堆叠 ndarrays 的最快方法

Question

Gist

Basically I want to perform an increase in dimension of two axes on a n-dimensonal tensor. For some reason this operation seems very slow on bigger tensors. If someone can give me a reason or better method I'd be very happy.

Goal

Going from (4, 8, 8, 4, 4, 4, 4, 4, 16, 8, 4, 4, 1) to (4, 32, 8, 4, 4, 4, 4, 4, 4, 8, 4, 4, 1) takes roughly 170 second . I'd like to improve on that. Below is an example, finding the correct indices is not necessary here.

Example Code

Increase dimension (0,2) of tensor

tensor = np.arange(16).reshape(2,2,4,1)
I = np.identity(4)

I tried 3 different methods:

np.kron

indices = [1,3,0,2]
result = np.kron(
            I, tensor.transpose(indices)
        ).transpose(np.argsort(indices))
print(result.shape) # should be (8,2,16,1)

---

manual stacking

col = []
for i in range(4):
    row  = [np.zeros_like(tensor)]*4
    row[i]=tensor
    col.append(a)
result = np.array(col).transpose(0,2,3,1,4,5).reshape(8,2,16,1)
print(result.shape) # should be (8,2,16,1)

---

np.einsum

result =np.einsum("ij, abcd -> iabjcd", I, tensor).reshape(8,2,16,1)
print(result.shape) # should be (8,2,16,1)

---

Results

On my machine they performed the following (on the big example with complex entries):

1. np.einsum ~ 170s
2. manual stacking ~ 185s
3. np.kron ~ 580s

Answer 1

As Jérôme pointed out:

all your operations seems to involve a transposition which is known to be very expensive on modern hardware.

I reworked my algorithm to not rely on the dimensional increase by doing certain preprocessing steps. This indeed speeds up the overall process substantially.