Just to be sure, you CAN'T expect sum to always have the same result in
float{32,64,128}. This is a properties of float numbers, but if you add
precission, you can hide this problem in some/most case.
Fred
On Sun, Aug 25, 2013 at 4:30 PM, Matthieu Brucher <
matthieu.bruc...@gmail.com> wrote:
> You can expect sum to always have the same result with this precision.
> Depending on how the sum is done (first to last or last to first for
> instance), you will have this kind of results.
> Le 25 août 2013 08:47, "Peter Prettenhofer" <peter.prettenho...@gmail.com>
> a écrit :
>
>
>>
>>
>> 2013/8/24 Lars Buitinck <l.j.buiti...@uva.nl>
>>
>>> 2013/8/24 Peter Prettenhofer <peter.prettenho...@gmail.com>:
>>> > can anybody help me understand why the output of the following code
>>> snippet
>>> > is different depending on 32bit or 64bit architecture::
>>> >
>>> > x = np.empty((10 ** 6,), dtype=np.float64)
>>> > x.fill(1e-9)
>>> > hash(x.mean())
>>> >
>>> > on 64bit I get: 2475364768
>>> > on 32bit I get: -1839780448
>>> >
>>> > I expected that given you set an explicit dtype the result would be
>>> equal on
>>> > whatever architecture.
>>>
>>> Is that Python's built-in hash function? That's not even guaranteed to
>>> be consistent between runs of the same Python interpreter on the same
>>> machine.
>>>
>>
>> yes - I used the builtin hash to show its different (if you print the
>> floating point you can see the difference as well).
>> If hash would not be consistent than pickling dicts would not be a good
>> idea... ?
>>
>>>
>>> > Even worse when I do the following::
>>> >
>>> > np.sum(x)
>>> >
>>> > I too get different results but when I do::
>>> >
>>> > sum(x)
>>> >
>>> > I get equal results... could it be that the temporary variables that
>>> numpy
>>> > uses in this routines are platform dependent or am I missing something
>>> here.
>>>
>>> How unequal? IIRC, x86 processors do all computations using 80-bit
>>> floating point numbers internally, and amd64 (x86-64) uses 128-bit.
>>>
>>
>> on 32bit machine:
>>
>> In [6]: print('%.16f' % sum(x))
>> 0.0010000000000082
>>
>> In [7]: print('%.16f' % np.sum(x))
>> 0.0010000000000000
>>
>>
>>
>>> --
>>> Lars Buitinck
>>> Scientific programmer, ILPS
>>> University of Amsterdam
>>>
>>>
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>>
>>
>>
>> --
>> Peter Prettenhofer
>>
>>
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