编程实现对率回归,并给出西瓜数据集 3.0α 上的结果.
1数据集编号,密度,含糖率,好瓜1,0.697,0.46,是2,0.774,0.376,是3,0.634,0.264,是4,0.608,0.318,是5,0.556,0.215,是6,0.403,0.237,是7,0.481,0.149,是8,0.437,0.211,是9,0.666,0.091,否10,0.243,0.267,否11,0.245,0.057,否12,0.343,0.099,否13
·
1数据集
编号,密度,含糖率,好瓜
1,0.697,0.46,是
2,0.774,0.376,是
3,0.634,0.264,是
4,0.608,0.318,是
5,0.556,0.215,是
6,0.403,0.237,是
7,0.481,0.149,是
8,0.437,0.211,是
9,0.666,0.091,否
10,0.243,0.267,否
11,0.245,0.057,否
12,0.343,0.099,否
13,0.639,0.161,否
14,0.657,0.198,否
15,0.36,0.37,否
16,0.593,0.042,否
17,0.719,0.103,否
2模型代码
import csv
import numpy as np
from matplotlib import pyplot as plt
class logodds_regress(object):
def sigmoid(self,z):
'''
@param z:beta * xi
'''
return 1/(1 + np.exp(-z))
def lossfunc(self,y,z):
'''
@param y:标签
@param z:beta * x_i
@return:返回目标函数值
'''
return np.sum(-y*z + np.log(1+np.exp(z)))
def dl_to_beta(self,xtrain,ytrain,beta):
'''
@param xtrain:(x,1) shape[N,d+1]
@param ytrain:label shape[N,1]
@param beta: (w,b) shape [1,d+1]
@return beta
'''
#shape [N, 1]
z = np.dot(xtrain,beta.T)
p1 = np.exp(z) / (1 + np.exp(z))
#shape [N, N]
p = np.diag((p1 * (1-p1)).reshape(-1))#生成对角阵
#shape [N, 1]
dl1 = -np.sum(xtrain * (ytrain - p1), 0, keepdims=True) #按列相加,保持矩阵的二维性
#shape [3, 3]
dl2 = xtrain.T .dot(p).dot(xtrain)
beta -= np.dot(dl1,np.linalg.inv(dl2))
return beta
def newton(self,xtrain, ytrain):
'''
牛顿迭代法求解beta
@param xtrain:(x,1) shape[N,d+1]
@param ytrain:label shape[N,1]
@return beta (w,b) shape [1,d+1]
'''
#initialization
beta = np.ones((1, 3))
#shape [N, 1]
z = np.dot(beta,xtrain.T)
#log-likehood
loss_current = 0
loss_next = self.lossfunc(ytrain,z)
err = 1e-5
while( np.abs(loss_current-loss_next) > err):
beta = self.dl_to_beta(xtrain,ytrain,beta)
z = np.dot(beta,xtrain.T)
loss_current = loss_next
loss_next = self.lossfunc(ytrain,z)
return beta
def gradient_descent(self,xtrain,ytrain):
'''
梯度下降法求解beta
@param xtrain:(x,1) shape[N,d+1]
@param ytrain:label shape[N,1]
@return beta (w,b) shape [1,d+1]
'''
beta = np.ones((1,3)) * 0.1
z = np.dot(xtrain,beta.T)
learn_rate = 0.05
iter_max = 2000
for i in range(iter_max):
p1 = np.exp(z) / (1 + np.exp(z))
#shape [N, N]
p = np.diag((p1 * (1-p1)).reshape(-1))#生成对角阵
#shape [N, 1] 一阶导数
dl1 = -np.sum(xtrain * (ytrain - p1), 0, keepdims=True) #按列相加,保持矩阵的二维性
beta -= dl1 * learn_rate
z = np.dot(xtrain,beta.T)
return beta
def solver_sklearn(self,xtrain,ytrain):
'''
sklearn 模块中的lbfgs方法求beta
@param xtrain:(x,1) shape[N,d+1]
@param ytrain:label shape[N,1]
@return beta (w,b) shape [1,d+1]
'''
from sklearn.linear_model import LogisticRegression
reg = LogisticRegression(solver='lbfgs', C=1000).fit(xtrain,ytrain)
beta = np.c_[reg.coef_,reg.intercept_]
return beta
def model(self,xtrain,ytrain,solver='newton'):
if solver == 'newton':
return self.newton(xtrain,ytrain)
elif solver == 'gradient_descent':
return self.gradient_descent(xtrain,ytrain)
elif solver == 'solver_sklearn':
xtrain = np.delete(xtrain,-1,axis=1)
return self.solver_sklearn(xtrain,ytrain)
else:
raise ValueError('Unknown method {}'.format(solver))
def predict(self,beta,xtest):
'''
#我们以0.5为界,预测值y大于0.5则判断为好瓜,赋值1;反之判断为不是好瓜,赋值0
'''
z = beta.dot(xtest.T)
ypredict = self.sigmoid(z)
ypredict[ypredict>0.5] = 1
ypredict[ypredict<=0.5] = 0
ypredict = ypredict.reshape(-1,1)
return ypredict
def read_waremelon_data(filename):
'''
读取西瓜数据并转换
@param filename:数据文件
'''
with open(filename,newline='',encoding='utf-8') as csvfile:
data = csv.DictReader(csvfile)
judge_to_num = {'是':1,'否':0}
density = [] #密度
sugar_rate = [] #含糖率
y = [] #标签
for item in data:
density.append(float(item['密度']))
sugar_rate.append(float(item['含糖率']))
y.append(judge_to_num[item['好瓜']])
density = np.array(density)
sugar_rate = np.array(sugar_rate)
xtrain = np.hstack((density.reshape(-1,1),sugar_rate.reshape(-1,1)))
return (xtrain,y)
if __name__=='__main__':
filename = 'table45.csv'
xtrain,y = read_waremelon_data(filename)
###绘制训练数据
y=np.array(y)
data_label1 = xtrain[y==1,:]
data_label0 = xtrain[y==0,:]
plt.scatter(data_label1[:, 0], data_label1[:, 1], c='y', marker='o')
plt.scatter(data_label0[:, 0], data_label0[:, 1], c='b', marker='+')
###数据w->beta,x->\hat{x}
#shape [N,3]
xtrain = np.hstack((xtrain,np.ones([len(y),1])))
#shape [N,1]
ytrain = y.reshape(-1,1)
### 建模
A = logodds_regress()
beta = A.model(xtrain,ytrain,solver='solver_sklearn')
ypredict = A.predict(beta,xtrain)
print(ypredict)
print('准确率',sum(ytrain==ypredict)/len(ytrain))
###绘图直线方程w1x1+w2x2+b=0-->x2=-(w1x1+b)/w2
ymin = -( beta[0, 0]*0.1 + beta[0, 2] ) / beta[0, 1]
ymax = -( beta[0, 0]*0.9 + beta[0, 2] ) / beta[0, 1]
plt.plot([0.1, 0.9], [ymin, ymax], 'k-')
plt.xlabel('density')
plt.ylabel('sugar rate')
plt.title("logit regression")
plt.show()
3预测结果


DAMO开发者矩阵,由阿里巴巴达摩院和中国互联网协会联合发起,致力于探讨最前沿的技术趋势与应用成果,搭建高质量的交流与分享平台,推动技术创新与产业应用链接,围绕“人工智能与新型计算”构建开放共享的开发者生态。
更多推荐
所有评论(0)