Question

我正在尝试使用pyTorch在GPU模式下对某些数据进行一次热编码，但是，它一直给我一个例外。有人可以帮我吗？

这里是一个例子：

def char_OneHotEncoding(x):
    coded = torch.zeros(x.shape[0], x.shape[1], 101)
    for i in range(x.shape[1]):
        coded[:,i] = scatter(x[:,i])
    return coded


def scatter(x):
    return torch.zeros(x.shape[0], 101).scatter_(1, x.view(-1,1), 1)

所以如果我在GPU上给它一个张量，它会显示如下：

x_train = [[ 0,  0,  0,  0,  0],
       [ 0,  0,  0,  0,  0],
       [ 0,  0,  0,  0,  0],
       [14, 13, 83, 18, 14],
       [ 0,  0,  0,  0,  0]]
print(char_OneHotEncoding(torch.tensor(x_train, dtype=torch.long).cuda()).shape)

---------------------------------------------------------------------------
RuntimeError                              Traceback (most recent call last)
<ipython-input-62-95c0c4ade406> in <module>()
      4        [14, 13, 83, 18, 14],
      5        [ 0,  0,  0,  0,  0]]
----> 6 print(char_OneHotEncoding(torch.tensor(x_train, dtype=torch.long).cuda()).shape)
      7 x_train[:5, maxlen:maxlen+5]

<ipython-input-53-055f1bf71306> in char_OneHotEncoding(x)
      2     coded = torch.zeros(x.shape[0], x.shape[1], 101)
      3     for i in range(x.shape[1]):
----> 4         coded[:,i] = scatter(x[:,i])
      5     return coded
      6 

<ipython-input-53-055f1bf71306> in scatter(x)
      7 
      8 def scatter(x):
----> 9     return torch.zeros(x.shape[0], 101).scatter_(1, x.view(-1,1), 1)

RuntimeError: Expected object of backend CPU but got backend CUDA for argument #3 'index'

顺便说一句，如果简单地移除.cuda()这里，一切顺利一个阱

print(char_OneHotEncoding(torch.tensor(x_train, dtype=torch.long)).shape)

torch.Size([5, 5, 101])

Answer 1

是的，有可能。您必须注意所有张量都在GPU上。特别是，默认情况下，torch.zeros之类的构造函数会在CPU上分配，这将导致这种不匹配。代码可以通过用构建固定device=x.device，如下

import torch

 def char_OneHotEncoding(x):
     coded = torch.zeros(x.shape[0], x.shape[1], 101, device=x.device)
     for i in range(x.shape[1]):
         coded[:,i] = scatter(x[:,i])
     return coded


 def scatter(x):
     return torch.zeros(x.shape[0], 101, device=x.device).scatter_(1, x.view(-1,1), 1)

 x_train = torch.tensor([
     [ 0,  0,  0,  0,  0],
     [ 0,  0,  0,  0,  0],
     [ 0,  0,  0,  0,  0],
     [14, 13, 83, 18, 14],
     [ 0,  0,  0,  0,  0]
 ], dtype=torch.long, device='cuda')

 print(char_OneHotEncoding(x_train).shape)

另一种替代被称为构造xxx_like，例如zeros_like，尽管在这种情况下，由于需要比x，我发现不同的形状device=x.device的可读性。< / p>

我们可以在GPU上使用pytorch scatter_吗？

1 个答案: