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up vote 6 down vote favorite
2

Given two numpy.arrays a and b,

c = numpy.outer(a, b)

returns an two-dimensional array where c[i, j] == a[i] * b[j]. Now, imagine a having k dimensions.

  • Which operation returns an array c of dimension k+1 where c[..., j] == a * b[j]?

Additionally, let b have l dimensions.

  • Which operation returns an array c of dimension k+1 where c[..., i1, i2, i3] == a * b[i1, i2, i3]?
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up vote 4 down vote accepted

The outer method of NumPy ufuncs treats multidimensional input the way you want, so you could do

numpy.multiply.outer(a, b)

rather than using numpy.outer.

All solutions suggested here are equally fast; for small arrays, multiply.outer has a slight edge

enter image description here

Code for generating the image:

import numpy
import perfplot


def multiply_outer(data):
    a, b = data
    return numpy.multiply.outer(a, b)


def outer_reshape(data):
    a, b = data
    return numpy.outer(a, b).reshape((a.shape + b.shape))


def tensor_dot(data):
    a, b = data
    return numpy.tensordot(a, b, 0)


perfplot.show(
        setup=lambda n: (numpy.random.rand(n, n), numpy.random.rand(n, n)),
        kernels=[multiply_outer, outer_reshape, tensor_dot],
        n_range=[2**k for k in range(7)],
        logx=True,
        logy=True,
        )
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up vote 2 down vote

One approach would be using np.outer and then reshape -

np.outer(a,b).reshape((a.shape + b.shape))
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up vote 2 down vote

I think np.tensordot also works

c = np.tensordot(a, b, 0)

inds = np.reshape(np.indices(b.shape), (b.ndim, -1))
for ind in inds.T:
    ind = tuple(ind)
    assert np.allclose(a * b[ind], c[(...,) + ind])
else:
    print('no error')
# no error
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up vote 1 down vote

I think you're looking for kroneker product

for example

> np.kron(np.eye(2), np.ones((2,2)))

array([[ 1.,  1.,  0.,  0.],
       [ 1.,  1.,  0.,  0.],
       [ 0.,  0.,  1.,  1.],
       [ 0.,  0.,  1.,  1.]])
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up vote 1 down vote

np.einsum is what you are looking for.

c[..., j] == a * b[j]

should be

c = np.einsum('...i,j -> ...ij', a, b)

and c[..., i1, i2, i3] == a * b[i1, i2, i3] should be

c = np.einsum('i,...jkl -> ...ijkl', a, b)

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