What is the best way to compute the trace of a matrix product in numpy?

Question

If I have numpy arrays A and B, then I can compute the trace of their matrix product with:

tr = numpy.linalg.trace(A.dot(B))


        

Answer 1

You can improve on @Bill's solution by reducing intermediate storage to the diagonal elements only:

from numpy.core.umath_tests import inner1d

m, n = 1000, 500

a = np.random.rand(m, n)
b = np.random.rand(n, m)

# They all should give the same result
print np.trace(a.dot(b))
print np.sum(a*b.T)
print np.sum(inner1d(a, b.T))

%timeit np.trace(a.dot(b))
10 loops, best of 3: 34.7 ms per loop

%timeit np.sum(a*b.T)
100 loops, best of 3: 4.85 ms per loop

%timeit np.sum(inner1d(a, b.T))
1000 loops, best of 3: 1.83 ms per loop

Another option is to use np.einsum and have no explicit intermediate storage at all:

# Will print the same as the others:
print np.einsum('ij,ji->', a, b)

On my system it runs slightly slower than using inner1d, but it may not hold for all systems, see this question:

%timeit np.einsum('ij,ji->', a, b)
100 loops, best of 3: 1.91 ms per loop

Answer 2

Note that one slight variant is to take the dot product of the vectorized matrices. In python, vectorization is done using .flatten('F'). It's slightly slower than taking the sum of the Hadamard product, on my computer, so it's a worse solution than wflynny's , but I think it's kind of interesting, since it can be more intuitive, in some situations, in my opinion. For example, personally I find that for the matrix normal distribution, the vectorized solution is easier for me to understand.

Speed comparison, on my system:

import numpy as np
import time

N = 1000

np.random.seed(123)
A = np.random.randn(N, N)
B = np.random.randn(N, N)

tart = time.time()
for i in range(10):
    C = np.trace(A.dot(B))
print(time.time() - start, C)

start = time.time()
for i in range(10):
    C = A.flatten('F').dot(B.T.flatten('F'))
print(time.time() - start, C)

start = time.time()
for i in range(10):
    C = (A.T * B).sum()
print(time.time() - start, C)

start = time.time()
for i in range(10):
    C = (A * B.T).sum()
print(time.time() - start, C)

Result:

6.246593236923218 -629.370798672
0.06539678573608398 -629.370798672
0.057890892028808594 -629.370798672
0.05709719657897949 -629.370798672

Answer 3

From wikipedia you can calculate the trace using the hadamard product (element-wise multiplication):

# Tr(A.B)
tr = (A*B.T).sum()

I think this takes less computation than doing numpy.trace(A.dot(B)).

Edit:

Ran some timers. This way is much faster than using numpy.trace.

In [37]: timeit("np.trace(A.dot(B))", setup="""import numpy as np;
A, B = np.random.rand(1000,1000), np.random.rand(1000,1000)""", number=100)
Out[38]: 8.6434469223022461

In [39]: timeit("(A*B.T).sum()", setup="""import numpy as np;
A, B = np.random.rand(1000,1000), np.random.rand(1000,1000)""", number=100)
Out[40]: 0.5516049861907959