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这是我的 lstm 网络代码，我将其实例化并传递给 Cuda 设备，但仍然收到隐藏和输入不在同一设备中的错误

class LSTM_net(nn.Module):
def __init__(self, input_size, hidden_size, output_size):
    super(LSTM_net, self).__init__()
    self.hidden_size = hidden_size
    self.lstm_cell = nn.LSTM(input_size, hidden_size)
    self.h2o = nn.Linear(hidden_size, output_size)
    self.softmax = nn.LogSoftmax(dim=1)

def forward(self, input, hidden_0=None, hidden_1=None, hidden_2=None):
    input=resnet(input)
    input=input.unsqueeze(0)
    out_0, hidden_0 = self.lstm_cell(input, hidden_0)
    out_1, hidden_1 = self.lstm_cell(out_0+input, hidden_1)
    out_2, hidden_2 = self.lstm_cell(out_1+input, hidden_2)
    output = self.h2o(hidden_2[0].view(-1, self.hidden_size))
    output = self.softmax(output)
    return output,hidden_0,hidden_1, hidden_2 

def init_hidden(self, batch_size = 1):
    return (torch.zeros(1, batch_size, self.hidden_size), torch.zeros(1, batch_size, self.hidden_size))

net1=LSTM_net(input_size=1000,hidden_size=1000, output_size=100)

net1=net1.to(device)

我想建立的连接图片，请指导我实现它

单击此处查看错误按摩的图像

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2 回答 2

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确保您为 forward() 方法提供的 hidden_​​0 驻留在 GPU 内存中，或者理想地将其作为参数张量存储在模型中，以便优化器对其进行更新并通过 model.cuda() 移动到 gpu。

模型中带有 hidden_​​0 的第二个解决方案的示例（在 init 中添加并self.hidden_0在 forward() 中使用）：

class LSTM_net(nn.Module):
    def __init__(self, input_size, hidden_size, output_size):
        super(LSTM_net, self).__init__()
        self.hidden_size = hidden_size
        self.lstm_cell = nn.LSTM(input_size, hidden_size)
        self.h2o = nn.Linear(hidden_size, output_size)
        self.softmax = nn.LogSoftmax(dim=1)
        self.hidden_0 = torch.nn.parameter.Parameter(torch.zeros(1, batch_size, self.hidden_size)) #taken from init_hidden, assuming that's the intended shape

    def forward(self, input, hidden_0=None, hidden_1=None, hidden_2=None):
        input=resnet(input)
        input=input.unsqueeze(0)
        out_0, hidden_0 = self.lstm_cell(input, self.hidden_0)
        out_1, hidden_1 = self.lstm_cell(out_0+input, hidden_1)
        out_2, hidden_2 = self.lstm_cell(out_1+input, hidden_2)
        output = self.h2o(hidden_2[0].view(-1, self.hidden_size))
        output = self.softmax(output)
        return output,hidden_0,hidden_1, hidden_2
于 2020-08-17T08:49:41.080 回答
-1

编辑：我想我现在看到了问题。尝试改变

    def init_hidden(self, batch_size = 1):
        return (torch.zeros(1, batch_size, self.hidden_size), torch.zeros(1, batch_size, self.hidden_size))

至

    def init_hidden(self, batch_size = 1):
        return (torch.zeros(1, batch_size, self.hidden_size).cuda(), torch.zeros(1, batch_size, self.hidden_size).cuda())

这是因为 init_hidden 方法创建的每个张量都不是函数父对象中的数据属性。因此，当您将 cuda() 应用于模型对象的实例时，它们没有应用 cuda() 。

尝试在所有涉及的张量/变量和模型上调用 .cuda() 。

net1.cuda() # net1.to(device) for device == cuda:0 works fine also 
            # cuda() is more succinct, though
input.cuda()

# now, calling net1 on a tensor named input should not produce the error.
out = net1(input)
于 2020-08-17T08:11:56.943 回答