题目太长啦！文档下载【传送门】
第1题
简述：实现逻辑回归。
此处使用了minimize函数代替Matlab的fminunc函数，参考了该博客【传送门】。
import numpy as np
import matplotlib.pyplot as plt
import scipy.optimize as op #S函数
def sigmoid(z):
g = 1/(1+np.exp(-z))
return g #cost计算函数
def costFunction(theta, X, y):
theta = np.array(theta).reshape((np.size(theta),1))
m = np.size(y)
h = sigmoid(np.dot(X, theta))
J = 1/m*(-np.dot(y.T, np.log(h)) - np.dot((1-y.T), np.log(1-h)))
return J.flatten() def gradient(theta, X, y):
theta = np.array(theta).reshape((np.size(theta), 1))
m = np.size(y)
h = sigmoid(np.dot(X, theta))
grad = 1/m*np.dot(X.T, h - y)
return grad.flatten() #读取数据，第一列是成绩1，第二列是成绩2，第三列是yes/no
data = np.loadtxt('ex2data1.txt', delimiter=',', dtype='float')
m = np.size(data[:, 0])
# print(data) #绘制样本点
X = data[:, 0:2]
y = data[:, 2:3]
pos = np.where(y == 1)[0]
neg = np.where(y == 0)[0]
X1 = X[pos, 0:2]
X0 = X[neg, 0:2]
plt.plot(X1[:, 0], X1[:, 1], 'k+')
plt.plot(X0[:, 0], X0[:, 1], 'yo')
plt.xlabel('Exam 1 score')
plt.ylabel('Exam 2 score') #求解最优解
one = np.ones(m)
X = np.insert(X, 0, values=one, axis=1)
initial_theta = np.zeros(np.size(X, 1))
result = op.minimize(fun=costFunction, x0=initial_theta, args=(X, y), method='TNC', jac=gradient)
# print(result)
theta = result.x
cost = result.fun
print('theta:', theta)
print('cost:', cost) #绘制决策边界
plot_x = np.array([np.min(X[:, 1]), np.max(X[:, 2])])
# print(plot_x)
plot_y = (-1/theta[2])*(theta[1]*plot_x+theta[0])
# print(plot_y)
plt.plot(plot_x,plot_y)
plt.legend(labels=['Admitted', 'Not admitted'])
plt.axis([30, 100, 30, 100])
plt.show() #预测[45 85]成绩的学生，并计算准确率
theta = np.array(theta).reshape((np.size(theta),1))
z = np.dot([1, 45, 85], theta)
prob = sigmoid(z)
print('For a student with scores 45 and 85, we predict an admission probability of ', prob)
p = np.round(sigmoid(np.dot(X,theta)))
acc = np.mean(p==y)*100
print('Train Accuracy: ',acc,'%')
运行结果：
第2题
简述：通过正规化实现逻辑回归。
import numpy as np
import matplotlib.pyplot as plt
import scipy.optimize as op #S函数
def sigmoid(z):
g = 1/(1+np.exp(-z))
return g #cost计算函数
def costFunction(theta, X, y, lamb):
theta = np.array(theta).reshape((np.size(theta), 1))
m = np.size(y)
h = sigmoid(np.dot(X, theta))
J = 1/m*(-np.dot(y.T, np.log(h)) - np.dot((1-y.T), np.log(1-h)))
# 添加项
theta2 = theta[1:, 0]
Jadd = lamb/(2*m)*np.sum(theta2**2)
J = J + Jadd
return J.flatten() #求梯度
def gradient(theta, X, y, lamb):
theta = np.array(theta).reshape((np.size(theta), 1))
m = np.size(y)
h = sigmoid(np.dot(X, theta))
grad = 1/m*np.dot(X.T, h - y)
#添加项
theta[0,0] = 0
gradadd = lamb/m*theta
grad = grad + gradadd
return grad.flatten() #求特征矩阵
def mapFeature(X1, X2):
degree = 6
out = np.ones((np.size(X1),1))
for i in range(1, degree+1):
for j in range(0, i+1):
res = np.multiply(np.power(X1, i-j), np.power(X2, j))
out = np.insert(out, np.size(out[0, :]), values=res, axis=1)
return out #读取数据，第一列是成绩1，第二列是成绩2，第三列是yes/no
data = np.loadtxt('ex2data2.txt', delimiter=',', dtype='float')
m = np.size(data[:, 0]) #绘制样本点
X = data[:, 0:2]
y = data[:, 2:3]
pos = np.where(y == 1)[0]
neg = np.where(y == 0)[0]
X1 = X[pos, 0:2]
X0 = X[neg, 0:2]
plt.plot(X1[:, 0], X1[:, 1], 'k+')
plt.plot(X0[:, 0], X0[:, 1], 'yo')
plt.xlabel('Microchip Test 1')
plt.ylabel('Microchip Test 2')
plt.legend(labels=['y = 1', 'y = 0']) #数据初始化
X = mapFeature(X[:, 0], X[:, 1])
# print(X)
lamb = 1
initial_theta = np.zeros(np.size(X, 1)) #求解最优解
result = op.minimize(fun=costFunction, x0=initial_theta, args=(X, y, lamb), method='TNC', jac=gradient)
# print(result)
cost = result.fun
theta = result.x
print('theta:', theta)
print('cost:', cost) #绘制决策边界
u = np.linspace(-1, 1.5, 50)
v = np.linspace(-1, 1.5, 50)
z = np.zeros((np.size(u),np.size(v)))
theta = np.array(theta).reshape((np.size(theta), 1))
for i in range(0, np.size(u)):
for j in range(0, np.size(v)):
z[i, j] = np.dot(mapFeature(u[i], v[j]), theta)
# print(z)
plt.contour(u, v, z.T, [0])
plt.show() #计算准确率
p = np.round(sigmoid(np.dot(X,theta)))
acc = np.mean(p==y)*100
print('Train Accuracy: ',acc,'%')
运行结果：