Matlab equality problem. Please help.

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Hi, this is my code.
theta(k) = 45
l = x.*cosd(theta(k))+y.*sind(theta(k));
l(1,1) == l(3,3);
x(1,1)*cosd(45)+ y(1,1)*sind(45) == x(3,3)*cosd(45)+ y(3,3)*sind(45);
x(1,1)+ y(1,1)==x(3,3)+ y(3,3);
However, matlab it saying
l(1,1) == l(3,3)
is false,
x(1,1)*cosd(45)+ y(1,1)*sind(45) == x(3,3)*cosd(45)+ y(3,3)*sind(45)
is false, and
x(1,1)+ y(1,1)==x(3,3)+ y(3,3)
is true. I am wondering why because I believe l(1,1) == l(3,3) is true.
  2 个评论
Rik
Rik 2018-10-3
Your code is not self-contained (i.e. we cannot run it independently of the code you have before this), and I don't understand what you mean with your code/notation.
A wild guess would be that you are checking equality on a non-integer data type, which means you might be encountering float rounding errors. If you run the code below, you can clearly see that 1.0000 does not equal 1.0000
a=1+eps*2;%very close to 1
b=sqrt(a);b=b^2;%do an operation and reverse it
clc,isequal(a,b)
disp(a)
disp(b)
Jenny Chen
Jenny Chen 2018-10-3
编辑:Stephen23 2018-10-4
Hello, sorry, this should be enough to run.
theta = [45];
[Pr,Pc] = size(P);
nt = length(theta);
x = zeros(Pr,Pc);
y = zeros(Pr,Pc);
for i = 1:Pr
for j = 1:Pc
x(i,j) = j - 1;
y(i,j) = Pr - i;
end
end
for k = 1:nt
l = x.*cosd(theta(k))+y.*sind(theta(k));
l(1,1) == l(3,3)
x(1,1)*cosd(45)+ y(1,1)*sind(45) == x(3,3)*cosd(45)+ y(3,3)*sind(45)
x(1,1)+ y(1,1)==x(3,3)+ y(3,3)
end

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采纳的回答

Stephen23
Stephen23 2018-10-4
编辑:Stephen23 2018-10-4
You have to learn about floating point numbers, and why you should not test them for equality like that. Always compare the absolute difference against a tolerance, like this:
abs(A-B)<tol
Start by reading these:
This is worth reading as well:
To see the "real" value download James Tursa's FEX submission:
"I am wondering why because I believe l(1,1) == l(3,3) is true"
If you use num2strExact on both of those values, you will find that they are not the same, and MATLAB's output is correct. That is why you should compare the absolute difference against a tolerance.

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Image Analyst
Image Analyst 2018-10-3

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