Ch3.3 对数几率回归

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问题：编程实现对数几率回归，并给出西瓜数据集3.0a上的结果

方法一：使用sklearn实现对数几率回归

import numpy as np
import matplotlib.pyplot as plt

from sklearn import model_selection
from sklearn.linear_model import LogisticRegression
from sklearn import metrics

data = np.array([[0.697, 0.460, 1],
        [0.774, 0.376, 1],
        [0.634, 0.264, 1],
        [0.608, 0.318, 1],
        [0.556, 0.215, 1],
        [0.403, 0.237, 1],
        [0.481, 0.149, 1],
        [0.437, 0.211, 1],
        [0.666, 0.091, 0],
        [0.243, 0.267, 0],
        [0.245, 0.057, 0],
        [0.343, 0.099, 0],
        [0.639, 0.161, 0],
        [0.657, 0.198, 0],
        [0.360, 0.370, 0],
        [0.593, 0.042, 0],
        [0.719, 0.103, 0]])
X=data[:,0:2]
Y=data[:,-1]

#绘制数据集
f1=plt.figure(1)
plt.title("watermelon_3a")
plt.xlabel("密度")
plt.ylabel("含糖量")
plt.scatter(X[Y==0,0],X[Y==0,1],marker='o',color='k',s=100,label='bad')
plt.scatter(X[Y==1,0],X[Y==1,1],marker='o',color='g',s=100,label='good')
plt.legend(loc='upper right')
plt.show()

#使用sklearn

X_train,X_test,Y_train,Y_test=model_selection.train_test_split(X,Y,test_size = 0.5,random_state = 0)

log_model=LogisticRegression()
log_model.fit(X_train,Y_train)

Y_pred=log_model.predict(X_test)

print(metrics.confusion_matrix(Y_test,Y_pred))
print(metrics.classification_report(Y_test,Y_pred))


log_model.fit(X,Y)
Y_pred=log_model.predict(X)
print(Y_pred)
right=0
for i in range(17):
        if Y_pred[i]==Y[i]:
                right+=1
print("accuect is :",right/17*100)

方法二：自己用python实现对数几率回归

import numpy as np
import matplotlib.pyplot as plt

from sklearn import model_selection
from sklearn.linear_model import LogisticRegression
from sklearn import metrics

data = np.array([[0.697, 0.460, 1],
        [0.774, 0.376, 1],
        [0.634, 0.264, 1],
        [0.608, 0.318, 1],
        [0.556, 0.215, 1],
        [0.403, 0.237, 1],
        [0.481, 0.149, 1],
        [0.437, 0.211, 1],
        [0.666, 0.091, 0],
        [0.243, 0.267, 0],
        [0.245, 0.057, 0],
        [0.343, 0.099, 0],
        [0.639, 0.161, 0],
        [0.657, 0.198, 0],
        [0.360, 0.370, 0],
        [0.593, 0.042, 0],
        [0.719, 0.103, 0]])
X=data[:,0:2]
Y=data[:,-1]

#绘制数据集
f1=plt.figure(1)
plt.title("watermelon_3a")
plt.xlabel("密度")
plt.ylabel("含糖量")
plt.scatter(X[Y==0,0],X[Y==0,1],marker='o',color='k',s=100,label='bad')
plt.scatter(X[Y==1,0],X[Y==1,1],marker='o',color='g',s=100,label='good')
plt.legend(loc='upper right')
plt.show()



# 批量梯度下降法
def logit(beta,x,y):
        #print(np.shape(beta))
        #print(np.shape(x))
        l=-y*(np.dot(beta.T,x))+np.math.log(1+np.math.exp(np.dot(beta.T,x)))
        return l

def logitFunction(beta,X,Y):
        m,n=np.shape(X)
        sum=0
        for i in range(m):
                sum+=logit(beta,X[i],Y[i])
        return sum

def gradDescent(X,Y):
        h=0.1
        max_iters=500
        m,n=np.shape(X)

        beta=np.zeros(n)
        deta_beta=np.ones(n)
        llh=0
        llh_temp=0
        old_beta=beta

        for i in range(max_iters):
                beta_temp=beta

                for j in range(n):
                        beta[j]+=deta_beta[j]
                        llh_temp=logitFunction(beta,X,Y)
                        deta_beta[j]=-h*(llh_temp-llh)/deta_beta[j]
                        beta[j]=beta_temp[j]

                beta+=deta_beta
                llh=logitFunction(beta,X,Y)
                old_beta=np.row_stack((old_beta,beta))
        return beta,old_beta


if __name__=='__main__':
        X=np.column_stack((X,np.ones(np.shape(X)[0])))
        #X = np.mat(X)
       # Y = np.mat(Y)
        #print(X)
        #print(Y)
        beta,old_beta=gradDescent(X,Y)
        print(beta)

        #绘制beta[0],beta[1],beta[2](=b)的迭代图像
        f2=plt.figure(2)
        m,n=np.shape(old_beta)
        print(m,n)
        print(old_beta)
        plt.subplot(311)
        plt.plot(np.arange(0,m-1,1),old_beta[1:,0])
        plt.xlabel("deta1")
        plt.ylabel("iters")

        plt.subplot(312)
        plt.plot(np.arange(0,m-1,1),old_beta[1:,1])
        plt.xlabel("deta2")
        plt.ylabel("iters")

        plt.subplot(313)
        plt.plot(np.arange(0,m-1,1),old_beta[1:,2])
        plt.xlabel("b")
        plt.ylabel("iters")
        plt.show()

        #测试对数几率回归模型，计算精度
        Y_pre=np.zeros((17,1))
        for i in range(17):
                Y_pre[i]=1/(1+np.math.exp(-1*(np.dot(beta,X[i]))))
                if(Y_pre[i]>0.5):
                        Y_pre[i]=1
                else:
                        Y_pre[i]=0
        print(Y_pre)
        right=0
        for i in range(17):
                if Y_pre[i]==Y[i]:
                 right+=1
        print("accuect is :",right/17*100)


        #画出决策边界，最佳拟合直线
        #令 beta[0]*x1+beta[1]*x2+b=0,得出x1和x2的关系，在图中显示
        # 由sigmod函数的性质，1/(1+exp(-z)), z=0是两个类别的分界处
        f3 = plt.figure(3)
        plt.title("watermelon_3a")
        plt.xlabel("密度")
        plt.ylabel("含糖量")
        plt.scatter(X[Y == 0, 0], X[Y == 0, 1], marker = 'o', color = 'k', s = 100, label = 'bad')
        plt.scatter(X[Y == 1, 0], X[Y == 1, 1], marker = 'o', color = 'g', s = 100, label = 'good')
        plt.legend(loc = 'upper right')
        xcord=np.arange(0,1,0.001)
        ycord= (-beta[0]*X - beta[2])/beta[1]
        plt.plot(X, ycord)
        plt.xlabel("x1")
        plt.ylabel("x2")
        plt.show()

方法二的运行结果：

beta的值如下：beta[0]、beta[1]、beta[2]（=参数b）