python实现逻辑回归的示例

代码

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

def initialize_params(dims):
 w = np.zeros((dims, 1))
 b = 0
 return w, b

def sigmoid(x):
 z = 1 / (1 + np.exp(-x))
 return z

def logistic(X, y, w, b):
 num_train = X.shape[0]
 y_hat = sigmoid(np.dot(X, w) + b)
 loss = -1 / num_train * np.sum(y * np.log(y_hat) + (1-y) * np.log(1-y_hat))
 cost = -1 / num_train * np.sum(y * np.log(y_hat) + (1 - y) * np.log(1 - y_hat))
 dw = np.dot(X.T, (y_hat - y)) / num_train
 db = np.sum(y_hat - y) / num_train
 return y_hat, cost, dw, db

def linear_train(X, y, learning_rate, epochs):
 # 参数初始化
 w, b = initialize_params(X.shape[1])

loss_list = []
 for i in range(epochs):
   # 计算当前的预测值、损失和梯度
   y_hat, loss, dw, db = logistic(X, y, w, b)
   loss_list.append(loss)

# 基于梯度下降的参数更新
   w += -learning_rate * dw
   b += -learning_rate * db

# 打印迭代次数和损失
   if i ％ 10000 == 0:
     print("epoch ％d loss ％f" ％ (i, loss))

# 保存参数
   params = {
     'w': w,
     'b': b
   }

# 保存梯度
   grads = {
     'dw': dw,
     'db': db
   }

return loss_list, loss, params, grads

def predict(X, params):
 w = params['w']
 b = params['b']
 y_pred = sigmoid(np.dot(X, w) + b)
 return y_pred

if __name__ == "__main__":
 # 生成数据
 X, labels = make_classification(n_samples=100,
                 n_features=2,
                 n_informative=2,
                 n_redundant=0,
                 random_state=1,
                 n_clusters_per_class=2)
 print(X.shape)
 print(labels.shape)

# 生成伪随机数
 rng = np.random.RandomState(2)
 X += 2 * rng.uniform(size=X.shape)

# 划分训练集和测试集
 offset = int(X.shape[0] * 0.9)
 X_train, y_train = X[:offset], labels[:offset]
 X_test, y_test = X[offset:], labels[offset:]
 y_train = y_train.reshape((-1, 1))
 y_test = y_test.reshape((-1, 1))
 print('X_train=', X_train.shape)
 print('y_train=', y_train.shape)
 print('X_test=', X_test.shape)
 print('y_test=', y_test.shape)

# 训练
 loss_list, loss, params, grads = linear_train(X_train, y_train, 0.01, 100000)
 print(params)

# 预测
 y_pred = predict(X_test, params)
 print(y_pred[:10])

来源：https://www.cnblogs.com/chenxiangzhen/p/10395231.html