How to store the non zero value from each column of a matrix as a column vector (but if there is only zeros on the column store zero) in each iteration?

How to store the non zero value from each column of an output matrix which its size(5*3) as a clounm vector with a size(3*1) (but if there is only zeros on the column store zero) in each iteration? Then combine all the 3*1 vector in one matrix call it Outputs.

For example : This output from an iteration of the following program

y =

0.38 0 0

0 0 0

0 0.58 0.71

0 0 0

0 0 0

in each iteration there are different loctions for the non zerovalue and in each column there is just only one non zerovalue.

Can anyone help me in this problem?

This is the program:

mf1=[fismf("trimf" , [0 0 2], 'Name', "ZE") ; 
     fismf("trimf" , [2 6 10], 'Name', "PS") ;
     fismf("trimf" , [10 14 18], 'Name', "M") ;
     fismf("trimf" , [18 22 26], 'Name', "PL") ;
     fismf("trapmf" ,  [26 27 60 60], 'Name', "PVL") ]
 
 for i = 1:127
     y = evalmf(mf1,RNF(:,i));
     
 end

[M]

Hi @Stephen , Can you help in this problem please?

[Arif Hoq]

do you want this result ?

output =

    0.3800
    0.5800
    0.7100

[M]

@Arif Hoq yes , but from each iteration there are different loctions for the non zerovalue in columns of the matrix y

[M]

@Arif Hoq and if there is only zeros in the cloumn, I want to store the zero

[M]

Is that possible to achieve?

Answer 1

[Stephen23]

The simplest solution is probably to use MAX:

M = [0.38,0,0;0,0,0;0,0.58,0.71;0,0,0;0,0,0]

M = 5×3

0.3800 0 0 0 0 0 0 0.5800 0.7100 0 0 0 0 0 0

V = max(M,[],1).'

V = 3×1

0.3800 0.5800 0.7100

An example where one column contains only zeros:

M = [0.38,0,0;0,0,0;0,0,0.71;0,0,0;0,0,0]

M = 5×3

0.3800 0 0 0 0 0 0 0 0.7100 0 0 0 0 0 0

V = max(M,[],1).'

V = 3×1

0.3800 0 0.7100

If the values can be negative:

M = [0.38,0,0;0,0,0;0,-0.58,0.71;0,0,0;0,0,0]

M = 5×3

0.3800 0 0 0 0 0 0 -0.5800 0.7100 0 0 0 0 0 0

V = (max(M,[],1)+min(M,[],1)).'

V = 3×1

0.3800 -0.5800 0.7100

" I want to combine all the column vectors in matrix at the end of the program"

Store them in a cell array, then concatenate them together after the loop:

N = number of iterations

C = cell(1,N)

for k = 1:N

V = the output of your calculation

C{k} = V;

end

M = horzcat(C{:}) % or VERTCAT

[M]

@Stephen , thank you so much.

[Stephen23]

"but how can I combine your solution in for loop and incroporate it with my code? and at the end of the iterations store all the column vectors in one matrix?"

Have a look at the end of my answer. I showed you how.

Answer 2

[Arif Hoq]

if there is only 1 zero in a column

A=[0.38 0 0
    0 0 5
    0 0.58 0.71
    0 0 5
    0 0 5];
idx=A==0;
for n=1:size(A,2)
   if  sum(idx(:,n))>1 % if there is more than 1 zero
       Output=A(A>0);
   elseif sum(idx(:,n))==1 % if there is only 1 zero
       Output=[A(A>0);0]
   end
end

Or if more than 1 zero in the column

A=[0.38 0 0
    0 0 0
    0 0.58 0.71
    0 0 0
    0 0 0];
idx=A==0;
for n=1:size(A,2)
   if  sum(idx(:,n))>1 % if there is more than 1 zero
       Output=A(A>0)
   else sum(idx(:,n))==1 % if there is only 1 zero
       Output=[A(A>0);0]
   end
end

[M]

@Arif Hoq , The output from y is diffrenet in each iteration. The code that you wrote is just for one iteration

[M]

Also, I want to combine all the column vectors in matrix at the end of the program

[M]

As well as your program doesnt store zero if the there is only zero the column of the matix.

I mean if the output from an iteration as the following:

y =

0.38 0 0

0 0 0

0 0 0.71

0 0 0

0 0 0

The output should be as the following:

output =

0.3800

0

0.7100