python - 矩阵-矩阵乘法的函数 numpy.dot()、@ 和方法 .dot() 之间有什么区别？

Question

有区别吗？如果不是，按惯例首选什么？性能似乎几乎相同。

a=np.random.rand(1000,1000)
b=np.random.rand(1000,1000)
%timeit a.dot(b)     #14.3 ms ± 374 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit np.dot(a,b)  #14.7 ms ± 315 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit a @ b        #15.1 ms ± 779 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 1

除了少数异常(exception)，它们几乎相同。

a.dot(b) 和 np.dot(a, b) 完全一样。参见 numpy.dot和 ndarray.dot .

但是，查看numpy.dot的文档:

If both a and b are 2-D arrays, it is matrix multiplication, but using matmul or a @ b is preferred.

a@b对应numpy.matmul(a, b) . dot 和 matmul 的区别如下:

matmul differs from dot in two important ways:

• Multiplication by scalars is not allowed, use * instead.
• Stacks of matrices are broadcast together as if the matrices were elements, respecting the signature (n,k),(k,m)->(n,m):

>>> a = np.ones([9, 5, 7, 4])
>>> c = np.ones([9, 5, 4, 3])
>>> np.dot(a, c).shape (9, 5, 7, 9, 5, 3)
>>> np.matmul(a, c).shape (9, 5, 7, 3)
>>> # n is 7, k is 4, m is 3