An aggregation is some kind of summary of a set of data. This can be the sum, length, minimum, etc. It is quite common to want to calculate such a summary repeatedly, e.g. “the maximum noise level in dB for the past 30 seconds” for a nuisance detector. In such a case we say there is a sliding window over our data, and we want to aggregate over our window.

If our aggregation is a binary operator with an inverse, like integer sums, there is a very easy solution using a double-ended queue:

from collections import deque

class SlidingWindowSum:
    def __init__(self):
        self.sum = 0
        self.elems = deque()
    
    def push(self, x):
        self.sum += x
        self.elems.append(x)
    
    def pop(self):
        self.sum -= self.elems.popleft()
    
    def eval(self):
        return self.sum

But what if our operator has no inverse? This is actually the case for most interesting summaries such as minimum, quantile, approximate unique count (for example using HyperLogLog), etc. In fact, even something as simple as a floating-point sum suffers from the fact that floating-point addition is not invertible. For example, if you ever have a NaN in your input data with the above naive algorithm your sum will forever remain NaN, even long after the bad value has left your window.

Six years ago I came up with an algorithm for maintaining just the minimum/maximum in a sliding window and posted it to cs.stackexchange. I now consider this algorithm pointless, because it turns out there is a simple and efficient algorithm that solves this problem for a very wide class of aggregations. I’m writing this blog post to spread the word, because I feel it should be more widely known.

I came across this algorithm while reading a far more advanced paper, Low-Latency Sliding-Window Aggregation in Worst-Case Constant Time by Tangwongsan et al. Why is this paper titled low-latency? Because it does the same as what I’m about to describe, but in O(1) time for each step. However, in it they also described a “two-stack” algorithm, which does it in amortized O(1), and is far, far simpler.

Funnily enough that paper attributes this algorithm to “adamax” from a 2011 Stack Overflow post. They in turn credit a 2001 lecture note by D. Sleator for the inspiration. However, this lecture note does not describe a sliding window aggregate, it describes the classical two-stack algorithm for implementing a FIFO queue and does amortized analysis on it. Ultimately I would not be surprised to find that this algorithm was already described in an obscure paper from the 1970s, seeing how simple and brilliant it is.

Like the authors of the paper, I will generalize the two-stack algorithm to arbitrary associative aggregation functions. By abstracting the aggregation as a set of functions, empty(), unit(x), combine(x, y) and finalize(x), you can describe many possible aggregations, for example a mean:

empty = lambda: (0, 0)
unit = lambda x: (x, 1)
combine = lambda x, y: (x[0] + y[0], x[1] + y[1])
finalize = lambda x: x[0] / x[1] if x[1] else None

I’d like to note here that these functions have the following signatures:

fn empty() -> Agg;
fn unit(x: Value) -> Agg;
fn combine(x: Agg, y: Agg) -> Agg;
fn finalize(x: Agg) -> Out;

I’m making a distinction here between Value, Agg and Out because while they seem superficially similar for something like an integer sum, for an approximate unique count on strings you would have (Value, Agg, Out) = (String, HyperLogLogSketch, u64), three wildly different types.

Without further ado, the algorithm:

class TwoStackAgg:
    def __init__(self):
        self.values = []
        self.values_agg = empty()
        self.cum_aggs = []
    
    def push(self, x):
        self.values.append(x)
        self.values_agg = combine(self.values_agg, unit(x))
    
    def pop(self):
        if not self.cum_aggs:
            cum_agg = empty()
            while self.values:
                cum_agg = combine(unit(self.values.pop()), cum_agg)
                self.cum_aggs.append(cum_agg)
            self.values_agg = empty()
        self.cum_aggs.pop()
    
    def eval(self):
        return finalize(
            combine(self.cum_aggs[-1], self.values_agg)
            if self.cum_aggs else self.values_agg
        )

That’s it, the entire algorithm. There’s two stacks (values and cum_aggs) and one more aggregate, values_agg. At any point in time values_agg holds the aggregate of values, and cum_aggs contains the cumulative aggregates of all values in our window that aren’t in values, in reverse order. From this we can get the aggregate over our entire window in constant time by by combining the last value of cum_aggs with values_agg.

The neat part is that (assuming w is our window size) every wth operation we drain all of values and maintain a running aggregate while pushing the partial cumulative aggregates onto cum_aggs. This is what makes it amortized O(1), doing O(w) internal operations every wth pop bounds the total amount of work per element to O(1), even though a singular operation might not be constant time.

I think this is best visualized. Suppose we sum [1, 2, ..., 10] with a fixed-size sliding window of four elements, then the state on each eval() call would look like this (values_agg not shown as it is simply the aggregate of the values):

             cum_aggs values        out
                   [] []            = 0
                   [] [1]           = 1
                   [] [1, 2]        = 1 + 2
                   [] [1, 2, 3]     = 1 + 2 + 3
                   [] [1, 2, 3, 4]  = 1 + 2 + 3 + 4
[4, 3 + 4, 2 + 3 + 4] [5]           = 2 + 3 + 4 + 5
           [4, 3 + 4] [5, 6]        = 3 + 4 + 5 + 6
                  [4] [5, 6, 7]     = 4 + 5 + 6 + 7
                   [] [5, 6, 7, 8]  = 5 + 6 + 7 + 8
[8, 7 + 8, 6 + 7 + 8] [9]           = 6 + 7 + 8 + 9
           [8, 7 + 8] [9, 10]       = 7 + 8 + 9 + 10
                  [8] [9, 10]       = 8 + 9 + 10
                   [] [9, 10]       = 9 + 10
                 [10] []            = 10
                   [] []            = 0

In total the memory usage is O(w), where w is your maximum window size. Note that for simplicity of analysis and the example I assumed a fixed-size window w, but there is nothing about the two-stack algorithm that requires this. You can call push(x) and pop() as many times as you’d like between each eval(), growing and shrinking the window size as needed.

Note that we required above that our aggregate combine is associative, meaning:

combine(combine(x, y), z) = combine(x, combine(y, z))

Technically speaking, floating-point addition doesn’t respect this. Nevertheless, the above algorithm is still very useful because the results closely match the expected outcome, even more so if you use a compensated summation algorithm like Kahan summation.

However, there is a second very useful property of the above algorithm. Each aggregate is strictly a combination of the elements in the window, and none outside the window. This means if your window contains a NaN or infinity (or some other outlier), that value only poisons the windows that contain it rather than the rest of your computation.

But even without NaN or infinity it is useful, due to not propagating errors endlessly. E.g. if your sliding window starts with [1e20, 1], this is what would happen with a naive rolling sum:

>>> 1e20 + 1 - 1e20 - 1
-1.0

Compensated summation will reduce these effects, but not making your result depend on values outside of the window will eliminate long-term error accumulation entirely.