On 08/19/2010 04:20 PM, josef.p...@gmail.com wrote:
On Thu, Aug 19, 2010 at 4:03 PM, John Salvatier
wrote:
Precise in what sense? Numerical accuracy? If so, why is that?
I don't remember where I ran into this example, maybe integer
underflow (?) with addition.
NIST ANOVA test cases have some
On Thu, Aug 19, 2010 at 4:03 PM, John Salvatier
wrote:
> Precise in what sense? Numerical accuracy? If so, why is that?
I don't remember where I ran into this example, maybe integer
underflow (?) with addition.
NIST ANOVA test cases have some nasty badly scaled variables
but I have problems crea
Word size.
imagine 2.123456789 vs 22.12345679
the more that can stored to the right of the decimal, the more precise.
That larger the number is on the left, the less that can stored on the right.
On Thu, Aug 19, 2010 at 4:03 PM, John Salvatier
wrote:
> Precise in what sense? Numerical accuracy
Precise in what sense? Numerical accuracy? If so, why is that?
On Thu, Aug 19, 2010 at 12:13 PM, wrote:
> On Thu, Aug 19, 2010 at 11:29 AM, Joe Harrington
> wrote:
> > On Thu, 19 Aug 2010 09:06:32 -0500, G?khan Sever
> wrote:
> >
> >>On Thu, Aug 19, 2010 at 9:01 AM, greg whittier wrote:
> >>
On Thu, Aug 19, 2010 at 11:29 AM, Joe Harrington wrote:
> On Thu, 19 Aug 2010 09:06:32 -0500, G?khan Sever
> wrote:
>
>>On Thu, Aug 19, 2010 at 9:01 AM, greg whittier wrote:
>>
>>> I frequently deal with 3D data and would like to sum (or find the
>>> mean, etc.) over the last two axes. I.e. su
On Thu, 19 Aug 2010 09:06:32 -0500, G?khan Sever wrote:
>On Thu, Aug 19, 2010 at 9:01 AM, greg whittier wrote:
>
>> I frequently deal with 3D data and would like to sum (or find the
>> mean, etc.) over the last two axes. I.e. sum a[i,j,k] over j and k.
>> I find using .sum() really convenient f
greg whittier gmail.com> writes:
>
> a = np.arange(27).reshape(3,3,3)
>
> # sum over axis 1 and 2
> result = a.reshape((a.shape[0], a.shape[1]*a.shape[2])).sum(axis=1)
>
> Is there a cleaner way to do this? I'm sure I'm missing something obvious.
>
Only slightly more convenient is:
a.reshape
On Thu, Aug 19, 2010 at 10:12 AM, Angus McMorland wrote:
> Another rank-generic approach is to use apply_over_axes (you get a
> different shape to the result this way):
>
> a = np.random.randint(20, size=(4,3,5))
> b = np.apply_over_axes(np.sum, a, [1,2]).flat
> assert( np.all( b == a.sum(axis=2).
On 19 August 2010 10:01, greg whittier wrote:
> I frequently deal with 3D data and would like to sum (or find the
> mean, etc.) over the last two axes. I.e. sum a[i,j,k] over j and k.
> I find using .sum() really convenient for 2d arrays but end up
> reshaping 2d arrays to do this. I know there
On Thu, Aug 19, 2010 at 9:01 AM, greg whittier wrote:
> I frequently deal with 3D data and would like to sum (or find the
> mean, etc.) over the last two axes. I.e. sum a[i,j,k] over j and k.
> I find using .sum() really convenient for 2d arrays but end up
> reshaping 2d arrays to do this. I kn
Hi,
Couldn't you do it with several sum steps? E.g.:
result = array.sum(axis=1).sum(axis=2)
Ian
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I frequently deal with 3D data and would like to sum (or find the
mean, etc.) over the last two axes. I.e. sum a[i,j,k] over j and k.
I find using .sum() really convenient for 2d arrays but end up
reshaping 2d arrays to do this. I know there has to be a more
convenient way. Here's what I'm doing
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