Question

我正在尝试将seq2seq模型用于我自己的任务https://github.com/spro/practical-pytorch/blob/master/seq2seq-translation/seq2seq-translation.ipynb

我在解码器阶段有两个张量

rnn_output: (1, 1, 256)       # time_step x batch_size x hidden_dimension
encoder_inputs: (10, 1, 256)  # seq_len   x batch_size x hidden_dimension

它们应该被乘以得到形状的注意力得分（在softmax之前）

attn_score: (10, 1, 1)

最好的方法是什么？笔记本似乎使用for循环，有没有更好的矩阵乘法运算？

Answer 1

没有使用pytorch的经验，但是可以这样做吗？

torch.einsum('ijk,abk->abc', (rnn_output, encoder_inputs))

将点积乘在最后一个轴上并添加几个空轴。

使用纯粹的numpy可以实现类似的功能（pytorch 0.4还没有...符号）

np.einsum('...ik,...jk', rnn_output.numpy(), encoder_inputs.numpy())

或np.tensordot

np.tensordot(rnn_output.numpy(), encoder_inputs.numpy(), axes=[2,2])

但是在这里你将获得输出形状：(1, 1, 10, 1)

你可以通过挤压和重新扩展来解决这个问题（几乎可以肯定必须有一些更清洁的方法来执行此操作）

 np.tensordot(rnn_output.numpy(), encoder_inputs.numpy(), axes=[2,2]).squeeze()[..., None, None]

Answer 2

使用torch.bmm()的示例：

import torch
from torch.autograd import Variable
import numpy as np

seq_len = 10
rnn_output = torch.rand((1, 1, 256))
encoder_outputs = torch.rand((seq_len, 1, 256))

# As computed in the tutorial:
attn_score = Variable(torch.zeros(seq_len))
for i in range(seq_len):
    attn_score[i] = rnn_output.squeeze().dot(encoder_outputs[i].squeeze())
    # note: the code would fail without the "squeeze()". I would assume the tensors in
    # the tutorial are actually (,256) and (10, 256)

# Alternative using batched matrix multiplication (bmm) with some data reformatting first:
attn_score_v2 = torch.bmm(rnn_output.expand(seq_len, 1, 256),
                          encoder_outputs.view(seq_len, 256, 1)).squeeze()

# ... Interestingly though, there are some numerical discrepancies between the 2 methods:
np.testing.assert_array_almost_equal(attn_score.data.numpy(), 
                                     attn_score_v2.data.numpy(), decimal=5)
# AssertionError: 
# Arrays are not almost equal to 5 decimals
# 
# (mismatch 30.0%)
#  x: array([60.32436, 69.04288, 72.04784, 70.19503, 71.75543, 67.45459,
#        63.01708, 71.70189, 63.07552, 67.48799], dtype=float32)
#  y: array([60.32434, 69.04287, 72.0478 , 70.19504, 71.7554 , 67.4546 ,
#        63.01709, 71.7019 , 63.07553, 67.488  ], dtype=float32)

在PyTorch和Numpy中快速获得形状（1,1,256）和（10,1,256）的多个3D张量

2 个答案: