loop matrix to new matrix
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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]
采纳的回答
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
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
And if you really need zeros and ones of class 'double' rather than falses and trues of class 'logical':
A_new = double(A > 30)
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