1 问题描述

在训练lstm网络过程中出现如下错误：

Traceback (most recent call last):File "D:\code\lstm_emotion_analyse\text_analyse.py", line 82, in <module>loss.backward()File "C:\Users\lishu\anaconda3\envs\pt2\lib\site-packages\torch\_tensor.py", line 487, in backwardtorch.autograd.backward(File "C:\Users\lishu\anaconda3\envs\pt2\lib\site-packages\torch\autograd\__init__.py", line 200, in backwardVariable._execution_engine.run_backward(  # Calls into the C++ engine to run the backward pass
RuntimeError: Trying to backward through the graph a second time (or directly access saved tensors after they have already been freed). Saved intermediate values of the graph are freed when you call .backward() or autograd.grad(). Specify retain_graph=True if you need to backward through the graph a second time or if you need to access saved tensors after calling backward.

2 问题分析

按照错误提示查阅相关资料了解到，实际上在大多数情况下retain_graph都应采用默认的False，除了几种特殊情况：

• 一个网络有两个output分别执行backward进行回传的时候: output1.backward(), output2.backward().
• 一个网络有两个loss需要分别执行backward进行回传的时候: loss1.backward(), loss2.backward().

但本项目的LSTM训练模型不属于以上情况，再次查找资料，在在pytorch的官方论坛上找到了真正的原因：

如截图中的描述，只要我们对变量进行运算了，就会加进计算图中。所以本项目的问题在于在for循环梯度反向传播中，使用了循环外部的变量h，如下所示：

epochs = 128step = 0model.train()  # 开启训练模式for epoch in range(epochs):h = model.init_hidden(batch_size)  # 初始化第一个Hidden_statefor data in tqdm(train_loader):x_train, y_train = datax_train, y_train = x_train.to(device), y_train.to(device)step += 1  # 训练次数+1x_input = x_train.to(device)model.zero_grad()output, h = model(x_input, h)# 计算损失loss = criterion(output, y_train.float().view(-1))loss.backward()nn.utils.clip_grad_norm_(model.parameters(), max_norm=5)optimizer.step()if step % 10 == 0:print("Epoch: {}/{}...".format(epoch + 1, epochs),"Step: {}...".format(step),"Loss: {:.6f}...".format(loss.item()))

3 问题解决

代码修改如下：

epochs = 128step = 0model.train()  # 开启训练模式for epoch in range(epochs):h = model.init_hidden(batch_size)  # 初始化第一个Hidden_statefor data in tqdm(train_loader):x_train, y_train = datax_train, y_train = x_train.to(device), y_train.to(device)step += 1  # 训练次数+1x_input = x_train.to(device)model.zero_grad()h = tuple([e.data for e in h])output, h = model(x_input, h)# 计算损失loss = criterion(output, y_train.float().view(-1))loss.backward()nn.utils.clip_grad_norm_(model.parameters(), max_norm=5)optimizer.step()if step % 10 == 0:print("Epoch: {}/{}...".format(epoch + 1, epochs),"Step: {}...".format(step),"Loss: {:.6f}...".format(loss.item()))

增加for循环内部变量，对外部变量进行复制，内部变量参与梯度传播，问题解决。