How to convert 3D matrix to 1D?

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Tan Wen Kun
Tan Wen Kun 2015-12-3
编辑： Guillaume 2015-12-4
I got a color image and store into a variable A.
A =
(22,10,10) (22,10,10) (39,40,50)
(89,11,14) (23,11,11) (99,10,10)
(89,11,14) (69,10,10) (99,10,10)
x=1 when loop horizontally I wish to register the first color I seem and label it into x.
when loop same color then label it into same label.
when new color is seem then register the color and label into x+1.
I wish to get a table like
color label
(22,10,10) 1
(39,40,50) 2
(89,11,14) 3
(23,11,11) 4
(99,10,10) 5
(69,10,10) 6
so result finally get is
B=
1 1 2
3 4 5
3 6 5
How to get the label table and how to get the result?
  2 个评论
Walter Roberson
Walter Roberson 2015-12-3
编辑：Walter Roberson 2015-12-3
What data type is A? The notation you use is not MATLAB syntax, and
A = {
[22,10,10], [22,10,10], [39,40,50];
[89,11,14], [23,11,11], [99,10,10];
[89,11,14], [69,10,10], [99,10,10] }
would not be a 3D matrix.
Tan Wen Kun
Tan Wen Kun 2015-12-3
编辑：Tan Wen Kun 2015-12-3
I should say I got a 3D image.
I got a color image like this so what I want to do is loop horizontal and label the color seem and register it.
I just show my problem in a simple form to see.
so I did not use
A(:,:,1) =
22 22 39
89 23 99
89 69 99
A(:,:,2)=
10 10 40
11 11 10
11 10 10
A(:,:,3)=
10 10 50
14 11 10
14 10 10

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采纳的回答

Guillaume
Guillaume 2015-12-3
The way to do this is to reshape your image into an nx3 matrix where each row correspond to the three colours of a pixel. You can then apply unique with the 'rows' option on this to get your labels. It's then just a matter of reshaping the output.
Note that doing the scanning by rows rather than by columns complicate the code somewhat.
Also, please, post valid matlab code rather than leaving it to us to generate valid matrix
A = cat(3, [22 22 39;89 23 99; 89 69 99], ... it would have been great
[10 10 40; 11 11 10; 11 10 10], ... if I could just have
[10 10 50; 14 11 10; 14 10 10]) %copy/pasted that from your question
[~, ~, labels] = unique(reshape(permute(A, [2 1 3]), [], 3), 'rows', 'stable');
B = reshape(labels, size(A, 2), size(A, 1))'
The same code if you do the labeling by column:
[~, ~, labels] = unique(reshape(A, [], 3), 'rows', 'stable'); %no need to transpose anymore
B = reshape(labels, size(A, 1), size(A, 2)) %no need to transpose either.

更多回答（1 个）

Walter Roberson
Walter Roberson 2015-12-3
If you do not need the labels to start from 1 at the upper left, then
[color, ~, idx] = unique(reshape(A, [], 3), 'rows');
B = reshape(idx, size(A,1), size(A,2));
  9 个评论
显示 7更早的评论
Tan Wen Kun
Tan Wen Kun 2015-12-3
Is that other method can do like stable? I hope to get 1 at upper left
Columns 1 through 10
152 152 152 152 152 152 152 152 152 152
Columns 11 through 20
152 152 152 152 152 152 152 152 152 152
Columns 21 through 30
152 152 152 152 152 152 152 152 152 152
Guillaume
Guillaume 2015-12-3
Without 'stable', it's not possible to have a row or column ordering of labels with unique. You would have to use sort with the stable option (I'm fairly certain that did exist in 2011b) and the find the duplicate yourself. It's possible but it's worthy of a question in itself.
As for your new question, please start a new question as it is barely related to the current one. When you ask the question, please explain clearly what your definition of connected is. In your above example, I consider 4 and 5 to be connected orthogonally, and 3 and 5 to be connected diagonally.

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