How to get find() results as a matrix?

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Queila Martins 2015-10-30

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评论： dpb 2015-11-3

Hi,

I have a binary image that was obtained from a color image (both were attached here). The background of the segmented image is black and the main object is white. I'm trying to use that binary image to get only the main object of the color image by using find() function.

I was able to do it using the following code:

I = imread(<binaryImage>);
I2 = imread(<color image>);
IMNONZERO = find(I); % this code returns only indexes of non-zero values
I3 = I2(IMNONZERO); % this gets all indexes of IMNONZERO in the I2 color image
imshow(I3);

This is my question: I saw that find() returns linear indexes. Also, I tried [row, col] = find(I) and it brings me the indexes of non-zero values as 2 separated arrays (one for row and another for column). However, I would like to know if it's possible to obtain those indexes as a regular matrix.

Is it possible?

Thank you in advance,

Queila

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Answer 1

dpb 2015-10-30

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Yes, if you wrap the find call inside ind2sub--

ijNONZERO = ind2sub(find(I),size(I));

But, you really don't need them for any purpose shown above.

Also, you can do w/o find entirely --

I3=I2(I~=0); % get all locations not zero in I from I2

However, I'm guessing this isn't going to work as you're hoping unless the image I has very special properties such that it returns another square/rectangular array. I'd guess imshow isn't happy with it has you're still going to get a vector of returned values, NOT a rectangular grid.

That's not doable by such logic and it's not a syntax issue it's a fundamental property of the characteristic of the image--there's no guarantee at all it will return such a set that can be arranged to be a valid image array.

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Queila Martins 2015-11-3

Hi dpb

First of all, thank you very much for your answer. I have another question related to this: how do I get a regular matrix with the values, not the indexes?

for example, I have a matrix a:

[0 1 1 0; 0 1 1 0; 0 1 1 0]

I want to remove all zeros, so my result should be this:

[1 1 1; 1 1 1; 1 1 1]

Then, as the second matrix would be similar (like the images), I would "crop" the second matrix to get only the values that are in the positions I got only "1" in the first matrix. Is it possible?

I need it this way because I have a code to run against non-zero values of the third matrix that should be generated(I3).

dpb 2015-11-3

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How do you get three ones/row from the original if you "want to remove all zeros"? That would leave a 3x2 array, not 3x3.

If it's regular as above, then

A=reshape(A(A~=0),size(A,1),[]);

does what I think you're asking.

If the area is not regular such that there are zeros scattered within the area that is otherwise nonzero, then you'll have to do something like --

>> A=[A(:,1:3) A]; A(2,4)=5  % another dummy array...
A =
   0     1     1     0     1     1     0
   0     2     2     5     2     2     0
   0     3     3     0     3     3     0
>> A(:,all(A==0))=[]  % remove all columns fully zero remaining rectangular
A =
   1     1     0     1     1
   2     2     5     2     2
   3     3     0     3     3
>>

If there are also rows; then use the second optional DIM option with all for them...

A(all(A==0,2),:)=[] % remove all rows fully zero

You might, depending on the need, want to check that there isn't a row/column in the middle rather than just edges first, that I don't know about whether would be a need or not...

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Answer 2

Philip Borghesani 2015-10-30

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How about:

[ind(:,1),ind(:,2)]=find(I);

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Queila Martins 2015-11-3

Hi, Philip

First of all, thank you very much for your answer. I have another question related to this: how do I get a regular matrix with the values, not the indexes?

for example, I have a matrix a:

[0 1 1 0; 0 1 1 0; 0 1 1 0]

I want to remove all zeros, so my result should be this:

[1 1 1; 1 1 1; 1 1 1]

Then, as the second matrix would be similar (like the images), I would "crop" the second matrix to get only the values that are in the positions I got only "1" in the first matrix. Is it possible?

I need it this way because I have a code to run against non-zero values of the third matrix that should be generated(I3).

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