On Tue, 25 Aug 2009 08:26:44 -0700, Scott David Daniels
wrote:
>Matjaz Bezovnik wrote:
>
>If you are using numpy (which it sounds like you are):
>
>IDLE 2.6.2
> >>> import numpy as np
> >>> v = np.array([[0,1,2],[3,4,5],[6,7,8]], dtype=float)
> >>> v
>array([[ 0., 1., 2.],
>[ 3., 4., 5.],
>[ 6., 7., 8.]])
> >>> w = np.array([[10,11,12],[13,14,15],[16,17,18]], dtype=float)
> >>> w
>array([[ 10., 11., 12.],
>[ 13., 14., 15.],
>[ 16., 17., 18.]])
> >>> r = np.zeros((6,6))
> >>> r
>array([[ 0., 0., 0., 0., 0., 0.],
>[ 0., 0., 0., 0., 0., 0.],
>[ 0., 0., 0., 0., 0., 0.],
>[ 0., 0., 0., 0., 0., 0.],
>[ 0., 0., 0., 0., 0., 0.],
>[ 0., 0., 0., 0., 0., 0.]])
> >>> r[:3,:3] = v
> >>> r
>array([[ 0., 1., 2., 0., 0., 0.],
>[ 3., 4., 5., 0., 0., 0.],
>[ 6., 7., 8., 0., 0., 0.],
>[ 0., 0., 0., 0., 0., 0.],
>[ 0., 0., 0., 0., 0., 0.],
>[ 0., 0., 0., 0., 0., 0.]])
> >>> r[3:,3:] = w
> >>> r
>array([[ 0., 1., 2., 0., 0., 0.],
>[ 3., 4., 5., 0., 0., 0.],
>[ 6., 7., 8., 0., 0., 0.],
>[ 0., 0., 0., 10., 11., 12.],
>[ 0., 0., 0., 13., 14., 15.],
>[ 0., 0., 0., 16., 17., 18.]])
> >>>
>
>In general, make the right-sized array of zeros, and at various points:
>and you can ssign to subranges of the result array:
>
> N = 3
> result = np.zeros((len(parts) * N, len(parts) * N), dtype=float)
> for n, chunk in enumerate(parts):
> base = n * 3
> result[base : base + 3, base : base + 3] = chunk
>
>--Scott David Daniels
>scott.dani...@acm.org
Scott, thank you very much for the snippet.
It is exactly what I looked for; simple to read and obvious as to what
it does even a month later to a non-pythonist!
Matjaz
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