本文参考,对后面将公式转化为矩阵的形式加了些解释,便于大家理解。
梯度下降是用来求函数最小值点的一种方法,所谓梯度在一元函数中是指某一点的斜率,在多元函数中可表示为一个向量,可由偏导求得,此向量的指向是函数值上升最快的方向。公式表示为:
\[ \nabla J(\Theta) = \langle \frac{\overrightarrow{\partial J}}{\partial \theta_1}, \frac{\overrightarrow{\partial J}}{\partial \theta_2}, \frac{\overrightarrow{\partial J}}{\partial \theta_3} \rangle \]比如
\[ \begin{aligned} J(\theta) &= \theta^2 \qquad \nabla J(\theta) = J'(\theta) = 2\theta\\ J(\theta) &= 2\theta_1+3\theta_2 \quad \nabla J(\theta) = \langle 2,3 \rangle \end{aligned} \]
由于梯度是函数值上升最快的方向,那么我们一直沿着梯度的反方向走,就能找到函数的局部最低点。
因此有梯度下降的递推公式:\[ \Theta^1 = \Theta^0 - \alpha \nabla J(\Theta^0) \] 其中\(\alpha\)称为学习率,直观得讲就是每次沿着梯度反方向移动的距离,学习率过高可能会越过最低点,学习率过低又会使得学习速度偏慢。 为了更好的理解上面的公式 我们举例:\[ \quad J(\theta) = \theta^2 \Longrightarrow \nabla J(\theta) = J'(\theta) = 2\theta \] 则梯度下降的迭代过程有:\[ \begin{aligned} \theta^0 &= 1 \\ \theta^1 &= \theta^0 - \alpha \times J'(\theta^0) \\ &= 1-0.1 \times 2 \\ &= 0.2 \\ \theta^2 &= \theta^1 - \alpha \times J'(\theta^1)\\ &= 0.04\\ \theta^3 &= 0.008\\ \theta^4 &= 0.0016\\ \end{aligned} \] 我们知道$J(\theta) = \theta^2 \(的极小值点在\)(0,0)\(处,而迭代四次后的结果\)(0.0016,0.00000256)$已相当接近。再举一个二元函数的例子:
\[J(\Theta) = \theta_1^2 + \theta_2^2 \Longrightarrow \nabla J(\Theta) = \langle 2\theta_1 + 2\theta_2 \rangle\] 令\(\alpha = 0.1\), 以初始点\((1,3)\)用梯度下降法求函数最低点\[ \begin{aligned} \Theta^0 &= (1,3) \\ \Theta^1 &= \Theta^0 - \alpha \nabla J(\Theta^0)\\ &= (1,3) - 0.1(2,6)\\ &= (0.8,2.4)\\ \Theta^2 &= (0.8,2.4) - 0.1(1.6,4.8)\\ &= (0.64,1.92) \\ \Theta^3 &= (0.512,1.536) \\ \end{aligned} \]值得注意的是,迭代的效率除了和步长(学习率)\(\alpha\)有关外,还与初始值的选取有关。
梯度下降做散点拟合
利用梯度下降法拟合出一条直线使得平均方差最小。
假设拟合后的直线为:\[h_\Theta(x) = \theta_0 + \theta_1 \times x \] 其中\(\Theta = (\theta_0, \theta_1)\) 把平均方差作为代价函数\[J(\Theta) = \frac{1}{2m}\sum_{i=1}^m \left\lbrace h_\Theta(x_i) - y_i \right\rbrace ^2\]\[ \begin{aligned} &\nabla J(\Theta) = \langle \frac{\delta J}{\delta \theta_0}, \frac{\delta J}{\delta\theta_1} \rangle \\ &\frac{\delta J}{\delta \theta_0} = \frac{1}{m} \sum_{i=1}^m \left\lbrace h_\Theta(x_i) - y_i \right\rbrace\\ &\frac{\delta J}{\delta \theta_1} = \frac{1}{m} \sum_{i=1}^m \left\lbrace h_\Theta(x_i) - y_i \right\rbrace x_i\\ \end{aligned} \]
为了便于利用python的矩阵运算,我们可以将公式稍微变形
\[ \begin{aligned} h_\Theta(x_i) &= \theta_0 \times 1 + \theta_1 \times x_i \\ &= \begin{bmatrix} 1 & x_0\\ 1 & x_1\\ \vdots & \vdots\\ 1 & x_{19}\\ \end{bmatrix} \times \begin{bmatrix} \theta_0\\ \theta_1 \end{bmatrix} \end{aligned} \]\(i\)的范围是\([1,20]\),因此最后得到
\[ h_\Theta(x_i)= \begin{bmatrix} h_0\\ h_1\\ \vdots\\ h_{19} \end{bmatrix} \]\[ 误差diff = \begin{bmatrix} h_0\\ h_1\\ \vdots\\ h_{19} \end{bmatrix}- \begin{bmatrix} y_0\\ y_1\\ \vdots\\ y_{19} \end{bmatrix} \]
\[ 方差和 \sum_{i=1}^m \left\lbrace h_\Theta(x_i) - y_i \right\rbrace ^2=diff^T \times diff \]
将公式变为矩阵向量相乘的形式后你应当更容易理解下面的python代码
import numpy as np# Size of the points dataset.m = 20# Points x-coordinate and dummy value (x0, x1).X0 = np.ones((m, 1))X1 = np.arange(1, m+1).reshape(m, 1)X = np.hstack((X0, X1))# Points y-coordinatey = np.array([ 3, 4, 5, 5, 2, 4, 7, 8, 11, 8, 12, 11, 13, 13, 16, 17, 18, 17, 19, 21]).reshape(m, 1)# The Learning Rate alpha.alpha = 0.01def error_function(theta, X, y): '''Error function J definition.''' diff = np.dot(X, theta) - y return (1./(2*m)) * np.dot(np.transpose(diff), diff)def gradient_function(theta, X, y): '''Gradient of the function J definition.''' diff = np.dot(X, theta) - y return (1./m) * np.dot(np.transpose(X), diff)def gradient_descent(X, y, alpha): '''Perform gradient descent.''' theta = np.array([1, 1]).reshape(2, 1) gradient = gradient_function(theta, X, y) while not np.all(np.absolute(gradient) <= 1e-5): theta = theta - alpha * gradient gradient = gradient_function(theta, X, y) return thetaoptimal = gradient_descent(X, y, alpha)print('optimal:', optimal)print('error function:', error_function(optimal, X, y)[0,0])