angdiff doesn't work as documented

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Eyal Itskovits
Eyal Itskovits 2016-8-31
编辑： Guillaume 2016-8-31
Hello,
I'm using matlab 2015b. When used with a single variable, angdiff is documented to return the angular difference (smallest distance) between two adjacent angles.
Practically, it returns the distance of all the angles from zero. For example, this - angdiff(0:(pi/6):2*pi)
suppose to return [pi/6, pi/6 ... ] but it returns, [0, pi/6, 2*pi/6 ..]
Thank you, Eyal.

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回答（2 个）

Guillaume
Guillaume 2016-8-31
编辑：Guillaume 2016-8-31
I don't have the robotics toolbox, so can't check. If it does truly behave as you say then it is a bug you should reports to mathworks.
In the meantime, the function is trivial to implement:
function delta = nonbuggy_angdiff(alpha, beta)
validateattributes(alpha, {'numeric'}, {'vector', 'nonsparse'}, 1);
if nargin > 1
validateattributes(beta, {'numeric'}, {'vector', 'nonsparse', 'size', size(alpha)}, 2);
alpha = beta - alpha;
else
alpha = diff(alpha);
end
delta = mod(alpha + pi, 2*pi) - pi; %constrain to [-pi, pi[. will prefer -pi over +pi
delta(delta == -pi & alpha >= 0) = pi; %so force -pi to +pi (only when the original angle was positive)
end
edit: bugfix. There may still be bugs lurking about.
  2 个评论
Thorsten
Thorsten 2016-8-31
编辑：Thorsten 2016-8-31
This functions returns pi for angdiff(0,pi) and angdiff(pi, 0), which is wrong, or at least a bit strange.
And it does not work properly on arrays.
It return only a single pi if all values are pi:
angdiff([0 pi], [pi 0])
ans =
3.1416
and it gives inconsistent results if not all elements equal -pi:
angdiff([0 pi 0 ], [pi 0 0.1])
ans =
-3.1416 -3.1416 0.1000
because with single values angdiff(0,pi) and angdiff(pi,0) both return pi, not -pi.
And it does not work on some single matrices, as expected:
>> angdiff([0 pi 0])
ans =
3.1416
Should be pi, -pi.
Guillaume
Guillaume 2016-8-31
The first issue was a bug. Now fixed.
As to what angle should be returned for a pi or -pi angle, I find it puzzling that the official angdiff returns both since they're the same mod 2pi.
In any case, that was trivial to adapt to, so now my function should behave the same as the official angdiff.
As said, I don't have the robotics toolbox to check.

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Thorsten
Thorsten 2016-8-31
Have a look which angdiff you use
which angdiff
it may be function different from the angdiff in the Robotics Toolbox.

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