逻辑回归练习1

继续进行代码实战，对向量计算有了更进一步的理解，对scipy.optimize还不是很明白。

#导入包
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
path = 'ex2data1.txt'
data = pd.read_csv(path, header=None, names=['Exam 1', 'Exam 2', 'Admitted'])
#print(data.head())
#导入数据并绘图
'''positive = data[data['Admitted'].isin([1])]
negative = data[data['Admitted'].isin([0])]

fig, ax = plt.subplots(figsize=(12,8))
ax.scatter(positive['Exam 1'], positive['Exam 2'], s=50, c='b', marker='o', label='Admitted')
ax.scatter(negative['Exam 1'], negative['Exam 2'], s=50, c='r', marker='x', label='Not Admitted')
ax.legend()
ax.set_xlabel('Exam 1 Score')
ax.set_ylabel('Exam 2 Score')
plt.show()
'''
def sigmoid(z):#定义假设函数
    return 1 / (1 + np.exp(-z))

def cost(theta, X, y):#定义代价函数J
    theta = np.matrix(theta)
    X = np.matrix(X)
    y = np.matrix(y)
    first = np.multiply(-y, np.log(sigmoid(X * theta.T)))#np.log(X)是分别对X中每一个元素求log
    second = np.multiply((1 - y), np.log(1 - sigmoid(X * theta.T)))
    return np.sum(first - second) / (len(X))

#将数据输入并转换为矩阵
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)#数据转换
y = np.array(y.values)
theta = np.zeros(3)

#向量化计算梯度
def gradient(theta, X, y):
    theta = np.matrix(theta)
    X = np.matrix(X)
    y = np.matrix(y)

    parameters = int(theta.ravel().shape[1])
    grad = np.zeros(parameters)

    error = sigmoid(X * theta.T) - y

    for i in range(parameters):
        term = np.multiply(error, X[:, i])
        grad[i] = np.sum(term) / len(X)

    return grad
#自动寻找步长的高级优化，scipy.optimize是优化和拟合
import scipy.optimize as opt
result = opt.fmin_tnc(func=cost, x0=theta, fprime=gradient, args=(X, y))
result

#print(cost(result[0], X, y))
#预测函数：当h大于等于0.5时，预测 y=1，否则 y=0
def predict(theta, X):
    probability = sigmoid(X * theta.T)
    return [1 if x >= 0.5 else 0 for x in probability]
#验证精准度
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))

)