吴恩达cs229|编程作业第一周（Python）

 

练习一：线性回归

 

目录

1.包含的文件

2.单元线性回归

3.多元线性回归

1.包含的文件

文件名	含义
ex1.py	单元线性回归
ex1_multi.py	多元线性回归
ex1data1.txt	单变量线性回归数据集
ex1data2.txt	多变量线性回归数据集
plotData.py	数据可视化
computeCost.py	损失函数
gradientDescent.py	梯度下降
featureNormalize.py	特征归一化
normalEqn.py	正规方程求解线性回归程序

 

红色部分需要自己填写。

2.单元线性回归

项目需要的包

import matplotlib.pyplot as plt
import numpy as np
from matplotlib.colors import LogNorm
from mpl_toolkits.mplot3d import axes3d, Axes3D
from computeCost import *
from gradientDescent import *
from plotData import *
import matplotlib.pyplot as plt

2.1可视化数据集

• 主程序

# ===================== Part 1: Plotting =====================
print('Plotting Data...')


data = np.loadtxt('ex1data1.txt', delimiter=',', usecols=(0, 1))
X = data[:, 0]
y = data[:, 1]
m = y.size


plot_data(X, y)

input('Program paused. Press ENTER to continue')

• 可视化函数plotData.py

import matplotlib.pyplot as plt


def plot_data(x, y):
    # ===================== Your Code Here =====================
    # Instructions : Plot the training data into a figure using the matplotlib.pyplot
    #                using the "plt.scatter" function. Set the axis labels using
    #                "plt.xlabel" and "plt.ylabel". Assume the population and revenue data
    #                have been passed in as the x and y.

    # Hint : You can use the 'marker' parameter in the "plt.scatter" function to change the marker type (e.g. "x", "o").
    #        Furthermore, you can change the color of markers with 'c' parameter.
    plt.scatter(x,y,marker='o',s=50,cmap='Blues',alpha=0.3)  #绘制散点图
    plt.xlabel('population')  #设置x轴标题
    plt.ylabel('profits')   #设置y轴标题
    
    # ===========================================================

    plt.show()

• 运行结果：

2.2梯度下降

• 单变量线性回归的目标是使损失函数最小化：

• 单变量线性模型是：

• 编写计算线性回归代价函数的程序computeCost.py

import numpy as np


def compute_cost(X, y, theta):
    # Initialize some useful values
    m = y.size
    cost = 0

    # ===================== Your Code Here =====================
    # Instructions : Compute the cost of a particular choice of theta.
    #                You should set the variable "cost" to the correct value.
    h_out = theta*X#回归输出
    error = (h_out.sum(axis=1)-y)*(h_out.sum(axis=1)-y)#先按行求和  再与Y按列求差 再内积 
    error_all = error.sum(axis=0)#按列求和 得到总误差
    cost = 1/(2*m)*error_all

    # ==========================================================

    return cost

• 运行结果为：

Initial cost : 32.0727338775 (This value should be about 32.07)

• 梯度下降为：

• 编写梯度下降程序gradientDescent.py

import numpy as np
from computeCost import *


def gradient_descent(X, y, theta, alpha, num_iters):
    # Initialize some useful values
    m = y.size
    J_history = np.zeros(num_iters)

    for i in range(0, num_iters):
        # ===================== Your Code Here =====================
        # Instructions : Perform a single gradient step on the parameter vector theta
        #
        # Hint: X.shape = (97, 2), y.shape = (97, ), theta.shape = (2, )
        h_out = theta * X#回归输出
        error = h_out.sum(axis=1) - y#先按行求和 a1x1+a0 再与y求偏差
        error_2d =  error.reshape(m,1)#转换成2维数组进行乘法
        error_all = error_2d*X#是一个 m*2大小的数组
        theta = theta -  alpha*(1/m)*error_all.sum(axis = 0)#按列求和成 1*2大小数组更新
        # ===========================================================
        # Save the cost every iteration
        J_history[i] = compute_cost(X, y, theta)

    return theta, J_history

• 运行结果为：

Theta found by gradient descent: [-3.63029144  1.16636235]

• 主程序

# ===================== Part 2: Gradient descent =====================
print('Running Gradient Descent...')
X = np.c_[np.ones(m), X]  # Add a column of ones to X
theta = np.zeros(2)  # initialize fitting parameters

# Some gradient descent settings
iterations = 1500
alpha = 0.01
# Compute and display initial cost
print('Initial cost : ' + str(compute_cost(X, y, theta)) + ' (This value should be about 32.07)')
theta, J_history = gradient_descent(X, y, theta, alpha, iterations)
print('Theta found by gradient descent: ' + str(theta.reshape(2)))

# Plot the linear fit
plt.figure(0)
line1, = plt.plot(X[:, 1], np.dot(X, theta), label='Linear Regression')
plt.legend(handles=[line1])
input('Program paused. Press ENTER to continue')

# Predict values for population sizes of 35,000 and 70,000
predict1 = np.dot(np.array([1, 3.5]), theta)
print('For population = 35,000, we predict a profit of {:0.3f} (This value should be about 4519.77)'.format(predict1*10000))
predict2 = np.dot(np.array([1, 7]), theta)
print('For population = 70,000, we predict a profit of {:0.3f} (This value should be about 45342.45)'.format(predict2*10000))

input('Program paused. Press ENTER to continue')

• 利用训练好的参数进行预测，与期望值进行比较，验证我们的程序是正确的：

For population = 35,000, we predict a profit of 4519.768 (This value should be about 4519.77)
For population = 70,000, we predict a profit of 45342.450 (This value should be about 45342.45)

• 拟合的直线：

2.3可视化代价函数

# ===================== Part 3: Visualizing J(theta0, theta1) =====================
print('Visualizing J(theta0, theta1) ...')
 
theta0_vals = np.linspace(-10, 10, 100)  #参数1的取值
theta1_vals = np.linspace(-1, 4, 100)    #参数2的取值
 
xs, ys = np.meshgrid(theta0_vals, theta1_vals)   #生成网格
J_vals = np.zeros(xs.shape)
 
 
for i in range(0, theta0_vals.size):
    for j in range(0, theta1_vals.size):
        t = np.array([theta0_vals[i], theta1_vals[j]])
        J_vals[i][j] = compute_cost(X, y, t)    #计算每个网格点的代价函数值
 
J_vals = np.transpose(J_vals)
 
fig1 = plt.figure(1)      #绘制3d图形
ax = fig1.gca(projection='3d')
ax.plot_surface(xs, ys, J_vals)
plt.xlabel(r'$\theta_0$')
plt.ylabel(r'$\theta_1$')
 
#绘制等高线图 相当于3d图形的投影
plt.figure(2)
lvls = np.logspace(-2, 3, 20)
plt.contour(xs, ys, J_vals, levels=lvls, norm=LogNorm())
plt.plot(theta[0], theta[1], c='r', marker="x")

• 可视化效果

3.多元线性回归

3.1特征归一化

• 编写归一化代码featureNormalize.py

import numpy as np


def feature_normalize(X):
    # You need to set these values correctly
    n = X.shape[1]  # the number of features
    X_norm = X
    mu = np.zeros(n)
    sigma = np.zeros(n)

    # ===================== Your Code Here =====================
    # Instructions : First, for each feature dimension, compute the mean
    #                of the feature and subtract it from the dataset,
    #                storing the mean value in mu. Next, compute the
    #                standard deviation of each feature and divide
    #                each feature by its standard deviation, storing
    #                the standard deviation in sigma
    #
    #                Note that X is a 2D array where each column is a
    #                feature and each row is an example. You need
    #                to perform the normalization separately for
    #                each feature.
    #
    # Hint: You might find the 'np.mean' and 'np.std' functions useful.
    #       To get the same result as Octave 'std', use np.std(X, 0, ddof=1)
    #
    mu = np.mean(X_norm,axis=0)#按列求平均值
    sigma = np.std(X_norm, axis=0) # axis=0计算每一列的标准差
    X_norm = (X-mu)/sigma

    # ===========================================================

    return X_norm, mu, sigma

•  测试函数：

# ===================== Part 1: Feature Normalization =====================
print('Loading Data...')
data = np.loadtxt('ex1data2.txt', delimiter=',', dtype=np.int64)
X = data[:, 0:2]
y = data[:, 2]
m = y.size

# Print out some data points
print('First 10 examples from the dataset: ')
for i in range(0, 10):
    print('x = {}, y = {}'.format(X[i], y[i]))

input('Program paused. Press ENTER to continue')

# Scale features and set them to zero mean
print('Normalizing Features ...')

X, mu, sigma = feature_normalize(X)
X = np.c_[np.ones(m), X]  # Add a column of ones to X

 

3.2梯度下降

• 多元线性回归，要以向量的形式：

• 编写多元参数梯度下降程序gradientDescent.py

def gradient_descent_multi(X, y, theta, alpha, num_iters):
    # Initialize some useful values
    m = y.size
    J_history = np.zeros(num_iters)

    for i in range(0, num_iters):
        # ===================== Your Code Here =====================
        # Instructions : Perform a single gradient step on the parameter vector theta
        #
        h_out = theta * X#回归输出
        error = h_out.sum(axis=1) - y#先按行求和 a1x1+a0 再与y求偏差
        error_2d =  error.reshape(m,1)#转换成2维数组进行乘法
        error_all = error_2d*X#是一个 m*2大小的数组
        theta = theta -  alpha*(1/m)*error_all.sum(axis = 0)#按列求和成 1*2大小数组更新
        
        # ===========================================================
        # Save the cost every iteration
        J_history[i] = compute_cost(X, y, theta)

    return theta, J_history

• 测试代码： 

# ===================== Part 2: Gradient Descent =====================

# ===================== Your Code Here =====================
# Instructions : We have provided you with the following starter
#                code that runs gradient descent with a particular
#                learning rate (alpha).
#
#                Your task is to first make sure that your functions -
#                computeCost and gradientDescent already work with
#                this starter code and support multiple variables.
#
#                After that, try running gradient descent with
#                different values of alpha and see which one gives
#                you the best result.
#
#                Finally, you should complete the code at the end
#                to predict the price of a 1650 sq-ft, 3 br house.
#
# Hint: At prediction, make sure you do the same feature normalization.
#

print('Running gradient descent ...')

# Choose some alpha value
alpha = 0.03

num_iters = 500

# Init theta and Run Gradient Descent
theta = np.zeros(3)
theta, J_history = gradient_descent_multi(X, y, theta, alpha, num_iters)

# Plot the convergence graph
plt.figure()
plt.plot(np.arange(J_history.size), J_history)
plt.xlabel('Number of iterations')
plt.ylabel('Cost J')

# Display gradient descent's result
print('Theta computed from gradient descent : \n{}'.format(theta))

# Estimate the price of a 1650 sq-ft, 3 br house
# ===================== Your Code Here =====================
# Recall that the first column of X is all-ones. Thus, it does
# not need to be normalized.
a1 = np.array([1650, 3])
a1_nromal = (a1 - mu)/sigma
X1 = np.c_[np.ones(1), a1_nromal.reshape(1,2)]#扩充一列
price = (theta*X1).sum(axis=1)[0] # You should change this


# ==========================================================

print('Predicted price of a 1650 sq-ft, 3 br house (using gradient descent) : {:0.3f}'.format(price))

input('Program paused. Press ENTER to continue')

• 梯度下降法求解的最优参数，样例的预测价格以及代价函数随迭代次数的变化曲线

3.3正规方程直接求解 

 

•  正规方程：

• 编写代码normalEqn.py：

import numpy as np

def normal_eqn(X, y):
    theta = np.zeros((X.shape[1], 1))

    # ===================== Your Code Here =====================
    # Instructions : Complete the code to compute the closed form solution
    #                to linear regression and put the result in theta
    #
    X_mat = np.mat(X)
    y_mat = np.mat(y)
    A = np.dot(X_mat.T, X_mat)
    B = np.dot(A.I,X.T)
    theta = np.dot(B,y)
    #theta = np.dot(np.dot(np.dot(X_mat.T, X_mat).I,X_mat.T), y).T#这样写是错误的
    theta = theta.getA()#矩阵转换成数组
    return theta

• 测试程序： 

# ===================== Part 3: Normal Equations =====================

print('Solving with normal equations ...')

# ===================== Your Code Here =====================
# Instructions : The following code computes the closed form
#                solution for linear regression using the normal
#                equations. You should complete the code in
#                normalEqn.py
#
#                After doing so, you should complete this code
#                to predict the price of a 1650 sq-ft, 3 br house.
#

# Load data
data = np.loadtxt('ex1data2.txt', delimiter=',', dtype=np.int64)
X = data[:, 0:2]
y = data[:, 2]
m = y.size

# Add intercept term to X
X = np.c_[np.ones(m), X]

theta = normal_eqn(X, y)

# Display normal equation's result
print('Theta computed from the normal equations : \n{}'.format(theta))

# Estimate the price of a 1650 sq-ft, 3 br house
# ===================== Your Code Here =====================
a2 = np.array([1650, 3])
X2 = np.c_[np.ones(1), a2.reshape(1,2)]#扩充一列
price = (theta*X2).sum(axis=1)[0]  # You should change this


# ==========================================================

print('Predicted price of a 1650 sq-ft, 3 br house (using normal equations) : {:0.3f}'.format(price))

input('ex1_multi Finished. Press ENTER to exit')

 

 

注：所有代码及说明PDF在全部更新完后统一上传。