我希望以特定的方式组合 3 个矩阵，我相信这对于习惯使用 numpy 的人来说都很容易：
w = np.array([[-0.46733567, 0.38864732],
[-0.42436867, -1.08760098],
[-1.01118741, 0.99096466]])
a = array([[ 0.63127368, 0.00167775, 0.97812284]])
d = [[-0.43252997]] # although d is length 1 in this example, its often >1 in length and code needs to be able to generalize, there will always be the same length of cols and and arrays between a and w, i.e in this example there are 3 arrays in w and 1 of length 3 in a (please excuse my terminology, I am new to linear algebra).
我正在尝试将它们组合起来以找到 dx，以便：
dx1 = [-0.46733567, 0.38864732] * 0.63127368 * -0.43252997
dx2 = [-0.42436867, -1.08760098] * 0.00167775 * -0.43252997
dx3 = [-1.01118741, 0.99096466] * 0.97812284 * -0.43252997
如果 d>长度为 1，例如 d= np.array（[[-0.43252997]，[0.87500009]]），则：
dx1_1 = [-0.46733567, 0.38864732] * 0.63127368 * -0.43252997
dx2_1 = [-0.42436867, -1.08760098] * 0.00167775 * -0.43252997
dx3_1 = [-1.01118741, 0.99096466] * 0.97812284 * -0.43252997
dx1_2 = [-0.46733567, 0.38864732] * 0.63127368 * 0.87500009
dx2_2 = [-0.42436867, -1.08760098] * 0.00167775 * 0.87500009
dx3_2 = [-1.01118741, 0.99096466] * 0.97812284 * 0.87500009
尝试了所有这些，但似乎没有成功，除非我错过了什么？
out = np.dot(d, w) * a
out = np.dot(d, w) * a.T
out = np.dot(d, w.T) * a
out = np.dot(a, w) * d
out = np.dot(a, w.T) * d
在很多情况下我收到错误：
ValueError: shapes (3,1) and (3,2) not aligned: 1 (dim 1) != 3 (dim 0)
任何有助于解决此问题的指针将不胜感激
你不需要任何点积，因为你不想对任何东西求和。
我在第三个轴上重复你的数组（如果 d 长于 1，这很重要）。
然后乘以相应的轴...：
w = np.array([[-0.46733567, 0.38864732],
[-0.42436867, -1.08760098],
[-1.01118741, 0.99096466]])
a = np.array([[ 0.63127368, 0.00167775, 0.97812284]])
d = np.array([[-0.43252997]])
(dim1, dim2), dim3 = w.shape, len(d[0])
wnew = w.reshape(dim1, dim2, 1)
anew = a.reshape(dim1, 1, 1)
dnew = d.reshape(1, 1, dim3)
product = wnew * anew * dnew
i, j = 0, 0
print product[i, :, j]
最后一条语句打印您所说的dx_ij
按照输入数组的格式，特别是所有输入数组的形状都是2D的，假设你想将输出存储在3D数组中，这里有一种使用 broadcasting 的方法 -
(w.T)[:,:,None]*(a[0,:,None]*d[:,0])
示例运行 -
In [48]: # Input arrays
...: w = np.array([[-0.46733567, 0.38864732],
...: [-0.42436867, -1.08760098],
...: [-1.01118741, 0.99096466]])
...:
...: a = np.array([[ 0.63127368, 0.00167775, 0.97812284]])
...:
...: d= np.array([[-0.43252997],[0.87500009]])
...:
...: dx1_1 = np.array([-0.46733567, 0.38864732]) * 0.63127368 * -0.43252997
...: dx2_1 = np.array([-0.42436867, -1.08760098]) * 0.00167775 * -0.43252997
...: dx3_1 = np.array([-1.01118741, 0.99096466]) * 0.97812284 * -0.43252997
...: dx1_2 = np.array([-0.46733567, 0.38864732]) * 0.63127368 * 0.87500009
...: dx2_2 = np.array([-0.42436867, -1.08760098]) * 0.00167775 * 0.87500009
...: dx3_2 = np.array([-1.01118741, 0.99096466]) * 0.97812284 * 0.87500009
...:
...: # Let's store these in a 3D array
...: p1 = np.column_stack((dx1_1,dx2_1,dx3_1))
...: p2 = np.column_stack((dx1_2,dx2_2,dx3_2))
...: out = np.dstack((p1,p2))
...:
验证的原始和拟议的基于广播的方法的打印输出 -
In [49]: out # Output from original approach
Out[49]:
array([[[ 1.27603568e-01, -2.58139646e-01],
[ 3.07954650e-04, -6.22986533e-04],
[ 4.27800472e-01, -8.65432403e-01]],
[[ -1.06118124e-01, 2.14674993e-01],
[ 7.89247187e-04, -1.59663239e-03],
[ -4.19244884e-01, 8.48124609e-01]]])
In [50]: (w.T)[:,:,None]*(a[0,:,None]*d[:,0]) # Output from proposed solution
Out[50]:
array([[[ 1.27603568e-01, -2.58139646e-01],
[ 3.07954650e-04, -6.22986533e-04],
[ 4.27800472e-01, -8.65432403e-01]],
[[ -1.06118124e-01, 2.14674993e-01],
[ 7.89247187e-04, -1.59663239e-03],
[ -4.19244884e-01, 8.48124609e-01]]])
如果输入数组a并且d是1D数组：
a = np.array([ 0.63127368, 0.00167775, 0.97812284])
d = np.array([-0.43252997,0.87500009])
，该解决方案将简化为 -
(w.T)[:,:,None]*(a[:,None]*d)
当你多个两个矩阵时，它们应该有一个共享的维度。在您的示例中：
>>> out = np.dot(a, w)
>>> print out
[[-1.28479419 1.21280327]]
欲了解更多信息： https://en.wikipedia.org/wiki/Matrix_multiplication如果 A 是 n × m 矩阵，B 是 m × p 矩阵，则矩阵乘积 AB（不带乘号或点表示）定义为 n × p 矩阵。
此致敬意亚龙