7.6非线性回归应用：logistic regretion application

python实现：

import numpy as np
import random

def gradientDescent(x,y,theta,alpha,m,numIterations):
    xTrans = x.transpose()
    for i in range(0,numIterations):
        hypothesis = np.dot(x,theta)
        loss = hypothesis - y
        cost = np.sum(loss ** 2)/(2 * m)
        print("Iteration %d / Cost: %f" %(i,cost))
        gradient = np.dot(xTrans,loss)/m
        theta = theta - alpha*gradient
    return theta

def genData(numPoints,bias,variance):
    x = np.zeros(shape=(numPoints,2))
    y = np.zeros(shape=numPoints)

    for i in range(0,numPoints):
        x[i][0] = 1
        x[i][1] = i

        y[i] = (i+bias)+random.uniform(0,1)*variance
    return x,y

x,y = genData(100,25,10)
print("x:")
print(x)
print("y:")
print(y)
m,n = np.shape(x)
n_y = np.shape(y)

print("x shape:",str(m)," ",str(n))
print("y shape:",str(n_y))

numIterations = 100000
alpha = 0.0005
theta = np.ones(n)
theta = gradientDescent(x,y,theta,alpha,m,numIterations)
print(theta)

结果：

.......

Iteration 99988 / Cost: 4.024027
Iteration 99989 / Cost: 4.024027
Iteration 99990 / Cost: 4.024027
Iteration 99991 / Cost: 4.024027
Iteration 99992 / Cost: 4.024027
Iteration 99993 / Cost: 4.024027
Iteration 99994 / Cost: 4.024027
Iteration 99995 / Cost: 4.024027
Iteration 99996 / Cost: 4.024027
Iteration 99997 / Cost: 4.024027
Iteration 99998 / Cost: 4.024027
Iteration 99999 / Cost: 4.024027
[29.89292353  0.99333989]

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