Use numpy.tensordot to replace a nested loop

Question

I have a piece of code, but I want to pull up the performance. My code is:

lis = []
for i in range(6):
    for j in range(6):
        for k in range(6):
             


        

Answer 1

You can just use a jit-compiler

Your solution isn't bad at all. The only thing I have changed is the indexing and variable loop ranges. If you have numpy arrays and excessive looping you can use a compiler (Numba), which is a really simple thing to do.

import numba as nb
import numpy as np
#The function is compiled only at the first call (with using same datatypes)
@nb.njit(cache=True) #set cache to false if copying the function to a command window
def almost_your_solution(matrix1,matrix2):
  lis = np.zeros(matrix1.shape,np.float64)
  for i in range(matrix2.shape[0]):
      for j in range(matrix2.shape[1]):
          for k in range(matrix2.shape[2]):
              for l in range(matrix2.shape[3]):
                  lis[i,j] += matrix1[k,l] * (2 * matrix2[i,j,k,l] - matrix2[i,k,j,l])

  return lis

Regarding code simplicity I would prefer the einsum solution from hpaulj over the solution shown above. The tensordot solution isn't that easy to understand to my opinion. But that's a a matter of taste.

Comparing performance

The function from hpaulj i used for comparison:

def hpaulj_1(matrix1,matrix2):
  matrix3 = 2*matrix2-matrix2.transpose(0,2,1,3)
  return np.einsum('kl,ijkl->ij', matrix1, matrix3)

def hpaulj_2(matrix1,matrix2):
  matrix3 = 2*matrix2-matrix2.transpose(0,2,1,3)
  (matrix1*matrix3).sum(axis=(2,3))
  return np.tensordot(matrix1, matrix3, [[0,1],[2,3]])

Very short arrays gives:

matrix1=np.random.rand(6,6)
matrix2=np.random.rand(6,6,6,6)

Original solution:    2.6 ms
Compiled solution:    2.1 µs
Einsum solution:      8.3 µs
Tensordot solution:   36.7 µs

Larger arrays gives:

matrix1=np.random.rand(60,60)
matrix2=np.random.rand(60,60,60,60)

Original solution:    13,3 s
Compiled solution:    18.2 ms
Einsum solution:      115  ms
Tensordot solution:   180  ms

Conclusion

Compilation speeds up the computation by about 3 orders of magnitude and outperforms all other solutions by quite a margin.

Answer 2

Test setup:

In [274]: lis = np.zeros((6,6),int)
In [275]: matrix1 = np.arange(36).reshape(6,6)
In [276]: matrix2 = np.arange(36*36).reshape(6,6,6,6)
In [277]: for i in range(6):
     ...:     for j in range(6):
     ...:         for k in range(6):
     ...:             for l in range(6):
     ...:                 lis[i,j] += matrix1[k,l] * (2 * matrix2[i,j,k,l] - mat
     ...: rix2[i,k,j,l])
     ...:                 
In [278]: lis
Out[278]: 
array([[-51240,  -9660,  31920,  73500, 115080, 156660],
       [ 84840, 126420, 168000, 209580, 251160, 292740],
       [220920, 262500, 304080, 345660, 387240, 428820],
       [357000, 398580, 440160, 481740, 523320, 564900],
       [493080, 534660, 576240, 617820, 659400, 700980],
       [629160, 670740, 712320, 753900, 795480, 837060]])

right?

I'm not sure that tensordot is the right tool; at least may not be the simplest. It certainly can't handle the matrix2 difference.

Let's start with an obvious substitution:

In [279]: matrix3 = 2*matrix2-matrix2.transpose(0,2,1,3)
In [280]: lis = np.zeros((6,6),int)
In [281]: for i in range(6):
     ...:     for j in range(6):
     ...:         for k in range(6):
     ...:             for l in range(6):
     ...:                 lis[i,j] += matrix1[k,l] * matrix3[i,j,k,l]

tests ok - same lis.

Now it is easy to express this with einsum - just replicate the indices

In [284]: np.einsum('kl,ijkl->ij', matrix1, matrix3)
Out[284]: 
array([[-51240,  -9660,  31920,  73500, 115080, 156660],
       [ 84840, 126420, 168000, 209580, 251160, 292740],
       [220920, 262500, 304080, 345660, 387240, 428820],
       [357000, 398580, 440160, 481740, 523320, 564900],
       [493080, 534660, 576240, 617820, 659400, 700980],
       [629160, 670740, 712320, 753900, 795480, 837060]])

elementwise product plus summation on two axes also works; and an equivalent tensordot (specifying which axes to sum over)

(matrix1*matrix3).sum(axis=(2,3))
np.tensordot(matrix1, matrix3, [[0,1],[2,3]])