Non zero min 3D matrix

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VISHNUPRIYA M S
VISHNUPRIYA M S on 12 Mar 2019
Edited: madhan ravi on 12 Mar 2019
I have 3D matrix of A(k,l,t) where t is time. I want to find the non zero min of each row for every t.
for t =1:10
for k = 1:10
f(k,1,t) = max(A(k,:,t)(A(k,:,t)>0));
end
end
it shows error ()-indexing must appear last in an index expression.
I was trying to use f = min(A(A>0))

Accepted Answer

KSSV
KSSV on 12 Mar 2019
A = rand(10,10,10) ;
for t =1:10
for k = 1:10
D = A(k,:,t) ;
D = D(D>0) ;
f(k,1,t) = max(D);
end
end

More Answers (1)

Raghunandan V
Raghunandan V on 12 Mar 2019
You can remove all the for loops and make it more effecient
A= rand(10,10,10);
A(A==0) = inf;
min(A)
Here I am just replacing the 0 with inf and then finding the minimum. This code even works for matrix with negative integers. :)
  5 Comments
madhan ravi
madhan ravi on 12 Mar 2019
+1, Ah didn’t examine the code because OP mentioned row in the statement but in the code turns out to find for columns, good proposal then ;-) .In your code what will happen if a column is all zeros? Why do you say if that the second solution proposed by me will not work?, it was the continuation of your code after replacing zeros with infs. By the way when you said vice versa matrix has to be transposed that’s what permute() does , by the way min() function can be used to find the minimum along the specific dimension.If the dimension of A is dynamic then
permute(A,[2 1 3:ndims(A)])

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