A better way is to use Welford's Algorithm, which is numerically stable.
Below is a gist for doing just that.
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import math | |
class Welford(object): | |
""" Implements Welford's algorithm for computing a running mean | |
and standard deviation as described at: | |
http://www.johndcook.com/standard_deviation.html | |
can take single values or iterables | |
Properties: | |
mean - returns the mean | |
std - returns the std | |
meanfull- returns the mean and std of the mean | |
Usage: | |
>>> foo = Welford() | |
>>> foo(range(100)) | |
>>> foo | |
<Welford: 49.5 +- 29.0114919759> | |
>>> foo([1]*1000) | |
>>> foo | |
<Welford: 5.40909090909 +- 16.4437417146> | |
>>> foo.mean | |
5.409090909090906 | |
>>> foo.std | |
16.44374171455467 | |
>>> foo.meanfull | |
(5.409090909090906, 0.4957974674244838) | |
""" | |
def __init__(self,lst=None): | |
self.k = 0 | |
self.M = 0 | |
self.S = 0 | |
self.__call__(lst) | |
def update(self,x): | |
if x is None: | |
return | |
self.k += 1 | |
newM = self.M + (x - self.M)*1./self.k | |
newS = self.S + (x - self.M)*(x - newM) | |
self.M, self.S = newM, newS | |
def consume(self,lst): | |
lst = iter(lst) | |
for x in lst: | |
self.update(x) | |
def __call__(self,x): | |
if hasattr(x,"__iter__"): | |
self.consume(x) | |
else: | |
self.update(x) | |
@property | |
def mean(self): | |
return self.M | |
@property | |
def meanfull(self): | |
return self.mean, self.std/math.sqrt(self.k) | |
@property | |
def std(self): | |
if self.k==1: | |
return 0 | |
return math.sqrt(self.S/(self.k-1)) | |
def __repr__(self): | |
return "<Welford: {} +- {}>".format(self.mean, self.std) |