问题描述
假设我们有两个矩阵A
和B
,并且让矩阵C
为A*B
(矩阵乘法不是元素方式的).我们希望仅获取C
的对角线条目,可以通过np.diagonal(C)
完成.但是,这会导致不必要的时间开销,因为即使我们只需要将A
中每一行与具有相同"id"的B
列(即第1行)相乘,我们就将A与B相乘. A
的第1列为B
,A
的第2列为B
的第2列,依此类推:构成C
的对角线的乘法.有没有办法使用Numpy有效地实现这一目标?我想避免使用循环来控制哪一行乘以哪一列,而是希望使用内置的numpy方法来执行这种操作以优化性能.
Let's say we have two matrices A
and B
and let matrix C
be A*B
(matrix multiplication not element-wise). We wish to get only the diagonal entries of C
, which can be done via np.diagonal(C)
. However, this causes unnecessary time overhead, because we are multiplying A with B even though we only need the the multiplications of each row in A
with the column of B
that has the same 'id', that is row 1 of A
with column 1 of B
, row 2 of A
with column 2 of B
and so on: the multiplications that form the diagonal of C
. Is there a way to efficiently achieve that using Numpy? I want to avoid using loops to control which row is multiplied with which column, instead, I wish for a built-in numpy method that does this kind of operation to optimize performance.
先谢谢了.
推荐答案
我可能在这里使用einsum
:
>>> a = np.random.randint(0, 10, (3,3))
>>> b = np.random.randint(0, 10, (3,3))
>>> a
array([[9, 2, 8],
[5, 4, 0],
[8, 0, 6]])
>>> b
array([[5, 5, 0],
[3, 5, 5],
[9, 4, 3]])
>>> a.dot(b)
array([[123, 87, 34],
[ 37, 45, 20],
[ 94, 64, 18]])
>>> np.diagonal(a.dot(b))
array([123, 45, 18])
>>> np.einsum('ij,ji->i', a,b)
array([123, 45, 18])
对于较大的数组,它比直接进行乘法要快得多:
For larger arrays, it'll be much faster than doing the multiplication directly:
>>> a = np.random.randint(0, 10, (1000,1000))
>>> b = np.random.randint(0, 10, (1000,1000))
>>> %timeit np.diagonal(a.dot(b))
1 loops, best of 3: 7.04 s per loop
>>> %timeit np.einsum('ij,ji->i', a, b)
100 loops, best of 3: 7.49 ms per loop
[注意:起初我是做元素级版本ii,ii->i
,而不是矩阵乘法.相同的einsum
技巧起作用.]
[Note: originally I'd done the elementwise version, ii,ii->i
, instead of matrix multiplication. The same einsum
tricks work.]
这篇关于(Python)如何在不执行A * B的情况下获得对角线(A * B)?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!