一:线性logistic 回归

代码如下:

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.optimize as opt
import seaborn as sns #读取数据集
path = 'ex2data1.txt'
data = pd.read_csv(path, header=None, names=['Exam 1', 'Exam 2', 'Admitted']) #将正负数据集分开
positive = data[data['Admitted'].isin([1])]
negative = data[data['Admitted'].isin([0])] '''
#查看分布
fig, ax = plt.subplots(figsize=(12, 8))
ax.scatter(positive['Exam 1'], positive['Exam 2'], s=60, c='b', marker='o', label='Admitted')
ax.scatter(negative['Exam 1'], negative['Exam 2'], s=50, c='r', marker='x', label='UnAdmitted')
ax.legend()
ax.set_xlabel('Exam 1 Score')
ax.set_ylabel('Exam 2 Score')
plt.show()
''' #sigmoid函数实现
def sigmoid(h):
return 1 / (1 + np.exp(-h)) '''
#测试sigmoid函数
nums = np.arange(-10, 11, step=1)
fig, ax = plt.subplots(figsize=(12, 8))
ax.plot(nums, sigmoid(nums), 'k')
plt.show()
''' #计算损失函数值
def cost(theta, X, y):
theta = np.matrix(theta)
X = np.matrix(X)
y = np.matrix(y) part1 = np.multiply(-y, np.log(sigmoid(X * theta.T)))
part2 = np.multiply((1-y), np.log(1-sigmoid(X * theta.T)))
return np.sum(part1-part2) / len(X) #在原矩阵第1列前加一列全1
data.insert(0, 'ones', 1) cols = data.shape[1] X = data.iloc[:, 0:cols-1]
y = data.iloc[:, cols-1:cols] X = np.array(X.values)
y = np.array(y.values)
theta = np.zeros(3) #这里是一个行向量 #返回梯度向量,注意是向量
def gradient(theta, X, y):
theta = np.matrix(theta)
X = np.matrix(X)
y = np.matrix(y) parameters = theta.ravel().shape[1]
grad = np.zeros(parameters) error = sigmoid(X * theta.T) - y grad = error.T.dot(X)
grad = grad / len(X)
return grad #通过高级算法计算出最好的theta值
result = opt.fmin_tnc(func=cost, x0=theta, fprime=gradient, args=(X, y)) #print(cost(result[0], X, y)) #测试所得theta的性能
#计算原数据集的预测情况
def predict(theta, X):
theta = np.matrix(theta)
X = np.matrix(X) probability = sigmoid(X * theta.T)
return [1 if i > 0.5 else 0 for i in probability] theta_min = result[0]
predictions = predict(theta_min, X) correct = [1 if((a == 1 and b == 1) or(a == 0 and b == 0)) else 0 for(a, b) in zip(predictions, y)]
accuracy = (sum(map(int, correct)) % len(correct))
print('accuracy = {0}%'.format(accuracy))#训练集测试准确度89% # 作图
theta_temp = theta_min
theta_temp = theta_temp / theta_temp[2] x = np.arange(130, step=0.1)
y = -(theta_temp[0] + theta_temp[1] * x)
#画出原点
sns.set(context='notebook', style='ticks', font_scale=1.5)
sns.lmplot('Exam 1', 'Exam 2', hue='Admitted', data=data,
size=6,
fit_reg=False,
scatter_kws={"s": 25}
)
#画出分界线
plt.plot(x, y, 'grey')
plt.xlim(0, 130)
plt.ylim(0, 130)
plt.title('Decision Boundary')
plt.show()

二:非线性logistic 回归(正则化)

代码如下:

import pandas as pd
import numpy as np
import scipy.optimize as opt
import matplotlib.pyplot as plt path = 'ex2data2.txt'
data = pd.read_csv(path, header=None, names=['Test 1', 'Test 2', 'Accepted']) positive = data[data['Accepted'].isin([1])]
negative = data[data['Accepted'].isin([0])] '''
#显示原始数据的分布
fig, ax = plt.subplots(figsize=(12, 8))
ax.scatter(positive['Test 1'], positive['Test 2'], s=50, c='b', marker='o', label='Accepted')
ax.scatter(negative['Test 1'], negative['Test 2'], s=50, c='r', marker='x', label='Unaccepted')
ax.legend() #显示右上角的Accepted 和 Unaccepted标签
ax.set_xlabel('Test 1 Score')
ax.set_ylabel('Test 2 Score')
plt.show()
'''
degree = 5
x1 = data['Test 1']
x2 = data['Test 2']
#在data的第三列插入一列全1
data.insert(3, 'Ones', 1) #创建多项式特征值,最高阶为4
for i in range(1, degree):
for j in range(0, i):
data['F' + str(i) + str(j)] = np.power(x1, i-j) * np.power(x2, j) #删除原数据中的test 1和test 2两列
data.drop('Test 1', axis=1, inplace=True)
data.drop('Test 2', axis=1, inplace=True) #sigmoid函数实现
def sigmoid(h):
return 1 / (1 + np.exp(-h)) def cost(theta, X, y, learnRate):
theta = np.matrix(theta)
X = np.matrix(X)
y = np.matrix(y) first = np.multiply(-y, np.log(sigmoid(X * theta.T)))
second = np.multiply((1 - y), np.log(1 - sigmoid(X * theta.T)))
reg = (learnRate / (2 * len(X))) * np.sum(np.power(theta[:, 1:theta.shape[1]], 2))
return np.sum(first - second) / len(X) + reg learnRate = 1
cols = data.shape[1] X = data.iloc[:, 1:cols]
y = data.iloc[:, 0:1] X = np.array(X)
y = np.array(y)
theta = np.zeros(X.shape[1]) #计算原数据集的预测情况
def predict(theta, X):
theta = np.matrix(theta)
X = np.matrix(X) probability = sigmoid(X * theta.T)
return [1 if i > 0.5 else 0 for i in probability] def gradientReg(theta, X, y, learnRate):
theta = np.matrix(theta)
X = np.matrix(X)
y = np.matrix(y) paramates = int(theta.ravel().shape[1])
grad = np.zeros(paramates) grad = (sigmoid(X * theta.T) - y).T * X / len(X) + (learnRate / len(X)) * theta[:, i]
grad[0] = grad[0] - (learnRate / len(X)) * theta[:, i]
return grad result = opt.fmin_tnc(func=cost, x0=theta, fprime=gradientReg, args=(X, y, learnRate))
print(result) theta_min = np.matrix(result[0])
predictions = predict(theta_min, X)
correct = [1 if((a == 1 and b == 1) or(a == 0 and b == 0)) else 0 for(a, b) in zip(predictions, y)]
accuracy = (sum(map(int, correct)) % len(correct)) print('accuracy = {0}%'.format(accuracy))

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