python - pytorch - 使用 'with statement' 内的设备

Question

有没有办法在特定 (GPU) 设备的上下文中运行 pytorch(无需为每个新张量指定设备，例如 .to 选项)？

类似于 tensorflow with tf.device('/device:GPU:0'):..

好像默认的设备是cpu(除非我做错了):

with torch.cuda.device('0'):
   a = torch.zeros(1)
   print(a.device)

>>> cpu

Answer 1

不幸的是，在当前的实现中，with-device 语句不能以这种方式工作，它只能用于在 cuda 设备之间切换。

---

您仍然必须使用 device 参数来指定使用哪个设备(或 .cuda() 将张量移动到指定的 GPU)，带有像这样的术语:

# allocates a tensor on GPU 1
a = torch.tensor([1., 2.], device=cuda)

因此要访问 cuda:1:

cuda = torch.device('cuda')

with torch.cuda.device(1):
    # allocates a tensor on GPU 1
    a = torch.tensor([1., 2.], device=cuda)

并访问 cuda:2:

cuda = torch.device('cuda')

with torch.cuda.device(2):
    # allocates a tensor on GPU 2
    a = torch.tensor([1., 2.], device=cuda)

但是没有 device 参数的张量仍然是 CPU 张量:

cuda = torch.device('cuda')

with torch.cuda.device(1):
    # allocates a tensor on CPU
    a = torch.tensor([1., 2.])

总结一下:

No - unfortunately it is in the current implementation of the with-device statement not possible to use in a way you described in your question.

---

以下是来自 documentation 的更多示例:

cuda = torch.device('cuda')     # Default CUDA device
cuda0 = torch.device('cuda:0')
cuda2 = torch.device('cuda:2')  # GPU 2 (these are 0-indexed)

x = torch.tensor([1., 2.], device=cuda0)
# x.device is device(type='cuda', index=0)
y = torch.tensor([1., 2.]).cuda()
# y.device is device(type='cuda', index=0)

with torch.cuda.device(1):
    # allocates a tensor on GPU 1
    a = torch.tensor([1., 2.], device=cuda)

    # transfers a tensor from CPU to GPU 1
    b = torch.tensor([1., 2.]).cuda()
    # a.device and b.device are device(type='cuda', index=1)

    # You can also use ``Tensor.to`` to transfer a tensor:
    b2 = torch.tensor([1., 2.]).to(device=cuda)
    # b.device and b2.device are device(type='cuda', index=1)

    c = a + b
    # c.device is device(type='cuda', index=1)

    z = x + y
    # z.device is device(type='cuda', index=0)

    # even within a context, you can specify the device
    # (or give a GPU index to the .cuda call)
    d = torch.randn(2, device=cuda2)
    e = torch.randn(2).to(cuda2)
    f = torch.randn(2).cuda(cuda2)
    # d.device, e.device, and f.device are all device(type='cuda', index=2)