Extracting only real numbers from a vector containing both real and complex values

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Hi, I have a matrix that contains both real and complex elements in it. How can I transfer all the real elements (NOT the real parts of all elements) to another matrix?
For example:
My orginal matrix is A:
A = [ -0.4406 - 1.5696i , -0.4406 + 1.5696i, 1.8812 + 0.0000i]
Here we see that A has 3 elements, the first two of which are complex while the third element is real. How do I write code that extracts only the real element ie, the third element (1.8812)?
I do NOT want to extract the real parts of all three variables.

Respuesta aceptada

Daniele Mascali
Daniele Mascali el 19 de Mzo. de 2021
Try this:
real(A(imag(A) == 0))
  2 comentarios
Rik
Rik el 19 de Mzo. de 2021
If you run into any issues with float rounding errors:
A = [ -0.4406 - 1.5696i , -0.4406 + 1.5696i, 1.8812 + 0.0000i];
A(4)= 1 + 1i*(sqrt(2)^2-2); % imaginary part should be 0 but isn't quite
B1=real(A(imag(A) == 0));
B2=real(A(abs(imag(A))<(10*eps)));
disp(A),disp(B1),disp(B2)
-0.4406 - 1.5696i -0.4406 + 1.5696i 1.8812 + 0.0000i 1.0000 + 0.0000i 1.8812 1.8812 1.0000
Mathieu NOE
Mathieu NOE el 19 de Mzo. de 2021
that's why I suggested to make a test with a given tolerance and not exact zero match with risks of wrong answer due to rounding errors

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Más respuestas (2)

Mathieu NOE
Mathieu NOE el 19 de Mzo. de 2021
hello
see code below :
A = [ -0.4406 - 1.5696i , -0.4406 + 1.5696i, 1.8812 + 0.0000i];
tol = eps;
ind = find(abs(imag(A))<tol);
B = A(ind);
  1 comentario
Stephen23
Stephen23 el 19 de Mzo. de 2021
Editada: Stephen23 el 19 de Mzo. de 2021
+1 A slightly larger tolerance might also be more robust, e.g. 1e-10
find is not required, logical indexing is simpler and more efficient:
tol = 1e-10;
idx = abs(imag(A))<tol;
B = A(idx)
B = 1.8812

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Ahmadreza Torabi
Ahmadreza Torabi el 15 de Dic. de 2021
Editada: Ahmadreza Torabi el 15 de Dic. de 2021
Hello
You can use something like this:
A = [1 2 7+i 3 8+i 4 9+i 5 6];
m = 1;
n = length(A);
for k = 1:n
if isreal(A(k))
B(m) = A(k);
m = m+1;
end
end
B
B = 1×6
1 2 3 4 5 6
Then B is just include real numbers of A.
  1 comentario
Rik
Rik el 15 de Dic. de 2021
This will dynamically grow the array. It is better to create a logical vector of the same size as A where you mark wether it should be removed. Then you only need to modify the B once: when you create it.

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