Simplification of double einsum

Question

I am trying to use numpy.einsumto simplify the loop that I have in my code.

My code currently looks something like this:

k = 100 
m = 50
n = 10
A = np.arange(k*m*n).reshape(k, m, n)
B = np.arange(m*m).reshape(m, m)

T = np.zeros((n, n)) 
for ind in xrange(k):
    T += np.dot(A[ind,:,:].T, np.dot(B, A[ind,:,:]))

I am trying to use numpy.einsumas an alternative to this loop:

Tp = np.einsum('nij,njk->ik', np.einsum('nij,jk->nik', A.transpose(0,2,1), B), A)
print np.allclose(T, Tp)

Is it possible to use one numpy.einsuminstead of two?

Answer 1

Really reasonable approaches that you got in another answer. I would like to add a numpy.dotbased approach to the mix, which also uses some reshaping. Here is one way to do this -

k,m,n = A.shape
((A.transpose(2,0,1).reshape(-1,m).dot(B.T)).reshape(n,-1)).dot(A.reshape(-1,n)).T

Runtime Tests -

, other answer numpy.dot, .

In [130]: k = 100 
     ...: m = 50
     ...: n = 10
     ...: A = np.arange(k*m*n).reshape(k, m, n)
     ...: B = np.arange(m*m).reshape(m, m)
     ...: 

In [131]: %timeit np.einsum('nij,njk->ik', np.einsum('nij,jk->nik', A.transpose(0,2,1), B), A)
100 loops, best of 3: 10.7 ms per loop

In [132]: %timeit np.einsum('nij, il, kln ->jk', A, B, A.T)
10 loops, best of 3: 105 ms per loop

In [133]: %timeit np.tensordot(A, np.tensordot(A, B, axes=(1, 1)), axes=((0, 1), (0, 2)))
100 loops, best of 3: 6.22 ms per loop

In [134]: %timeit ((A.transpose(2,0,1).reshape(-1,m).dot(B.T)).reshape(n,-1)).dot(A.reshape(-1,n)).T
100 loops, best of 3: 5.3 ms per loop

Answer 2

:

np.einsum('nij,njk->ik', np.einsum('nij,jk->nik', A.transpose(0,2,1), B), A)
# 100 loops, best of 3: 4.55 ms per loop

:

T2 = np.einsum('nij, il, kln ->jk', A, B, A.T)
#  10 loops, best of 3: 51.9 ms per loop

np.tensordot():

T3 = np.tensordot(A, np.tensordot(A, B, axes=(1, 1)), axes=((0, 1), (0, 2)))
# 100 loops, best of 3: 2.73 ms per loop

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