How to extract diagonal elements of multidimensional array ?

2023 3 23
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更新时间：2023 3 24
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If I have a m-order n-dimensional tensor. How should I extract the diagonal elements ?
For example
% Generate a 3-order 4-dimensional tensor
rng('default')
A = rand(4,4,4)
A =
A(:,:,1) = 0.8147 0.6324 0.9575 0.9572 0.9058 0.0975 0.9649 0.4854 0.1270 0.2785 0.1576 0.8003 0.9134 0.5469 0.9706 0.1419 A(:,:,2) = 0.4218 0.6557 0.6787 0.6555 0.9157 0.0357 0.7577 0.1712 0.7922 0.8491 0.7431 0.7060 0.9595 0.9340 0.3922 0.0318 A(:,:,3) = 0.2769 0.6948 0.4387 0.1869 0.0462 0.3171 0.3816 0.4898 0.0971 0.9502 0.7655 0.4456 0.8235 0.0344 0.7952 0.6463 A(:,:,4) = 0.7094 0.6551 0.9597 0.7513 0.7547 0.1626 0.3404 0.2551 0.2760 0.1190 0.5853 0.5060 0.6797 0.4984 0.2238 0.6991
The diagonal elements are A(1,1,1) = 0.8147, A(2,2,2) = 0.0357, A(3,3,3) = 0.7655 and A(4,4,4) = 0.6991.
I was hoping to have a tensor_diag function that takes a tensor A as an input parameter and returns a vector consisting of its diagonal elements.

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Mattia Marsetti
Mattia Marsetti 2023-3-23
Ciao shuang Yang
I don't know of any native Matlab function for this purpose. You can try to use the following custom function to solve your problem:
function out=get_tensor(v)
size_v=size(v);
if sum(size_v == size_v(1))<numel(size_v)
error('the input vector is not a sqare matrix');
end
N = size_v(1);
out=zeros(N,1);
for i=1:N
str='v(';
for s=1:N
str=[str 'i,'];
end
str(end:end+1)=');';
out(i)=eval(str);
end
end
Stephen23
Stephen23 2023-3-23
编辑：Stephen23 2023-3-23
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@Mattia Marsetti: your code throws an error on the example array:
rng('default')
A = rand(4,4,4)
A =
A(:,:,1) = 0.8147 0.6324 0.9575 0.9572 0.9058 0.0975 0.9649 0.4854 0.1270 0.2785 0.1576 0.8003 0.9134 0.5469 0.9706 0.1419 A(:,:,2) = 0.4218 0.6557 0.6787 0.6555 0.9157 0.0357 0.7577 0.1712 0.7922 0.8491 0.7431 0.7060 0.9595 0.9340 0.3922 0.0318 A(:,:,3) = 0.2769 0.6948 0.4387 0.1869 0.0462 0.3171 0.3816 0.4898 0.0971 0.9502 0.7655 0.4456 0.8235 0.0344 0.7952 0.6463 A(:,:,4) = 0.7094 0.6551 0.9597 0.7513 0.7547 0.1626 0.3404 0.2551 0.2760 0.1190 0.5853 0.5060 0.6797 0.4984 0.2238 0.6991
get_tensor(A)
Index in position 4 exceeds array bounds. Index must not exceed 1.

Error in solution>get_tensor (line 19)
out(i)=eval(str);
function out=get_tensor(v)
size_v=size(v);
if sum(size_v == size_v(1))<numel(size_v)
error('the input vector is not a sqare matrix');
end
N = size_v(1);
out=zeros(N,1);
for i=1:N
str='v(';
for s=1:N
str=[str 'i,'];
end
str(end:end+1)=');';
out(i)=eval(str);
end
end
Note that you could easily replace the evil EVAL with a cell array and a comma-separated list.
Mattia Marsetti
Mattia Marsetti 2023-3-24
Thanks for the hint Stephen23, I'll try that
Regards

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the cyclist
the cyclist 2023-3-23
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rng('default')
N = 4;
A = rand(N,N,N);
A(1:N^2+N+1:end)
ans = 1×4
0.8147 0.0357 0.7655 0.6991

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Bruno Luong
Bruno Luong 2023-3-23
编辑：Bruno Luong 2023-3-23
Note that cyclist's indexing (1:N^2+N+1:end) is
sub2ind([N,N,N],1:N,1:N,1:N)
the cyclist
the cyclist 2023-3-23
Nice -- and more intuitive about what is going on
shuang Yang
shuang Yang 2023-3-23
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Thank you very much for your answer, although this is not the correct answer, but I should have guessed how to write it, the correct version of it should be like this
rng('default')
%% order = 2
N = 3;
A = rand(N,N)
A = 3×3
0.8147 0.9134 0.2785 0.9058 0.6324 0.5469 0.1270 0.0975 0.9575
A(1:N+1:end)
ans = 1×3
0.8147 0.6324 0.9575
%% order = 3
N = 4;
A = rand(N,N,N) %
A =
A(:,:,1) = 0.9649 0.4854 0.9157 0.0357 0.1576 0.8003 0.7922 0.8491 0.9706 0.1419 0.9595 0.9340 0.9572 0.4218 0.6557 0.6787 A(:,:,2) = 0.7577 0.1712 0.0462 0.3171 0.7431 0.7060 0.0971 0.9502 0.3922 0.0318 0.8235 0.0344 0.6555 0.2769 0.6948 0.4387 A(:,:,3) = 0.3816 0.4898 0.7547 0.1626 0.7655 0.4456 0.2760 0.1190 0.7952 0.6463 0.6797 0.4984 0.1869 0.7094 0.6551 0.9597 A(:,:,4) = 0.3404 0.2551 0.9593 0.2575 0.5853 0.5060 0.5472 0.8407 0.2238 0.6991 0.1386 0.2543 0.7513 0.8909 0.1493 0.8143
A(1:N^2+N+1:end)
ans = 1×4
0.9649 0.7060 0.6797 0.8143
%% order = 4
N = 4;
A = rand(N,N,N,N);
A(1:N^3+N^2+N+1:end)
ans = 1×4
0.2435 0.3692 0.0292 0.6569
A(1)
ans = 0.2435
A(2,2,2,2)
ans = 0.3692
A(3,3,3,3)
ans = 0.0292
%% order = 5
N = 7;
A = rand(N,N,N,N,N);
A(1:N^4+N^3+N^2+N+1:end)
ans = 1×7
0.6280 0.7118 0.6596 0.5435 0.6933 0.9827 0.9083
the cyclist
the cyclist 2023-3-23
Ah, I read your question too quickly, and didn't make my solution general enough. Glad you found it.
Stephen23
Stephen23 2023-3-23
@shuang Yang: you could generalize that:
A(1:sum(N.^(0:ndims(A)-1):end)

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Bruno Luong
Bruno Luong 2023-3-23
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N = 7;
A = rand(N,N,N,N,N);
p=ndims(A);
N=length(A);
% Method 1: generalization of cyclist's answer
step = polyval(ones(1,p),N);
idx = 1:step:N^p;
A(idx)
ans = 1×7
0.4728 0.2804 0.8097 0.9442 0.5680 0.1666 0.9129
% Method 2
c = repmat({1:N}, [1,p]);
idx = sub2ind(size(A), c{:});
A(idx)
ans = 1×7
0.4728 0.2804 0.8097 0.9442 0.5680 0.1666 0.9129

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