On Thu, Aug 11, 2011 at 7:15 PM, Fernando Perez <fperez....@gmail.com>wrote:
> On Thu, Aug 11, 2011 at 4:43 PM, Jose Borreguero <borregu...@gmail.com> > wrote: > > a = random.randn(3,3) > > b = a.reshape(1,3,3).repeat(50,axis=0) > > scipy.linalg.block_diag( *b ) > > > > slightly simpler, but equivalent, code: > > b = [a]*50 > scipy.linalg.block_diag( *b) > > The following is unnecessarily complicated--using block_diag is fine--but it can be fun to stretch out into the fourth dimension with stride tricks: from numpy import array, zeros from numpy.lib.stride_tricks import as_strided # N is the number of 3x3 blocks. # N = 50 N = 4 a = array([[1,2,3],[4,5,6],[7,8,9]]) # b will be the block-diagonal array. b = zeros((3*N, 3*N), dtype=a.dtype) bstr = b.strides c = as_strided(b, shape=(N,N,3,3), strides=(3*bstr[0], 3*bstr[1], bstr[0], bstr[1])) # Assign a to the diagonal blocks. c[range(N), range(N)] = a print b Output: [[1 2 3 0 0 0 0 0 0 0 0 0] [4 5 6 0 0 0 0 0 0 0 0 0] [7 8 9 0 0 0 0 0 0 0 0 0] [0 0 0 1 2 3 0 0 0 0 0 0] [0 0 0 4 5 6 0 0 0 0 0 0] [0 0 0 7 8 9 0 0 0 0 0 0] [0 0 0 0 0 0 1 2 3 0 0 0] [0 0 0 0 0 0 4 5 6 0 0 0] [0 0 0 0 0 0 7 8 9 0 0 0] [0 0 0 0 0 0 0 0 0 1 2 3] [0 0 0 0 0 0 0 0 0 4 5 6] [0 0 0 0 0 0 0 0 0 7 8 9]] Warren Cheers, > > f > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >
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