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

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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.
  3 Kommentare
Image Analyst
Image Analyst am 25 Feb. 2018
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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Akzeptierte Antwort

John BG
John BG am 24 Feb. 2018
Bearbeitet: John BG am 24 Feb. 2018
Hi Zara Khan
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:
start image
clear all;clc;close all
the for loop with the counting of pixels for each quadrant:
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
for k=1:1:d1*d2
[nk1 nk2]=ind2sub([d1 d2],k);
if nk1<vert_bord && nk2<horz_bord
if A1(nk1,nk2)==255
if nk1<vert_bord && nk2>horz_bord
if A1(nk1,nk2)==255
if nk1>vert_bord && nk2<horz_bord
if A1(nk1,nk2)==255
if nk1>vert_bord && nk2>horz_bord
if A1(nk1,nk2)==255
resulting quadrants, coloring just to check the counting is correct
the amount of white pixels in each quadrant is
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.
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
  15 Kommentare
Zara Khan
Zara Khan am 26 Feb. 2018
Bearbeitet: Zara Khan am 26 Feb. 2018
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 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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Weitere Antworten (2)

Guillaume am 23 Feb. 2018
Bearbeitet: Guillaume am 23 Feb. 2018
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!
  4 Kommentare
Zara Khan
Zara Khan am 26 Feb. 2018
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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sumaiya khan
sumaiya khan am 7 Dez. 2018
How can I diagnolly divide the image into 4 quadrants using the centroid of the blob ?

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