Oops, the subject line was somehow cut off. Please use this one if you follow up. - Huaiyu On Tue, 17 Sep 2002, Huaiyu Zhu wrote: Some of the choices between rank-0 arrays and new scalar types might be resolved by enumerating the properties desired of them. Most properties of rank-0 arrays could be fixed by consistency requirements alone, using operations that reduce array dimensions. Let a = ones((2,3,4)) b = sum(a) c = sum(b) d = sum(c) Property 1: the shape of an array is a tuple of integers a.shape == (2, 3, 4) b.shape == (3, 4) c.shape == (4,) d.shape == () Property 2: rank(a) == len(a.shape) rank(a) == 3 == len(a.shape) rank(b) == 2 == len(b.shape) rank(c) == 1 == len(c.shape) rank(d) == 0 == len(d.shape) Property 3: len(a) == a.shape[0] len(a) == 2 == a.shape[0] len(b) == 3 == b.shape[0] len(c) == 4 == c.shape[0] len(d) == Exception == d.shape[0] # Currently the last is wrong? Property 4: size(a) == product(a.shape) size(a) == 24 == product(a.shape) size(b) == 12 == product(b.shape) size(c) == 4 == product(c.shape) size(d) == 1 == product(d.shape) # Currently the last is wrong Property 5: rank-0 array behaves as mutable numbers when used as value array(2) is similar to 2 array(2.0) is similar to 2.0 array(2j) is similar to 2j # This is a summary of many concrete properties. Property 6: Indexing reduces rank. Slicing preserves rank. a[:,:,:].shape = (2, 3, 4) a[1,:,:].shape = (3, 4) a[1,1,:].shape = (4,) a[1,1,1].shape = () Property 7: Indexing by tuple of ints gives scalar. a[1,1,1] == 1 b[1,1] == 2 c[1,] == 6 d[()] == 24 # So rank-0 array indexed by empty tuple should be scalar. # Currently the last is wrong Property 8: Indexing by tuple of slices gives array. a[:,:,:] == ones((2,3,4)) b[:,:] == ones((3,4)) * 2 c[:] == ones((,4)) * 6 d[()] == ones(()) * 24 # So rank-0 array indexed by empty tuple should be rank-0 array. # Currently the last is wrong Property 9: Indexing as lvalues a[1,1,1] = 2 b[1,1] = 2 c[1,] = 2 d[()] = 2 Property 10: Indexing and slicing as lvalues a[:,:,:] = ones((2, 3, 4)) a[1,:,:] = ones((3, 4)) a[1,1,:] = ones((4,)) a[1,1,1] = ones(()) # But the last is wrong. Conclusion 1: rank-0 arrays are equivalent to scalars. See properties 7 and 8. Conclusion 2: rank-0 arrays are mutable. See property 9. Conclusion 3: shape(scalar), size(scalar) are all defined, but len(scalar) should not be defined. See conclusion 1 and properties 1, 2, 3, 4. Conclusion 4: A missing axis is similar to having dimension 1. See property 4. Conclusion 5: rank-0 int arrays should be allowed to act as indices. See property 5. Conclusion 6: rank-0 arrays should not be hashable except by object id. See conclusion 2. Discussions: - These properties correspond to the current implementation quite well, except a few rough edges. - Mutable scalars are useful in their own rights. - Is there substantial difference in overhead between rank-0 arrays and scalars? - How to write literal values? array(1) is too many characters. - For rank-1 and rank-0 arrays, Python notation distinguishes: c[1] vs c[1,] d[] vs d[()] Should these be used to handle semantic difference between indexing and slicing? Should d[] be syntactically allowed? Hope these observations help. Huaiyu