Logical with complex numbers in a vector
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I have a complex number in a vector,
e =
-0.0000
3.0000
9.0000
>> logical(e) %checking if the the first entry is actually 0. Results show it is not
ans =
3×1 logical array
1
1
1
>> t = sqrt(e) % sqauring the entries shows all the three entries are complex numbers
t =
0.0000 + 0.0000i
1.7321 + 0.0000i
3.0000 + 0.0000i
L = logical(t) % In Matlab documentation it not possible to use logical for complex numbers
Error using logical
Complex values cannot be converted to logicals.
>> t(1,1)==0 % checking if the first entry is actually 0?
ans =
logical
0
>> t(1,1)<0 % checking if the first entry is actually less than 0?
ans =
logical
0
>> t(1,1)>0 % checking if the first entry is actually greater than 0?
ans =
logical
0
% QUESTION: is there a way to compare t(1,1) with 0 to know whether is it 0 or not?
% Alternatively: What is the genral way of using logicals for complex numbers?
采纳的回答
e = [-1e-9;3;9;0]
t = sqrt(e);
[real(t), imag(t)]
t == 0
format long g
e
[real(t), imag(t)]
You are being mislead by your default output format. If you switch output formats you will be able to see quickly that your e(1) is a small negative value.
You can see that t == 0 is enough to compare both parts of a complex number to 0, and correspondingly,
Lt = t ~= 0
is the equivalent of logical()
更多回答（2 个）
Image Analyst
2021-2-27
Perhaps this:
complexVector =[
0.0000 + 0.1000i
1.7321 + 0.0010i
3.0000 + 0.0000i]
% Define a tolerance:
tolerance = 0.01
% Find out which elements have an imaginary magnitude less than tolerance:
belowTolerance = imag(complexVector) < tolerance
% For those elements with an imaginary value below the tolerance,
% turn them into a real number with no imaginary component.
complexVector(belowTolerance) = real(complexVector(belowTolerance))
You get:
complexVector =
0 + 0.1i
1.7321 + 0.001i
3 + 0i
tolerance =
0.01
belowTolerance =
3×1 logical array
0
1
1
complexVector =
0 + 0.1i
1.7321 + 0i
3 + 0i
Note the second and third elements are real, not complex, numbers because the imaginary part was stripped off.
David Hill
2021-2-23
%depends on what you are trying to compare
real(t)==0;
imag(t)==0;
norm(t)==0;
7 个评论
"If the real part is 0 and the imaginary part is not 0, can we then conclude that the complex number is 0?"
The answer to that question is clearly "no", because the imaginary part is non-zero.
You seem to be under the mistaken impression that it is meaningful to compare complex numbers with non-complex numbers. Until you define what that means, the answer will be "no".
Image Analyst
2021-2-26
Let's take a step back and ask why you think you want to do that. What is your use case?
Let's say you had a general complex number a + bi. Now, how were those numbers generated in the first place (what do they mean), and under what cases of a and b would you want a true logical value and under what cases of a and b would you want a false logical variable?
Walter Roberson
2021-2-27
Your E, U, V are not complex.
You can test value ~= 0 and that will work for complex values as well.
A = ([-1 1 0; -1 0 1; 0 -1 1; -2 1 1]);
e=eig(A'*A)
f = sqrt(e);
format short
[real(e), imag(e)]
[real(f), imag(f)]
f ~= 0
format long g
[real(e), imag(e)]
[real(f), imag(f)]
f ~= 0
You can see from this that the first entry in e is truly not 0... it just looks that way when you have "format short" in effect.
You can also see that comparing ~= 0 is successful in detecting that the imaginary component is not 0, which is what you want.
As = sym(A)
es = eig(As'*As)
Working with symbolic values shows us that the ideal value for the first eigenvalue is exact 0; the -1e-16 is due to numeric round-off.
Hmm!
2021-2-27
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