[Numpy-discussion] longdouble (float96) literals
David Cournapeau
cournape@gmail....
Wed Aug 18 10:07:35 CDT 2010
On Wed, Aug 18, 2010 at 11:52 PM, <josef.pktd@gmail.com> wrote:
> On Wed, Aug 18, 2010 at 10:36 AM, Charles R Harris
> <charlesr.harris@gmail.com> wrote:
>>
>>
>> On Wed, Aug 18, 2010 at 8:25 AM, Colin Macdonald <macdonald@maths.ox.ac.uk>
>> wrote:
>>>
>>> On 08/18/10 15:14, Charles R Harris wrote:
>>> > However, the various constants supplied by numpy, pi and such, are
>>> > full precision.
>>>
>>> no, they are not. My example demonstrated that numpy.pi is only
>>> double precision.
>>>
>>
>> Hmm, the full precision values are available internally but it looks like
>> they aren't available to the public. I wonder what the easiest way to
>> provide them would be? Maybe they should be long double types by default?
>
> playing with some examples, I don't seem to be able to do anything
> with longdouble on win32, py2.5
For all practical purposes, double == long double on windows: although
some compilers support the intel long format or even quadd precision,
the MSVC runtime does not (so you cannot print, scan, etc... making
them useless).
cheers,
David
>
>>>> np.array([3141592653589793238L], np.int64).astype(np.longdouble)[0]
> 3141592653589793300.0
>>>> np.array([3141592653589793238L], np.int64).astype(float)[0]
> 3.1415926535897933e+018
>>>> 1./np.array([np.pi],np.longdouble)[0] - 1/np.pi
> 0.0
>>>> 1./np.array([np.pi],np.longdouble)[0]
> 0.31830988618379069
>
> and it doesn't look like it's the print precision
>>>> 1./np.array([np.pi],np.longdouble)[0]*1e18
> 318309886183790720.0
>>>> 1./np.array([np.pi],float)[0]*1e18
> 3.1830988618379072e+017
>
>
> type conversion and calculations seem to go through float
>
> Josef
>
>>
>> Chuck
>>
>>
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