定义损失函数以使用外部数组

Question

在我的神经网络 (RNN) 中，我定义了损失函数，以便神经网络的输出用于查找索引（二进制），然后使用索引从数组中提取所需的元素这又将用于计算 MSELoss。

但是，程序给出了 parameter().grad = None 错误，这主要是因为图形在某处中断了。错误函数定义有什么问题

框架：Pytorch

代码如下：神经网络：

class RNN(nn.Module):
  def __init__(self):
    super(RNN, self).__init__()
    self.hidden_size = 8
    # self.input_size = 2
    self.h2o = nn.Linear(self.hidden_size, 1)
    self.h2h = nn.Linear(self.hidden_size, self.hidden_size)
    self.sigmoid = nn.Sigmoid()
  def forward(self,hidden):
    output = self.h2o(hidden)
    output = self.sigmoid(output)
    hidden = self.h2h(hidden)
    return output, hidden
  def init_hidden(self):
    return torch.zeros(1, self.hidden_size)

损失函数、训练步骤和训练

rnn = RNN()
criterion = nn.MSELoss()

def loss_function(previous, output, index):
  code = 2*(output > 0.5).long()
  current = Q_m2[code:code+2, i]
  return criterion(current, previous), current

def train_step():
  hidden = rnn.init_hidden()
  rnn.zero_grad()
  # Q_m2.requires_grad = True
  # Q_m2.create_graph = True 
  loss = 0
  previous = Q_m[0:2, 0]
  for i in range(1, samples):
    output, hidden = rnn(hidden)
    l, previous = loss_function(previous, output, i)
    loss+=l
  loss.backward()
  # Q_m2.retain_grad()
  for p in rnn.parameters():
    p.data.add_(p.grad.data, alpha=-0.05)
  return output, loss.item()/(samples - 1)

def training(epochs):
  running_loss = 0
  for i in range(epochs):
    output, loss = train_step()
    print(f'Epoch Number: {i+1}, Loss: {loss}')
    running_loss +=loss

Q_m2

Q_m = np.zeros((4, samples))
for i in range(samples):
  Q_m[:,i] = q_x(U_m[:,i])
Q_m = torch.FloatTensor(Q_m)
Q_m2 = Q_m
Q_m2.requires_grad = True
Q_m2.create_graph = True

错误：

<ipython-input-36-feefd257c97a> in train_step()
     21   # Q_m2.retain_grad()
     22   for p in rnn.parameters():
---> 23     p.data.add_(p.grad.data, alpha=-0.05)
     24   return output, loss.item()/(samples - 1)
     25 

AttributeError: 'NoneType' object has no attribute 'data'

Answer 1

这是 K. Frank at discuss.pytorch.org

向我建议的可能解决方案

As I read it, code is calculated to be either 0 or 2. You could instead interpret output (processed appropriately, as necessary) to be the probability that code should be 0 vs. 2, and then use that probability to form a weighted average of the 0 and 2 entries in your Q_m2 array.

定义损失函数以使用外部数组

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