python实现多变量线性回归(Linear Regression with Multiple Variables)

python基础

浏览数:248

2019-9-5

本文介绍如何使用python实现多变量线性回归,文章参考NG的视频和黄海广博士的笔记

现在对房价模型增加更多的特征,例如房间数楼层等,构成一个含有多个变量的模型,模型中的特征为( x1,x2,…,xn)

表示为:

引入 x0=1,则公式
转化为:

 

1、加载训练数据

数据格式为:

X1,X2,Y

2104,3,399900

1600,3,329900

2400,3,369000

1416,2,232000

将数据逐行读取,用逗号切分,并放入np.array

#加载数据

#加载数据
def load_exdata(filename):
    data = []
    with open(filename, 'r') as f:
        for line in f.readlines(): 
            line = line.split(',')
            current = [int(item) for item in line]
            #5.5277,9.1302
            data.append(current)
    return data

data = load_exdata('ex1data2.txt');
data = np.array(data,np.int64)

x = data[:,(0,1)].reshape((-1,2))
y = data[:,2].reshape((-1,1))
m = y.shape[0]

# Print out some data points
print('First 10 examples from the dataset: \n')
print(' x = ',x[range(10),:],'\ny=',y[range(10),:])

  

 

First 10 examples from the dataset:

x = [[2104 3]

[1600 3]

[2400 3]

[1416 2]

[3000 4]

[1985 4]

[1534 3]

[1427 3]

[1380 3]

[1494 3]]

y= [[399900]

[329900]

[369000]

[232000]

[539900]

[299900]

[314900]

[198999]

[212000]

[242500]]

2、通过梯度下降求解theta

 (1)在多维特征问题的时候,要保证特征具有相近的尺度,这将帮助梯度下降算法更快地收敛。

解决的方法是尝试将所有特征的尺度都尽量缩放到-1 到 1 之间,最简单的方法就是(X – mu) / sigma,其中mu是平均值, sigma 是标准差。

(2)损失函数和单变量一样,依然计算损失平方和均值

我们的目标和单变量线性回归问题中一样,是要找出使得代价函数最小的一系列参数。多变量线性回归的批量梯度下降算法为:

求导数后得到:

(3)向量化计算

向量化计算可以加快计算速度,怎么转化为向量化计算呢?

在多变量情况下,损失函数可以写为:

对theta求导后得到:

(1/2*m) * (X.T.dot(X.dot(theta) y))

因此,theta迭代公式为:

theta = theta (alpha/m) * (X.T.dot(X.dot(theta) y))

(4)完整代码如下:

  

#特征缩放
def featureNormalize(X):
    X_norm = X;
    mu = np.zeros((1,X.shape[1]))
    sigma = np.zeros((1,X.shape[1]))
    for i in range(X.shape[1]):
        mu[0,i] = np.mean(X[:,i]) # 均值
        sigma[0,i] = np.std(X[:,i])     # 标准差
#     print(mu)
#     print(sigma)
    X_norm  = (X - mu) / sigma
    return X_norm,mu,sigma

#计算损失
def computeCost(X, y, theta):
    m = y.shape[0]
#     J = (np.sum((X.dot(theta) - y)**2)) / (2*m) 
    C = X.dot(theta) - y
    J2 = (C.T.dot(C))/ (2*m)
    return J2

#梯度下降
def gradientDescent(X, y, theta, alpha, num_iters):
    m = y.shape[0]
    #print(m)
    # 存储历史误差
    J_history = np.zeros((num_iters, 1))
    for iter in range(num_iters):
        # 对J求导,得到 alpha/m * (WX - Y)*x(i), (3,m)*(m,1)  X (m,3)*(3,1) = (m,1)
        theta = theta - (alpha/m) * (X.T.dot(X.dot(theta) - y))
        J_history[iter] = computeCost(X, y, theta)
    return J_history,theta
    

iterations = 10000  #迭代次数
alpha = 0.01    #学习率
x = data[:,(0,1)].reshape((-1,2))
y = data[:,2].reshape((-1,1))
m = y.shape[0]
x,mu,sigma = featureNormalize(x)
X = np.hstack([x,np.ones((x.shape[0], 1))])
# X = X[range(2),:]
# y = y[range(2),:]

theta = np.zeros((3, 1))

j = computeCost(X,y,theta)
J_history,theta = gradientDescent(X, y, theta, alpha, iterations)


print('Theta found by gradient descent',theta)

  

Theta found by gradient descent [[ 109447.79646964]

[ -6578.35485416]

[ 340412.65957447]]

绘制迭代收敛图

 

plt.plot(J_history)

plt.ylabel(‘lost’);

plt.xlabel(‘iter count’)

plt.title(‘convergence graph’)

使用模型预测结果

 

def predict(data):
    testx = np.array(data)
    testx = ((testx - mu) / sigma)
    testx = np.hstack([testx,np.ones((testx.shape[0], 1))])
    price = testx.dot(theta)
    print('price is %d ' % (price))

predict([1650,3])

  

price is 293081

no bb,上代码,代码下载

作者:JadePeng