Is there a way to access array elements returned by a function directly

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Leos Pohl
Leos Pohl el 14 de Jun. de 2021
Comentada: Leos Pohl el 17 de Jun. de 2021
Assume this code:
a=magic(4);
then I can do:
a(1,2)
to access the elements of the matrix a. Is there a way to do it directly without associating the variable a with the result of the function magic, that is, something like:
magic(4)(1,2)
?

Respuestas (2)

Steven Lord
Steven Lord el 14 de Jun. de 2021
What you've written is not valid MATLAB syntax. You could do something close to what you want using a helper function.
valueAt = @(A, varargin) A(varargin{:});
valueAt(magic(4), 1, 2)
ans = 2
M = magic(4) % so you can see what element (1, 2) is
M = 4×4
16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1
  4 comentarios
Steven Lord
Steven Lord el 14 de Jun. de 2021
It's easy to say how the syntax you described should work in the simple cases that interest you.
It's not nearly so easy to say how it should work in the general case for which we would have to design that feature.
As an example, the eig function has two outputs. The first is a matrix of eigenvectors and the second the corresponding eigenvalues. Suppose I wanted the third eigenvector and eigenvalue? How would you do that with your syntax?
format longg
A = magic(5);
[V, D] = eig(A)
V = 5×5
-0.447213595499958 0.0976400709329798 -0.632976910516004 0.67800972395606 -0.261860649935008 -0.447213595499958 0.352538557500108 0.589480807113999 0.322279530272056 -0.173165093661346 -0.447213595499958 0.550110625795028 -0.391529640194349 -0.550110625795028 0.391529640194349 -0.447213595499958 -0.322279530272056 0.173165093661345 -0.352538557500107 -0.589480807113999 -0.447213595499958 -0.678009723956059 0.261860649935008 -0.0976400709329798 0.632976910516003
D = 5×5
65.0000000000001 0 0 0 0 0 -21.2767654714738 0 0 0 0 0 -13.1262809307092 0 0 0 0 0 21.2767654714738 0 0 0 0 0 13.1262809307092
check = A*V(:, 3)-V(:, 3)*D(3, 3)
check = 5×1
-1.77635683940025e-14 -8.88178419700125e-16 -7.105427357601e-15 -5.32907051820075e-15 -2.66453525910038e-15
Now what would you do to get the third eigenvector and eigenvalue using this call to eig?
[V, D] = eig(A, 'vector')
V = 5×5
-0.447213595499958 0.0976400709329798 -0.632976910516004 0.67800972395606 -0.261860649935008 -0.447213595499958 0.352538557500108 0.589480807113999 0.322279530272056 -0.173165093661346 -0.447213595499958 0.550110625795028 -0.391529640194349 -0.550110625795028 0.391529640194349 -0.447213595499958 -0.322279530272056 0.173165093661345 -0.352538557500107 -0.589480807113999 -0.447213595499958 -0.678009723956059 0.261860649935008 -0.0976400709329798 0.632976910516003
D = 5×1
65.0000000000001 -21.2767654714738 -13.1262809307092 21.2767654714738 13.1262809307092
Leos Pohl
Leos Pohl el 17 de Jun. de 2021
Like this (the first says all from the first dimensions and the second says only third element in the second dimension):
eig(A, 'vector')(:,3)
Look, i am not criticizing but rather trying to find the way these things are done properly and, most importnatly, efficiently in Matlab especially when it comes to memory. I come from Mathematica and such thing is simple there but that is a different concept.

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Matt J
Matt J el 15 de Jun. de 2021
In theory, it is possible to make a function directly indexable using OOP trickery,
>> Magic=IndexableFunction(@magic);
>> Magic{4}(1,2)
ans =
2
It's not worth it...

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