Creating 3rd order tensors with python and numpy

Question

I have a two 1 dimensional arrays, a such that np.shape(a) == (n,) and b such that np.shape(b) == (m,).

I want to make a (3rd order) tensor c such that np.shape(c) == (n,n,m,)by doing c = np.outer(np.outer(a,a),b).

But when I do this, I get:

>> np.shape(c) 

(n*n,m)

which is just a rectangular matrix. How can I make a 3D tensor like I want?

np.outer flattens the input as described in the docs... are you maybe searching for np.kron? — Jörn Hees
– Jörn Hees, Commented May 19, 2015 at 11:32
So it appears, through experimentation in ipython, that np.kron(b,np.kron(a,a)).reshape(m,n,n) gives what I wanted, although indeces are reversed. But that's clean so I'll use this I think. — user27886
– user27886, Commented May 19, 2015 at 11:41

Alex Riley · Accepted Answer · 2015-05-19 11:39:45Z

3

You could perhaps use np.multiply.outer instead of np.outer to get the required outer product:

>>> a = np.arange(4)
>>> b = np.ones(5)
>>> mo = np.multiply.outer

Then we have:

>>> mo(mo(a, a), b).shape
(4, 4, 5)

A better way could be to use np.einsum (this avoids creating intermediate arrays):

>>> c = np.einsum('i,j,k->ijk', a, a, b)
>>> c.shape
(4, 4, 5)

answered May 19, 2015 at 11:39

Alex Riley

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1 Comment

I like this better than using np.kron as it retains the shape information.