注意
转到末尾 下载完整的示例代码。
PyTorch: 控制流 + 权重共享#
创建日期:2017年3月24日 | 最后更新:2021年12月28日 | 最后验证:2024年11月05日
为了展示PyTorch动态图的强大功能,我们将实现一个非常奇怪的模型:一个三到五次的多项式,在每次前向传播时,它会选择一个介于4到5之间的随机数,并使用该数量的次数,重复使用相同的权重来计算四次和五次。
import random
import torch
import math
class DynamicNet(torch.nn.Module):
def __init__(self):
"""
In the constructor we instantiate five parameters and assign them as members.
"""
super().__init__()
self.a = torch.nn.Parameter(torch.randn(()))
self.b = torch.nn.Parameter(torch.randn(()))
self.c = torch.nn.Parameter(torch.randn(()))
self.d = torch.nn.Parameter(torch.randn(()))
self.e = torch.nn.Parameter(torch.randn(()))
def forward(self, x):
"""
For the forward pass of the model, we randomly choose either 4, 5
and reuse the e parameter to compute the contribution of these orders.
Since each forward pass builds a dynamic computation graph, we can use normal
Python control-flow operators like loops or conditional statements when
defining the forward pass of the model.
Here we also see that it is perfectly safe to reuse the same parameter many
times when defining a computational graph.
"""
y = self.a + self.b * x + self.c * x ** 2 + self.d * x ** 3
for exp in range(4, random.randint(4, 6)):
y = y + self.e * x ** exp
return y
def string(self):
"""
Just like any class in Python, you can also define custom method on PyTorch modules
"""
return f'y = {self.a.item()} + {self.b.item()} x + {self.c.item()} x^2 + {self.d.item()} x^3 + {self.e.item()} x^4 ? + {self.e.item()} x^5 ?'
# Create Tensors to hold input and outputs.
x = torch.linspace(-math.pi, math.pi, 2000)
y = torch.sin(x)
# Construct our model by instantiating the class defined above
model = DynamicNet()
# Construct our loss function and an Optimizer. Training this strange model with
# vanilla stochastic gradient descent is tough, so we use momentum
criterion = torch.nn.MSELoss(reduction='sum')
optimizer = torch.optim.SGD(model.parameters(), lr=1e-8, momentum=0.9)
for t in range(30000):
# Forward pass: Compute predicted y by passing x to the model
y_pred = model(x)
# Compute and print loss
loss = criterion(y_pred, y)
if t % 2000 == 1999:
print(t, loss.item())
# Zero gradients, perform a backward pass, and update the weights.
optimizer.zero_grad()
loss.backward()
optimizer.step()
print(f'Result: {model.string()}')