在numpy中计算矩阵乘积的迹线的最佳方法是什么? [英] What is the best way to compute the trace of a matrix product in numpy?

问题描述

如果我有numpy数组A和B，则可以使用以下公式计算其矩阵乘积的迹线:

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))

但是，当轨迹中仅使用对角线元素时，矩阵乘法A.dot(B)不必要地计算矩阵乘积中的所有非对角线条目.相反，我可以做类似的事情:

However, the matrix multiplication A.dot(B) unnecessarily computes all of the off-diagonal entries in the matrix product, when only the diagonal elements are used in the trace. Instead, I could do something like:

tr = 0.0
for i in range(n):
    tr += A[i, :].dot(B[:, i])

但是这会在Python代码中执行循环，并且不如numpy.linalg.trace那样明显.

but this performs the loop in Python code and isn't as obvious as numpy.linalg.trace.

是否有更好的方法来计算numpy数组的矩阵乘积的轨迹?最快或最惯用的方法是什么?

Is there a better way to compute the trace of a matrix product of numpy arrays? What is the fastest or most idiomatic way to do this?

推荐答案

您可以通过将中间存储空间仅减少到对角线元素来改进@Bill的解决方案:

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

另一种选择是使用np.einsum并且根本没有显式的中间存储:

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)

在我的系统上，它的运行速度比使用inner1d慢一些，但可能不适用于所有系统，请参见

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

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