我有三个numpy矩阵x,r和r。谁的价值观是:
x = np.array([[4,2],
[0,-1],
[-2,5],
[2,6]])
y = np.array([[1,7],
[2,6],
[5,2]])
r = np.array([[2,2,1],
[2,3,1],
[9,5,1],
[2,0,4]])
我要做的是:(用文字描述很难,所以我用代码来表达我想做的事情)
K = r.shape[1]
D = x.shape[1]
v = np.zeros((K, D, D))
for k in range(K):
v[k] = (r[:, k] * (x - y[k]).transpose() @ (x - y[k]))
print(v)
最后的v是我需要的,v等于
[[[103. 38.]
[ 38. 216.]]
[[100. 46.]
[ 46. 184.]]
[[111. -54.]
[-54. 82.]]]
有没有优雅或pythonic方式来实现这个没有for循环?
谢谢
这应该适合你:
A = x[np.newaxis,...]-y[:,np.newaxis,:] # equivalent to (x-y[k]) for all k
B = A.swapaxes(1,2) # equivalent to (x-y[k]).transpose() for all k
C = r.T[:,np.newaxis,:]*B # equivalent to r[:, k] * (x - y[k]).transpose()
D = C@A # equivalent to r[:, k] *(x - y[k]).transpose() @ (x - y[k])
或者是怪物不可读的形式
((r.T[:,np.newaxis,:]*(x[np.newaxis,...]
-y[:,np.newaxis,:]).swapaxes(1,2))@
(x[np.newaxis,...]-y[:,np.newaxis,:]))
证明:
>>> (v==((r.T[:,np.newaxis,:]*(x[np.newaxis,...]
-y[:,np.newaxis,:]).swapaxes(1,2))@
(x[np.newaxis,...]-y[:,np.newaxis,:]))).all()
True