用线性模型逼近平方函数时,PyTorch不收敛

时间:2019-04-30 01:55:20

标签: python machine-learning neural-network regression pytorch

我正在尝试学习PyTorch,并引用了此讨论here

作者提供了最少的代码片段,说明了如何使用PyTorch求解已被随机噪声污染的未知线性函数。

这段代码对我来说很好。

但是,当我更改函数以使我希望t = X ^ 2时,参数似乎没有收敛。

import torch
import torch.nn as nn
import torch.optim as optim
from torch.autograd import Variable

# Let's make some data for a linear regression.
A = 3.1415926
b = 2.7189351
error = 0.1
N = 100 # number of data points

# Data
X = Variable(torch.randn(N, 1))

# (noisy) Target values that we want to learn.
t = X * X + Variable(torch.randn(N, 1) * error)

# Creating a model, making the optimizer, defining loss
model = nn.Linear(1, 1)
optimizer = optim.SGD(model.parameters(), lr=0.05)
loss_fn = nn.MSELoss()

# Run training
niter = 50
for _ in range(0, niter):
    optimizer.zero_grad()
    predictions = model(X)
    loss = loss_fn(predictions, t)
    loss.backward()
    optimizer.step()

    print("-" * 50)
    print("error = {}".format(loss.data[0]))
    print("learned A = {}".format(list(model.parameters())[0].data[0, 0]))
    print("learned b = {}".format(list(model.parameters())[1].data[0]))

当我执行此代码时,新的A和b参数似乎是随机的,因此不会收敛。我认为这应该收敛,因为您可以使用斜率和偏移函数近似任何函数。我的理论是我没有正确使用PyTorch。

有人能确定我的t = X * X + Variable(torch.randn(N, 1) * error)代码行有问题吗?

1 个答案:

答案 0 :(得分:3)

您无法将二阶多项式与线性函数拟合。您不能期望超过随机数(因为您有多项式中的随机样本)。
您可以做的就是尝试并输入两个输入xx^2并从它们中进行拟合:

model = nn.Linear(2, 1)  # you have 2 inputs now
X_input = torch.cat((X, X**2), dim=1)  # have 2 inputs per entry
# ...

    predictions = model(X_input)  # 2 inputs -> 1 output
    loss = loss_fn(predictions, t)
    # ...
    # learning t = c*x^2 + a*x + b
    print("learned a = {}".format(list(model.parameters())[0].data[0, 0]))
    print("learned c = {}".format(list(model.parameters())[0].data[0, 1])) 
    print("learned b = {}".format(list(model.parameters())[1].data[0]))