Reputation: 151
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):
for l in range(6):
lis[i][j] += matrix1[k][l] * (2 * matrix2[i][j][k][l] - matrix2[i][k][j][l])
print(lis)
matrix2 is a 4-dimensional np-array, and matrix1 is a 2d-array.
I want to speed up this code by using np.tensordot(matrix1, matrix2), but then I'm lost.
Upvotes: 5
Views: 587
Reputation: 231385
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]])
The newer np.matmul/@
can also be used, but requires some reshaping
In [111]: (matrix1.ravel()[None,None,None,:]@matrix3.reshape(6,6,-1,1)).squeeze(
...: )
Out[111]:
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]])
This reduces the kl
dimensions down to one, and does 'broadcasting' on the ij
dimensions.
Upvotes: 1
Reputation: 6482
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.
Upvotes: 2