Please help - get average

[Frohnhofer, James]

> Robin Becker wrote
> ...... interestingly the standard algorithms for means/deviations are 
> numerically poor. As an example
> 
>  >>> (1.e30+1-1.0e30)/3
> 0.0
> 
> Where we have total loss of information in one of the terms.
> 
> 
> There are better algorithms involving recursion,
> 
> eg
> 
>     mu[n] = (data[n]+(n-1)*mu[n-1])/n
> The error in the existing estimate, mu[n-1], gets multiplied by the 
> number (n-1)/n which is less than one (ie the errors are damped).
> 

I'm not sure I believe this.  It's been a while since I studied this, but I
think I remember maintaining precision at the Frankenstein level:
"Multiplication bad.  Addition good."  Keep the multiplications and divisions
to a minimum, and try to do them at the end.

> However, in the case of widely differing magnitudes I think 
> you need to 
> accumulate the sums specially.


Well, its not even that hard:

 >>> data=[1.e30,1.,-1.e30]
 >>> sum(data)/len(data)
 0.0
 >>> data.sort(lambda x,y:-cmp(abs(x),abs(y)))
 >>> data
 [-1e+030, 1e+030, 1.0]
 >>> sum(data)/len(data)  #Assuming sum works the way I think it should . . .
 0.33333333333333331

But using the recursive 
 >>> def mu(lst):
         n = len(lst)
         if n==1:
             return lst[0]
         else:
             return (lst[0] + ((n-1)*mu(lst[1:])))/n

 >>> data=[1.e30,1.,-1.e30]
 >>> mu(data)
 0.0
 >>> data.sort(lambda x,y:-cmp(abs(x),abs(y)))
 >>> mu(data)
 0.0

it seems we lose even the ability to maintain the precision by sorting on the
magnitude. 

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