Here’s a little lunch-time diversionary math. Suppose you want a function that takes a number, and returns the average of all the numbers it’s been called with so far. Handy for continuously updated displays, that kind of thing. Here’s a method that will return this averaging function.

private static Func MakeAverager() { float sum = 0; int count = 0; return delegate(float x) { sum += x; count += 1; return sum/count; }; }

It creates `sum`

and `count`

variables for the function to close over. The function takes a number, `x`

, and adds it to `sum`

. It increments `count`

, and divides. Pretty standard.

Now, let’s get crazy, and pretend this code is going on Voyager, and it’ll be running for ever. `sum`

will get pretty high, right? We’ll blow through 2^{31}, the upper-bound for .Net 32-bit ints. Sure, we could make it a `long`

, and go up to 2^{63}, but that’s not the point. The point is, it’ll eventually run out, because `sum`

is too high.

I’ve been chewing on this brain-teaser for a while. I knew there must be a way to calculate a long-running average without storing the sum and the count; it seems the average so far, and the count, should be enough, but I don’t want to resort to `((average * count) + x) / count++`

, because that’s the exact same problem. (Of course, `count`

could still get so high it overflows, but that’s somewhat less likely. Hush, you.)

I finally sat down and figured it out. The trick is, each successive x tugs your average up or down — first by a lot, but by less over time. With each x, the average gets harder to move: the effect each new x has on the average is inversely proportionate to the count. We can put it like this:

average += (x - average) / count

We tug average by `x - average`

, the distance between them, scaled down by count. Then, add that on to average (of course, if x < average, then x – average is negative, so it’ll tug the average down).

Let’s make a new averager.

private static Func MakeNewAverager() { float average = 0; int count = 0; return delegate(float x) { average += (x - average) / ++count; return average; }; }

It works the same, but it’ll take a lot longer for `count`

to overflow than `sum`

.

For the record, here’s the ruby I used to sketch this idea out. Of course, in ruby, this problem is even more bogus, because ruby’s Bignum can handle any number that your machine has enough free RAM to store. But still.

def make_averager sum, count = 0, 0 lambda { |x| sum += x count += 1 sum.to_f / count } end def make_sum_free_averager avg = 0.0 count = 0 lambda { |x| count += 1 avg += (x - avg) / count } end

Very cool solution to the problem – thanks for sharing your thoughts.

Out of curiosity to see how exact this method is I figured I would do a very simple test. Both examples have an average of 2 after 200 iterations but the 201st iteration passes in a value of x=4 – the two resultant averages are very, close to each other (I imagine the difference is due to the calculator I used forcing a rounding on me) – I just figured I would post it here for consideration:

Traditional Result: (404/201)

2.0099502487562189054726368159204

Your Algorithim: 2 + (2/201)

2.009950248756218905472636815920398

Again, thanks for sharing. I like it and it seems like an easier way, in general, to compute a running average.

Whoa! That aspect of the problem didn’t occur to me, but I don’t think I’d have expected that result, even if it had. I guess I can deal with only-accurate-to-30-decimal-places. Thanks!

If you replace your count with a weight factor you’ve got an exponential average (http://en.wikipedia.org/wiki/Exponential_average#Exponential_moving_average). No chance for overflow but current values have more weight in the average that prior values.

Paul, thanks for pointing that out…especially since it links to Triangle numbers, which I’d thought about, but didn’t know had a name.

Every time I drive my Prius, I want it to give me a weighted running average of my mileage, instead of the straight running average. I care more about my mileage over this tank of gas, or the past few weeks, than I do about 4 months ago. I’ll send this to Toyota. :)

I had thought of doing this, but I was always worried that in a long-running system you’d reach a point where a (x – average)/count = 0 due to precision problems. Essentially each iteration becomes less precise until you’ve lost all your precision. Then it will never change even if the next (10 * count) values of x are (average + 1)

I suppose as long as this takes a few orders of magnitude longer than the sum version, then it is useful.

@Drew

You mean, as count gets really huge, (x-average)/count tends towards 0, and eventually the precision will bottom out, right? I hadn’t thought of that, good point. I tried it out in ruby — a really tiny x-average, and count is .Net’s long.MaxValue, 263, or ~9 quintillion.

1.0e-300 / (2 ** 63) #-> 1.08417765323403e-319

So there’s enough precision (in ruby, on WinXP 32, at least) to last through all that. I guess that’s the few orders of magnitude difference. :)

I felt drawn to work on this problem before seeing your solution. (I tend to like such challenges). As I worked on it, I realized that it was just an exercise in transferring the use of precision on the left side of the decimal point to the right side. Really, what is the difference between multiplying two numbers for a large result and dividing two numbers for a tiny result? The computer is just bit shifting in the opposite direction, now that I think about it.

So, in the end, it is all about precision, whether during multiplication or division. You started out trying to keep things within 32 bits, but notice how you resolved the division precision with 63 :) We will run into a similar situation if the count exceeds that as well.

In any case, thank you for the stimulating post!

You’re not summing an int, or a long, you are summing a float (http://steve.hollasch.net/cgindex/coding/ieeefloat.html). If you use a double-precision one, this will not top-out at 2^63, but roughly 2^1023. Your algorithm will top-out much faster than this, as it will break horribly when your integer ‘count’ overflows 2^32 and your reasoning for ignoring this depends on the values of x being larger than 1, and the value of sum also being an int. Finally, your algorithm is susceptible to ignoring any total contribution from small numbers that are not near the mean (summation underflow). Do everything in double precision, and be careful, and you’ll be able to make it run forever. See http://en.wikipedia.org/wiki/Kahan_summation_algorithm for some ideas.

Stoat.

Stoat, witness the sloppy thinking: I was actually thinking of summing ints, not floats, when I started the problem. Then, when integer division caused wicked rounding errors, I thought “oh, right,” and made them floats, forgetting about the change in range that meant. Maybe I should have phrased the whole thing as just a mental exercise, with silly constraints, and a proof of correctness in code. I trust no one’s going to drop this code into production soon.

This is a great way of keeping running averages. I’ve used this for computing cluster centroids, since the average *is* the centroid, it’s easier to keep the count, as you said. We worked with floats exclusively, as the values were multidimensional vectors converted from various data types. There’s a variation on this that can be used to recompute the standard deviation from the new mean by tracking the power sum average, but you do need to recompute the sums:

(python)

count=0

mean=0

pwrSumAvg=0

stdDev=0

def update(x):

count += 1

mean += (x-mean)/count

pwrSumAvg += ( x*x – pwrSumAvg) / count

stdDev = sqrt(pwrSumAvg*count – mean*count) / (count*(count-1))

Could you please explain how you came to the expression:

average += (x – average) / count ?

from the brute-force version:

average(n) = ( sum(n-1) + x ) / ( count(n-1) + 1 )

Thanks!

The math:

Let a := old average, a’ := new average

Let n := *new* count

Let x := value just inserted

From brute-force:

a’ = (a(n-1) + x) / n [a * (n-1) is the sum of the n-1 numbers]

= a(n-1)/n + x/n

= an/n – a/n + x/n

= a + (x-a)/n

which is average += (x – average)/n.

@Guy, I think jonathantan86 did it just fine. Though I have to admit, I didn’t go through all that rigor, going from brute force to new equation…I kind of felt my way through the problem, formalized the new version in code, and tested it against the brute force version. What can I say? Programming is a Good Medium for Expressing Poorly Understood and Sloppily Formulated Ideas.

Thanks for the solution

You can also use boost/accumulators with tag::mean(immediate):

#include

#include

#include

#include

#include

#include

namespace bacc = boost::accumulators;

void test1()

{

int i, j;

typedef int16_t mytype;

bacc::accumulator_set<mytype, bacc::stats > bm;

for(i = 0; i < 100; i++) {

for(j = 0; j < 99; j++) {

bm(10000);

}

std::cout << i << "%\n";

}

std::cout << bacc::mean(bm) << "\n";

}

void test2()

{

int i;

typedef char mytype;

bacc::accumulator_set<mytype, bacc::stats > bm;

for(i = 0; i < 100; i++) {

bm(i);

std::cout << "bm: " << bacc::mean(bm) << "\n";

}

}

void test3()

{

int i;

typedef int64_t mytype;

bacc::accumulator_set<mytype, bacc::stats > bm;

for(i = 0; i < 100; i++) {

bm(i);

std::cout << "bm: " << bacc::mean(bm) << "\n";

}

}

int main()

{

std::cout << "TEST 1\n";

test1();

std::cout << "TEST 2\n";

test2();

std::cout << "TEST 3\n";

test3();

return 0;

}

Gah… Includes got removed…

Here are the ones I provided:

iostream

boost/cstdint.hpp

boost/accumulators/accumulators.hpp

boost/accumulators/statistics/mean.hpp

boost/accumulators/statistics/stats.hpp

You still have a failure area, and that is the count can overflow, you can do this by faking a running average.

double avg;

void update(double val) {

avg = (avg + val / 9) * .9;

}

That will give a nice running average for a fast updating system, a nice “health monitor” of some procedure or temperature with jittery inputs.

Play with 9 and .9 till you get the responsiveness you desire :)

Just a note about that routine….

avg = (avg + newval/A) * B;

Where B = A / (A+1); Choose A

This will amplify certain frequencies generated in the signal. For fast update use A = .1; For long use A=10 or more;

Apparantly you have discovered that part of Welford’s algorithm to online compute the mean: http://en.wikipedia.org/wiki/Algorithms_for_calculating_variance#On-line_algorithm

Thanks… I figured there was a name behind that one, I noticed it back in the early 1990′s when writing some audio processing software that needed a frequency filter. Random wave samples (white/pink noise) in, and you get a nice roar/boom around a certain frequency out.