Computing a third order tensor efficiently in python

Question

If x denotes a d dimensional column vector and e_i denotes the ith standard basis in the Euclidean space R^d, I would like to compute the following third-order tensor in python in an efficient way:

Here, the circled-cross refers to the tensor outer product. For now I have been relying on the following rather inefficient code:

import numpy as np
from sktensor import ktensor,dtensor

d=5
x= np.random.normal(0,1,(d,1))

z= np.zeros((d,1))
I= np.identity(d)

T1= ktensor([x,x,x])
T2= ktensor([z,z,z])
T3= ktensor([z,z,z])
T4= ktensor([z,z,z])

for j in range(d):
    T2 = T2+ ktensor([I(:,j),I(:,j),x]
    T3 = T3+ ktensor([I(:,j),x,I(:,j)]
    T4 = T4+ ktensor([x,I(:,j),I(:,j)]

T= T1-T2-T3-T4

I'd suggest moving away from sktensor for a start, seems poorly maintained (I can't seem to get it running to compare). You could write this in pure numpy or maybe have a look at tensorflow. — piman314
– piman314, Commented Dec 18, 2017 at 16:40
@ncfirth: What would be the implementation of this quantity in numpy? — pikachuchameleon
– pikachuchameleon, Commented Dec 19, 2017 at 2:51
I'll write an answer in numpy, but I can't test whether it's faster as I can't install sktensor. — piman314
– piman314, Commented Dec 19, 2017 at 10:05
@ncfirth: It's possible to install sktensor in Ubuntu. Windows I am not sure. You need a distribution called nose and afterwards it's straightforward. — pikachuchameleon
– pikachuchameleon, Commented Dec 19, 2017 at 16:32

piman314 · Accepted Answer · 2017-12-19 10:07:05Z

1

numpy based answer as request by OP's comment. Not tested against sktensor

import numpy as np
np.random.seed(42)
d = 5
x = np.random.normal(0,1, size=(d,1))
I = np.identity(d)

ans = np.outer(x, np.outer(x, x))
for i in range(d):
    ans -= np.outer(I[:, i], np.outer(I[:, i], x))
    ans -= np.outer(I[:, i], np.outer(x, I[:, i]))
    ans -= np.outer(x, np.outer(I[:, i], I[:, i]))

answered Dec 19, 2017 at 10:07

piman314

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