Using Python, assume I’m running through a known quantity of items
I, and have the ability to time how long it takes to process each one
t, as well as a running total of time spent processing
T and the number of items processed so far
c. I’m currently calculating the average on the fly
A = T / c but this can be skewed by say a single item taking an extraordinarily long time to process (a few seconds compared to a few milliseconds).
I would like to show a running Standard Deviation. How can I do this without keeping a record of each
I use Welford’s Method, which gives more accurate results. This link points to John D. Cook’s overview. Here’s a paragraph from it that summarizes why it is a preferred approach:
This better way of computing variance goes back to a 1962 paper by B. P. Welford and is presented in Donald Knuth’s Art of Computer Programming, Vol 2, page 232, 3rd edition. Although this solution has been known for decades, not enough people know about it. Most people are probably unaware that computing sample variance can be difficult until the first time they compute a standard deviation and get an exception for taking the square root of a negative number.
As outlined in the Wikipedia article on the standard deviation, it is enough to keep track of the following three sums:
s0 = sum(1 for x in samples) s1 = sum(x for x in samples) s2 = sum(x*x for x in samples)
These sums are easily updated as new values arrive. The standard deviation can be calculated as
std_dev = math.sqrt((s0 * s2 - s1 * s1)/(s0 * (s0 - 1)))
Note that this way of computing the standard deviation can be numerically ill-conditioned if your samples are floating point numbers and the standard deviation is small compared to the mean of the samples. If you expect samples of this type, you should resort to Welford’s method (see the accepted answer).
Based on Welford’s algorithm:
import numpy as np class OnlineVariance(object): """ Welford's algorithm computes the sample variance incrementally. """ def __init__(self, iterable=None, ddof=1): self.ddof, self.n, self.mean, self.M2 = ddof, 0, 0.0, 0.0 if iterable is not None: for datum in iterable: self.include(datum) def include(self, datum): self.n += 1 self.delta = datum - self.mean self.mean += self.delta / self.n self.M2 += self.delta * (datum - self.mean) def variance(self): return self.M2 / (self.n - self.ddof) def std(self): return np.sqrt(self.variance)
Update the variance with each new piece of data:
N = 100 data = np.random.random(N) ov = OnlineVariance(ddof=0) for d in data: ov.include(d) std = ov.std print(std)
Check our result against the standard deviation computed by numpy:
assert np.allclose(std, data.std())