Re: [Numpy-discussion] What does Fortran order mean?

[Stefan v. d. W. <st...@su...>]

On Tue, Oct 17, 2006 at 10:01:51AM -0600, Travis Oliphant wrote:
> Charles R Harris wrote:
>=20
> > Travis,
> >
> > I note that
> >
> > >>> a =3D arange(6).reshape(2,3,order=3D'F')
> > >>> a
> > array([[0, 1, 2],
> >        [3, 4, 5]])
> >
> > Shouldn't that be 3x2? Or maybe [[0,2,4],[1,3,5]]? Reshape is making =
a=20
> > copy, but flat, flatten, and tostring all show the elements in 'C'=20
> > order. I ask because I wonder if changing the order can be used to=20
> > prepare arrays for input into the LaPack routines.
>=20
>=20
> The order argument to reshape means (how should the big-chunk of data b=
e=20
> interpreted when you reshape).  So, yes, this should be=20
> [[0,2,4],[1,3,5]].  It is a bug that it does not do the right thing in=20
> this case.

A bit counter-ituitive (I somehow expect reshape to return an array
that satisfies all the constraints specified as parameters -- i.e.
shape and order in memory), but consistent with Numeric.

What confuses me is that, if you call the array constructor, you get

In [2]: N.array(N.array([[1,2,3],[4,5,6]]),order=3D'F')
Out[2]:=20
array([[1, 2, 3],
       [4, 5, 6]])

so here 'order' means something else.

So then you do

x =3D N.array([[0,1,2],[3,4,5]],order=3D'F')
x.reshape((2,3),order=3D'C')

and you get

array([[0, 1, 2],
       [3, 4, 5]])

Hmm.

Cheers
St=E9fan