激活梯度惩罚

Question

这是一个简单的神经网络,我试图惩罚激活梯度的规范:

class Net(nn.Module):
    def __init__(self):
        super(Net, self).__init__()
        self.conv1 = nn.Conv2d(3, 32, kernel_size=5)
        self.conv2 = nn.Conv2d(32, 64, kernel_size=5)
        self.pool = nn.MaxPool2d(2, 2)
        self.relu = nn.ReLU()
        self.linear = nn.Linear(64 * 5 * 5, 10)

    def forward(self, input):
        conv1 = self.conv1(input)
        pool1 = self.pool(conv1)
        self.relu1 = self.relu(pool1)
        self.relu1.retain_grad()
        conv2 = self.conv2(relu1)
        pool2 = self.pool(conv2)
        relu2 = self.relu(pool2)
        self.relu2 = relu2.view(relu2.size(0), -1)
        self.relu2.retain_grad()
        return self.linear(relu2)

model = Net()
optimizer = torch.optim.SGD(model.parameters(), lr=0.001)

for i in range(1000):
    output = model(input)
    loss = nn.CrossEntropyLoss()(output, label)
    optimizer.zero_grad()
    loss.backward(retain_graph=True)

    grads = torch.autograd.grad(loss, [model.relu1, model.relu2], create_graph=True)

    grad_norm = 0
    for grad in grads:
        grad_norm += grad.pow(2).sum()

    grad_norm.backward()

    optimizer.step()

但是,它不会产生所需的正则化效果.如果我对重量(而不是激活)做同样的事情,它运作良好.我做得对吗(就火炬机械而言)？具体来说,grad_norm.backward()调用会发生什么？我只想确保更新重量梯度,而不是激活渐变.目前,当我在该行之前和之后立即打印出权重和激活的渐变时,两者都会改变 - 所以我不确定发生了什么.

Answer 1

我认为您的代码最终会在每个步骤中计算一些梯度两次。我还怀疑它实际上永远不会将激活梯度归零，因此它们会跨步骤累积。

一般来说：

• x.backward()计算xwrt 的梯度。计算图叶子（例如权重张量和其他变量），以及wrt。节点明确标记为retain_grad(). 它累积张量属性中计算的梯度.grad。

• autograd.grad(x, [y, z])返回xwrt 的梯度。y并且z无论他们通常是否会保留毕业资格。默认情况下，它还会在所有叶子的.grad属性中累积梯度。您可以通过传递来防止这种情况发生only_inputs=True。

我更喜欢仅用于backward()优化步骤，并且autograd.grad()每当我的目标是获得“具体化”梯度作为另一次计算的中间值时。.grad这样，我可以确保在完成处理后，张量的属性中不会残留任何不需要的梯度。

import torch
from torch import nn
class Net(nn.Module):
    def __init__(self):
        super(Net, self).__init__()
        self.conv1 = nn.Conv2d(3, 32, kernel_size=5)
        self.conv2 = nn.Conv2d(32, 64, kernel_size=5)
        self.pool = nn.MaxPool2d(2, 2)
        self.relu = nn.ReLU()
        self.linear = nn.Linear(64 * 5 * 5, 10)

    def forward(self, input):
        conv1 = self.conv1(input)
        pool1 = self.pool(conv1)
        self.relu1 = self.relu(pool1)
        conv2 = self.conv2(self.relu1)
        pool2 = self.pool(conv2)
        self.relu2 = self.relu(pool2)
        relu2 = self.relu2.view(self.relu2.size(0), -1)
        return self.linear(relu2)


model = Net()
optimizer = torch.optim.SGD(model.parameters(), lr=0.001)
grad_penalty_weight = 10.

for i in range(1000000):
    # Random input and labels; we're not really learning anything
    input = torch.rand(1, 3, 32, 32)
    label = torch.randint(0, 10, (1,))

    output = model(input)
    loss = nn.CrossEntropyLoss()(output, label)

    # This is where the activation gradients are computed
    # only_inputs is optional here, since we're going to call optimizer.zero_grad() later
    # But it makes clear that we're *only* interested in the activation gradients at this point
    grads = torch.autograd.grad(loss, [model.relu1, model.relu2], create_graph=True, only_inputs=True)

    grad_norm = 0
    for grad in grads:
        grad_norm += grad.pow(2).sum()

    optimizer.zero_grad()
    loss = loss + grad_norm * grad_penalty_weight
    loss.backward()
    optimizer.step()

这段代码似乎有效，因为激活梯度确实变小了。我无法评论这种技术作为正则化方法的可行性。