Hello, I'm sure there's a simple solution, but I'm not seeing it so any hints would be greatly appreciated. I have an array "A" of shape (NX, NY, NZ), and then I have a second array "B" of shape (NX, NY) that ranges from 0 to NZ in value. I want to create a third array "C" of shape (NX, NY) that holds the "B"-th slice for each (NX, NY). For example: A = numpy.zeros((NX,NY,NZ)) A[:,:,0] = numpy.arange(0, NX*NY).reshape(NX,NY) A[:,:,1] = numpy.arange(1, NX*NY+1).reshape(NX,NY) and so on B = numpy.zeros((NX,NY)) B[0,0] = 0 B[0,1] = 1 so C[0,0] = A[0,0,B[0,0]] C[0,1] = A[0,1,B[0,1]] ... C[NX-1,NY-1] = A[NX-1,NY-1,B[NX-1,NY-1]] and C has shape(NX,NY) How do I accomplish this without loops? Thank-you for any advice, Catherine
Hi Catherine
On 2014-12-04 01:12:30, Moroney, Catherine M (398E)
I have an array "A" of shape (NX, NY, NZ), and then I have a second array "B" of shape (NX, NY) that ranges from 0 to NZ in value.
I want to create a third array "C" of shape (NX, NY) that holds the "B"-th slice for each (NX, NY)
Those two arrays can broadcast if you expand the dimensions of B: A: (NX, NY, NZ) B: (NX, NY, 1) Your result would be B = B[..., np.newaxis] # now shape (NX, NY, 1) C = A[B] For more information on this type of broadcasting manipulation, see http://nbviewer.ipython.org/github/stefanv/teaching/blob/master/2014_assp_sp... and http://wiki.scipy.org/EricsBroadcastingDoc Stéfan
Posting in the correct thread now...
A slightly different way to look at it (I don't think it is exactly the
same problem, but the description reminded me of it):
http://mail.scipy.org/pipermail/numpy-discussion/2013-April/066269.html
(and I think there are some things that can be done to make that faster,
but I don't recall it right now)
Ben Root
On Wed, Dec 3, 2014 at 7:02 PM, Stefan van der Walt
Hi Catherine
On 2014-12-04 01:12:30, Moroney, Catherine M (398E) < Catherine.M.Moroney@jpl.nasa.gov> wrote:
I have an array "A" of shape (NX, NY, NZ), and then I have a second array "B" of shape (NX, NY) that ranges from 0 to NZ in value.
I want to create a third array "C" of shape (NX, NY) that holds the "B"-th slice for each (NX, NY)
Those two arrays can broadcast if you expand the dimensions of B:
A: (NX, NY, NZ) B: (NX, NY, 1)
Your result would be
B = B[..., np.newaxis] # now shape (NX, NY, 1) C = A[B]
For more information on this type of broadcasting manipulation, see
http://nbviewer.ipython.org/github/stefanv/teaching/blob/master/2014_assp_sp...
and
http://wiki.scipy.org/EricsBroadcastingDoc
Stéfan _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
On Wed, Dec 3, 2014 at 4:02 PM, Stefan van der Walt
Hi Catherine
On 2014-12-04 01:12:30, Moroney, Catherine M (398E) < Catherine.M.Moroney@jpl.nasa.gov> wrote:
I have an array "A" of shape (NX, NY, NZ), and then I have a second array "B" of shape (NX, NY) that ranges from 0 to NZ in value.
I want to create a third array "C" of shape (NX, NY) that holds the "B"-th slice for each (NX, NY)
Those two arrays can broadcast if you expand the dimensions of B:
A: (NX, NY, NZ) B: (NX, NY, 1)
Your result would be
B = B[..., np.newaxis] # now shape (NX, NY, 1) C = A[B]
For more information on this type of broadcasting manipulation, see
http://nbviewer.ipython.org/github/stefanv/teaching/blob/master/2014_assp_sp...
and
I don't think this would quite work... Even though it now has 3 dimensions (Nx, Ny, 1), B is still a single array, so when fancy indexing A with it, it will only be applied to the first axis, so the return will be of shape B.shape + A.shape[1:], that is (Nx, Ny, 1, Ny, Nz). What you need to have is three indexing arrays, one per dimension of A, that together broadcast to the desired shape of the output C. For this particular case, you could do: nx = np.arange(A.shape[0])[:, np.newaxis] ny = np.arange(A.shape[1]) C = A[nx, ny, B] To show that this works:
A = np.arange(2*3*4).reshape(2, 3, 4) A array([[[ 0, 1, 2, 3], [ 4, 5, 6, 7], [ 8, 9, 10, 11]],
[[12, 13, 14, 15], [16, 17, 18, 19], [20, 21, 22, 23]]])
B = np.random.randint(4, size=(2, 3)) B array([[2, 1, 2], [2, 0, 3]]) nx = np.arange(A.shape[0])[:, None] ny = np.arange(A.shape[1]) A[nx, ny, B] array([[ 2, 5, 10], [14, 16, 23]])
Jaime -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial.
On 2014-12-04 03:41:35, Jaime Fernández del Río
nx = np.arange(A.shape[0])[:, np.newaxis] ny = np.arange(A.shape[1]) C = A[nx, ny, B]
That's the correct answer--in my answer I essentially wrote C = A[B] (== A[B, :, :]) which broadcasts the shape of B against the second and third dimensions of A (it's almost always a bad idea to combine index broadcasting and slicing). The notes I linked to are correct, though, and explain Jamie's answer in more detail (search for "Jack's Dilemma"). Regards Stéfan
participants (4)
-
Benjamin Root -
Jaime Fernández del Río -
Moroney, Catherine M (398E) -
Stefan van der Walt