3D matrix with various chain!!

17 Aug 2012
4 Answers
Answer Accepted
3 Views (30 days)

0 votes

If I have U3 matrix as:
U3(:,:,1) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,2) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0
]
U3(:,:,3) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
I assumed that these coordinates can be represented as an individual chain U3(3,2,1) and U3(3,2,2)and U3(4,2,2) and U3(5,2,2)and U3(3,2,3). and the other coordinates as another individual chain U3(3,4,1) and U3(3,4,2) and U3(3,4,3).
How can I give each chain a similar individual number? EX.
U3(3,2,1) and U3(3,2,2)and U3(4,2,2) and U3(5,2,2)and U3(3,2,3)=2
U3(3,4,1) and U3(3,4,2) and U3(3,4,3)=3

Sign in to answer this question.

 Accepted Answer

Image Analyst
Image Analyst on 17 Aug 2012
Edited: Image Analyst on 17 Aug 2012

0 votes

You can use bwconncomp. See this demo I wrote for you:
U3(:,:,1) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,2) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0
]
U3(:,:,3) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
% Method 1
% Make the list for how you want the objects labeled:
desiredNumberList = [2 3 4 42 69 123 73]; % Whatever you want.
% Make an object for the output
labeledImage = zeros(size(U3), 'int16')
% Do connected components labeling.
connectivityObject = bwconncomp(U3)
% Get the number of object it found (optional).
numberOfObjects = connectivityObject.NumObjects
% Now go through the blobs, reassiging them with the labels Hisham wants.
for blob = 1 : numberOfObjects
% Get the linear indices of the pixels that
% identify this particular blob.
pixelIndexes = connectivityObject.PixelIdxList{blob}
% Make those pixels have the desired number.
labeledImage(pixelIndexes) = desiredNumberList(blob);
end
% Print out the output image to the command window.
labeledImage
% Method 2
% Use labelmatrix but cast to uint16 if you ever expect to have more than 255 blobs.
labeledImage = uint16(labelmatrix(connectivityObject))
% But you didn't want 1,2,3, etc. You wanted custom numbers.
% For that we need to remap the default label numbers using intlut().
% Right now the blobs are labeled 1,2,3, etc.
% Make those pixels have the desired number with intlut().
% Make the list for how you want the objects labeled:
% But first one has to be zero because zero must remain as zero.
desiredNumberList = zeros(1, int32(intmax('uint16'))+1, 'uint16');
desiredNumberList(1:8) = [0 2 3 4 42 69 123 73]; % Whatever you want.
labeledImage = intlut(labeledImage, desiredNumberList)

More Answers (3)

Oleg Komarov
Oleg Komarov on 17 Aug 2012
Edited: Oleg Komarov on 17 Aug 2012

0 votes

If you have the Image Processing Toolbox:
CC = bwconncomp(U3);
labelmatrix(CC)
Hisham
Hisham on 20 Aug 2012
Edited: Hisham on 20 Aug 2012

0 votes

It is very nice. The first part of the problem has been solved.
The second part of the problem is:
How can I stop my "while" loop when the beginning of any chain start from the first row of the matrix, and its end is on the last row of the matrix?
for example:
the following matrix is inside the "while" loop and I want stop the loop at the previous condition "when any chain terminals are on the first row and the last row of the matrix as in the next example:
U3(:,:,1) = [
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,2) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,3) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0
]
Image Analyst
Image Analyst on 20 Aug 2012

0 votes

topPlane = squeeze(U3(1, :, :));
bottomPlane = squeeze(U3(end, :, :));
if any(topPlane(:)) || any(bottomPlane(:))
break; % Bail out of the while loop.
end

6 Comments
Show 4 older comments Hide 4 older comments

Hisham
Hisham on 20 Aug 2012
Edited: Hisham on 20 Aug 2012
Actually, I tried the code that you provided here but it is not work and I checked whether any chain's terminals on the first row and the last row is connected or not, and It showed me the same message. See the example: -----------------
Firstly
-----------------
U3(:,:,1) = [
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,2) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,3) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0
]
topPlane = squeeze(U3(1, :, :));
bottomPlane = squeeze(U3(end, :, :));
if any(topPlane(:)) any(bottomPlane(:))
disp('Ok')
else
disp('No')
end
-------------------
Secondly
-------------------
U3(:,:,1) = [
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,2) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,3) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0
]
topPlane = squeeze(U3(1, :, :));
bottomPlane = squeeze(U3(end, :, :));
if any(topPlane(:)) any(bottomPlane(:))
disp('Ok')
else
disp('No')
end
Image Analyst
Image Analyst on 20 Aug 2012
Perhaps I don't know what you mean. My code will enter the "if" if there is a 1 in the top row or bottom row of the three U3 planes you show. Both of your two examples have that so that's why both will display OK (in my code it would have hit the break line). If that's not what you want then explain what you want. For some irregularly shaped grouping of 1's, there is no "chain terminal" that I know of. Let's say the blob is shaped like a star. What is/are the terminals?
Hisham
Hisham on 20 Aug 2012
Edited: Hisham on 20 Aug 2012
I would like to thank you for your helps. Secondly, I mean that if I have a chain which starts from any point on the top row of the matrix and finishes at any point of the bottom row of the matrix. When it occurs, the loop will stop. For example: the chain in the following matrix starts from the top row of the matrix and end at the lowest row of the matrix
U3(:,:,1) = [
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,2) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,3) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0
]
whereas, the chain in the following matrix does not start from the top row of the matrix but ends at the lowest row of the matrix.
U3(:,:,1) = [
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,2) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,3) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0
]
Image Analyst
Image Analyst on 20 Aug 2012
Change from OR to AND:
if any(topPlane(:)) && any(bottomPlane(:))
Hisham
Hisham on 20 Aug 2012
Thaaaaaaaaaaaaaaaaaaaaanks so much. It is work now.
Hisham
Hisham on 21 Aug 2012
Edited: Hisham on 21 Aug 2012
Sorry it is not the one that I want ):
Actually, the code that you provided is going to check whether the top row and the bottom row contain number one, and then break the "while" loop.
However, the code that I want it is check for the chain starting and ending. If the same chain starts from the first row and finishes at the bottom row, then stop the "while" loop working.
Example of a chain starts from the stopped row and finishes at the bottom row:
U3(:,:,1) = [
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,2) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0
]
U3(:,:,3) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
whereas the next example shows there is no chain starts from the topped row and finishes at the bottom row.
U3(:,:,1) = [
0 , 0 , 0 , 1 , 0;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]
U3(:,:,2) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 1 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 0
]
U3(:,:,3) = [
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0;
0 , 1 , 0 , 0 , 1;
0 , 0 , 0 , 0 , 0;
0 , 0 , 0 , 0 , 0
]

Sign in to comment.

Categories

Find more on Labels and Annotations in Help Center and File Exchange

Products

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!