loop matrix to new matrix

Question

Emilia 2022-7-24

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1766540-loop-matrix-to-new-matrix

评论： Voss 2022-7-24

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]

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Answer 1

Voss 2022-7-24

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1766540-loop-matrix-to-new-matrix#answer_1013925

在 MATLAB Online 中打开

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