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 Another interesting avenue to is aim for highest possible accuracy. Consider using math.fsum() to avoid rounding errors in the summation of large numbers of nearly equal values. Raymond ------------- follow my python tips on twitter: @raymondh -- http://mail.python.org/mailman/listinfo/python-list
