> >> On Tue, Mar 29, 2016 at 1:46 PM, Benjamin Root <ben.v.r...@gmail.com> > >> wrote: > >> > Is there a quick-n-easy way to reflect a NxM array that represents a > >> > quadrant into a 2Nx2M array? Essentially, I am trying to reduce the size > >> > of > >> > an expensive calculation by taking advantage of the fact that the first > >> > part > >> > of the calculation is just computing gaussian weights, which is radially > >> > symmetric. > >> > > >> > It doesn't seem like np.tile() could support this (yet?). Maybe we could > >> > allow negative repetitions to mean "reflected"? But I was hoping there > >> > was > >> > some existing function or stride trick that could accomplish what I am > >> > trying. > >> > > >> > x = np.linspace(-5, 5, 20) > >> > y = np.linspace(-5, 5, 24) > >> > z = np.hypot(x[None, :], y[:, None]) > >> > zz = np.hypot(x[None, :int(len(x)//2)], y[:int(len(y)//2), None]) > >> > zz = some_mirroring_trick(zz) > >> > > You can avoid the allocation with preallocation: > > nx = len(x) // 2 > ny = len(y) // 2 > zz = np.zeros((len(y), len(x))) > zz[:ny,-nx:] = np.hypot.outer(y[:ny], x[:nx]) > zz[:ny, :nx] = zz[:ny,:-nx-1:-1] > zz[-ny:, :] = zz[ny::-1, :] > > if nx * 2 != len(x): > zz[:ny, nx] = y[::-1] > zz[-ny:, nx] = y > if ny * 2 != len(y): > zz[ny, :nx] = x[::-1] > zz[ny, -nx:] = x > > All of the steps after the call to `hypot.outer` create views. This is > untested, so you may need to tweak the indices a little. >
A couple of months ago I wrote a C-code with ctypes to do this sort of mirroring trick on an (N,N,N) numpy array of fft weights (where you can exploit the 48-fold symmetry of using interchangeable axes), which was pretty useful since I had N^3 >> 1e9 and the weight function was quite expensive. Obviously the (N,M) case doesn't allow quite so much optimization but if it could be interesting then PM me. Best, Peter _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org https://mail.scipy.org/mailman/listinfo/numpy-discussion
