【机器学习】【线性回归】梯度下降

文章目录

@[toc]
数据集
实际值
估计值
估计误差
代价函数
学习率
参数更新
`Python`实现
线性拟合结果
代价结果

数据集

$\left(x^{(i)} , y^{(i)}\right) , i = 1 , 2 , \cdots , m$

实际值

$y^{(i)}$

估计值

$h_{\theta}{\left(x^{(i)}\right)} = \theta_{0} + \theta_{1}{x^{(i)}}$

估计误差

$h_{\theta}{\left(x^{(i)}\right)} - y^{(i)}$

代价函数

$J(\theta) = J(\theta_{0} , \theta_{1}) = \cfrac{1}{2m} \displaystyle\sum\limits_{i = 1}^{m}{\left(h_{\theta}{\left(x^{(i)}\right)} - y^{(i)}\right)^{2}} = \cfrac{1}{2m} \displaystyle\sum\limits_{i = 1}^{m}{\left(\theta_{0} + \theta_{1}{x^{(i)}} - y^{(i)}\right)^{2}}$

学习率

$\alpha$ 是学习率，一个大于 $0$ 的很小的经验值，决定代价函数下降的程度

参数更新

$\Delta{\theta_{j}} = \cfrac{\partial}{\partial{\theta_{j}}} J(\theta_{0} , \theta_{1})$

$\theta_{j} := \theta_{j} - \alpha \Delta{\theta_{j}} = \theta_{j} - \alpha \cfrac{\partial}{\partial{\theta_{j}}} J(\theta_{0} , \theta_{1})$

$\left[ \begin{matrix} \theta_{0} \\ \theta_{1} \end{matrix} \right] := \left[ \begin{matrix} \theta_{0} \\ \theta_{1} \end{matrix} \right] - \alpha \left[ \begin{matrix} \cfrac{\partial{J(\theta_{0} , \theta_{1})}}{\partial{\theta_{0}}} \\ \cfrac{\partial{J(\theta_{0} , \theta_{1})}}{\partial{\theta_{1}}} \end{matrix} \right]$

$\left[ \begin{matrix} \cfrac{\partial{J(\theta_{0} , \theta_{1})}}{\partial{\theta_{0}}} \\ \cfrac{\partial{J(\theta_{0} , \theta_{1})}}{\partial{\theta_{1}}} \end{matrix} \right] = \left[ \begin{matrix} \cfrac{1}{m} \displaystyle\sum\limits_{i = 1}^{m}{\left(h_{\theta}{\left(x^{(i)}\right)} - y^{(i)}\right)} \\ \cfrac{1}{m} \displaystyle\sum\limits_{i = 1}^{m}{\left(h_{\theta}{\left(x^{(i)}\right)} - y^{(i)}\right) x^{(i)}} \end{matrix} \right] = \left[ \begin{matrix} \cfrac{1}{m} \displaystyle\sum\limits_{i = 1}^{m}{e^{(i)}} \\ \cfrac{1}{m} \displaystyle\sum\limits_{i = 1}^{m}{e^{(i)} x^{(i)}} \end{matrix} \right] \kern{2em} e^{(i)} = h_{\theta}{\left(x^{(i)}\right)} - y^{(i)}$

$\begin{aligned} \left[ \begin{matrix} \cfrac{\partial{J(\theta_{0} , \theta_{1})}}{\partial{\theta_{0}}} \\ \cfrac{\partial{J(\theta_{0} , \theta_{1})}}{\partial{\theta_{1}}} \end{matrix} \right] &= \left[ \begin{matrix} \cfrac{1}{m} \displaystyle\sum\limits_{i = 1}^{m}{e^{(i)}} \\ \cfrac{1}{m} \displaystyle\sum\limits_{i = 1}^{m}{e^{(i)} x^{(i)}} \end{matrix} \right] = \left[ \begin{matrix} \cfrac{1}{m} \left(e^{(1)} + e^{(2)} + \cdots + e^{(m)}\right) \\ \cfrac{1}{m} \left(e^{(1)} + e^{(2)} + \cdots + e^{(m)}\right) x^{(i)} \end{matrix} \right] \\ &= \cfrac{1}{m} \left[ \begin{matrix} 1 & 1 & \cdots & 1 \\ x^{(1)} & x^{(2)} & \cdots & x^{(m)} \end{matrix} \right] \left[ \begin{matrix} e^{(1)} \\ e^{(2)} \\ \vdots \\ e^{(m)} \end{matrix} \right] = \cfrac{1}{m} X^{T} e = \cfrac{1}{m} X^{T} (X \theta - y) \end{aligned}$

由上述推导得

$\Delta{\theta} = \cfrac{1}{m} X^{T} e$

$\theta := \theta - \alpha \Delta{\theta} = \theta - \alpha \cfrac{1}{m} X^{T} e$

`Python`实现

import numpy as np
import matplotlib.pyplot as plt

x = np.array([4, 3, 3, 4, 2, 2, 0, 1, 2, 5, 1, 2, 5, 1, 3])
y = np.array([8, 6, 6, 7, 4, 4, 2, 4, 5, 9, 3, 4, 8, 3, 6])

m = len(x)

x = np.c_[np.ones([m, 1]), x]
y = y.reshape(m, 1)  # 转成列向量
theta = np.zeros([2, 1])

alpha = 0.01
iter_cnt = 1000  # 迭代次数
cost = np.zeros([iter_cnt])  # 代价数据

for i in range(iter_cnt):
    h = x.dot(theta)  # 估计值
    error = h - y  # 误差值
    cost[i] = 1 / 2 * m * error.T.dot(error)  # 代价值

    # 更新参数
    delta_theta = 1 / m * x.T.dot(error)
    theta -= alpha * delta_theta

# 线性拟合结果
plt.scatter(x[:, 1], y, c='blue')
plt.plot(x[:, 1], h, 'r-')
plt.show()

# 代价结果
plt.plot(cost)
plt.show()