I have the following function that calculates the eucledian distance between all combinations of the vectors in Matrix A and Matrix B
def distance_matrix(A,B):
n=A.shape[1]
m=B.shape[1]
C=np.zeros((n,m))
for ai, a in enumerate(A.T):
for bi, b in enumerate(B.T):
C[ai][bi]=np.linalg.norm(a-b)
return C
This works fine and creates an n*m-Matrix from a d*n-Matrix and a d*m-Matrix containing the eucledian distance between all combinations of the column vectors.
>>> print(A)
[[-1 -1 1 1 2]
[ 1 -1 2 -1 1]]
>>> print(B)
[[-2 -1 1 2]
[-1 2 1 -1]]
>>> print(distance_matrix(A,B))
[[2.23606798 1. 2. 3.60555128]
[1. 3. 2.82842712 3. ]
[4.24264069 2. 1. 3.16227766]
[3. 3.60555128 2. 1. ]
[4.47213595 3.16227766 1. 2. ]]
I spent some time looking for a numpy or scipy function to achieve this in a more efficient way. Is there such a function or what would be the vecotrized way to do this?
CodePudding user response：
You can use:
np.linalg.norm(A[:,:,None]-B[:,None,:],axis=0)
or (totaly equivalent but without in-built function)
((A[:,:,None]-B[:,None,:])**2).sum(axis=0)**0.5
We need a 5x4 final array so we extend our array this way:
A[:,:,None] -> 2,5,1
↑ ↓
B[:,None,:] -> 2,1,4
A[:,:,None] - B[:,None,:] -> 2,5,4
and we apply our sum over the axis 0 to finally get a 5,4 ndarray.
CodePudding user response：
Yes, you can broadcast your vectors:
A = np.array([[-1, -1, 1, 1, 2], [ 1, -1, 2, -1, 1]])
B = np.array([[-2, -1, 1, 2], [-1, 2, 1, -1]])
C = np.linalg.norm(A.T[:, None, :] - B.T[None, :, :], axis=-1)
print(C)
array([[2.23606798, 1. , 2. , 3.60555128],
[1. , 3. , 2.82842712, 3. ],
[4.24264069, 2. , 1. , 3.16227766],
[3. , 3.60555128, 2. , 1. ],
[4.47213595, 3.16227766, 1. , 2. ]])
You can get an explanation of how it works here: