Non zero min 3D matrix

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))

Answer 1

[KSSV]

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

Answer 2

[Raghunandan V]

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. :)

[madhan ravi]

I suspect it should be

min(permute(A,[2 1 3])) 
% or simply
min(A,[],2)

[Raghunandan V]

The first solution wont work if the size of matrix is dynamic and

both of your solutions will not solve if there are multiple zeros in the same row. Please check.

[madhan ravi]

I am not certain with yours too because your solution gives min of each column instead of each row in each page.

[Raghunandan V]

Please check the question. the formula written there is max(A(k,:,t). this actually means they are trying to find the minimum of a column as 2nd argument is :. If you need the minimum or rows you can transpose the matrix and then use minimum.

:)

[madhan ravi]

+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)])