How to replace zeros in a matrix by the elements of an array?

Hi everybody! I have a matrix (A) that contains zeros but it may contain also ones.

    R=1:8;     A=[1 0 0 0;
                  0 0 0 0; 
                  1 0 1 1]

How to replace all the ones by zeros and all the zeros by elements of an array (R) going column-wise so that the resulting matrix (B) would look like:

               B=[0 2 5 7;
                  1 3 6 8;
                  0 4 0 0]

[Thomas]

I think this might be homework.. can you show what you have done so far?

[Julius]

Yes, I've done this:

D=3;

N=4;

A=zeros(D,N);

equations=sum(A(:)==0);

R=1:equations;

for j=1:N

for i=1:D

if A(i,j)==1

A(i,j)=0;

else

ID(i,j)=?????;

end

end

end

B=A

....but I don't know how to continue

Answer 1

[Thomas]

You are on the right track, though it can be done in much simpler way but since you are a beginner this will suffice..

use

Your input matrix was zeros, just add a count for R values

Editing your code:

A=[1 0 0 0;
                  0 0 0 0; 
                  1 0 1 1]  
D=3;
N=4;
% A=zeros(D,N);
equations=sum(A(:)==0);
R=1:equations;
count=1;
for j=1:N
for i=1:D
if A(i,j)==1
A(i,j)=0;
else
ID(i,j)=R(count);
count=count+1;
end
end
end 

ID

[Julius]

''count'' was exactly the command I've been looking for. But I didn't find it so I thought it didn't exist. Thank you!

[Thomas]

count is a variable.. I'm incrementing it every iteration..

[Julius]

I see.....that's why I didn't find such command. Very clever!

Answer 2

[Image Analyst]

R=1:8;    
A=[1 0 0 0;
   0 0 0 0; 
   1 0 1 1]
linearIndices = A == 0
B = A; % Make a copy.
B(A==1) = 0
B(linearIndices) = R;

This is a vectorized way of doing it rather than your for loop method.

[Julius]

I just wanted to write down that the answer for your question is: B(A==0)=R.Meanwhile, you have done it instead of me. I just tried it and it worked. It wouldn't even cross my mind to do it that way. And it seems to be a faster code than the mine one. Thank you!