When I go and define 'theta' outside of the if statment, Matlab just bypasses the if statement entirely and only uses the defined value.
"theta" is defined inside of the statement, but when I run my code it gives me an error saying that " 'theta' is not a recognized function or variable'. Any solutions?
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x = [2 2 0 -3 -2 -1 0 0 2];
y = [0 1 3 1 0 -2 0 -2 2];
r = (sqrt((x.^2)+(y.^2)));
if x > 0
theta = atan(y./x);
elseif x < 0
theta = atan(y./x);
if y > 0
theta = atan(y./x) + (pi);
elseif y < 0
theta = atan(y./x) - (pi);
elseif y == 0
theta = pi;
end
elseif x == 0
if y > 0
theta = (pi./2);
elseif y < 0
theta = (-pi./2);
elseif y == 0
theta = 0;
end
end
theta = rad2deg(theta); % Changing theta from radians to degrees
w = [x' y' r' theta']; % A matrix containing all of the values
2 Kommentare
Stephen23
am 28 Jan. 2023
""theta" is defined inside of the statement"
Nope. As the IF documentation states, "An expression is true when its result is nonempty and contains only nonzero elements". Lets check which of your logical expressions are considered true:
x = [2,2,0,-3,-2,-1,0,0,2];
y = [0,1,3,1,0,-2,0,-2,2];
all(x>0)
all(x<0)
all(x==0)
all(y>0)
all(y<0)
all(y==0)
So none of them are true, none of your IF/ELSEIF will run, and THETA is never defined inside them.
In any case, you should be using indexing for this task, not IF/ELSEIF. Or ATAN2.
Akzeptierte Antwort
cdawg
am 28 Jan. 2023
x is a vector so when you say "if x > 0" it is returning a logic vector of the locations where that is the case. You need to loop through:
X = [2 2 0 -3 -2 -1 0 0 2];
Y = [0 1 3 1 0 -2 0 -2 2];
R = (sqrt((X.^2)+(Y.^2)));
for ii = 1:length(X)
x = X(ii);
y = Y(ii);
r = R(ii);
if x > 0
theta = atan(y./x);
elseif x < 0
theta = atan(y./x);
if y > 0
theta = atan(y./x) + (pi);
elseif y < 0
theta = atan(y./x) - (pi);
elseif y == 0
theta = pi;
end
elseif x == 0
if y > 0
theta = (pi./2);
elseif y < 0
theta = (-pi./2);
elseif y == 0
theta = 0;
end
end
THETA(ii) = theta;
end
THETA = rad2deg(THETA); % Changing theta from radians to degrees
w = [X' Y' R' THETA']; % A matrix containing all of the values
3 Kommentare
Stephen23
am 28 Jan. 2023
Bearbeitet: Stephen23
am 28 Jan. 2023
"You need to loop through"
No, you don't need a loop. This is easy to solve using exactly one command and no loop. Lets try:
X = [2,2,0,-3,-2,-1,0,0,2];
Y = [0,1,3,1,0,-2,0,-2,2];
R = (sqrt((X.^2)+(Y.^2)));
for ii = 1:length(X)
x = X(ii);
y = Y(ii);
r = R(ii);
if x > 0
theta = atan(y./x);
elseif x < 0
theta = atan(y./x);
if y > 0
theta = atan(y./x) + (pi);
elseif y < 0
theta = atan(y./x) - (pi);
elseif y == 0
theta = pi;
end
elseif x == 0
if y > 0
theta = (pi./2);
elseif y < 0
theta = (-pi./2);
elseif y == 0
theta = 0;
end
end
THETA(ii) = theta;
end
T1 = rad2deg(THETA) % Changing theta from radians to degrees
T2 = atan2d(Y,X) % the simple MATLAB approach
There you are: one simple function that gives the same output (to within floating point precision).
No loops or complex IF/ELSEIF required.
cdawg
am 28 Jan. 2023
Thanks Stephen. I guess I shouldn’t have said you *need* to loop through. In fact that’s something I’m trying to work on as it’s usually my “goto” method, but is often unnecessary :-)
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