Logistic回归

 

逻辑斯蒂回归

对于目标值是离散变量的两类分类问题，假设目标值为{0,1}，所以先改变模型使其预测值在[0,1]之间，我们选择这样一个函数：

分类概率值：

似然函数：

对数似然函数：

更新规则：

 # coding:utf-8
import matplotlib.pyplot as plt
import numpy as np

def dataN(length):#生成数据
    x = np.ones(shape = (length,3))
    y = np.zeros(length)
    for i in np.arange(0,length/100,0.02):
        x[100*i][1]=i
        x[100*i][2]=i + 1 + np.random.uniform(0,1.2)
        y[100*i]=1
        x[100*i+1][1]=i+0.01
        x[100*i+1][2]=i+0.01 + np.random.uniform(0,1.2)
    return np.mat(x),np.mat(y).T

def sigmoid(x): #simoid 函数
    return 1.0/(1+np.exp(-x))

def alphA(x,y): #选取前20次迭代cost最小的alpha
    c=float("inf")
    for k in range(1,1000):
            a=1.0/k**3
            f=gD(x,y,20,a)[1][-1]
            if f>c:
                break
            c=f
            alpha=a
    return alpha

def gD(x,y,iter,alpha):#梯度下降
    theta=np.ones((3,1))

    cost=[]
    for i in range(iter):
        hypothesis =sigmoid(np.dot(x,theta))
        loss = hypothesis-y
        cost.append(np.sum(loss[0]**2))
        gradient = np.dot(x.transpose(),loss)
        theta = theta -alpha * gradient
    return theta,cost

def tesT(theta, x, y):#准确率
    length=len(x)
    count=0
    for i in xrange(length):
        predict = sigmoid(x[i, :] * theta)[0, 0] > 0.5
        if predict == bool(y[i, 0]):
            count+= 1
    accuracy = float(count)/length
    return accuracy

length=200
iter=1000
x,y=dataN(length)
theta,cost=gD(x,y,iter,alphA(x,y))
print tesT(theta, x, y) #0.92

plt.figure(1)
plt.plot(range(iter),cost)
plt.figure(2)
color=['or','ob']
for i in xrange(length):
    plt.plot(x[i, 1], x[i, 2],color[int(y[i])])
theta = theta.getA()
plt.plot([0,length/100],[-theta[0],-theta[0]-theta[1]*length/100]/theta[2])
plt.show()