Hi Robert 2008/7/17 Robert Kern <robert.kern@gmail.com>:
In [42]: smallcube = cube[idx_i,idx_j,idx_k]
Fantastic -- a good way to warm up the brain-circuit in the morning! Is there an easy-to-remember rule that predicts the output shape of the operation above? I'm trying to imaging how the output would change if I altered the dimensions of idx_i or idx_j, but it's hard. It looks like you can do all sorts of interesting things by manipulation the indices. For example, if I take In [137]: x = np.arange(12).reshape((3,4)) I can produce either In [138]: x[np.array([[0,1]]), np.array([[1, 2]])] Out[138]: array([[1, 6]]) or In [140]: x[np.array([[0],[1]]), np.array([[1], [2]])] Out[140]: array([[1], [6]]) and even In [141]: x[np.array([[0],[1]]), np.array([[1, 2]])] Out[141]: array([[1, 2], [5, 6]]) or its transpose In [143]: x[np.array([[0,1]]), np.array([[1], [2]])] Out[143]: array([[1, 5], [2, 6]]) Is it possible to separate the indexing in order to understand it better? My thinking was cube_i = cube[idx_i,:,:].squeeze() cube_j = cube_i[:,idx_j,:].squeeze() cube_k = cube_j[:,:,idx_k].squeeze() Not sure what would happen if the original array had single dimensions, though. Back to the original problem: In [127]: idx_i.shape Out[127]: (10, 1, 1) In [128]: idx_j.shape Out[128]: (1, 15, 1) In [129]: idx_k.shape Out[129]: (10, 15, 7) For the constant slice case, I guess idx_k also have been (1, 1, 7)? The construction of the cube could probably be done using only cube.flat = np.arange(nk) Fernando is right: this is good food for thought and excellent cookbook material! Regards Stéfan