numpy.outer — NumPy v1.13 Manual (original) (raw)

Compute the outer product of two vectors.

Given two vectors, a = [a0, a1, ..., aM] andb = [b0, b1, ..., bN], the outer product [R60] is:

[[a0b0 a0b1 ... a0bN ] [a1b0 . [ ... . [aMb0 aMbN ]]

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)