How do I get the sum of a dot product efficiently
如何有效地获得点积的总和
A = np.random.rand(20,200)
x= np.random.rand(20)
y= np.random.rand(20)
num= np.zeros(20)
for i in range (A.shape[0]):
num[i] = np.sum(A.T[i,:].dot(x[i]+y[i]))
print num
Is there a way to find num without a for loop
有没有办法在没有for循环的情况下找到num
1 个解决方案
#1
1
You could use np.einsum -
你可以使用np.einsum -
num = np.einsum('ji,i->i',A[:,:20],x+y)
That slicing of [:,:20] is needed because even though you are iterating along the rows of A.T with A.T[i,:] i.e. columns of A, you are not iterating through all of those columns.
需要切片[:,:20],因为即使您使用A.T [i,:](即A的列)沿A.T的行迭代,您也不会遍历所有这些列。
#1
1
You could use np.einsum -
你可以使用np.einsum -
num = np.einsum('ji,i->i',A[:,:20],x+y)
That slicing of [:,:20] is needed because even though you are iterating along the rows of A.T with A.T[i,:] i.e. columns of A, you are not iterating through all of those columns.
需要切片[:,:20],因为即使您使用A.T [i,:](即A的列)沿A.T的行迭代,您也不会遍历所有这些列。