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

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

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