[Numpy-discussion] Unexpected float96 precision loss

Michael Gilbert michael.s.gilbert@gmail....
Wed Sep 1 16:30:08 CDT 2010


On Wed, 1 Sep 2010 21:15:22 +0000 (UTC), Pauli Virtanen wrote:
> Wed, 01 Sep 2010 16:26:59 -0400, Michael Gilbert wrote:
> > I've been using numpy's float96 class lately, and I've run into some
> > strange precision errors.
> [clip]
> >   >>> x = numpy.array( [0.01] , numpy.float96 )
> [clip]
> > I would expect the float96 calculation to also produce 0.0 exactly as
> > found in the float32 and float64 examples.  Why isn't this the case?
> 
> (i) It is not possible to write long double literals in Python.
>     "float96(0.0001)" means in fact "float96(float64(0.0001))"
> 
> (ii) It is not possible to represent numbers 10^-r, r > 1 exactly
>      in base-2 floating point.
> 
> So if you write "float96(0.0001)", the result is not the float96 number 
> closest to 0.0001, but the 96-bit representation of the 64-bit number 
> closest to 0.0001. Indeed,
> 
> >>> float96(0.0001), float96(1.0)/1000
> (0.00010000000000000000479, 0.00099999999999999999996)

Interesting.  float96( '0.0001' ) also seems to evaluate to the first
result. I assume that it also does a float64( '0.0001' ) conversion
first. I understand that you can't change how python passes in floats,
but wouldn't it be better to exactly handle strings since those can be
converted exactly, which is what the user wants/expects?

Thanks so much for the quick responses.

Best wishes,
Mike


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