[Numpy-discussion] numpy.fft.rfftn -> non-contiguous
Pierre Thibault
thibault at physics.cornell.edu
Fri Dec 1 22:09:32 CST 2006
Hello!
I'm a little confused about what rfftn is doing: It seems to me that
the best would be for it to return a C-contiguous array with the first
element reduced by a half (plus one), so that one can easily obtain
the non-repeated slices.
What I get is the following:
In [1]: from numpy import *
In [2]: a = random.rand(8,8,8)
In [3]: fa = fft.rfftn(a)
In [4]: fa.shape
Out[4]: (8, 8, 5)
In [5]: fa.flags
Out[5]:
CONTIGUOUS : False
FORTRAN : False
OWNDATA : False
WRITEABLE : True
ALIGNED : True
UPDATEIFCOPY : False
In [6]: fa.swapaxes(-1,-2).flags
Out[6]:
CONTIGUOUS : True
FORTRAN : False
OWNDATA : False
WRITEABLE : True
ALIGNED : True
UPDATEIFCOPY : False
What I would like to have is
fa.shape: (5,8,8)
fa.flags: CONTIGUOUS = True
So, now fa[0] is a contiguous block containing the data that is not
supposed to appear twice in the complex fft (that is, it would be nice
if fft.rfftn(a)[0] == fft.fftn(a)[0]).
I tried playing with the axes argument in rfftn, to no avail. I can do
without, but at least it looks kind of ugly to me that the array
returned by a built-in function is not contiguous.
By the way, I am using the official debian unstable numpy package. It
doesn't look like it is using fftw, and I don't know if any of this
behaviour would be different if I had compiled numpy by myself - I am
not even sure using fftw is an option with numpy.
Thanks for any answer/comments on that!
Pierre
--
Pierre Thibault
616 Clark Hall, Cornell University
(607) 255-5522
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