pixels[] Question

Programming Questions

subatomic..

in Programming Questions • 1 year ago

Howdy All,

At the end is the code snippet from the reference manual for pixels[].

So, you can set the color using an english word describing the color (in the case below, "pink"). My question is can you feed it a set of RGB values and get the english equivalent word back? So, (255, 255, 0) would return "yellow." Essentially, query pixels[] and instead of the RGB values, it returns the english equivalent word.

Is this possible? I am working on some code (will have other questions about it soon!) that goes through a series of images and counts the number of pixels in each that are red, yellow, and green. For example, to classify a pixel as red the R value has to be above 200 and the other two values have to be below 50. But I am testing this on "perfect" test images and the actual working images will have varying shades of each color and I will need to account for that. I'm playing around with ways to compare each of the values to each other, but that can get rather tedious...

If one cannot get the english color name back, are there any functions that can classify various shades of a given color as the color? Or will I have to "brute force" it?

Thanks!

james

color pink = color(255, 102, 204); loadPixels(); for (int i = 0; i < (width*height/2)-width/2; i++) { pixels[i] = pink; } updatePixels();

Replies(9)

Chrisir

Re: pixels[] Question

1 year ago

better readable;

color pink = color(255, 102, 204);
loadPixels();
for (int i = 0; i < (width*height/2)-width/2; i++)
{
pixels[i] = pink;
}
updatePixels();

benja

Re: pixels[] Question

1 year ago

In this example "pink" is just the name of a variable storing a color-value.
I dont't think that i've seen pre defined consants for colours, so you would have to create a list yourself and then test every colour, to find out, wich is the closest from your list.

subatomic..

Re: Re: pixels[] Question

1 year ago

Oh, crap, I'm an idiot!

Sorry!

Chrisir

Re: pixels[] Question

1 year ago

I think so, too

You could make a class and then an array of this class and then store the values in it :

class ColorName {
String nameEnglish;
color bestColor;
int valueRedLowerBound;
int valueRedUpperBound;
int valueGreenLowerBound;
int valueGreenUpperBound;
int valueBlueLowerBound;
int valueBlueUpperBound;
}
ColorName [] colornames;

then in a function

String colorNameFromColor(color myCol) ...

for-loop over colornames
check

if (red(myCol)>colornames[i].valueRedLowerBound && red(myCol)<colornames[i].valueRedUpperBound&&
green(myCol)>colornames[i].valueGreenLowerBound && green(myCol)<colornames[i].valueGreenUpperBound&&
....)
{
return colornames[i].nameEnglish;
} // if
} // for

Greetings, Chrisir

subatomic..

Re: Re: pixels[] Question

1 year ago

Thanks, Chrisir, that looks like a tidier way of going about it than what I was thinking of.

Thank you!

benja

Re: pixels[] Question

1 year ago

My idea was to compute the difference between colors and just compare wich one from you colorNames-array fits best.

class ColorName {
String name;
int r, g, b;
ColorName(int value, String name) {
r = (value >> 16) & 0xFF;
g = (value >> 8) & 0xFF;
b = value & 0xFF;
this.name = name;
}
}
String colorNameFromColor(color testColor) {
ColorName currCol;
int diff = 765; // set to maximum difference (255*3)
int bestIndex = 0;
// loop through the arraylist of colornames
for (int i= 0 ; i< colorNames.size(); i++) {
currCol = colorNames.get(i);
// Extract the red, green, and blue components of the test color
int testR = (testColor >> 16) & 0xFF;
int testG = (testColor >> 8) & 0xFF;
int testB = testColor & 0xFF;
// Compute the difference of the red, green, and blue values
int diffR = abs(currCol.r - testR);
int diffG = abs(currCol.g - testG);
int diffB = abs(currCol.b - testB);
// see if its closer to the color than previous ones
if(diff >= diffR+diffG+diffB){
diff = diffR+diffG+diffB;
bestIndex = i;
}
}
return colorNames.get(bestIndex).name;
}

This way you could easily add new colours without thinking about boundaries.
If you're not only looking for basic colors, there are plenty of lists in the web with colornames and there hex-values.
And this is an example in javascript that does pretty much what this thread is about.

subatomic..

Re: Re: pixels[] Question

1 year ago

Hi Guys,

I've implemented two methods, including the RGB distance method. The first method (non-RGB distance method) gives me what I expect but the RGB distance method doesn't, but based on the logic, it should. Please don't laugh at the lack of an rgbDistance class...I haven't done java in many, many years and am coming from a perl background.

Here is the input image:

And here is the output of the non-rgbDistance method (this is what I expected):

Here is the code:

PImage a;
void setup() {
String folderPath = selectFolder("Choose a folder containing images");
println(folderPath);
File folder = new File (folderPath);
String[] names = folder.list();
int foundCount = 0;
if (names != null) {
for (int w = 0; w < names.length; w++) {
if (names[w].charAt(0) == '.') continue;
if (names[w].toLowerCase().endsWith(".jpg")) {
println("Found JPG image: " + names[w]);
foundCount++;
String pathFile = folderPath + "/" + names[w];
println(pathFile);
a = loadImage(pathFile); // Load image
size(a.width, a.height);
//a.filter(BLUR, 3);
a.loadPixels();
int yellow_pix = 0;
int red_pix = 0;
int green_pix = 0;
int blue_pix = 0;
for (int i= 0; i < a.width*a.height; i++) {
color p = a.pixels[i];
float r = red(p);
float g = green(p);
float b = blue(p);
//if (r > 200 && g < 50 && b < 50) {
if (r > (g+10) && r > (b+10)) {
red_pix++;
a.pixels[i] = color(255,0,0);
}
//if (g > 200 && r < 50 && b < 50) {
if (g > (r+10) && g > (b+10)) {
green_pix++;
a.pixels[i] = color(0,255,0);
}
//if (r >200 && g > 200 && b < 50) {
if (r > (b+10) && g > (b+10)) {
yellow_pix++;
a.pixels[i] = color(255,255,0);
}
//if (r > 200 && g > 200 && b > 200) {
if (b > (r+10) && b > (g+10)) {
//white_pix
blue_pix++;
a.pixels[i] = color(0,0,255);
}
}
println ("Blue = " + blue_pix + " Red = " + red_pix + " Green = " + green_pix + " Yellow = " + yellow_pix);
a.updatePixels();
}
}
println("Found " + foundCount + " image(s).");
int[] numbers = new int[foundCount];
}
else {
println("Could not access images.");
}
}//void setup end
void draw() {
image(a, 0, 0);
}

Here is the output of the rgbDistance method (not what I expected):

And the code (try not to laugh ;) ):

PImage a;
void setup() {
String folderPath = selectFolder("Choose a folder containing images");
println(folderPath);
File folder = new File (folderPath);
String[] names = folder.list();
int foundCount = 0;
/*
int greenR = 0;
int greenG = 255;
int greenB = 0;
int redR = 255;
int redG = 0;
int redB = 0;
int blueR = 0;
int blueG = 0;
int blueB = 255;
int yellowR = 255;
int yellowG = 255;
int yellowB = 0;
*/
if (names != null) {
for (int w = 0; w < names.length; w++) {
if (names[w].charAt(0) == '.') continue;
if (names[w].toLowerCase().endsWith(".jpg")) {
println("Found JPG image: " + names[w]);
foundCount++;
String pathFile = folderPath + "/" + names[w];
println(pathFile);
a = loadImage(pathFile); // Load image
size(a.width, a.height);
//a.filter(BLUR, 3);
a.loadPixels();
int yellow_pix = 0;
int red_pix = 0;
int green_pix = 0;
int blue_pix = 0;
int white_pix = 0;
int black_pix = 0;
int grey_pix = 0;
for (int i= 0; i < a.width*a. height; i++) {
color p = a.pixels[i];
float r = red(p);
float g = green(p);
float b = blue(p);
float rgbDistanceG = Math.abs(r - 0) + Math.abs(g - 255) + Math.abs(b - 0);
float rgbDistanceR = Math.abs(r - 255) + Math.abs(g - 0) + Math.abs(b - 0);
float rgbDistanceB = Math.abs(r - 0) + Math.abs(g - 0) + Math.abs(b - 255);
float rgbDistanceY = Math.abs(r - 255) + Math.abs(g - 255) + Math.abs(b - 0);
float rgbDistanceW = Math.abs(r - 255) + Math.abs(g - 255) + Math.abs(b - 255);
float rgbDistanceBlack = Math.abs(r - 0) + Math.abs(g - 0) + Math.abs(b - 0);
float rgbDistanceGrey = Math.abs(r - 128) + Math.abs(g - 128) + Math.abs(b - 128);
if (rgbDistanceG < rgbDistanceR && rgbDistanceG < rgbDistanceB && rgbDistanceG < rgbDistanceY && rgbDistanceG < rgbDistanceW && rgbDistanceG < rgbDistanceBlack && rgbDistanceG < rgbDistanceGrey) {
green_pix++;
a.pixels[i] = color(0,255,0);
} else if (rgbDistanceR < rgbDistanceG && rgbDistanceR < rgbDistanceB && rgbDistanceR < rgbDistanceY && rgbDistanceR < rgbDistanceW && rgbDistanceR < rgbDistanceBlack && rgbDistanceR < rgbDistanceGrey) {
red_pix++;
a.pixels[i] = color(255,0,0);
} else if (rgbDistanceB < rgbDistanceG && rgbDistanceB < rgbDistanceR && rgbDistanceB < rgbDistanceY && rgbDistanceB < rgbDistanceW && rgbDistanceB < rgbDistanceBlack && rgbDistanceB < rgbDistanceGrey) {
blue_pix++;
a.pixels[i] = color(0,0,255);
} else if (rgbDistanceY < rgbDistanceG && rgbDistanceY < rgbDistanceB && rgbDistanceY < rgbDistanceR && rgbDistanceY < rgbDistanceW && rgbDistanceY < rgbDistanceBlack && rgbDistanceY < rgbDistanceGrey) {
yellow_pix++;
a.pixels[i] = color(255,255,0);
} else if (rgbDistanceW < rgbDistanceG && rgbDistanceW < rgbDistanceB && rgbDistanceW < rgbDistanceR && rgbDistanceW < rgbDistanceY && rgbDistanceW < rgbDistanceBlack && rgbDistanceW < rgbDistanceGrey) {
white_pix++;
a.pixels[i] = color(255,255,255);
}
else if (rgbDistanceBlack < rgbDistanceG && rgbDistanceBlack < rgbDistanceB && rgbDistanceBlack < rgbDistanceR && rgbDistanceBlack < rgbDistanceY && rgbDistanceBlack < rgbDistanceW && rgbDistanceBlack < rgbDistanceGrey) {
black_pix++;
a.pixels[i] = color(0,0,0);
}
else if (rgbDistanceGrey < rgbDistanceG && rgbDistanceGrey < rgbDistanceB && rgbDistanceGrey < rgbDistanceR && rgbDistanceGrey < rgbDistanceY && rgbDistanceGrey < rgbDistanceW && rgbDistanceGrey < rgbDistanceBlack) {
grey_pix++;
a.pixels[i] = color(255,255,255);
} //else { a.pixels[i] = color(255,128,100); }
}
println ("Blue = " + blue_pix + " Red = " + red_pix + " Green = " + green_pix + " Yellow = " + yellow_pix + " White: " + white_pix + " Black: " + black_pix + " Grey: " + grey_pix);
a.updatePixels();
}
}
println("Found " + foundCount + " image(s).");
int[] numbers = new int[foundCount];
}
else {
println("Could not access images.");
}
}//void setup end
void draw() {
image(a, 0, 0);
}

Any idea of why the differences in output?

Also, any tips on cleaning up the rgbDistance code would be welcome.

Thanks so much, guys!

james

benja

Re: pixels[] Question

1 year ago

As far as i can see, both codes work as i would expect it.
In the second code you use more colors, including a "grey". Of course a lot of pixel are more similiar to grey, than to pure red, green, blue or yellow. Thats why you have a different result.

Regarding code-improvements... i don't know where to start.
Use arrays, loops and classes. Makes the code much cleaner and saves a lot of work/time.

subatomic..

Re: Re: pixels[] Question

1 year ago

Thanks, iron13.

Now what I want to do is to graph the color counts in a line graph with the images going out on the x-axis (so, if there are 10 images, x will go from 1 to 10) and the pixel color (r, g, b, y) count on y.

Any tips would be appreciated...I am thinking about creating ArrayLists for each color and then going through them to draw the graph.

Figured I'd ask the question as I begin work on it...

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