机器学习笔记：线性回归

Posted 2020-10-12 菩提树下的杨过

下列代码来自 https://zh.gluon.ai/chapter_supervised-learning/linear-regression-scratch.html

 1 from mxnet import ndarray as nd
 2 from mxnet import autograd
 3 import random
 4 
 5 num_inputs = 2
 6 num_examples = 1000
 7 
 8 true_w = [2, -3.4]
 9 true_b = 4.2
10 
11 X = nd.random_normal(shape=(num_examples, num_inputs)) #1000行，2列的数据集
12 y = true_w[0] * X[:, 0] + true_w[1] * X[:, 1] + true_b #已知答案的结果
13 y += .01 * nd.random_normal(shape=y.shape) #加入噪音
14 
15 batch_size = 10
16 def data_iter():
17     #产生一个随机索引列表
18     idx = list(range(num_examples))
19     random.shuffle(idx)
20     for i in range(0, num_examples, batch_size):
21         j = nd.array(idx[i:min(i+batch_size,num_examples)])
22         yield nd.take(X, j), nd.take(y, j) #每次随机取出10行，2列的数据
23 
24 #初始化模型参数（即：需要求解的参数变量）
25 w = nd.random_normal(shape=(num_inputs, 1))
26 b = nd.zeros((1,))
27 params = [w, b]
28 
29 #创建梯度
30 for param in params:
31     param.attach_grad()
32     
33 #定义线性回归模型
34 def net(X):
35     return nd.dot(X, w) + b
36 
37 #定义损失函数
38 def square_loss(yhat, y):
39     # 注意这里我们把y变形成yhat的形状来避免自动广播
40     return (yhat - y.reshape(yhat.shape)) ** 2
41 
42 #随机梯度下降法
43 def SGD(params, lr):
44     for param in params:
45         param[:] = param - lr * param.grad
46         
47 
48 #训练
49 epochs = 5
50 learning_rate = .001
51 for e in range(epochs):
52     total_loss = 0
53     for data, label in data_iter():
54         with autograd.record():
55             output = net(data)
56             loss = square_loss(output, label)
57         loss.backward()
58         SGD(params, learning_rate)
59 
60         total_loss += nd.sum(loss).asscalar()
61     print("Epoch %d, average loss: %f" % (e, total_loss/num_examples))
62 
63 print(true_w) #打印答案
64 print(w) #打印求解结果
65 
66 print(true_b) #打印答案
67 print(b) #打印求解结果

Epoch 0, average loss: 6.012281
Epoch 1, average loss: 0.102830
Epoch 2, average loss: 0.001909
Epoch 3, average loss: 0.000133
Epoch 4, average loss: 0.000101 #5次迭代后，已经快速收敛
[2, -3.4] #已知答案

[[ 2.00017834] #求解结果
 [-3.40006614]]
<NDArray 2x1 @cpu(0)>
4.2

[ 4.19863892]
<NDArray 1 @cpu(0)>