块状（n，1，m）至（n，m）

Question

I am working on a problem which involves a batch of 19 tokens each with 400 features. I get the shape (19,1,400) when concatenating two vectors of size (1, 200) into the final feature vector. If I squeeze the 1 out I am left with (19,) but I am trying to get (19,400). I have tried converting to list, squeezing and raveling but nothing has worked.

Is there a way to convert this array to the correct shape?

def attn_output_concat(sample):
  out_h, state_h = get_output_and_state_history(agent.model, sample)
  attns = get_attentions(state_h)
  inner_outputs = get_inner_outputs(state_h)
  if len(attns) != len(inner_outputs):
    print 'Length err'
  else:
    tokens = [np.zeros((400))] * largest
    print(tokens.shape)
    for j, (attns_token, inner_token) in enumerate(zip(attns, inner_outputs)):
      tokens[j] = np.concatenate([attns_token, inner_token], axis=1)
    print(np.array(tokens).shape)
    return tokens

Answer 1

The easiest way would be to declare tokens to be a numpy.shape=(19,400) array to start with. That's also more memory/time efficient. Here's the relevant portion of your code revised...

import numpy as np
attns_token = np.zeros(shape=(1,200))
inner_token = np.zeros(shape=(1,200))
largest = 19
tokens = np.zeros(shape=(largest,400))
for j in range(largest):
    tokens[j] = np.concatenate([attns_token, inner_token], axis=1)
print(tokens.shape)

BTW... It makes it difficult for people to help you if you don't include a self-contained and runnable segment of code (which is probably why you haven't gotten a response on this yet). Something like the above snippet is preferred and will help you get better answers because there's less guessing at what your trying to accomplish.