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

98 Ansichten (letzte 30 Tage)
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.

Akzeptierte Antwort

Daniele Mascali
Daniele Mascali am 19 Mär. 2021
Try this:
real(A(imag(A) == 0))
  2 Kommentare
Rik
Rik am 19 Mär. 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 am 19 Mär. 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

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (2)

Mathieu NOE
Mathieu NOE am 19 Mär. 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 Kommentar
Stephen23
Stephen23 am 19 Mär. 2021
Bearbeitet: Stephen23 am 19 Mär. 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

Melden Sie sich an, um zu kommentieren.


Ahmadreza Torabi
Ahmadreza Torabi am 15 Dez. 2021
Bearbeitet: Ahmadreza Torabi am 15 Dez. 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 Kommentar
Rik
Rik am 15 Dez. 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.

Melden Sie sich an, um zu kommentieren.

Produkte


Version

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by