The NumPy Fortran-ordering quiz
Charles R Harris
charlesr.harris at gmail.com
Tue Oct 17 21:21:02 CDT 2006
On 10/17/06, Travis Oliphant <oliphant.travis at ieee.org> wrote:
>
> Charles R Harris wrote:
> >
> >
> > On 10/17/06, *Lisandro Dalcin* <dalcinl at gmail.com
> > <mailto:dalcinl at gmail.com>> wrote:
> >
> > I was surprised by this
> >
> > In [14]: array([[1,2,3],[4,5,6]]).reshape((3,2),order='F')
> > Out[14]:
> > array([[1, 5],
> > [4, 3],
> > [2, 6]])
> >
> >
> > This one still looks wrong.
> >
> > In [15]: array([1,2,3,4,5,6]).reshape((3,2),order='F')
> > Out[15]:
> > array([[1, 2],
> > [3, 4],
> > [5, 6]])
> >
> >
> >
> > This one is fixed,
> >
> > In [3]: array([[1,2,3,4,5,6]]).reshape((3,2),order='F')
> > Out[3]:
> > array([[1, 4],
> > [2, 5],
> > [3, 6]])
> >
> > I also don't understand why a copy is returned if 'F' just fiddles
> > with the indices and strides; the underlying data should be the same,
> > just the view changes. FWIW, I think both examples should be returning
> > views.
>
> You are right, it doesn't need to. My check is not general enough.
>
> It can be challenging to come up with a general way to differentiate the
> view-vs-copy situation and I struggled with it. In this case, it's the
> fact that while self->nd > 1, the other dimensions are only of shape 1
> and so don't really matter. If you could come up with some kind of
> striding check that would distinguish the two cases, I would appreciate
> it.
I suppose the problem is mostly in discontiguous arrays. Hmmm..., this isn't
too different that reshaping the transpose.
a.reshape((m,n),order='F' ~ a.reshape((n,m)).T.reshape(m,n)
for instance:
In [26]: a
Out[26]:
array([[1, 2, 3],
[4, 5, 6]])
In [27]: a.reshape((3,2)).T.reshape((2,3))
Out[27]:
array([[1, 3, 5],
[2, 4, 6]])
In [28]: a.reshape((2,3), order='F')
Out[28]:
array([[1, 2, 3],
[4, 5, 6]])
Where I actually think the second reshape is correct and the third
incorrect. This has the advantage that *all* the steps return views.
Chuck
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