[Numpy-discussion] please change mean to use dtype=float
Robert Kern
robert.kern at gmail.com
Wed Sep 20 12:22:16 CDT 2006
David M. Cooke wrote:
> On Wed, Sep 20, 2006 at 03:01:18AM -0500, Robert Kern wrote:
>> Let me offer a third path: the algorithms used for .mean() and .var() are
>> substandard. There are much better incremental algorithms that entirely avoid
>> the need to accumulate such large (and therefore precision-losing) intermediate
>> values. The algorithms look like the following for 1D arrays in Python:
>>
>> def mean(a):
>> m = a[0]
>> for i in range(1, len(a)):
>> m += (a[i] - m) / (i + 1)
>> return m
>
> This isn't really going to be any better than using a simple sum.
> It'll also be slower (a division per iteration).
With one exception, every test that I've thrown at it shows that it's better for
float32. That exception is uniformly spaced arrays, like linspace().
> You do avoid
> accumulating large sums, but then doing the division a[i]/len(a) and
> adding that will do the same.
Okay, this is true.
> Now, if you want to avoid losing precision, you want to use a better
> summation technique, like compensated (or Kahan) summation:
>
> def mean(a):
> s = e = a.dtype.type(0)
> for i in range(0, len(a)):
> temp = s
> y = a[i] + e
> s = temp + y
> e = (temp - s) + y
> return s / len(a)
>
> Some numerical experiments in Maple using 5-digit precision show that
> your mean is maybe a bit better in some cases, but can also be much
> worse, than sum(a)/len(a), but both are quite poor in comparision to the
> Kahan summation.
>
> (We could probably use a fast implementation of Kahan summation in
> addition to a.sum())
+1
>> def var(a):
>> m = a[0]
>> t = a.dtype.type(0)
>> for i in range(1, len(a)):
>> q = a[i] - m
>> r = q / (i+1)
>> m += r
>> t += i * q * r
>> t /= len(a)
>> return t
>>
>> Alternatively, from Knuth:
>>
>> def var_knuth(a):
>> m = a.dtype.type(0)
>> variance = a.dtype.type(0)
>> for i in range(len(a)):
>> delta = a[i] - m
>> m += delta / (i+1)
>> variance += delta * (a[i] - m)
>> variance /= len(a)
>> return variance
>
> These formulas are good when you can only do one pass over the data
> (like in a calculator where you don't store all the data points), but
> are slightly worse than doing two passes. Kahan summation would probably
> also be good here too.
Again, my tests show otherwise for float32. I'll condense my ipython log into a
module for everyone's perusal. It's possible that the Kahan summation of the
squared residuals will work better than the current two-pass algorithm and the
implementations I give above.
--
Robert Kern
"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco
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