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 >>> s.send(2) 0.7071067811865476 >>> s.send(5) 1.5275252316519465 >>> s.send(5) 1.4999999999999998 The non-running calculation of stdev gives this: >>> stats.stdev([3, 2, 5, 5]) 1.5 http://pypi.python.org/pypi/stats/ http://code.google.com/p/pycalcstats/ Be warned that the version on Google Code is unstable, and currently broken. Feedback is welcome! -- Steven -- http://mail.python.org/mailman/listinfo/python-list