Steven D'Aprano wrote:
On Fri, 03 Jun 2011 13:09:43 -0700, Raymond Hettinger wrote:
On Jun 3, 10:55 am, Billy Mays <no...@nohow.com> wrote:
I'm trying to shorten a one-liner I have for calculating the standard
deviation of a list of numbers. I have something so far, but I was
wondering if it could be made any shorter (without imports).
Here's my function:
a=lambda d:(sum((x-1.*sum(d)/len(d))**2 for x in
d)/(1.*(len(d)-1)))**.5
The functions is invoked as follows:
>>> a([1,2,3,4])
1.2909944487358056
Besides trying to do it one line, it is also interesting to write an
one-pass version with incremental results:
http://mathcentral.uregina.ca/QQ/database/QQ.09.06/h/murtaza2.html
I'm not convinced that's a good approach, although I haven't tried it. In
general, the so-called "computational formula" for variance is optimized
for pencil and paper calculations of small amounts of data, but is
numerically unstable.
See
http://www.johndcook.com/blog/2008/09/26/comparing-three-methods-of-
computing-standard-deviation/
http://en.wikipedia.org/wiki/Algorithms_for_calculating_variance
I'll also take this opportunity to plug my experimental stats package,
which includes coroutine-based running statistics, including standard
deviation:
--> s = stats.co.stdev()
--> s.send(3)
nan
Look! A NaN in the wild! :)
~Ethan~
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