[Numpy-discussion] please change mean to use dtype=float
Tim Hochberg
tim.hochberg at ieee.org
Thu Sep 21 13:34:42 CDT 2006
Tim Hochberg wrote:
> Robert Kern wrote:
>
>> 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.
>>
>>
> This is what my tests show as well var_knuth outperformed any simple two
> pass algorithm I could come up with, even ones using Kahan sums.
> Interestingly, for 1D arrays the built in float32 variance performs
> better than it should. After a bit of twiddling around I discovered that
> it actually does most of it's calculations in float64. It uses a two
> pass calculation, the result of mean is a scalar, and in the process of
> converting that back to an array we end up with float64 values. Or
> something like that; I was mostly reverse engineering the sequence of
> events from the results.
>
Here's a simple of example of how var is a little wacky. A shape-[N]
array will give you a different result than a shape-[1,N] array. The
reason is clear -- in the second case the mean is not a scalar so there
isn't the inadvertent promotion to float64, but it's still odd.
>>> data = (1000*(random.random([10000]) - 0.1)).astype(float32)
>>> print data.var() - data.reshape([1, -1]).var(-1)
[ 0.1171875]
I'm going to go ahead and attach a module containing the versions of
mean, var, etc that I've been playing with in case someone wants to mess
with them. Some were stolen from traffic on this list, for others I
grabbed the algorithms from wikipedia or equivalent.
-tim
> -tim
>
>
>
>
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