Can i use Color map to find which color is the most used color in an image ?

How can i find dominant color from an image( Yellow, green , red , blue, orange , black , white etc..)? What is the basic method benind this ??

[Walter Roberson]

We do not know what "dominant color" means. (Seriously, we do not. See past discussions.)

https://www.mathworks.com/matlabcentral/answers/?term=tag%3A%22dominant+color%22

[sam CP]

I meant the majority used color in an image.

[Walter Roberson]

Do you mean the single exact RGB combination that occurs most often? If so, then suppose there is an image that is lots of shades of red, but no one red-ish RGB has more than 9 copies, but scattered around the image there are 10 pixels with the exact same blue. Is blue the dominant color?

[sam CP]

Yellow is the most used color in this image. Which method can find this ?

[Walter Roberson]

No, black is the most used color. Run the image through https://www.mathworks.com/matlabcentral/fileexchange/28164-color-frequency-image

[sam CP]

How can i display the color map and color bar of this image ?

Answer 1

[Image Analyst]

You can display the 3-D color gamut with colorcloud(). Call that on your image and see what you think the dominant color is.

You might want too look at the Color Frequency Image. See this link

To give a good answer we need to know what you plan on doing assuming you had the "dominant color". If we knew that we might be able to suggest something. Post your image(s) and explain what you want to know about them or do with them assuming you knew the dominant color.

Finally, look at these fun web sites:

https://www.npr.org/sections/thetwo-way/2010/12/10/131960390/color-picker-sorts-flickr-photos-for-fun

https://labs.tineye.com/multicolr/

http://colorthesaurus.epfl.ch/

[sam CP]

How can i find which is the majority used color in this image ? Which method can be used ?

[Image Analyst]

In the color frequency image it would be the one with the greatest area (largest histogram bin value).

[sam CP]

if true
 IM = imread('demo.jpg');
IM=imresize(IM,[256,256]);
figure, imshow(IM),title('Input Image');
imhist(IM);
end

How can i find that area by using largest histogram bin value?

[Image Analyst]

With the color frequency image, you do

[maxCount, linearIndex] = max(gamut3D(:))
[r, c, z] = ind2sub(size(gamut3D), linearIndex)

[sam CP]

How can i apply attached image into this mentioned code?

[Image Analyst]

Did you run the demo? It asks you to pick the image and then it uses it. If you had run it, you couldn't have missed that.

[sam CP]

But I can’t understand that where can i apply the input image in your codes?

[Walter Roberson]

When you ran the code, did it not pop up a file selector with title "Select a color image file" ?

Answer 2

[Jan]

Walter is right. What is the dominant color of the French flag? Imagine a photo of a traffic light in the dusk. Most of the pixels are dark gray, but if you ask a person about the color, you get the answer green (or yellow, or red). Of course there are some rare examples, which allows a unique decision about the most frequent RGB or Lab color. But even then there is no unique relation to a name of a color. Do you remember the discussion about the blue/black or gold/white dress (link) ? Even human could not decide which color the image contain.

 What is the basic method behind this ?

There is no basic method. If you have a clear and unique definition, and an exactly defined workflow considering the color profiles of the camera, monitor, etc., it can be implemented in Matlab.

Currently the best method is to show the image printed on paper or shown on a calibrated monitor to 10 persons. If they agree, there is a high chance that other human have the same impression.

[Jan]

Did you even read my answer? Start with defining, what a "dominant color" is.

close all; clear all; clc; warning off;

This is a really bad idea: Killing everything and disabling all warnings is brute and provokes errors.

[sam CP]

Okay

Answer 3

[Walter Roberson]

topN = 50;
img = imread('demo.jpg');
rimg = reshape(img, [], 3);
[urgb, ~, uidx] = unique(rimg, 'rows');
labs = reshape(uint32(uidx),size(img,1),size(img,2));
recolored = uint8(ind2rgb(labs-1, urgb));
subplot(2,2,1)
image(img);
title('original')
counts = accumarray(uidx,1);
[sortedcounts, sortidx] = sort(counts,'descend');
mcc1 = urgb(sortidx(1),:);
mcc2 = urgb(sortidx(2),:);
subplot(2,2,3)
image(reshape(mcc1,1,1,3))
title(sprintf('Most common color [%d,%d,%d]\n%d occurrences', mcc1, sortedcounts(1)));
subplot(2,2,4)
image(reshape(mcc2,1,1,3))
title(sprintf('Second most common color [%d,%d,%d]\n%d occurrences', mcc2, sortedcounts(2)));
onlytop = double(ismember(labs,sortidx(1:topN)));
subplot(2,2,2)
image(recolored, 'alphadata', onlytop);
title(sprintf('showing only top %d colors', topN));
set(gca, 'color', 'k')

You have to go to the 47th most common RGB combination to get one that is not essentially black. No yellow-ish RGB combination occupies more than 96 pixels.

[sam CP]

Please explain the steps that are used in this code

[Walter Roberson]

First we display the original.

The RGB values for any pixel occur along the third dimension. If you reshape the image to contain just rows like this, then each row will contain the RGB combination of one pixel.

Then, find the unique RGB combinations in sorted numeric order (first output of unique). Also, for each original row, find the index of which of the unique rows the pixel was. For example [0 0 0] is always going to sort first in the list, which corresponds to value 1 (first in unique list), and then any entries in the uidx list that come out 1 do so because the corresponding row in rimg was [0 0 0]

Reshape this list of indices to be 2D. This is now an array of color numbers suitable for use in pseudocolor work. All of the pixels that matched the RGB triple urgb(1,:) originally will have value 1, all the pixels that matched the RGB triple urgb(2,:) originally will have value 2, and so on. So now urgb is effectively a color map and labs is the index into the color map for each original pixel.

For testing purpose, we then put the image back together in the array "recolored". This should be the same as the original. Subtracting 1 from labs is done for technical reasons involving whether 0 or 1 is to be used to designate the first color in the list.

We then count how many of each unique color there were by counting the number of times each index occurred. That is done by using accumarray().

Then we sort the counts in descending order, producing also the list of where in the original order the count came from. So sortedindex(1) tells you which counts() entry had the highest count and sortedcounts(1) tells you what the corresponding count was.

Then we extract the unique RGB triple that was associated with the most common color and also the second most common color. And we put up two images showing samples of those two colors and the count of each one. These are intended just to be solid color images.

Then we take the 50 most commonly occurring colors, examining their indices in the urgb list, and matching that to the label indices. We want to know, for each pixel, whether the pixel was one of those 50 most common colors -- just a 0 (false) or a 1 (yes it was.) We convert that to class double for technical reasons. Then we put up the color image, but we make it transparent (alphadata = 0) everywhere that the pixel was not one of the most common colors, and making it not transparent (alphadata = 1) everywhere the pixel was one of the most common colors.

Then because we made the image transparent in most places, we set the background color (which will show through) black.

You might notice that most of the image is black. It probably looks like just 2 colors, black and a bit of yellow, but really it is 50 different colors, 49 of which are different shades of black -- because the first 46 most common RGB triples in the image are all shades of black (and the 48th, 49th and 50th are too.)

Therefore the dominant color of the image is black. Not yellow, black.

It is possible to keep redefining "dominant color" until you get to the point where the count for a color approaching yellow exceeds the count for the colors approaching black, such as by doing color quantizing. But you have to be careful, because when you do, you will probably end up quantizing to a shade of yellow that does not actually exist at all in the image !

[sam CP]

can i find this by using hsv ?

[Walter Roberson]

Each rgb corresponds to a unique hsv, so if you were to convert to hsv first, you would get exactly the same results. Shades of black would still be the most common 46 colors.

But if you need that proved..

topN = 50;
rgbimg = imread('demo.jpg');
img = rgb2hsv(rgbimg);
rimg = reshape(img, [], 3);
[uhsv, ~, uidx] = unique(rimg, 'rows');
labs = reshape(uint32(uidx),size(img,1),size(img,2));
recoloredhsv = uint8(ind2rgb(labs-1, uhsv));
subplot(2,2,1)
image(rgbimg);
title('original')
counts = accumarray(uidx,1);
[sortedcounts, sortidx] = sort(counts,'descend');
mcc1hsv = uhsv(sortidx(1),:);
mcc2hsv = uhsv(sortidx(2),:);
subplot(2,2,3)
mcc1rgb = hsv2rgb(reshape(mcc1hsv,1,1,3));
image(mcc1rgb)
title(sprintf('Most common hsv [%d,%d,%d]\n%d occurrences', mcc1hsv, sortedcounts(1)));
subplot(2,2,4)
mcc2rgb = hsv2rgb(reshape(mcc2hsv,1,1,3));
image(mcc2rgb)
title(sprintf('Second most common hsv [%d,%d,%d]\n%d occurrences', mcc2hsv, sortedcounts(2)));
onlytop = double(ismember(labs,sortidx(1:topN)));
subplot(2,2,2)
image(rgbimg, 'alphadata', onlytop);
title(sprintf('showing only top %d colors', topN));
set(gca, 'color', 'k')

[sam CP]

Please explain 5,6,7, 11,12,13,14 lines of codes

[Walter Roberson]

5. Then, find the unique hsv combinations in sorted numeric order (first output of unique). Also, for each original row, find the index of which of the unique rows the pixel was. For example [0 0 0] is always going to sort first in the list, which corresponds to value 1 (first in unique list), and then any entries in the uidx list that come out 1 do so because the corresponding row in rimg was [0 0 0]

6. Reshape this list of indices to be 2D. This is now an array of color numbers suitable for use in pseudocolor work. All of the pixels that matched the hsv triple uhsv(1,:) originally will have value 1, all the pixels that matched the hsv triple uhsv(2,:) originally will have value 2, and so on. So now uhsv is effectively a color map and labs is the index into the color map for each original pixel.

7. For testing purpose, we then put the image back together in the array "recoloredhsv". Subtracting 1 from labs is done for technical reasons involving whether 0 or 1 is to be used to designate the first color in the list. This might not be the same as the hsv image because I did not pay attention to whether rgb2hsv produces floating point or integer values; probably recoloredhsv is not useful without a correction to that line. I did not use the array in the later code.

11. We then count how many of each unique color there were by counting the number of times each index occurred. That is done by using accumarray().

The line

counts = accumarray(uidx,1);

is equivalent to (but much more efficient than)

counts = [];
for IDX = 1 : numel(uidx)
  temp = uidx(IDX);
  if size(counts,1) < temp
    counts(temp,1) = 1;
  else
    counts(temp,1) = counts(temp,1) + 1;
  end
end

That is, it counts the number of times each value in uidx occurs.

12. Then we sort the counts in descending order, producing also the list of where in the original order the count came from. So sortedindex(1) tells you which counts() entry had the highest count and sortedcounts(1) tells you what the corresponding count was. If you need more information about what the sort index means, see https://www.mathworks.com/help/matlab/ref/sort.html#bt8nojg-1-I

13 and 14. Then we extract the unique hsv triple that was associated with the most common color and also the second most common color.