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I looked at some solutions here but none provide what I need, so:

I need to average an array of angles(0 to 359.9, no negatives) (A1 + A2 + A3 + An) / n

The issue is when you get an array {1, 359, 2, 358} the average if you use the formula above is 180, but actually it is supposed to be 0.

Any thoughts?

dseibert
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Miro J.
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    The problem is that the mean of angles is not a well-defined operation; with your formula it depends on where you start to measure them, for example, you would obtain the result you imagine if you used a (-180, 180] angle representation. You should explain better what you expect to obtain in general. – Matteo Italia Mar 17 '11 at 18:40
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    Why is it supposed to be 0? – Jon Skeet Mar 17 '11 at 18:40
  • What output would you want if you had {0, 180}? Is the "right" answer 90 or 270? – Nathan S. Mar 17 '11 at 18:59
  • The issue is that we need to average the angles regardless of what sector they are in and for us 0 = 360 = 720 and so on. Even if we shift the 0 with 180 there will be another point that would have the same problem. in the case of -180 to 180, it will be averaging {-179, 179}. It is supposed to give you 180 and not the arithmetic average 0. – Miro J. Mar 17 '11 at 20:14
  • @Jon, if you look at your hand-watch and someone tells you they will be at your place between 11:59 and 12:01, most likely you will think that they spoke of the smaller sector between these lines, which includes 12:00 and not 6:00. Some may say 6:00 is between the 2 lines 12:01 and 11:59. Now, because time has a direction, but angles don't, I need the smaller sector. – Miro J. Apr 25 '14 at 19:04
  • @Miro: It would really have helped if you'd talked about that in the question then. It's a bit late to give more information 3 years after asking the question. In other contexts, 180 *would* make more sense as the average. – Jon Skeet Apr 25 '14 at 19:08
  • The proper solution is already marked. I am just giving more information for all the people that will visit this place later. Problem was solved! See Mark's answer. – Miro J. Apr 25 '14 at 22:06

4 Answers4

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Add unit vectors of each angle, and convert the resulting vector back into an angle. If the result vector is of zero length, the inputs cancelled each other out and the result is indeterminate.

A unit vector has a length of 1, and its x and y lengths are given by the cosine and sine of the angle. Thus you average your examples as in the following pseudo-code:

x = cos(radians(1)) + cos(radians(359)) + cos(radians(2)) + cos(radians(358));
y = sin(radians(1)) + sin(radians(359)) + sin(radians(2)) + sin(radians(358));
angle = degrees(atan2(y, x));
Mark Ransom
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  • Sorry, I don't understand what do you mean by unit vectors. Can you explain in an example with the numbers I gave above? – Miro J. Mar 17 '11 at 18:50
  • @Miroslav, I've updated the answer. Unfortunately I don't know the proper syntax for .NET, but this should be close enough for you to understand. – Mark Ransom Mar 17 '11 at 19:01
  • It's ok, you were a great help. We are implementing it now. – Miro J. Mar 17 '11 at 20:40
  • Thank you very much, that saved me a _lot_ of hassle! – Echelon Mar 23 '11 at 08:19
  • This is a clever and easy solution, and for most purposes it works great. However not all values gets equal weight. e.g. what should the average of 0,60,180 be? Normally one would say 80, but since 0 and 180 cancel out you would get 60. See my answer below for an alternative. – Martin Vatshelle Oct 24 '14 at 21:15
0

Averaging degrees is a little trickier than it sounds. I had to do this at work the other day and want to share my findings.

  1. Adding 360 to any of your values should not change your average.
    e.g. avg(2 4 18) = avg(362,4,18) = 8
  2. Shifting all values by a constant value should shift the average equally much.
    e.g. avg(2-3,4-3,18-3) = avg(359,1,15) = 8-3
  3. If your degrees "cancel out", e.g. 0,120,240, the average is undefined
    (i.e. there are several equally good answers).
    I see no way to deal with this other than giving an error message.

After falling into several traps, I ended up with the following definition: Average is the value which minimizes the variance.

Once given the average, variance can be computed by:

  1. Shifting all values by the same angle such that the average is at 180.
    (This is equivalent to "cutting" the circle at the opposite side of the average.)
  2. Normalizing all the angles to be between 0 and 360.
  3. Computing the normal average.
  4. Shifting the average back by the angle you shifted the values before (reversing step 1).

This takes O(n^2) time.

However, if all your values lie within a span of 180 degree, you can shift the values such that 0 is contained in the biggest value gap and then compute the normal average. This takes O(n) time.

Martin Vatshelle
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0

A ray leaving a circle can be represented in more than one manner. It could be 0 degrees, 360 degrees, 720 degrees, etc. You need to determine what is the correct acceptable answer for you case, and covert intermediate answers to the final answer before presenting it.

1 + 359 + 2 + 358 = 360 + 360 = 720 degrees in total
720 / 4 = 360 / 2 = 180 degrees on average

It's not that the answer should be zero, it's that 359 degrees is not equivalent to -1 degrees because the angle goes around the circle "the other way".

Edwin Buck
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-1

Do you want to round it off i.e. 0 should come after 359 degree?

take a modulus of the result with 360.

avg = ( (a1 + a2 + an) /n ) % 360
Unmesh Kondolikar
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