在本demo中，我们使用的二次函数为
[begin{aligned} f(x) &= left(x - 4 right) left(x - 8right) rand \ &= x ^ 2 - 12 x 32 rand end{aligned} ]
其中(rand)表示一个满足标准正态分布(Nleft(0, 1right))的随机数（平均值为0，方差为1）
基于Parameter手动构建
我们可以尝试用类似的方式，从参数（Parameter）开始构建一个二次函数。
一个二次函数的基本结构如下
[gleft(xright)=ax^2 bx c ]
其中包含三个参数：(a)、(b)、(c)。
代码如下：
代码语言：javascript复制import matplotlib.pyplot as plt
import torch
from torch import nn
from torch.autograd import Variable
class SquareRegression(nn.Module):
def __init__(self):
nn.Module.__init__(self)
self.a = nn.Parameter(torch.randn(1, 1), requires_grad=True) # 1 x 1
self.b = nn.Parameter(torch.randn(1, 1), requires_grad=True) # 1 x 1
self.c = nn.Parameter(torch.randn(1, 1), requires_grad=True) # 1 x 1
def forward(self, x_):
p_ = (x_ ** 2).mm(self.a) # n x 1
q_ = x_.mm(self.b) # n x 1
t_ = self.c # 1 x 1
return p_ q_ t_.expand_as(p_) # n x 1
if __name__ == "__main__":
n = 100
x = torch.linspace(-2, 12, n).resize_((n, 1)) # n x 1 tensor
y = (x - 4) * (x - 8) torch.randn(x.size()) # n x 1 tensor
model = SquareRegression()
criterion = nn.MSELoss()
optimizer = torch.optim.SGD(model.parameters(), lr=2e-5)
num_epochs = 500000
for epoch in range(num_epochs):
inputs, targets = Variable(x), Variable(y)
out = model(inputs)
loss = criterion(out, targets)
optimizer.zero_grad()
loss.backward()
optimizer.step()
if (epoch 1) % 100 == 0:
print('Epoch[{}/{}], loss:{:.6f}'.format(epoch 1, num_epochs, loss.item()))
for name, param in model.named_parameters():
print(name, param.data)
predict = model(x)
plt.plot(x.numpy(), y.numpy(), 'ro', label='Original Data')
plt.plot(x.numpy(), predict.data.numpy(), label='Fitting Line')
plt.show()输出结果如下：
代码语言：javascript复制Epoch[100/500000], loss:386.833191
Epoch[200/500000], loss:385.401001
Epoch[300/500000], loss:383.982391
Epoch[400/500000], loss:382.576904
Epoch[500/500000], loss:381.184418
Epoch[600/500000], loss:379.804657
Epoch[700/500000], loss:378.437256
Epoch[800/500000], loss:377.082153
Epoch[900/500000], loss:375.738983
Epoch[1000/500000], loss:374.407532
Epoch[1100/500000], loss:373.087585
Epoch[1200/500000], loss:371.778870
Epoch[1300/500000], loss:370.481293
Epoch[1400/500000], loss:369.194580
Epoch[1500/500000], loss:367.918549
... ... ... ...
Epoch[499400/500000], loss:0.985296
Epoch[499500/500000], loss:0.985296
Epoch[499600/500000], loss:0.985296
Epoch[499700/500000], loss:0.985296
Epoch[499800/500000], loss:0.985296
Epoch[499900/500000], loss:0.985296
Epoch[500000/500000], loss:0.985296
a tensor([[1.0048]])
b tensor([[-12.0305]])
c tensor([[31.8826]])生成图像如下：
可以发现，我们生成的拟合解如下
[begin{cases} a &= 1.0048 \ b &= -12.0305 \ c &= 31.8826 end{cases} ]
所得到的函数为
[begin{aligned} gleft(xright) &= a x^2 b x c \ &= 1.0048 x^2 -12.0305 x 31.8826 end{aligned} ]
和原公式(f(x) = x^2 - 12 x 32)已经十分接近，图像上的拟合结果也与期望结果基本一致。
基于Linear叠加构建
除此之外，我们还可以通过叠加Linear来构建一个二次函数。
本次使用的公式如下：
[begin{aligned} hleft(xright) &= left(w_1 x b_1right) left(w_2 x b_2right) \ &= w_1 w_2 x left(w_1 b_2 w_2 b_1right)x b_1 b_2 end{aligned} ]
包含四个参数(w_1)、(b_1)、(w_2)、(b_2)。
代码如下：
代码语言：javascript复制import matplotlib.pyplot as plt
import torch
from torch import nn
from torch.autograd import Variable
class SquareRegression(nn.Module):
def __init__(self):
nn.Module.__init__(self)
self.linear1 = nn.Linear(1, 1)
self.linear2 = nn.Linear(1, 1)
def forward(self, x_):
return self.linear2(x_) * self.linear1(x_) # n x 1
if __name__ == "__main__":
n = 100
x = torch.linspace(-2, 12, n).resize_((n, 1)) # n x 1 tensor
y = (x - 4) * (x - 8) torch.randn(x.size()) # n x 1 tensor
model = SquareRegression()
criterion = nn.MSELoss()
optimizer = torch.optim.SGD(model.parameters(), lr=2e-5)
num_epochs = 20000
for epoch in range(num_epochs):
inputs, targets = Variable(x), Variable(y)
out = model(inputs)
loss = criterion(out, targets)
optimizer.zero_grad()
loss.backward()
optimizer.step()
if (epoch 1) % 100 == 0:
print('Epoch[{}/{}], loss:{:.6f}'.format(epoch 1, num_epochs, loss.item()))
for name, param in model.named_parameters():
print(name, param.data)
predict = model(x)
plt.plot(x.numpy(), y.numpy(), 'ro', label='Original Data')
plt.plot(x.numpy(), predict.data.numpy(), label='Fitting Line')
plt.show()输出结果如下：
代码语言：javascript复制Epoch[100/20000], loss:448.569519
Epoch[200/20000], loss:444.616638
Epoch[300/20000], loss:441.241089
Epoch[400/20000], loss:438.343903
Epoch[500/20000], loss:435.845673
Epoch[600/20000], loss:433.682007
Epoch[700/20000], loss:431.799835
Epoch[800/20000], loss:430.154999
Epoch[900/20000], loss:428.709595
Epoch[1000/20000], loss:427.431366
Epoch[1100/20000], loss:426.291718
Epoch[1200/20000], loss:425.265594
Epoch[1300/20000], loss:424.330353
Epoch[1400/20000], loss:423.465973
Epoch[1500/20000], loss:422.654541
... ... ... ...
Epoch[19400/20000], loss:1.044041
Epoch[19500/20000], loss:1.044041
Epoch[19600/20000], loss:1.044041
Epoch[19700/20000], loss:1.044041
Epoch[19800/20000], loss:1.044041
Epoch[19900/20000], loss:1.044041
Epoch[20000/20000], loss:1.044041
linear1.weight tensor([[0.7155]])
linear1.bias tensor([-5.7249])
linear2.weight tensor([[1.4002]])
linear2.bias tensor([-5.6002])生成图像如下：
可以发现，我们生成的拟合解如下
[begin{cases} w_1 &= 0.7155 \ b_1 &= -5.7249 \ w_2 &= 1.4002 \ b_2 &= -5.6002 end{cases} ]
所得到的函数为
[begin{aligned} h(x) &= left( w_1 x b_1 right) left( w_2 x b_2 right) \ &= left( 0.7155 x - 5.7249 right) left( 1.4002 x - 5.6002 right) \ &= 1.0018431 x^2 - 12.02294808 x 32.06058498 end{aligned} ]
和原公式(f(x) = x^2 - 12 x 32)已经十分接近，图像上的拟合结果也与期望结果基本一致。而且训练速度还得到了大幅度的提高（原本进行了500000次，这次在使用同样learning rate的情况下，只进行了20000次就达到了类似的效果。
其他
关于Linear
通过阅读源码可以发现，Linear内含两个最为关键的参数：weight和bias。一定程度上可以理解为所构建的线性函数公式为
[f(x) = wx b ]
以此类推并扩展，在2输入（in_features == 2）3输出（out_features == 3）的时候，公式为：
[begin{cases} f_1(x) &= w_{1, 1} x_1 w_{1, 2} x_2 b_1 \ f_2(x) &= w_{2, 1} x_1 w_{2, 2} x_2 b_2 \ f_3(x) &= w_{3, 1} x_1 w_{3, 2} x_2 b_3 end{cases} ]
其中(w)和(b)都将以Tensor的格式，作为weight和bias进行存储，前向传播的时候则会进行矩阵运算并生成计算图。
