read each line of 3 values of a ( n*3 matrix) as a x,y,z positions

2014 5 26
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更新时间：2014 5 26
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Hi there! Hi have this k =
1 1 2
2 1 2
2 2 2
n1 n2 n3
n4 n5 n6
n7 n8 n9
n11 n12 n13
(...) (...) (...)
And I want for each line to be read as a position xyz, .
For example:
1st line: x=1, y=1, z=2
Afterwards I want to create a new 2*2*2 matrix of zeros, where the xyz positions that I got above are written with a 1 on the new matrix.
ANy help? I need to solve these asap!
Regards

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André Sousa
André Sousa 2014-5-26
following the same question: This is my full code:
aa=rand(2,2,2)
aa(1,1,1)=5;
aa(2,1,1)=8;
aa(1,2,1)=3;
aa(2,2,1)=3;
aa(1,1,2)=7;
aa(2,1,2)=8; aa(1,2,2)=3;
aa(2,2,2)=3;
dimention_aa=size(aa); aareshaped=reshape(aa,[2,4])
ii=rand(3,1);
ii(1,1)=8;
ii(2,1)=3;
ii(3,1)=7;
for( l=(ii(1):ii(end)))
ind_reshaped=find(aareshaped==l)
[R1, C1, S1]=ind2sub(dimention_aa,find(aareshaped==l)
k=[R1, C1, S1]
x1=k(:,1)
y1=k(:,2)
z1=k(:,3)
newMatrix = zeros(dimention_aa)
for j = 1 : size(k, 1)
newMatrix( x1(j) , y1(j) ,z1(j)) = 1
end
% roi_mat(l,:)=x;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%% Basically the k matrix(that has lines correspondent to xyz positions) is genereated by
[R1, C1, S1]=ind2sub(dimention_aa,find(aareshaped==l)
k=[R1, C1, S1]
And l is a vector with diferent numbers. How do I do the same procedure to go through every element of the l? The code I wrote above, doesn't work quite well.. Regards

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Image Analyst
Image Analyst 2014-5-26

0 个投票

Try this:
k =[...
1 1 1
2 1 2
2 2 2]
x = k(:, 1);
y = k(:, 2);
z = k(:, 3);
newMatrix = zeros(2,2,2);
for j = 1 : size(k, 1)
% Note: y = row, x = column.
newMatrix(y(j), x(j) ,z(j)) = 1;
end
newMatrix % Print to command window.

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