【吴恩达机器学习】练习1-线性回归

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

path =  'ex1data1.txt'
data = pd.read_csv(path, header=None, names=['Population', 'Profit'])

data.head()

data.info

data.describe()

# 散点图
data.plot(kind='scatter', x='Population', y='Profit', figsize=(12,8))
plt.show()

def computeCost(X, y, theta):
    inner = np.power(((X * theta.T) - y), 2)
    return np.sum(inner) / (2 * len(X))

data.insert(0, 'Ones', 1)

data.insert(1, 'Twos', 1)

data

data.drop(['Twos'],axis=1,inplace=True)

data

# set X (training data训练数据集) and y (target variable目标变量)
cols = data.shape[1]
X = data.iloc[:,0:cols-1]#X是所有行，去掉最后一列
y = data.iloc[:,cols-1:cols]#y是所有行，取最后一列

X.head()#head()是观察前5行

y.head()

X = np.matrix(X.values)
y = np.matrix(y.values)
theta = np.matrix(np.array([0,0]))

theta

X.shape, theta.shape, y.shape

computeCost(X, y, theta)

def gradientDescent(X, y, theta, alpha, iters):
    temp = np.matrix(np.zeros(theta.shape))# 初始化theta.shape维的0向量
    parameters = int(theta.ravel().shape[1])
    cost = np.zeros(iters)# 迭代次数
    
    # 
    for i in range(iters):
        error = (X * theta.T) - y
        for j in range(parameters):
            term = np.multiply(error, X[:,j])
            temp[0,j] = theta[0,j] - ((alpha / len(X)) * np.sum(term))
            
        theta = temp
        cost[i] = computeCost(X, y, theta)
        
    return theta, cost

alpha = 0.01
iters = 1000

g, cost = gradientDescent(X, y, theta, alpha, iters)
g

computeCost(X, y, g)

x = np.linspace(data.Population.min(), data.Population.max(), 100)# 返回均匀间隔的100个数
f = g[0, 0] + (g[0, 1] * x)

fig, ax = plt.subplots(figsize=(12,8))
ax.plot(x, f, 'r', label='Prediction')
ax.scatter(data.Population, data.Profit, label='Traning Data')
ax.legend(loc=2)
ax.set_xlabel('Population')
ax.set_ylabel('Profit')
ax.set_title('Predicted Profit vs. Population Size')
plt.show()

fig, ax = plt.subplots(figsize=(12,8))
ax.plot(np.arange(iters), cost, 'r')
ax.set_xlabel('Iterations')
ax.set_ylabel('Cost')
ax.set_title('Error vs. Training Epoch')
plt.show()

path =  'ex1data2.txt'
data2 = pd.read_csv(path, header=None, names=['Size', 'Bedrooms', 'Price'])
data2.head()

data2 = (data2 - data2.mean()) / data2.std()
data2.head()

# add ones column
data2.insert(0, 'Ones', 1)

# set X (training data) and y (target variable)
cols = data2.shape[1]
X2 = data2.iloc[:,0:cols-1]
y2 = data2.iloc[:,cols-1:cols]

# convert to matrices and initialize theta
X2 = np.matrix(X2.values)
y2 = np.matrix(y2.values)
theta2 = np.matrix(np.array([0,0,0]))

# perform linear regression on the data set
g2, cost2 = gradientDescent(X2, y2, theta2, alpha, iters)

# get the cost (error) of the model
computeCost(X2, y2, g2)

fig, ax = plt.subplots(figsize=(12,8))
ax.plot(np.arange(iters), cost2, 'r')
ax.set_xlabel('Iterations')
ax.set_ylabel('Cost')
ax.set_title('Error vs. Training Epoch')
plt.show()

from sklearn import linear_model
model = linear_model.LinearRegression()
model.fit(X, y)

x = np.array(X[:, 1].A1)
f = model.predict(X).flatten()

fig, ax = plt.subplots(figsize=(12,8))
ax.plot(x, f, 'r', label='Prediction')
ax.scatter(data.Population, data.Profit, label='Traning Data')
ax.legend(loc=2)
ax.set_xlabel('Population')
ax.set_ylabel('Profit')
ax.set_title('Predicted Profit vs. Population Size')
plt.show()

# 正规方程
def normalEqn(X, y):
    theta = np.linalg.inv(X.T@X)@X.T@y#X.T@X等价于X.T.dot(X)
    return theta

final_theta2=normalEqn(X, y)#感觉和批量梯度下降的theta的值有点差距
final_theta2

#梯度下降得到的结果是matrix([[-3.24140214,  1.1272942 ]])