How to extract color parameters of each rice grain and then how should I compare them?

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Ankit Rathore 2022-2-10

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评论： Ankit Rathore 2022-2-18

50.jpg

I have an image named 50, I have segmented the iamge and obtained label of each object (rice grain). This image contains rice of different varieties and I want to separate them based on their colour properties because they are similar in shape and size. Can anyone explain me which parmateters will help me to distinguish them.

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KSSV 2022-2-10

You can go by area, perimeter. ???

Ankit Rathore 2022-2-10

I would have, but the grains are of same size and another problem is that both varities contains some grains of same size?

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Answer 1

DGM 2022-2-10

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编辑：DGM 2022-2-10

在 MATLAB Online 中打开

Consider what information is available. The exposure doesn't seem too bad. There are some grains which seem maybe a bit more colorful than others, but it's hard to tell what's meaningful.

First, strip out the background and adjust the levels so that the grain color is more pronounced.

A = imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/890025/50.jpg');

A = rot90(A); % i'm just rotating this so it fits better in the web-view

A = imadjust(A,[0.7 1]);

mask = bwareaopen(rgb2gray(A)>128,1000);

A(~mask(:,:,[1 1 1])) = 0;

imshow(A)

Is that enough information to identify which are which? I'm not sure.

Select only the pixels representing the rice grains. Out of this entire population of pixels, is there any salient group information?

mask = imerode(mask,strel('disk',5)); % erode the mask to ignore edge shading

Am = reshape(A(mask(:,:,[1 1 1])),[],1,3);

Amlab = rgb2lab(Am);

for c = 1:3

subplot(3,1,c)

histogram(Amlab(:,:,c))

end

Maybe? It's hard to tell. Of course, that's the entire population of pixels. What if we consider the average color of each grain?

Alab = rgb2lab(A);

[L N] = bwlabel(mask);

graincolor = zeros(N,3);

for b = 1:N

grainmk = L==b;

Agrain = Alab(grainmk(:,:,[1 1 1]));

graincolor(b,:) = mean(reshape(Agrain,[],3),1);

end

graincolor % average color [L a b] for each grain

graincolor = 50×3

69.0426 -9.3356 27.3126 80.1833 -7.5283 30.9029 76.8120 -8.9078 25.2107 78.9900 -7.3287 23.0446 75.2678 -6.4827 22.0959 77.4466 -8.9433 26.8382 82.7059 -8.4911 24.3281 88.5259 -6.4093 18.3634 85.4462 -7.5029 22.2742 87.3065 -9.6987 29.2395

figure

for c = 1:3

subplot(3,1,c)

histogram(graincolor(:,c))

end

Now it looks even less like there's a discernable pattern. What if we just look at where those color points are and try to find a pattern by eye?

figure

plot3(graincolor(:,2),graincolor(:,3),graincolor(:,1),'b.')

grid on

Well, I'm not sure if that really helps either. It's hard to say if I'm getting anywhere since I'm not sure which are which to begin with. Let's just say that there's a group defined by L>90. Let's just select that.

figure

grainmask = graincolor(:,1)>90;

S = regionprops(mask,'centroid');

C = vertcat(S.Centroid);

imshow(A); hold on

for b = 1:N

if grainmask(b)

plot(C(b,1),C(b,2),'bo','markersize',25);

else

plot(C(b,1),C(b,2),'ro','markersize',25);

end

Are those the right ones?

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DGM 2022-2-14

编辑：DGM 2022-2-14

在 MATLAB Online 中打开

This is what I got:

A = imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/893335/Main.jpg');

A = imresize(A,0.25); % i'm just going to work with a smaller copy for testing

A = imadjust(A,[0.7 1]);

mask = bwareaopen(rgb2gray(A)>128,1000);

A(~mask(:,:,[1 1 1])) = 0;

% get mean color of each grain

Alab = rgb2lab(A);

[L N] = bwlabel(mask);

graincolor = zeros(N,3);

for b = 1:N

grainmk = L==b;

Agrain = Alab(grainmk(:,:,[1 1 1]));

graincolor(b,:) = mean(reshape(Agrain,[],3),1);

end

% plot grain color points

r51marked = [5 6 24 30]; % indices for marked grains in rice51.jpg

mk = false(N,1);

mk(r51marked) = true;

plot3(graincolor(mk,2),graincolor(mk,3),graincolor(mk,1),'r.'); hold on

plot3(graincolor(~mk,2),graincolor(~mk,3),graincolor(~mk,1),'b.')

grid on

figure

% make a selection and plot

grainmask = graincolor(:,1)>93 ...

& graincolor(:,2)<-6 ...

& graincolor(:,3)>19 ...

& sum(graincolor(:,2:3),2)>12;

S = regionprops(mask,'centroid');

C = vertcat(S.Centroid);

imshow(A); hold on

for b = 1:N

if grainmask(b)

plot(C(b,1),C(b,2),'ro','markersize',15);

else

plot(C(b,1),C(b,2),'bo','markersize',15);

end

hold off

Note that while I was able to find those four grains, the mask needed to be fairly elaborate to exclude nearby color points. With a larger sample size, it may prove impractical to find a geometry which can reliably make this exclusion.

Note also that the mask thresholds depend on the exposure and the input levels given to imadjust().

DGM 2022-2-14

编辑：DGM 2022-2-15

在 MATLAB Online 中打开

I meant that rgb2ycbcr() would be quicker than the demonstrated method, which was rgb2lab(), which it is about 10x as fast for me (though that depends). I didn't think that HSV really offered much margin for error in a rectangular mask region. To be fair, the demonstrated method didn't either, hence the diagonal face.

One can of course extend the color point viewing routine to provide more information about the margin around a simple box selection in various models. As before, this plots the target grains in red, and the unselected grains in blue. Any false positives that are unintentionally selected with the target grains are in green. For a scale factor of 1, this creates the smallest axis-aligned box selection that can include the target points. By simply scaling the box geometry, one can get an idea how much wiggle room there is. I've included some numbers for a handful of tests.

EDIT: fixed box scaling factor, updated

A = imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/893335/Main.jpg');

A = imresize(A,0.25);

A = imadjust(A,[0.7 1]);

mask = bwareaopen(rgb2gray(A)>128,1000);

A(~mask(:,:,[1 1 1])) = 0;

% get mean color of each grain

Alab = rgb2hsv(A); % yeah yeah i didn't change the variable name

[L N] = bwlabel(mask);

graincolor = zeros(N,3);

for b = 1:N

grainmk = L==b;

Agrain = Alab(grainmk(:,:,[1 1 1]));

graincolor(b,:) = mean(reshape(Agrain,[],3),1);

end

% plot grain color points

% find roi grains

r51marked = [5 6 24 30]; % indices for marked grains in rice51

mk1 = false(N,1);

mk1(r51marked) = true;

% maximum allowable box size

% NATIVE TOOLS

% +8% for lab (rgb2lab(A))

% +13% for hsv (rgb2hsv(A))

% +44% for ycbcr (rgb2ycbcr(A))

% +47% for yiq (rgb2ntsc(A))

% MIMT TOOLS

% +56% for hsl (rgb2hsl(A))

% +16% for hsi (rgb2hsi(A))

% +45% for lchab (rgb2lch(A,'lab'))

% +32% for lchuv (rgb2lch(A,'luv'))

% +45% for lchsr (rgb2lch(A,'srlab')) (srlab2)

% +46% for lchok (rgb2lch(A,'oklab')) (oklab)

% +49% for ych (rgb2lch(A,'ypbpr')) (polar ypbpr)

% -26% for hsy (rgb2hsy(A)) (chroma-normalized ypbpr)

% find false matches, isolate masks

boxscale = 1.15; % scale selection box

roibox = graincolor(mk1,:);

roibox = [min(roibox,[],1); max(roibox,[],1)];

bwdelta = abs(diff(roibox,1,1))*(boxscale-1);

roibox = roibox + [-1; 1].*bwdelta/2;

mk3 = all(graincolor>=roibox(1,:) & graincolor<=roibox(2,:),2);

mk2 = ~mk1 & ~mk3;

mk3 = mk3 & ~mk1;

nnz(mk3) % number of false positives

ans = 1

% plot color points

ms = 20;

scatter3(graincolor(mk1,2),graincolor(mk1,3),graincolor(mk1,1),ms,'r','filled'); hold on

scatter3(graincolor(mk2,2),graincolor(mk2,3),graincolor(mk2,1),ms,'b','filled')

ph = scatter3(graincolor(mk3,2),graincolor(mk3,3),graincolor(mk3,1),ms,'g','filled');

ph.MarkerEdgeColor = 'k';

% plot selection box

fa = 0.1;

patch(roibox([1 2 2 1],2),roibox([1 1 2 2],3),roibox([1 1 1 1],1),'k','facealpha',fa)

patch(roibox([1 2 2 1],2),roibox([1 1 2 2],3),roibox([2 2 2 2],1),'k','facealpha',fa)

patch(roibox([1 1 1 1],2),roibox([1 2 2 1],3),roibox([1 1 2 2],1),'k','facealpha',fa)

patch(roibox([2 2 2 2],2),roibox([1 2 2 1],3),roibox([1 1 2 2],1),'k','facealpha',fa)

patch(roibox([1 1 2 2],2),roibox([1 1 1 1],3),roibox([1 2 2 1],1),'k','facealpha',fa)

patch(roibox([1 1 2 2],2),roibox([2 2 2 2],3),roibox([1 2 2 1],1),'k','facealpha',fa)

grid on

The first group of conversions can be done with regular MATLAB tools. The rest are done using tools from MIMT, though LCHab and LCHuv can be done with existing tools. It's just not as convenient. Generally, simple transformations like HSV/HSL/HSI or luma-chroma models like YCbCr/YPbPr/YIQ will be much faster than LAB/LUV/etc. I don't think anything fancy is needed here. I'm just including a bunch of things to demonstrate that something as simple as YCbCr will probably work fine, and that there aren't really any esoteric miracle solutions that do remarkably better.

The particular properties of the various model don't matter that much right now, so long as the color points are distributed in a way that makes them (more) easily separated from their neighbors using a simple 3D geometry (a rectangular prism in this example). RIght now, the strategy is still squarely in the realm of "trying to make the best of a bad situation".

We're still trying to make a low-tolerance estimation of an unknown 3D boundary with only a very sparse test set. The apparent margins provided by the various models should be considered very much subject to change as the number of samples increases. What seems best now may have little margins with some other image, or like HSY in this case, it may simply not be able to make the selection with a simple box.

Image Analyst 2022-2-18

Hyper spectral could help. Post those images. Take a picture with all one type of rice on the left and all the other type on the right.

I've been designing imaging systems for about 40 years so I know a little bit about it. 🙂 Other than that (using your hyperspectral images) you should have a light booth that excludes ambient light. You should also check the illumination level with a lux meter to make sure it's the same each time. Aim for somewhere between 1000 and 6000 lux. Take a "blank shot" of just the background so you know how the exposure varies with location in the image due to lens shading and light non-uniformities. Then divide your test images by that background image. See attached demo.

But the big problem is your sample presentation. Your rice you said is all mixed together and you don't know which is which. So, using image analysis, let's say you figure out which type each grain of rice is. Now you need to collect those into bags or buckets (each of one type) to separate them. I guess you'd have to do that by hand, which is subject to error. Plus it would be slow and tedious - definitely not a high throughput method where you could sort through hundreds of kg per day. So maybe you could have the rice on a moving belt and analyze just one at a time. Then, depending on what type of rice it is, a blower tube blasts each grain of rice to the right or left (off the belt) into the proper bucket. Then you'd need to have a bucket fill indicator so you can either automatically switch out the bucket or shutdown the system until the user can put in a new bucket.

Ankit Rathore 2022-2-18

@Image Analyst The attached demo is missing

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Answer 2

Benjamin Thompson 2022-2-10

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You will have to decide how to separate them based on your intended purpose. See the documentation on the regionprops function in the Image Processing Toolbox for a variety of shape measurements that can be calculated. Your image also is grayscale or very nearly so, which means your ability to separate by color will be difficult. Try running the colorThresholder tool on it to create some masks.

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Answer 3

Image Analyst 2022-2-10

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How many colors are you seeing there? Because most of us see one : white. In fact it looks like some of the grains might be saturated (overexposed) so they are clipped at 255. If there are two types of rice there you might try hyperspectral imaging. Maybe as some narrow band of light there will be some brightness differentiation.

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