我已经构建了一个带有注意力的RNN语言模型,并且通过关注所有先前的隐藏状态(只有一个方向),为输入的每个元素创建上下文向量。
在我看来,最直接的解决方案是在RNN输出上使用for循环,以便逐个计算每个上下文向量。
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:这里的forward
方法仅用于训练。
然而,这种解决方案效率不是很高,因为代码不能很好地并行化顺序地计算每个上下文向量。但由于上下文向量并不相互依赖,我想知道是否有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循环的情况下计算上下文向量,以便更多的计算可以并行化?
发布于 2018-12-11 06:27:32
好吧,为了清楚起见:我假设我们只关心向量化for
循环。x
的形状是什么?假设x
是二维的,我有以下代码,其中v1
执行您的循环,v2
是矢量化版本:
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()
它向量化了你的循环,但有一些注意事项。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,因此不会影响向量化。这是真的,但不幸的是,填充0的向量v
的softmax不等于未填充的v
的0填充的softmax。不过,这可以通过重整化来解决。请让我知道,如果我的假设是正确的,这是否是一个足够好的起点,你的工作。
https://stackoverflow.com/questions/53706462
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