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Splitting a binary image into 2 parts

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Ankit Sahay
Ankit Sahay on 14 Sep 2020
Commented: Ankit Sahay on 20 Sep 2020
HI everybody! I have a binary image matrix uploaded as A.mat:
White pixels correspond to 1, black regions correspond to 0. I would like to split the curve into left and right parts, something like the following:
Initially, I thought of finding the indices of midponts of elements having 1 at their positions, at each row. Using the indices, I can substitute ones on the left (and right) of the midpoints (at each row) using relational operators. Maybe there might be a better way to do this, but this method comes to my mind right now.
So I start from the bottommost row, and I keep on moving upwards. As evident from the A.mat figure (first figure), a row can have more than 2 cells having values one. To choose the appropriate cells for calculating midpoints at row "r", I compare the number of elements having one on left and right side of midpoint at row "r+1". I am attaching the code below:
[m,n] = size(A);
midpoint = zeros(m,n);
midpt_idx = zeros(m,1);
A_sep = A;
for r = m:-1:1
k = find(A(r,:));
if size(k,2)==2
midpt_idx(r) = ceil((k(1)+k(2))*0.5);
elseif sum(A(r,1:midpt_idx(r+1))) < sum(A(r,midpt_idx(r+1)+1:end))
midpt_idx(r) = ceil((k(1)+k(2))/2);
elseif sum(A(r,1:midpt_idx(r+1))) > sum(A(r,midpt_idx(r+1)+1:end))
midpt_idx(r) = ceil((k(end-1)+k(end))/2);
elseif sum(A(r,1:midpt_idx(r+1))) == sum(A(r,midpt_idx(r+1)+1:end))
midpt_idx(r) = ceil((k(size(k,2)/2)+k((size(k,2)/2)+1))*0.5);
A_sep(r,midpt_idx(r)) = 2; % Midpoints have value 2 in their cells, I change this to 1 for checking solution by using imshow(A_sep)
An error is dispayed when execution comes at row 64. There is only one cell having 1 in row 64. For now, this error can be ignored (I think!).
This code runs correctly for some parts, but fails at other parts:
If you run the code along with A.mat file, starting from row 416, it gives correct results till row 278. At row 277, the code fails because there are a larger number of points adjacent to each other on right side, compared to left side of the midpoint index at row 279.
So, basically my method fails in these situations. Can you please help me with this? Maybe an advice for my code, or some other method which is better than what I am doing right now.

Accepted Answer

Matt J
Matt J on 14 Sep 2020
Edited: Matt J on 14 Sep 2020
There is a small break in the curve which I assumed was supposed to indicate the desired dividing point between the right and left sides.
Under that assumption, I propose the following:
load A
for i=1:numel(C)
if ismember(flip(B(j,:)),C(i).PixelList,'rows')
C=fliplr(C(i).PixelList); break
Left=accumarray(C,1,size(A)); %the resultant images
Ankit Sahay
Ankit Sahay on 19 Sep 2020
Yeah, I get it now. I did not anticipate such a situation when I started the analysis. It is only after analysing a few more data did I come across a situation where the code failed. Let me reformulate the question such that no ambiguities arise I will take help from the 10X10 matrix that you have provided. Thanks for your input.

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More Answers (1)

Image Analyst
Image Analyst on 19 Sep 2020
Why not simply scan your image column by column with find() finding the topmost and bottommost row?
Ankit Sahay
Ankit Sahay on 20 Sep 2020
I apologize profusely for this mistake. I should have been more careful. I have attached the complete program along with the .mat file that the programs needs to run. Again, I am really sorry for wasting your time.

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