(QByteArray imageArray, const int width, const int height){
if (vertFlip){
/* Each pixel constist of four unisgned chars: Red Green Blue Alpha.
* The field is normally 640*480, this means that the whole picture is in fact 640*4 uChars wide.
* The whole ByteArray is onedimensional, this means that 640*4 is the red of the first pixel of the second row
* This function is EXTREMELY SLOW
*/
QByteArray tempArray = imageArray;
for (int h = 0; h < height; ++h){
for (int w = 0; w < width/2; ++w){
for (int i = 0; i < 4; ++i){
imageArray.data()[h*width*4 + 4*w + i] = tempArray.data()[h*width*4 + (4*width - 4*w) + i ];
imageArray.data()[h*width*4 + (4*width - 4*w) + i] = tempArray.data()[h*width*4 + 4*w + i];
}
}
}
}
return imageArray;
}
This is the code I use right now to vertically flip an image which is 640*480 (The image is actually not guaranteed to be 640*480, but it mostly is). The color encoding is RGBA, which means that the total array size is 640*480*4. I get the images with 30 FPS, and I want to show them on the screen with the same FPS.
On an older CPU (Athlon x2) this code is just too much: the CPU is racing to keep up with the 30 FPS, so the question is: can I do this more efficient?
I am also working with OpenGL, does that have a gimmic I am not aware of that can flip images with relativly low CPU/GPU usage?
According to this question, you can flip an image in OpenGL by scaling it by (1,-1,1). This question explains how to do transformations and scaling.
You can improve at least by doing it blockwise, making use of the cache architecture. In your example one of the accesses (either the read OR the write) will be off-cache.
For a start it can help to "capture scanlines" if you're using two loops to loop through the pixels of an image, like so:
for (int y = 0; y < height; ++y)
{
// Capture scanline.
char* scanline = imageArray.data() + y*width*4;
for (int x = 0; x < width/2; ++x)
{
const int flipped_x = width - x-1;
for (int i = 0; i < 4; ++i)
swap(scanline[x*4 + i], scanline[flipped_x*4 + i]);
}
}
Another thing to note is that I used swap instead of a temporary image. That'll tend to be more efficient since you can just swap using registers instead of loading pixels from a copy of the entire image.
But also it generally helps if you use a 32-bit integer instead of working one byte at a time if you're going to be doing anything like this. If you're working with pixels with 8-bit types but know that each pixel is 32-bits, e.g., as in your case, you can generally get away with a case to uint32_t*, e.g.
for (int y = 0; y < height; ++y)
{
uint32_t* scanline = (uint32_t*)imageArray.data() + y*width;
std::reverse(scanline, scanline + width);
}
At this point you might parellelize the y loop. Flipping an image horizontally (it should be "horizontal" if I understood your original code correctly) in this way is a little bit tricky with the access patterns, but you should be able to get quite a decent boost using the above techniques.
I am also working with OpenGL, does that have a gimmic I am not aware
of that can flip images with relativly low CPU/GPU usage?
Naturally the fastest way to flip images is to not touch their pixels at all and just save the flipping for the final part of the pipeline when you render the result. For this you might render a texture in OGL with negative scaling instead of modifying the pixels of a texture.
Another thing that's really useful in video and image processing is to represent an image to process like this for all your image operations:
struct Image32
{
uint32_t* pixels;
int32_t width;
int32_t height;
int32_t x_stride;
int32_t y_stride;
};
The stride fields are what you use to get from one scanline (row) of an image to the next vertically and one column to the next horizontally. When you use this representation, you can use negative values for the stride and offset the pixels accordingly. You can also use the stride fields to, say, render only every other scanline of an image for fast interactive half-res scanline previews by using y_stride=height*2 and height/=2. You can quarter-res an image by setting x stride to 2 and y stride to 2*width and then halving the width and height. You can render a cropped image without making your blit functions accept a boatload of parameters by just modifying these fields and keeping the y stride to width to get from one row of the cropped section of the image to the next:
// Using the stride representation of Image32, this can now
// blit a cropped source, a horizontally flipped source,
// a vertically flipped source, a source flipped both ways,
// a half-res source, a quarter-res source, a quarter-res
// source that is horizontally flipped and cropped, etc,
// and all without modifying the source image in advance
// or having to accept all kinds of extra drawing parameters.
void blit(int dst_x, int dst_y, Image32 dst, Image32 src);
// We don't have to do things like this (and I think I lost
// some capabilities with this version below but it hurts my
// brain too much to think about what capabilities were lost):
void blit_gross(int dst_x, int dst_y, int dst_w, int dst_h, uint32_t* dst,
int src_x, int src_y, int src_w, int src_h,
const uint32_t* src, bool flip_x, bool flip_y);
By using negative values and passing it to an image operation (ex: a blit operation), the result will naturally be flipped without having to actually flip the image. It'll end up being "drawn flipped", so to speak, just as with the case of using OGL with a negative scaling transformation matrix.