import numpy as np
from .metrics import accuracy_scoreclass LogisticRegression:def __init__(self):# 截距theta0self.intercept_ = None# 系数，theta1 ... thetaNself.coef_ = None# theta列向量self._theta = None# 定义Sigmoid私有函数def _sigmoid(self, t):return 1. / (1. + np.exp(-t))# 使用批量梯度下降法，根据训练数据集X_train，y_train训练LogisticRegression模型def fit(self, X_train, y_train, is_debug=False, eta=0.01, n_iters=1e4):assert X_train.shape[0] == y_train.shape[0], \"特征数据矩阵的行数要等于样本结果数据的行数"# 定义逻辑回归损失函数def L(theta, X_b, y):# 定义逻辑回归概率公式y_hat = self._sigmoid(X_train.dot(theta))try:return -np.sum(y*np.log(y_hat)+(1-y)*np.log(1-y_hat)) / len(X_b)except:return float('inf')# 定义逻辑回归梯度def dL(theta, X_b, y):return X_b.T.dot(self._sigmoid(X_b.dot(theta)) - y) / len(X_b)def dL_debug(theta, X_b, y, epsilon=0.01):# 开辟大小与theta向量一致的向量空间result = np.empty(len(theta))# 便利theta向量中的每一个thetafor i in range(len(theta)):# 复制一份theta向量theta_1 = theta.copy()# 将第i个theta加上一个距离，既求该theta正方向的thetatheta_1[i] += epsilon# 在复制一份theta向量theta_2 = theta.copy()# 将第i个theta减去同样的距离，既求该theta负方向的thetatheta_2[i] -= epsilon# 求出这两个点连线的斜率，既模拟该theta的导数result[i] = (L(theta_1, X_b, y) - L(theta_2, X_b, y)) / (2 * epsilon)return result# 实现批量梯度下降法def gradient_descent(X_b, y, initial_theta, eta, difference=1e-8):theta = initial_thetai_iter = 0while i_iter < n_iters:# 当is_debug为True时走debug的求梯度的方法，反之走梯度公式的方法if is_debug:gradient = dL_debug(theta, X_b, y)else:gradient = dL(theta, X_b, y)last_theta = thetatheta = theta - eta * gradientif (abs(L(theta, X_b, y) - L(last_theta, X_b, y)) < difference):breaki_iter += 1return theta# 构建X_bX_b = np.hstack([np.ones((len(X_train), 1)), X_train])# 初始化theta向量为元素全为0的向量initial_theta = np.zeros(X_b.shape[1])self._theta = gradient_descent(X_b, y_train, initial_theta, eta)self.intercept_ = self._theta[0]self.coef_ = self._theta[1:]return self# 计算概率，给定待预测数据集X_predict，返回表示X_predict的结果概率向量def predict_probability(self, X_predict):assert self.intercept_ is not None and self.coef_ is not None, \"截距和系数都不为空，表示已经经过了fit方法"assert X_predict.shape[1] == len(self.coef_), \"要预测的特征数据集列数要与theta的系数数量相等"X_b = np.hstack([np.ones((len(X_predict), 1)), X_predict])# 返回0，1之间的浮点数return self._sigmoid(X_b.dot(self._theta))# 给定待预测数据集X_predict，返回表示X_predict的结果向量def predict(self, X_predict):assert self.intercept_ is not None and self.coef_ is not None, \"截距和系数都不为空，表示已经经过了fit方法"assert X_predict.shape[1] == len(self.coef_), \"要预测的特征数据集列数要与theta的系数数量相等"probability = self.predict_probability(X_predict)# 将概率转换为0和1的向量，True对应1，False对应0return np.array(probability >= 0.5, dtype='int')# 根据测试数据集X_test和y_test确定当前模型的准确度def score(self, X_test, y_test):y_predict = self.predict(X_test)return accuracy_score(y_test, y_predict)def __repr__(self):return "LinearRegression()"

import numpy as np
import matplotlib.pyplot as plt
from sklearn import datasetsiris = datasets.load_iris()
X = iris.data
y = iris.target

# 只取前两类的鸢尾花数据
X = X[y<2, :2]
y = y[y<2]plt.scatter(X[y==0, 0], X[y==0, 1], color='red')
plt.scatter(X[y==1, 0], X[y==1, 1], color='blue')
plt.show()

from myML.LogisticRegression import LogisticRegression
from myML.modelSelection import train_test_splitX_train, y_train, X_test, y_test = train_test_split(X, y, seed=666)log_reg = LogisticRegression()
log_reg.fit(X_train, y_train)
log_reg.score(X_test, y_test)
# 结果
1.0log_reg.predict_probability(X_test)
# 概率结果
array([ 0.92972035,  0.98664939,  0.14852024,  0.17601199,  0.0369836 ,0.0186637 ,  0.04936918,  0.99669244,  0.97993941,  0.74524655,0.04473194,  0.00339285,  0.26131273,  0.0369836 ,  0.84192923,0.79892262,  0.82890209,  0.32358166,  0.06535323,  0.20735334])log_reg.predict(X_test)
# 分类结果
array([1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0])y_test
# 测试数据结果
array([1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0])