loop matrix to new matrix

Hello,

I am given a matrix A, I want to make a conditional loop if a number is over 30 then place 1 in a new matrix, if a value is less than 30 place 0. How do this.

I would appreciate help fixing my code.

Thank

A=[60 56 44 44 22 18 22 18; 60 56 44 40 8 8 8 8;56 44 12 12 8 4 8 12];
switch C
    case C>30
        C(x,y)=1
    case C<=30
        C(x,y)=0
end

I need to get a new binary matrix, like this answer

A_new=[1 1 1 1 0 0 0 0;1 1 1 1 0 0 0 0;1 1 0 0 0 0 0 0]

Answer 1

[Voss]

A=[60 56 44 44 22 18 22 18; 60 56 44 40 8 8 8 8;56 44 12 12 8 4 8 12];

Here's one way to write the loop you want:

% initialize A_new to a matrix of zeros the same size as A
A_new = zeros(size(A));

for ii = 1:numel(A)
    % if the ii-th element of A is greater than 30, then set the 
    % ii-th element of A_new to 1.
    % (otherwise A_new(ii) is already 0 so there's nothing to do)
    if A(ii) > 30
        A_new(ii) = 1;
    end
end

A_new

A_new = 3×8

1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 0 0 0 0 0 0

However, you don't need a loop at all; you can merely perform the comparison on the entire matrix A at once:

A_new = A > 30

A_new = 3×8 logical array
   1   1   1   1   0   0   0   0
   1   1   1   1   0   0   0   0
   1   1   0   0   0   0   0   0

And if you really need zeros and ones of class 'double' rather than falses and trues of class 'logical':

A_new = double(A > 30)

A_new = 3×8

1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 0 0 0 0 0 0

[Emilia]

Thank you!

[Voss]

You're welcome!