梯度下降法实现（Python语言描述）

提示：扫一扫查出行
【扫一扫了解最新限行尾号】
复制提示

原文地址：传送门

import numpy as np import matplotlib.pyplot as plt %matplotlib inline

plt.style.use(['ggplot'])

当你初次涉足机器学习时，你学习的第一个基本算法就是 梯度下降 (Gradient Descent), 可以说梯度下降法是机器学习算法的支柱。在这篇文章中，我尝试使用

python

python 解释梯度下降法的基本原理。一旦掌握了梯度下降法，很多问题就会变得容易理解，并且利于理解不同的算法。

如果你想尝试自己实现梯度下降法，你需要加载基本的

python

packages

packages ——

numpy

numpy and

matplotlib

首先，我们将创建包含着噪声的线性数据

# 随机创建一些噪声 X = 2 * np.random.rand(100, 1) y = 4 + 3 * X + np.random.randn(100, 1)

接下来通过 matplotlib 可视化数据

# 可视化数据 plt.plot(X, y, 'b.') plt.xlabel("$x$", fontsize=18) plt.ylabel("$y$", rotation=0, fontsize=18) plt.axis([0, 2, 0, 15])

显然，

y 与

x 具有良好的线性关系，这个数据非常简单，只有一个自变量

我们可以将其表示为简单的线性关系：

y = b + mx

y=b+mx

并求出

b ,

m。

这种被称为解方程的分析方法并没有什么不妥，但机器学习是涉及矩阵计算的，因此我们使用矩阵法（向量法）进行分析。

我们将

y 替换成

(

)

J(\theta)

J(θ)，

b 替换成

\theta_0

θ0，

m 替换成

\theta_1

θ1。
得到如下表达式：

(

)

J(\theta) = \theta_0 + \theta_1 x

J(θ)=θ0+θ1x

注意： 本例中

\theta_0= 4

θ0=4，

\theta_1= 3

θ1=3

求解

\theta_0

θ0 和

\theta_1

θ1 的分析方法，代码如下：

X_b = np.c_[np.ones((100, 1)), X] # 为X添加了一个偏置单位，对于X中的每个向量都是1 theta_best = np.linalg.inv(X_b.T.dot(X_b)).dot(X_b.T).dot(y) theta_best

array([[3.86687149], [3.12408839]])

不难发现这个值接近真实的

\theta_0

θ0，

\theta_1

θ1，由于我在数据中引入了噪声，所以存在误差。

X_new = np.array([[0], [2]]) X_new_b = np.c_[np.ones((2, 1)), X_new] y_predict = X_new_b.dot(theta_best) y_predict

array([[ 3.86687149], [10.11504826]])

梯度下降法（Gradient Descent）

Cost Function & Gradients

计算代价函数和梯度的公式如下所示。

注意：代价函数用于线性回归，对于其他算法，代价函数是不同的，梯度必须从代价函数中推导出来。

Cost

(

)

∑

(

)

(

)

−

(

)

J(\theta) = 1/2m \sum_{i=1}^{m} (h(\theta)^{(i)} - y^{(i)})^2

J(θ)=1/2mi=1∑m(h(θ)(i)−y(i))2

Gradient

∂

(

)

∂

∑

(

)

−

(

)

(

)

\frac{\partial J(\theta)}{\partial \theta_j} = 1/m\sum_{i=1}^{m}(h(\theta^{(i)} - y^{(i)}).X_j^{(i)}

∂θj∂J(θ)=1/mi=1∑m(h(θ(i)−y(i)).Xj(i)

Gradients

−

(

∑

(

)

−

(

)

(

)

\theta_0: = \theta_0 -\alpha . (1/m .\sum_{i=1}^{m}(h(\theta^{(i)} - y^{(i)}).X_0^{(i)})

θ0:=θ0−α.(1/m.i=1∑m(h(θ(i)−y(i)).X0(i))

−

(

∑

(

)

−

(

)

(

)

\theta_1: = \theta_1 -\alpha . (1/m .\sum_{i=1}^{m}(h(\theta^{(i)} - y^{(i)}).X_1^{(i)})

θ1:=θ1−α.(1/m.i=1∑m(h(θ(i)−y(i)).X1(i))

−

(

∑

(

)

−

(

)

(

)

\theta_2: = \theta_2 -\alpha . (1/m .\sum_{i=1}^{m}(h(\theta^{(i)} - y^{(i)}).X_2^{(i)})

θ2:=θ2−α.(1/m.i=1∑m(h(θ(i)−y(i)).X2(i))

−

(

∑

(

)

−

(

)

(

)

\theta_j: = \theta_j -\alpha . (1/m .\sum_{i=1}^{m}(h(\theta^{(i)} - y^{(i)}).X_0^{(i)})

θj:=θj−α.(1/m.i=1∑m(h(θ(i)−y(i)).X0(i))

def cal_cost(theta, X, y): ''' Calculates the cost for given X and Y. The following shows and example of a single dimensional X theta = Vector of thetas X = Row of X's np.zeros((2,j)) y = Actual y's np.zeros((2,1)) where: j is the no of features ''' m = len(y) predictions = X.dot(theta) cost = (1/2*m) * np.sum(np.square(predictions - y)) return cost

def gradient_descent(X, y, theta, learning_rate = 0.01, iterations = 100): ''' X = Matrix of X with added bias units y = Vector of Y theta=Vector of thetas np.random.randn(j,1) learning_rate iterations = no of iterations Returns the final theta vector and array of cost history over no of iterations ''' m = len(y) # learning_rate = 0.01 # iterations = 100 cost_history = np.zeros(iterations) theta_history = np.zeros((iterations, 2)) for i in range(iterations): prediction = np.dot(X, theta) theta = theta - (1/m) * learning_rate * (X.T.dot((prediction - y))) theta_history[i, :] = theta.T cost_history[i] = cal_cost(theta, X, y) return theta, cost_history, theta_history

# 从1000次迭代开始，学习率为0.01。从高斯分布的θ开始 lr =0.01 n_iter = 1000 theta = np.random.randn(2, 1) X_b = np.c_[np.ones((len(X), 1)), X] theta, cost_history, theta_history = gradient_descent(X_b, y, theta, lr, n_iter)print('Theta0: {:0.3f},\nTheta1: {:0.3f}'.format(theta[0][0],theta[1][0])) print('Final cost/MSE: {:0.3f}'.format(cost_history[-1]))

Theta0: 3.867, Theta1: 3.124 Final cost/MSE: 5457.747

# 绘制迭代的成本图 fig, ax = plt.subplots(figsize=(12,8))ax.set_ylabel('J(Theta)') ax.set_xlabel('Iterations') ax.plot(range(1000), cost_history, 'b.')

在大约 150 次迭代之后代价函数趋于稳定，因此放大到迭代200，看看曲线

fig, ax = plt.subplots(figsize=(10,8)) ax.plot(range(200), cost_history[:200], 'b.')

值得注意的是，最初成本下降得更快，然后成本降低的收益就不那么多了。我们可以尝试使用不同的学习速率和迭代组合，并得到不同学习率和迭代的效果会如何。

让我们建立一个函数，它可以显示效果，也可以显示梯度下降实际上是如何工作的。

def plot_GD(n_iter, lr, ax, ax1=None): ''' n_iter = no of iterations lr = Learning Rate ax = Axis to plot the Gradient Descent ax1 = Axis to plot cost_history vs Iterations plot ''' ax.plot(X, y, 'b.') theta = np.random.randn(2, 1) tr = 0.1 cost_history = np.zeros(n_iter) for i in range(n_iter): pred_prev = X_b.dot(theta) theta, h, _ = gradient_descent(X_b, y, theta, lr, 1) pred = X_b.dot(theta) cost_history[i] = h[0] if ((i % 25 == 0)): ax.plot(X, pred, 'r-', alpha=tr) if tr < 0.8: tr += 0.2 if not ax1 == None: ax1.plot(range(n_iter), cost_history, 'b.')

# 绘制不同迭代和学习率组合的图 fig = plt.figure(figsize=(30,25), dpi=200) fig.subplots_adjust(hspace=0.4, wspace=0.4)it_lr = [(2000, 0.001), (500, 0.01), (200, 0.05), (100, 0.1)] count = 0 for n_iter, lr in it_lr: count += 1 ax = fig.add_subplot(4, 2, count) count += 1 ax1 = fig.add_subplot(4, 2, count) ax.set_title("lr:{}" .format(lr)) ax1.set_title("Iterations:{}" .format(n_iter)) plot_GD(n_iter, lr, ax, ax1)

通过观察发现，以较小的学习速率收集解决方案需要很长时间，而学习速度越大，学习速度越快。

_, ax = plt.subplots(figsize=(14, 10)) plot_GD(100, 0.1, ax)

随机梯度下降法（Stochastic Gradient Descent）

随机梯度下降法，其实和批量梯度下降法原理类似，区别在与求梯度时没有用所有的

m 个样本的数据，而是仅仅选取一个样本

j 来求梯度。对应的更新公式是：

−

(

)

(

)

(

)

−

)

(

)

\theta_i = \theta_i - \alpha (h_\theta(x_0^{(j)}, x_1^{(j)}, ...x_n^{(j)}) - y_j)x_i^{(j)}

θi=θi−α(hθ(x0(j),x1(j),...xn(j))−yj)xi(j)

def stocashtic_gradient_descent(X, y, theta, learning_rate=0.01, iterations=10): ''' X = Matrix of X with added bias units y = Vector of Y theta=Vector of thetas np.random.randn(j,1) learning_rate iterations = no of iterations Returns the final theta vector and array of cost history over no of iterations ''' m = len(y) cost_history = np.zeros(iterations) for it in range(iterations): cost = 0.0 for i in range(m): rand_ind = np.random.randint(0, m) X_i = X[rand_ind, :].reshape(1, X.shape[1]) y_i = y[rand_ind, :].reshape(1, 1) prediction = np.dot(X_i, theta) theta -= (1/m) * learning_rate * (X_i.T.dot((prediction - y_i))) cost += cal_cost(theta, X_i, y_i) cost_history[it] = cost return theta, cost_history

lr = 0.5 n_iter = 50 theta = np.random.randn(2,1) X_b = np.c_[np.ones((len(X),1)), X] theta, cost_history = stocashtic_gradient_descent(X_b, y, theta, lr, n_iter)print('Theta0: {:0.3f},\nTheta1: {:0.3f}' .format(theta[0][0],theta[1][0])) print('Final cost/MSE: {:0.3f}' .format(cost_history[-1]))

Theta0: 3.762, Theta1: 3.159 Final cost/MSE: 46.964

fig, ax = plt.subplots(figsize=(10,8))ax.set_ylabel('$J(\Theta)$' ,rotation=0) ax.set_xlabel('$Iterations$') theta = np.random.randn(2,1)ax.plot(range(n_iter), cost_history, 'b.')

小批量梯度下降法（Mini-batch Gradient Descent）

小批量梯度下降法是批量梯度下降法和随机梯度下降法的折衷，也就是对于

m 个样本，我们采用x个样子来迭代，

1<x<m

1<x<m。一般可以取

x=10

x=10，当然根据样本的数据，可以调整这个

x 的值。对应的更新公式是：

−

∑

−

(

)

(

)

(

)

−

)

(

)

\theta_i = \theta_i - \alpha \sum\limits_{j=t}^{t+x-1}(h_\theta(x_0^{(j)}, x_1^{(j)}, ...x_n^{(j)}) - y_j)x_i^{(j)}

θi=θi−αj=t∑t+x−1(hθ(x0(j),x1(j),...xn(j))−yj)xi(j)

def minibatch_gradient_descent(X, y, theta, learning_rate=0.01, iterations=10, batch_size=20): ''' X = Matrix of X without added bias units y = Vector of Y theta=Vector of thetas np.random.randn(j,1) learning_rate iterations = no of iterations Returns the final theta vector and array of cost history over no of iterations ''' m = len(y) cost_history = np.zeros(iterations) n_batches = int(m / batch_size) for it in range(iterations): cost = 0.0 indices = np.random.permutation(m) X = X[indices] y = y[indices] for i in range(0, m, batch_size): X_i = X[i: i+batch_size] y_i = y[i: i+batch_size] X_i = np.c_[np.ones(len(X_i)), X_i] prediction = np.dot(X_i, theta) theta -= (1/m) * learning_rate * (X_i.T.dot((prediction - y_i))) cost += cal_cost(theta, X_i, y_i) cost_history[it] = cost return theta, cost_history

lr = 0.1 n_iter = 200 theta = np.random.randn(2, 1) theta, cost_history = minibatch_gradient_descent(X, y, theta, lr, n_iter)print('Theta0: {:0.3f},\nTheta1: {:0.3f}' .format(theta[0][0], theta[1][0])) print('Final cost/MSE: {:0.3f}' .format(cost_history[-1]))

Theta0: 3.842, Theta1: 3.146 Final cost/MSE: 1090.518

fig, ax = plt.subplots(figsize=(10,8))ax.set_ylabel('$J(\Theta)$', rotation=0) ax.set_xlabel('$Iterations$') theta = np.random.randn(2, 1)ax.plot(range(n_iter), cost_history, 'b.')