Compute the outer product of two vectors.
Given two vectors, a = [a0, a1, ..., aM] and b = [b0, b1, ..., bN], the outer product [1] is:
[[a0*b0 a0*b1 ... a0*bN ] [a1*b0 . [ ... . [aM*b0 aM*bN ]]
>>> rl = np.outer(np.ones((5,)), np.linspace(-2, 2, 5))
>>> rl
array([[-2., -1., 0., 1., 2.],
[-2., -1., 0., 1., 2.],
[-2., -1., 0., 1., 2.],
[-2., -1., 0., 1., 2.],
[-2., -1., 0., 1., 2.]])
>>> im = np.outer(1j*np.linspace(2, -2, 5), np.ones((5,)))
>>> im
array([[0.+2.j, 0.+2.j, 0.+2.j, 0.+2.j, 0.+2.j],
[0.+1.j, 0.+1.j, 0.+1.j, 0.+1.j, 0.+1.j],
[0.+0.j, 0.+0.j, 0.+0.j, 0.+0.j, 0.+0.j],
[0.-1.j, 0.-1.j, 0.-1.j, 0.-1.j, 0.-1.j],
[0.-2.j, 0.-2.j, 0.-2.j, 0.-2.j, 0.-2.j]])
>>> grid = rl + im
>>> grid
array([[-2.+2.j, -1.+2.j, 0.+2.j, 1.+2.j, 2.+2.j],
[-2.+1.j, -1.+1.j, 0.+1.j, 1.+1.j, 2.+1.j],
[-2.+0.j, -1.+0.j, 0.+0.j, 1.+0.j, 2.+0.j],
[-2.-1.j, -1.-1.j, 0.-1.j, 1.-1.j, 2.-1.j],
[-2.-2.j, -1.-2.j, 0.-2.j, 1.-2.j, 2.-2.j]])
>>> x = np.array(['a', 'b', 'c'], dtype=object)
>>> np.outer(x, [1, 2, 3])
array([['a', 'aa', 'aaa'],
['b', 'bb', 'bbb'],
['c', 'cc', 'ccc']], dtype=object)
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