How to save multiple output?

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Ammy
Ammy 2021 年 9 月 15 日
コメント済み: Ammy 2021 年 9 月 15 日
a =randperm(18);
b=reshape(a,9,2);
vals=b;
[m,n] =size(vals) ;
for i = 1:m
[A,x,y]=myfunction(I,vals(i,1), vals(i,2));
end
[A,x,y] is the output of myfunction.
How to save A, x and y where A is a square matrix, x and y both are single vaues.

採用された回答

Mathieu NOE
Mathieu NOE 2021 年 9 月 15 日
hello
I would recommend to use structures (indexed) like in this example
a =randperm(18);
b=reshape(a,9,2);
vals=b;
[m,n] =size(vals) ;
I = rand(3,5);
for i = 1:m
[A,x,y]=myfunction(I,vals(i,1), vals(i,2));
output{i}.A = A;
output{i}.x = x;
output{i}.y = y;
end
% [A,x,y] is the output of myfunction.
% How to save A, x and y where A is a square matrix, x and y both are single vaues.
%%%%%%%%%%%%%%%%%%
function [A,x,y] = myfunction(I,a,b)
A = I ;
x = a;
y = b;
end
so , for example , to access the data from the first iteration :
output{1}
ans =
struct with fields:
A: [3×5 double]
x: 14
y: 7
>> output{1}.A
ans =
0.6948 0.0344 0.7655 0.4898 0.7094
0.3171 0.4387 0.7952 0.4456 0.7547
0.9502 0.3816 0.1869 0.6463 0.2760
>> output{1}.x
ans = 14
  3 件のコメント
Mathieu NOE
Mathieu NOE 2021 年 9 月 15 日
My pleasure
I have to say that there is a better coding : rather than nesting lots of scalar structures inside a cell array, it would be simpler to just use one structure array
a =randperm(18);
b=reshape(a,9,2);
vals=b;
[m,n] =size(vals) ;
I = rand(3,5);
output = struct([]); % ensure that "output" is a structure
for ii = 1:m
[A,x,y]=myfunction(I,vals(ii,1), vals(ii,2));
[output(ii).A, output(ii).x, output(ii).y] = myfunction(I, vals(ii,1), vals(ii,2));
end
%%%%%%%%%%%%%%%%%%
function [A,x,y] = myfunction(I,a,b)
% [A,x,y] is the output of myfunction.
A = I ;
x = a;
y = b;
end
Ammy
Ammy 2021 年 9 月 15 日
well said, thank you very much.

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その他の回答 (1 件)

Steven Lord
Steven Lord 2021 年 9 月 15 日
In your code:
%{
for i = 1:m
[A,x,y]=myfunction(I,vals(i,1), vals(i,2));
end
%}
is A always going to be the same size? If so consider making a 3-dimensional array:
A = magic(3);
B = zeros([size(A), 5]);
for k = 1:5
B(:, :, k) = A.^k;
end
B(:, :, 3) % The cube of each element in A
ans = 3×3
512 1 216 27 125 343 64 729 8
Similarly for your scalar x and y values, store them in a vector or a 3-dimensional equivalent.

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