python怎么实现逻辑回归的梯度下降法

发布网友发布时间：2022-04-26 01:18

共2个回答

懂视网时间：2022-04-18 12:09

这篇文章主要介绍了python实现逻辑回归的方法示例，这是机器学习课程的一个实验，整理出来共享给大家，需要的朋友可以参考学习，下来要一起看看吧。

本文实现的原理很简单，优化方法是用的梯度下降。后面有测试结果。

先来看看实现的示例代码：

# coding=utf-8
from math import exp

import matplotlib.pyplot as plt
import numpy as np
from sklearn.datasets.samples_generator import make_blobs


def sigmoid(num):
 '''

 :param num: 待计算的x
 :return: sigmoid之后的数值
 '''
 if type(num) == int or type(num) == float:
 return 1.0 / (1 + exp(-1 * num))
 else:
 raise ValueError, 'only int or float data can compute sigmoid'


class logistic():
 def init(self, x, y): 
 if type(x) == type(y) == list:
 self.x = np.array(x)
 self.y = np.array(y)
 elif type(x) == type(y) == np.ndarray:
 self.x = x
 self.y = y
 else:
 raise ValueError, 'input data error'

 def sigmoid(self, x):
 '''

 :param x: 输入向量
 :return: 对输入向量整体进行simgoid计算后的向量结果
 '''
 s = np.frompyfunc(lambda x: sigmoid(x), 1, 1)
 return s(x)

 def train_with_punish(self, alpha, errors, punish=0.0001):
 '''

 :param alpha: alpha为学习速率
 :param errors: 误差小于多少时停止迭代的阈值
 :param punish: 惩罚系数
 :param times: 最大迭代次数
 :return:
 '''
 self.punish = punish
 dimension = self.x.shape[1]
 self.theta = np.random.random(dimension)
 compute_error = 100000000
 times = 0
 while compute_error > errors:
 res = np.dot(self.x, self.theta)
 delta = self.sigmoid(res) - self.y
 self.theta = self.theta - alpha * np.dot(self.x.T, delta) - punish * self.theta # 带惩罚的梯度下降方法
 compute_error = np.sum(delta)
 times += 1

 def predict(self, x):
 '''

 :param x: 给入新的未标注的向量
 :return: 按照计算出的参数返回判定的类别
 '''
 x = np.array(x)
 if self.sigmoid(np.dot(x, self.theta)) > 0.5:
 return 1
 else:
 return 0


def test1():
 '''
 用来进行测试和画图，展现效果
 :return:
 '''
 x, y = make_blobs(n_samples=200, centers=2, n_features=2, random_state=0, center_box=(10, 20))
 x1 = []
 y1 = []
 x2 = []
 y2 = []
 for i in range(len(y)):
 if y[i] == 0:
 x1.append(x[i][0])
 y1.append(x[i][1])
 elif y[i] == 1:
 x2.append(x[i][0])
 y2.append(x[i][1])
 # 以上均为处理数据，生成出两类数据
 p = logistic(x, y)
 p.train_with_punish(alpha=0.00001, errors=0.005, punish=0.01) # 步长是0.00001，最大允许误差是0.005，惩罚系数是0.01
 x_test = np.arange(10, 20, 0.01)
 y_test = (-1 * p.theta[0] / p.theta[1]) * x_test
 plt.plot(x_test, y_test, c='g', label='logistic_line')
 plt.scatter(x1, y1, c='r', label='positive')
 plt.scatter(x2, y2, c='b', label='negative')
 plt.legend(loc=2)
 plt.title('punish value = ' + p.punish.str())
 plt.show()


if name == 'main':
 test1()

运行结果如下图

总结

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热心网友时间：2022-04-18 09:17

import sys

#Training data set
#each element in x represents (x0,x1,x2)
x = [(1,0.,3) , (1,1.,3) ,(1,2.,3), (1,3.,2) , (1,4.,4)]
#y[i] is the output of y = theta0 * x[0] + theta1 * x[1] +theta2 * x[2]
y = [95.364,97.217205,75.195834,60.105519,49.342380]

epsilon = 0.0001
#learning rate
alpha = 0.01
diff = [0,0]
max_itor = 1000
error1 = 0
error0 =0
cnt = 0
m = len(x)

#init the parameters to zero
theta0 = 0
theta1 = 0
theta2 = 0

while True:

cnt = cnt + 1

#calculate the parameters
for i in range(m):

diff[0] = y[i]-( theta0 + theta1 * x[i][1] + theta2 * x[i][2] )

theta0 = theta0 + alpha * diff[0] * x[i][0]
theta1 = theta1 + alpha * diff[0]* x[i][1]
theta2 = theta2 + alpha * diff[0]* x[i][2]

#calculate the cost function
error1 = 0
for lp in range(len(x)):
error1 += ( y[i]-( theta0 + theta1 * x[i][1] + theta2 * x[i][2] ) )**2/2

if abs(error1-error0) < epsilon:
break
else:
error0 = error1

print ' theta0 : %f, theta1 : %f, theta2 : %f, error1 : %f'%(theta0,theta1,theta2,error1)

print 'Done: theta0 : %f, theta1 : %f, theta2 : %f'%(theta0,theta1,theta2)