YueZhengMeng
commented
2 months ago # 修改后
def train_epoch_ch8(net, train_iter, loss, updater, device, use_random_iter):
"""训练网络一个迭代周期（定义见第8章）"""
state, timer = None, d2l.Timer()
metric = d2l.Accumulator(2) # 训练损失之和,词元数量
for X, Y in train_iter:
X, Y = X.to(device), Y.to(device)
y = Y.T.reshape(-1)
state = net.begin_state(batch_size=X.shape[0], device=device)
y_hat, state = net(X, state)
l = loss(y_hat, y.long()).mean()
if isinstance(updater, torch.optim.Optimizer):
updater.zero_grad()
l.backward(retain_graph=True)
d2l.grad_clipping(net, 1)
updater.step()
else:
l.backward(retain_graph=True)
d2l.grad_clipping(net, 1)
# 因为已经调用了mean函数
updater(batch_size=1)
metric.add(l * y.numel(), y.numel())
return math.exp(metric[0] / metric[1]), metric[1] / timer.stop()
def train_ch8(net, train_iter, vocab, lr, num_epochs, device,
use_random_iter=False):
"""训练模型（定义见第8章）"""
loss = nn.CrossEntropyLoss()
animator = d2l.Animator(xlabel='epoch', ylabel='perplexity',
legend=['train'], xlim=[10, num_epochs])
# 初始化
if isinstance(net, nn.Module):
updater = torch.optim.SGD(net.parameters(), lr)
else:
updater = lambda batch_size: d2l.sgd(net.params, lr, batch_size)
predict = lambda prefix: d2l.predict_ch8(prefix, 50, net, vocab, device)
# 训练和预测
for epoch in range(num_epochs):
ppl, speed = train_epoch_ch8(
net, train_iter, loss, updater, device, use_random_iter)
if (epoch + 1) % 10 == 0:
print(predict('time traveller'))
animator.add(epoch + 1, [ppl])
print(f'困惑度 {ppl:.1f}, {speed:.1f} 词元/秒 {str(device)}')
print(predict('time traveller'))
print(predict('traveller'))
解答： 先对输入和标签进行设备（device）变换和形状（reshape）变换，再进行前向计算和反向传播，将隐状态的分离操作放在更新之前，避免了更新中对隐状态进行计算，这样无需对隐状态进行修改，即可实现了不会从计算图中分离隐状态。
但是给出的解答代码里并没有与分离梯度相关的detach_()函数