Tim Peters added the comment:
As I said, the last code I posted is "fast enough" - I can't imagine a real
application can't live with being able to do "only" tens of thousands of roots
per second. A geometric mean is typically an output summary statistic, not a
transformation applied billions of times to input data.
To get a 100% guarantee of 100% portability (although still confined to IEEE
754 format boxes), we can't use the libm pow() at all. Then you can kiss speed
goodbye. Mark's `nroot` is portable in that way, but can take tens of
thousands of times longer to compute a root.
Here's another way, but routinely 100 times slower (than the last code I
posted):
import decimal
c = decimal.DefaultContext.copy()
c.prec = 25
def slow(x, n,
D=decimal.Decimal,
pow=c.power,
div=c.divide,
d1=decimal.Decimal(1)):
return float(pow(D(x), div(d1, n)))
del decimal, c
Too slow for my tastes, but at least it's obvious ;-)
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<https://bugs.python.org/issue27761>
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