Averaging angles… Again

Question

I want to calculate the average of a set of angles, which represents source bearing (0 to 360 deg) - (similar to wind-direction)

I know it has been

Answer 1

Edit: Equivalent, but more robust algorithm (and simpler):

1. divide angles into 2 groups, [0-180) and [180-360)
2. numerically average both groups
3. average the 2 group averages with proper weighting
4. if wraparound occurred, correct by 180˚

This works because number averaging works "logically" if all the angles are in the same hemicircle. We then delay getting wraparound error until the very last step, where it is easily detected and corrected. I also threw in some code for handling opposite angle cases. If the averages are opposite we favor the hemisphere that had more angles in it, and in the case of equal angles in both hemispheres we return None because no average would make sense.

The new code:

def averageAngles2(angles):
    newAngles = [a % 360 for a in angles];
    smallAngles = []
    largeAngles = []
    # split the angles into 2 groups: [0-180) and [180-360)
    for angle in newAngles:
        if angle < 180:
            smallAngles.append(angle)
        else:
            largeAngles.append(angle)
    smallCount = len(smallAngles)
    largeCount = len(largeAngles)
    #averaging each of the groups will work with standard averages
    smallAverage = sum(smallAngles) / float(smallCount) if smallCount else 0
    largeAverage = sum(largeAngles) / float(largeCount) if largeCount else 0
    if smallCount == 0:
        return largeAverage
    if largeCount == 0:
        return smallAverage
    average = (smallAverage * smallCount + largeAverage * largeCount) / \
        float(smallCount + largeCount)
    if largeAverage < smallAverage + 180:
        # average will not hit wraparound
        return average
    elif largeAverage > smallAverage + 180:
        # average will hit wraparound, so will be off by 180 degrees
        return (average + 180) % 360
    else:
        # opposite angles: return whichever has more weight
        if smallCount > largeCount:
            return smallAverage
        elif smallCount < largeCount:
            return largeAverage
        else:
            return None

 

>>> averageAngles2([0, 0, 90])
30.0
>>> averageAngles2([30, 350])
10.0
>>> averageAngles2([0, 200])
280.0

Here's a slightly naive algorithm:

1. remove all oposite angles from the list
2. take a pair of angles
3. rotate them to the first and second quadrant and average them
4. rotate average angle back by same amount
5. for each remaining angle, average in same way, but with successively increasing weight to the composite angle

some python code (step 1 not implemented)

def averageAngles(angles):
    newAngles = [a % 360 for a in angles];
    average = 0
    weight = 0
    for ang in newAngles:
        theta = 0
        if 0 < ang - average <= 180:
            theta = 180 - ang
        else:
            theta = 180 - average
        r_ang = (ang + theta) % 360
        r_avg = (average + theta) % 360
        average = ((r_avg * weight + r_ang) / float(weight + 1) - theta) % 360
        weight += 1
    return average