Reputation: 3181
How to transform 100 of 8 element vectors into 10 16 element vectors using 1000 different (8,16) weight matrices? Each of the 10 output vectors is a sum of 100 dot products:
A = np.random.randn(100,8)
W = np.random.randn(1000,8,16)
B = []
for i in range(10):
sum = np.zeros((1,16))
for j in range(100):
sum += np.dot(A[j], W[i*100+j])
B.append(sum)
B = np.asarray(B) #B.shape=(10,16)
Is there a function in Numpy or TensorFlow for that? I looked at dot, tensordot, einsum, and matmul in Numpy, and I'm still not sure which one is the right option.
EDIT: I just realized that I actually want to produce an intermediate result before summing dot products: (100,8)x(10,100,8,16) -> (10,100,16).
I'm guessing this could be done with reshaping (100,8) to (1,100,1,8) and (1000,8,16) to (10,100,8,16), and doing np.einsum('ijkl,ijlm->ijm', A, B)
But I'm not sure if it will broadcast 1 to 10 correctly.
Per @Divakar comment, np.einsum('jk,ijkl->ijl', V, W.reshape(10,100,8,16))
does the trick.
Upvotes: 2
Views: 1166
Reputation: 221754
You can use tensor based multiplication, np.tensordot
-
def tensordot_app(A, W):
m,n,r = W.shape
Wr = W.reshape(-1,A.shape[0],n,r)
return np.tensordot(A,Wr, axes=((0,1),(1,2)))
Related post to understand tensordot
.
Runtime test -
In [62]: A = np.random.randn(100,8)
...: W = np.random.randn(1000,8,16)
In [63]: %%timeit
...: B = []
...: for i in range(10):
...: sum = np.zeros((1,16))
...: for j in range(100):
...: sum += np.dot(A[j], W[i*100+j])
...: B.append(sum)
...: B = np.asarray(B) #B.shape=(10,16)
1000 loops, best of 3: 1.81 ms per loop
# Other post's einsum soln
In [64]: %timeit np.einsum('ij,ikjl',A,np.reshape(W,(100,10,8,16), order='F'))
10000 loops, best of 3: 83.4 µs per loop
# Other post's einsum soln without fortran re-ordering
In [65]: %timeit np.einsum('jk,ijkl', A, np.reshape(W, (10, 100, 8, 16)))
10000 loops, best of 3: 83.3 µs per loop
In [66]: %timeit tensordot_app(A, W)
10000 loops, best of 3: 193 µs per loop
Upvotes: 2
Reputation:
In one line, it's
B1 = np.einsum('ij,ikjl', A, np.reshape(W, (100, 10, 8, 16), order='F'))
Test it with np.allclose(B.squeeze(), B1)
where you need .squeeze
because your B has an extra dimension of size 1.
Explanation: your W has ugly shape, its first dimension of size 1000 should really be separated in 10 blocks of size 100 (as you in effect do with index manipulation in a loop). This is what reshape is for. The Fortran-style order is needed because we want to take out elements of W by changing the first index the fastest.
After that it's straightforward Einstein summation: over j to have matrix multiplication, over i to add 100 results.
Upvotes: 4