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().
>
Here's version of mean based on the Kahan sum, including the
compensation term that seems to be consistently much better than builtin
mean It seems to be typically 4 orders or magnitude closer to the
"correct" answer than the builtin mean and 2 orders of magnitude better
than just naively using the Kahan sum. I'm using the sum performed with
dtype=int64 as the "correct" value.
def rawKahanSum(values): # Stolen from mathworld
if not input:
return 0.0
total = values[0]
c = type(total)(0.0) # A running compensation for lost
low-order bits.
for x in values[1:]:
y = x - c # So far, so good: c is zero.
t = total + y # Alas, total is big, y small, so
low-order digits of y are lost.
c = (t - total) - y # (t - total) recovers the high-order
part of y; subtracting y recovers -(low part of y)
total = t # Algebriacally, c should always be
zero. Beware eagerly optimising compilers!
# Next time around, the lost low part
will be added to y in a fresh attempt.
return total, c
def kahanSum(values):
total, c = rawKahanSum(values)
return total
def mean(values):
total, c = rawKahanSum(values)
n = float(len(values))
return total / n - c / n
for i in range(100):
data = random.random([10000]).astype(float32)
baseline = data.mean(dtype=float64)
delta_builtin_mean = baseline - data.mean()
delta_compensated_mean = baseline - mean(data)
delta_kahan_mean = baseline - (kahanSum(data) / len(data))
if not abs(delta_builtin_mean) >= abs(delta_kahan_mean) >=
abs(delta_compensated_mean):
print ">>>",
print delta_builtin_mean, delta_kahan_mean, delta_compensated_mean
-tim
> > 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.
>
>
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