Lists: Converting Double to Single

[Neil Cerutti]

On 2007-02-27, Steven D'Aprano <steve at REMOVE.THIS.cybersource.com.au> wrote:
> On Tue, 27 Feb 2007 10:06:29 +0000, Duncan Booth wrote:
>> Adding up a long list of values and then dividing by the
>> number of values is the classic computer science example of
>> how to get an inaccurate answer from a floating point
>> calculation.
>
> I'm not entirely ignorant when it comes to computational
> mathematics, but I must admit this is a new one to me.
>
> If the values vary greatly in magnitude, you probably want to add them
> from smallest to biggest; other than that, how else can you calculate the
> mean?
>
> The only alternative I thought of was to divide each value by the count
> before summing, but that would be horribly inaccurate.
>
> Or would it?
>
>>>> def mean1(*args):
> ...     return sum(args)/len(args)
> ...
>>>> def mean2(*args):
> ...     n = len(args)
> ...     return sum([x/n for x in args])
> ...
>>>> L = range(25, 597)  # 572 values
>>>> L = [x/3.3 for x in L]
>>>>
>>>> mean1(*L)
> 94.090909090909108
>>>> mean2(*L)
> 94.090909090909108
>
> The first calculation has 571 additions and one division; the
> second calculation has 571 additions and 572 divisions, but
> they both give the same result to 15 decimal places.

How about this?

>>> L = range(5, 10)
>>> L = [pow(2, -x) for x in L]
>>> "%40.40f" % mean1(*L)
'0.0121093750000000000000000000000000000000'
>>> "%40.40f" % mean2(*L)
'0.0121093750000000020000000000000000000000'

Offhand, I think the first is "righter". Weird!

-- 
Neil Cerutti