我尝试在PyTorch中进行多项式回归。首先，我仅尝试了线性回归（b + wx）。

model_1 = RegressionModel()W = torch.zeros(1, requires_grad=True)b = torch.zeros(1, requires_grad = True)optimizer_1 = torch.optim.SGD([W, b], lr = 0.001)x_train = torch.FloatTensor(dataset.x_data['LSTAT'])y_train = torch.FloatTensor(dataset.data['target'])nb_epochs = 10000for epoch in range(nb_epochs + 1):    hypothesis = x_train * W + b    cost = torch.nn.functional.mse_loss(hypothesis, y_train.float())    optimizer_1.zero_grad()    cost.backward()    optimizer_1.step()    print('Epoch {:4d}/{} W: {:.3f}, b: {:.3f}, Cost: {:.6f}'.format(epoch,     nb_epochs, W.item(), b.item(), cost.item()))

然后，我更改并添加了一些变量来进行多项式回归（b + w1x + w2x^2）

model_2 = RegressionModel()W1 = torch.zeros(1, requires_grad=True) W2 = torch.zeros(1, requires_grad=True) b = torch.zeros(1, requires_grad = True)optimizer_2 = torch.optim.SGD([W2, W1, b], lr = 0.0000099)x_train = torch.FloatTensor(dataset.x_data['LSTAT'])y_train = torch.FloatTensor(dataset.data['target'])nb_epochs = 10000for epoch in range(nb_epochs + 1):    hypothesis = b + x_train * W1 + x_train * x_train * W2    cost = torch.nn.functional.mse_loss(hypothesis, y_train.float())    optimizer_2.zero_grad()    cost.backward()    optimizer_2.step()    print('Epoch {:4d}/{} W1: {:.3f}, W2: {:.3f}, b: {:.3f}, Cost:     {:.6f}'.format(epoch, nb_epochs, W1.item(), W2.item(), b.item(),     cost.item()))

我可以像这样尝试多项式回归吗？如果不行，我会非常感激你告诉我。我对PyTorch真的是个新手…

回答：

你的代码应该可以工作。在处理较大数据时，如果你通过单一矩阵操作进行回归，会更有效率。为此，你需要首先预计算输入特征的多项式：

x_train_polynomial = torch.stack([x_train, x_train ** 2], dim=1)

为了节省一些代码行，你可以将投影重写为线性层：

在训练循环中，你可以调用：

hypothesis = projection(x_train_polynomial)