Question

我已经建立了一个注意力十足的RNN语言模型，并通过参与所有先前的隐藏状态（仅一个方向） ，为输入的每个元素创建了上下文向量。

在我看来，最直接的解决方案是在RNN输出上使用 for-loop ，这样，每个上下文向量都是一个接一个地计算的。

import torch
import torch.nn as nn
import torch.nn.functional as F

class RNN_LM(nn.Module):
    def __init__(self, hidden_size, vocab_size, embedding_dim=None, droprate=0.5):
        super().__init__()
        if not embedding_dim:
            embedding_dim = hidden_size
        self.embedding_matrix = nn.Embedding(vocab_size, embedding_dim)

        self.lstm = nn.LSTM(input_size=embedding_dim, hidden_size=hidden_size, batch_first=False)
        self.attn = nn.Linear(hidden_size, hidden_size)
        self.vocab_dist = nn.Linear(hidden_size, vocab_size)
        self.dropout = nn.Dropout(droprate)

    def forward(self, x):
        x = self.dropout(self.embedding_matrix(x.view(-1, 1)))
        x, states = self.lstm(x)
        #print(x.size())
        x = x.squeeze()
        content_vectors = [x[0].view(1, -1)]
        # for-loop over hidden states and attention
        for i in range(1, x.size(0)):
            prev_states = x[:i]
            current_state = x[i].view(1, -1)

            attn_prod = torch.mm(self.attn(current_state), prev_states.t())
            attn_weights = F.softmax(attn_prod, dim=1)
            context = torch.mm(attn_weights, prev_states)
            content_vectors.append(context)

        return self.vocab_dist(self.dropout(torch.cat(content_vectors)))

注意：此处的forward方法仅用于培训。

但是，由于该代码与随后计算每个上下文向量的并行性很差，因此该解决方案不是很有效。但是，由于上下文向量不相互依赖，所以我想知道是否存在一种非顺序的方法来计算它们。

那么有没有一种方法可以在没有 for循环的情况下计算上下文向量，从而使更多计算可以并行化？

Answer 1

好吧，为清楚起见：我想我们只在乎将for循环向量化。 x的形状是什么？假设x是二维的，我有以下代码，其中v1执行循环，而v2是向量化版本：

import torch
import torch.nn.functional as F

torch.manual_seed(0)

x = torch.randn(3, 6)

def v1():
    for i in range(1, x.size(0)):
        prev = x[:i]
        curr = x[i].view(1, -1)

        prod = torch.mm(curr, prev.t())
        attn = prod # same shape
        context = torch.mm(attn, prev)
        print(context)

def v2():
    # we're going to unroll the loop by vectorizing over the new,
    # 0-th dimension of `x`. We repeat it as many times as there
    # are iterations in the for loop
    repeated = x.unsqueeze(0).repeat(x.size(0), 1, 1)

    # we're looking to build a `prevs` tensor such that
    # prevs[i, x, y] == prev[x, y] at i-th iteration of the loop in v1,
    # up to 0-padding necessary to make them all the same size.
    # We need to build a higher-dimensional equivalent of torch.triu
    xs = torch.arange(x.size(0)).reshape(1, -1, 1)
    zs = torch.arange(x.size(0)).reshape(-1, 1, 1)
    prevs = torch.where(zs < xs, torch.tensor(0.), repeated)

    # this is an equivalent of the above iteration starting at 1
    prevs = prevs[:-1]
    currs = x[1:]

    # a batched matrix multiplication
    prod = torch.matmul(currs, prevs.transpose(1, 2))
    attn = prod # same shape
    context = torch.matmul(attn, prevs)
    # equivalent of a higher dimensional torch.diagonal
    contexts = torch.einsum('iij->ij', (context))
    print(contexts)

print(x)

print('\n------ v1 -------\n')
v1()
print('\n------ v2 -------\n')
v2()

通过一些警告来向量化您的循环。首先，我假设x是二维的。其次，我跳过了softmax，声称它不会改变输入的大小，因此不会影响矢量化。确实如此，但不幸的是，填充0的向量v的softmax不等于未填充v的0填充softmax。这可以通过重新规范化解决。请让我知道我的假设是否正确，以及这是否是您工作的足够好的起点。

PyTorch-有效地吸引注意力

1 个答案: