numpy.outer without flatten

Question

$x$ is an $N \times M$ matrix.

$y$ is a $1 \times L$ vector.

I want to return "outer product" between $x$ and $y$, let's call it $z$.

z[n,m,l] = x[n,m] * y[l]

I could probably do this using einsum.

np.einsum("ij,k->ijk", x[:, :, k], y[:, k])

or reshape afterwards.

 np.outer(x[:, :, k], y).reshape((x.shape[0],x.shape[1],y.shape[0]))

But I'm thinking of doing this in np.outer only or something seems simpler, memory efficient.

Is there a way?

$\begingroup$ What does the k variable contain in your code examples? $\endgroup$ – Federico Poloni Jul 15 at 7:52 — Federico Poloni, Jul 15 at 7:52
$\begingroup$ also cross-posted on StackOverflow with an accepted answer $\endgroup$ – Anton Menshov♦ Jul 15 at 17:30 — Anton Menshov♦, Jul 15 at 17:30

LedHead · Accepted Answer · 2019-07-15 06:29:00Z

I think numpy.tensordot does what you need.

import numpy as np

N=2
M=3
L=4

x=np.arange(N*M).reshape(N,M)
y=np.arange(L)
z=np.tensordot(x,y,axes=0)

print('x=',x)
print('y=',y)
print('z=',z)

x= [[0 1 2]
 [3 4 5]]
y= [0 1 2 3]
z= [[[ 0  0  0  0]
  [ 0  1  2  3]
  [ 0  2  4  6]]

 [[ 0  3  6  9]
  [ 0  4  8 12]
  [ 0  5 10 15]]]

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