How to ignore values in a matrix that is out of bound when performing column 'find' in an array?

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Z is a 1024X1024 matrix containing negative value in some pixels.
For each pixel in Z, I need to find the column from an array 0,5,10,15,20,25,30,35,40; Because of out of bound values existing in Z matrix, I get this error 'Subscript indices must either be real positive integers or logicals.' How do I set all the negative values in matrix Z to '0' or is there a way to get around this problem? Below is the code.
m=1024;
n=1024;
Th = (0,5,10,15,20,25,30,35,40);
i=1:length(Th)
ratioAB(:,:,i)=A./B % A and B are matrices for each Th
RF=ones(m,n);
for i=1:m
for j=1:n
Z2=Z(i,round(j));
col = find(Th > Z2,1,'first')-[1 0]
RF(i,j)= diff(cat(3,ratioAB(i,j,col)),[],3)/diff(Th(col))*(Z2 Th(col(1)))+ratioAB(i,j,col(1));
end
end
  2 件のコメント
Image Analyst
Image Analyst 2012 年 8 月 21 日
編集済み: Image Analyst 2012 年 8 月 21 日
Lots of syntax errors, like in the definition of Th and RF. Plus what's with the definition of col? You're defining it as a scalar minus a 1 by 2 array. That's not a column number like you said you want.
Yun Inn
Yun Inn 2012 年 8 月 22 日
I have eight thicknesses, Th=0,5,10,15,20,25,30,35,40. For each of the Th, I have matrix, ratioAB.
Then, I have Z that is a 2D matrix consisting of pixels with different thicknesses. For each of the pixel, I need to find the correct ratioAB depending on the thickness. The result, RF, should be a matrix of ratioAB.
Because the Th range from 0 to 40, when I have negative value or value more than 40 in the Z matrix, I get the error 'Subscript indices must either be real positive integers or logicals.'
How do I fix the code?

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採用された回答

Matt Fig
Matt Fig 2012 年 8 月 21 日

その他の回答 (1 件)

Azzi Abdelmalek
Azzi Abdelmalek 2012 年 8 月 21 日
編集済み: Azzi Abdelmalek 2012 年 8 月 21 日
Z(find(Z<0))=0

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