cart2pol angle sin 0 to 2pi range instead of -pi to pi range?

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Alexander Hannides Rajapakse
回答： theodore panagos 2018-11-7
I have a huge set of x and y coordinates that I have converted to polar coordinates using cart2pol, but I need to have my range of angles to be 0 to 2pi instead of -pi to pi. How can I do this without manually adding 2pi to each negative theta?

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回答（2 个）

John D'Errico
John D'Errico 2018-10-24
The most efficient solution is to use mod. Thus...
theta = mod(cart2pol(x,y),2*pi);
This works, because what you really want is to add 2*pi to the angle, whenever the angle is less than zero. mod does exactly that. And, yes, mod works for non-integer moduli, and of course mod is vectorized, so it does everything you need without even a need for a loop.
theodore panagos
theodore panagos 2018-11-7
You can use the formula:
atan(x,y)=pi()-pi()/2*(1+sign(x))*(1-sign(y^2))-pi()/4*(2+sign(x))*sign(y)
-sign(x*y)*atan((abs(x)-abs(y))/(abs(x)+abs(y)))
there is x=x2-x1 and y=y2-y1
The formula give the angle from 0 to2pi for all the values x and y
For x=y=0 the result is undefined.

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