How can I create a set of N diagnal matrices range from diag(1, 1, ... , 1) to diag(-1, -1, ... , -1) for testing purposes?

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Zac Minson
Zac Minson 2021-9-20
编辑： Matt J 2021-9-21
I am needing to create 16 4x4 diagnal matrices consisting of all of the combinations of "1" and "-1" and have them be able to be called for use in a loop. Besides creating each individual matrix, how can I go about accomplishing this?
for instance:
Z1 = diag(1,1,1,1);
.
Z8 = diag(1,-1,-1,1);
.
Z16 = diag(-1,-1,-1,-1);
Thank you for any guidance.
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Zac Minson
Zac Minson 2021-9-20
I ended up doing this, but was wondering if there is a more general method in the case of 2^N matrices.
Stephen23
Stephen23 2021-9-21
@Zac Minson: using a cell array is a very general method, which works for any number of matrices.

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Matt J
Matt J 2021-9-20
编辑：Matt J 2021-9-21
Do you really need all Z{i} at the same time?
[C{1:N}]=ndgrid([-1,1]);
C=reshape( cat(N+1,C{:}) ,[],N);
Z=zeros(N^2,2^N); %EDITED - removed loops
Z(1:N+1:N^2,:)=C;
Z=reshape(Z, N,N,[])
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Matt J
Matt J 2021-9-20
Ran in:
Here's a loop-free version:
N = 4; % Set this value
[C{1:N}]=ndgrid([-1,1]);
C=reshape( cat(N+1,C{:}) ,[],N).';
Z=zeros(N^2,2^N);
Z(1:N+1:N^2,:)=C;
Z=reshape(Z, N,N,[])
Z =
Z(:,:,1) = -1 0 0 0 0 -1 0 0 0 0 -1 0 0 0 0 -1 Z(:,:,2) = 1 0 0 0 0 -1 0 0 0 0 -1 0 0 0 0 -1 Z(:,:,3) = -1 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 -1 Z(:,:,4) = 1 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 -1 Z(:,:,5) = -1 0 0 0 0 -1 0 0 0 0 1 0 0 0 0 -1 Z(:,:,6) = 1 0 0 0 0 -1 0 0 0 0 1 0 0 0 0 -1 Z(:,:,7) = -1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 -1 Z(:,:,8) = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 -1 Z(:,:,9) = -1 0 0 0 0 -1 0 0 0 0 -1 0 0 0 0 1 Z(:,:,10) = 1 0 0 0 0 -1 0 0 0 0 -1 0 0 0 0 1 Z(:,:,11) = -1 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 1 Z(:,:,12) = 1 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 1 Z(:,:,13) = -1 0 0 0 0 -1 0 0 0 0 1 0 0 0 0 1 Z(:,:,14) = 1 0 0 0 0 -1 0 0 0 0 1 0 0 0 0 1 Z(:,:,15) = -1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 Z(:,:,16) = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
Zac Minson
Zac Minson 2021-9-21
Thank you so much. I was really wracking my brain over how to generalize this problem. I appreciate your time.

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更多回答（2 个）

Fangjun Jiang
Fangjun Jiang 2021-9-20
a=(dec2bin(15:-1:0)-48)*2-1;
z=zeros(4,4,16);
for k=1:16
z(:,:,k)=diag(a(k,:));
end
  1 个评论
the cyclist
the cyclist 2021-9-20
Here is a generalization of this solution:
N = 4; % Set this value
a=(dec2bin((2^N-1):-1:0)-48)*2-1;
z=zeros(N,N,2^N);
for k=1:2^N
z(:,:,k)=diag(a(k,:));
end

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Jan
Jan 2021-9-20
编辑：Jan 2021-9-21
z = repmat(eye(4), 1, 1, 16);
z(z==1) = 1 - rem(floor((0:15) ./ [1; 2; 4; 8]), 2) * 2; % Inlined DEC2BIN
Generalized:
n = 3;
z = repmat(eye(n), 1, 1, 2^n);
z(z==1) = 1 - rem(floor((0:2^n-1) ./ pow2(0:n-1).'), 2) * 2;
Now z(:,:,i) is the i.th matrix. This is more efficient than storing them in a cell. But if you want this:
zC = num2cell(z, [1,2]);

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