在您的情况下它可以在没有 retain_graph=True 的情况下工作的原因是您有一个非常简单的图形,可能没有内部中间缓冲区,反过来不会释放任何缓冲区,因此无需使用 retain_graph=True。
但是当向您的图表添加额外的计算时,一切都会发生变化:
代码:
x = torch.ones(2, 2, requires_grad=True)
v = x.pow(3)
y = v + 2
y.backward(torch.ones(2, 2))
print('Backward 1st time w/o retain')
print('x.grad:', x.grad)
print('Backward 2nd time w/o retain')
try:
y.backward(torch.ones(2, 2))
except RuntimeError as err:
print(err)
print('x.grad:', x.grad)
输出:
Backward 1st time w/o retain
x.grad: tensor([[3., 3.],
[3., 3.]])
Backward 2nd time w/o retain
Trying to backward through the graph a second time, but the buffers have already been freed. Specify retain_graph=True when calling backward the first time.
x.grad: tensor([[3., 3.],
[3., 3.]]).
在这种情况下,额外的内部v.grad 将被计算,但torch 不存储中间值(中间梯度等),并且retain_graph=False v.grad 将在第一个backward 之后被释放。
所以,如果你想第二次反向传播,你需要指定retain_graph=True 来“保留”图表。
代码:
x = torch.ones(2, 2, requires_grad=True)
v = x.pow(3)
y = v + 2
y.backward(torch.ones(2, 2), retain_graph=True)
print('Backward 1st time w/ retain')
print('x.grad:', x.grad)
print('Backward 2nd time w/ retain')
try:
y.backward(torch.ones(2, 2))
except RuntimeError as err:
print(err)
print('x.grad:', x.grad)
输出:
Backward 1st time w/ retain
x.grad: tensor([[3., 3.],
[3., 3.]])
Backward 2nd time w/ retain
x.grad: tensor([[6., 6.],
[6., 6.]])