Question

我正在尝试通过pytorch进行多项式回归。首先，我只尝试了线性回归（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 = 10000

for 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以进行多项式回归（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 = 10000

for 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()))

我可以这样尝试多项式回归吗？如果没有，如果您让我知道，我将不胜感激。我真的不是菜鸟...

Answer 1

您的代码应该可以使用。当使用较大的数据时，如果在单个矩阵运算中进行回归，则效率会更高。为此，您需要先预先计算输入特征的多项式：

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

要保存一些线，可以将投影重写为线性层：

import torch.nn as nn
projection = nn.Linear(2, 1, bias=True)

在训练循环中，您可以致电：

hypothesis = projection(x_train_polynomial)

我在pytorch中使用多项式回归吗？

1 个答案: