How a binary image can be divided into four equal parts using loop ?

Question

Zara Khan 2018-2-23

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回答： sumaiya khan 2018-12-7

I have a binary image . I want to divide this into 4 equal parts using a loop and want to store each part individually. later I want to find out the no of white pixels in each parts.

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Zara Khan 2018-2-24

II.png

Image Analyst : its just splitting into four equal quadrants. for each of the quadrants I have to find out f=total no. of pixels/no of object pixels. and this will be done using loop.

Image Analyst 2018-2-25

OK, but does this mean that the center of the quadrants will be located at the center of the blob, or at the center of the image? And I presume that edges parallel with the edges of the image are okay? And that it's okay if each quadrant does not have the same number of pixels in it?

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Answer 1

John BG 2018-2-24

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编辑：John BG 2018-2-24

在 MATLAB Online 中打开

im02.png

Hi Zara Khan

this is John BG <mailto:jgb2012@sky.com jgb2012@sky.com>

I use a for loop to answer your question, as requested, and I have added the variable nq to count the amount of pixels in each quadrant, please have a look, attached script and start image that I have reshaped to 16:9 format, to be able to tell what size was vertical and what horizontal:

1.

start image

clear all;clc;close all
A=imread('im02.png');imshow(A)

.

2.

the for loop with the counting of pixels for each quadrant:

d2=size(A,2);
d1=size(A,1);
vert_bord=floor(d1/2)
horz_bord=floor(d2/2)
nq=[0 0 0 0]; % 1st: top left quadrant red
                     % 2nd: top right quadrant green
                     % 3rd: bottom left quadrant blue
                     % 4th: bottom right quadrant magenta
hold all
A1=A(:,:,1);
for k=1:1:d1*d2
    [nk1 nk2]=ind2sub([d1 d2],k);
    if nk1<vert_bord && nk2<horz_bord 
        if A1(nk1,nk2)==255
            nq(1)=nq(1)+1; 
            plot(nk2,nk1,'r.');
        end
    end;
    if nk1<vert_bord && nk2>horz_bord 
        if A1(nk1,nk2)==255
            nq(2)=nq(2)+1; 
            plot(nk2,nk1,'g.');
        end
    end;
    if nk1>vert_bord && nk2<horz_bord
        if A1(nk1,nk2)==255
            nq(3)=nq(3)+1;
            plot(nk2,nk1,'b.');
        end
    end;
    if nk1>vert_bord && nk2>horz_bord 
        if A1(nk1,nk2)==255
            nq(4)=nq(4)+1;
            plot(nk2,nk1,'m.');
        end
    end;
end

.

3.

resulting quadrants, coloring just to check the counting is correct

.

4.

the amount of white pixels in each quadrant is

nq
   =
          1007        1084        1235        1666

.

nq(1): top left quadrant, red.

nq(2): top right quadrant, green.

nq(3): bottom left quadrant, blue.

nq(4): bottom right quadrant, magenta.

.

Zara

if you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance for time and attention

John BG

<mailto:jgb2012@sky.com jgb2012@sky.com>

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John BG 2018-2-24

编辑：John BG 2018-2-25

在 MATLAB Online 中打开

Zara

Ok, let's calculate the areas.

1.

Replacing the image I used with the image you attached to the question the results are as follows:

.

Now

2.

Area of 1 quadrant, any quadrant, in pixels, all pixels:

   area_quadrant=vert_bord*horz_bord
    =
           27224

2. Percentage of quadrant area against complete image area

pc_area_quadrant=100*round(area_quadrant/(d1*d2),2)
=
   25.0000000000000

as expected.

3.

Areas of white pixels in each quadrant? it's nq, directly

nq =
         468         421        2019        2403

4.

Percentage of quadrant area white pixels against quadrant area

pc_wp_per_quadrant=round(100*nq/area_quadrant,2)
=
   1.720000000000000   1.550000000000000
   7.420000000000000   8.830000000000000

.

Item oneThe 1st quadrant TOP LEFT, in RED, has 468 white pixels.

1.72% of this quadrant has white pixels.

Item twoThe 2nd quadrant TOP RIGHT, in GREEN, has 421 white pixels.

1.55% of this quadrant has white pixels.

Item threeThe 3rd quadrant BOTTOM LEFT, in BLUE, has 2019 white pixels.

7.42% of this quadrant has white pixels.

Item fourThe 4th quadrant BOTTOM RIGHT, in MAG, has 2403 white pixels.

8.83% of this quadrant has white pixels.

.

If you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance for time and attention

John BG

<mailto:jgb2012@sky.com jgb2012@sky.com>

John BG 2018-2-25

在 MATLAB Online 中打开

Hello again Zara

This is John BG <mailto:jgb2012@sky.com jgb2012@sky.com>

Ok you need

1.- to check whether input images need cropping

2.- image splitting into quadrants

1.- CROPPING

While the supplied image in the initial question has black pixels only wrapping the white pixels of interest, the 2 new images you have supplied have white pixels framing the black pixels that in turn surround the white pixels of interest.

Before applying the code supplied above, the input image has to be cropped, that can be easily done the following way:

clear all;clc;close all;
A=imread('img1.png');
imshow(A);
d1=size(A,1);
d2=size(A,2);
hold all;
A1=A(:,:,1);
p1=1;          % start 1st sweep top left corner
while A1(p1)>250
    p1=p1+1;
end
[n1p1,n2p1]=ind2sub(size(A1),p1);
plot(n2p1,n1p1,'r*');
p2=d1*d2;    % start 2nd sweep bottom right corner
while A1(p2)>250
    p2=p2-1;
end
[n1p2,n2p2]=ind2sub(size(A1),p2);
plot(n2p2,n1p2,'g*');
% cropping
B=A([n1p1:n1p2],[n2p1:n2p2],:);
figure;imshow(B)

.

on the right hand side BEFORE cropping, on the left hand side AFTER cropping.

.

2.- QUADRANTS SPLITTING

As requested

d01=floor(size(B,1)/2);
d02=floor(size(B,2)/2);
B1=B([1:d01],[1:d02],:);  % 1st quadrant: top left
B2=B([1:d01],[d02+1:end],:);  % 2nd  quadrant: top right
B3=B([d01+1:end],[1:d02],:);  % 1st quadrant: 
B4=B([d01+1:end],[d02+1:end],:);  % 1st quadrant: 
% check
figure;imshow(B1);
figure;imshow(B2);
figure;imshow(B3);
figure;imshow(B4);

.

Now run the code answering the initial question on each B2 B3 B3 B4 and you will obtain how many white pixels in each sub-quadrant, and so on so forth, you can carry on splitting while and image is 2x2 or larger, but there may not may many white pixels of interest in too small split images, may it not?

Zara

If you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link.

thanks in advance for time and attention

John BG <mailto:jgb2012@sky.com jgb2012@sky.com>

To Image Analyst: would you please be so kind to post your prospective answer in a different answer thread?

Zara Khan 2018-2-26

Image Analyst: It does not mean that all the quadrants need to have only foreground pixels only. The thing is that I have to scan from centroid column to right upto width to find out the first column which will consists off all black pixels,then from centroid column to left and finally from centroid row to top for the similar thing. now the distance from centroid to the right,left and top will be calculated. the max distance or length will be considered and have to take two square of sides equal to the length have to take from both sides means left side of centroid column and right side of centroid column. Again I need to divide these square to four equal quadrants. But these quadrants can be rectangular and you can divide by following any image spitting rules. After diving these two square will get 8 parts in total. now for these 8 parts I have to find out f=no. of white pixels/area of the part. These thing I have to repeat for a dataset of 1000.

I hope, I hope, I made it clear to you now.

Guillaume 2018-2-26

Zara,

We seem to have gone back to the same topic as your previous two questions, how-to-scan-an-image-to-find-out-the-row-or-column-that-will-be-the-first-background-pixel and how-to-divide-a-binary-image-into-equal-sized-square-considering-major-and-minor-axis-as-x-axis-or-y.

With your new question, that's now 4 on the same subject. What your questions seem to show is that you don't appear to know much about image processing, nor how image pixels are stored and accessed. I would suggest you spend more time learning about that from a book.

Zara Khan 2018-2-26

编辑：Zara Khan 2018-2-26

Yes am new to this. The fact is the solution you provided me for my previous problem was ok. But am unable to find out the upper right image. Moreover your code is good with some of my dataset but not for all.as I have 1000 of data so most of the cases am not getting top like left ,right and bottom. So am facing problem. So I asked again if anyone can provide me any other solution . I hope you can link to my previous problem that was the one I accepted your answer.

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Answer 2

Guillaume 2018-2-23

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https://ww2.mathworks.cn/matlabcentral/answers/384432-how-a-binary-image-can-be-divided-into-four-equal-parts-using-loop#answer_306749

编辑：Guillaume 2018-2-23

在 MATLAB Online 中打开

I'm not sure why or even how you'd use a loop.

[height, width, ncols] = size(yourimage);
splitimages = mat2cell(yourimage, [height height]/2, [width width]/2, ncols)

Your four images are splitimages{1}, splitimages{2}, splitimages{3}, and splitimages{4}.

To find the number of white pixels in each subimage:

numwhitepixels = cellfun(@(subimg) sum(sum(all(subimg == 1, 3))), splitimages);  %assuming images of type double, where white == 1

edit: stupidly forgot the image argument to mat2cell!

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Zara Khan 2018-2-24

Thanks for the help. Just have done it using a for loop. Then I have used your numwhitepixels. What will be the syntax to find the total no pixels in each part ?

Zara Khan 2018-2-26

Like no. of white pixels calculation , will I be able to find out area of each cell using a single syntax only ??

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Answer 3

sumaiya khan 2018-12-7

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How can I diagnolly divide the image into 4 quadrants using the centroid of the blob ?

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