From 0b9bd58554129fad5624c011a12be94dc1cebf59 Mon Sep 17 00:00:00 2001
From: Xionghaoran <719929980@qq.com>
Date: Thu, 31 Oct 2024 16:19:52 +0800
Subject: [PATCH] =?UTF-8?q?=E6=9B=B4=E6=96=B0=20function.py?=
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---
 function.py | 40 ++++++++++------------------------------
 1 file changed, 10 insertions(+), 30 deletions(-)

diff --git a/function.py b/function.py
index 6e0e038..9dc71f9 100644
--- a/function.py
+++ b/function.py
@@ -19,49 +19,29 @@ id = "202210423213"
 name = "熊浩燃"
 
 # 对率回归训练函数
-def train_logreg(X, y, learning_rate=0.01, iterations=1000):
-    """
-    对率回归训练函数。
-    Args:
-    - X (numpy array): 特征数据。
-    - y (numpy array): 标签数据。
-    - learning_rate (float): 学习率，默认为0.01。
-    - iterations (int): 迭代次数，默认为1000。
-    Returns:
-    - numpy array: 训练得到的权重。
-    """
-    # 初始化权重
+def train_logreg(X, y, learning_rate=0.001, iterations=1000):
+    # 标准化特征数据
+    X = (X - np.mean(X, axis=0)) / np.std(X, axis=0)
+    
     weights = np.zeros(X.shape[1])
-
+    
     for i in range(iterations):
-        # 预测值：通过对率回归模型得到
         z = np.dot(X, weights)
         predictions = 1 / (1 + np.exp(-z))
         
-        # 计算梯度并更新权重
+        # 防止 log(0) 引发 NaN
+        predictions = np.clip(predictions, 1e-10, 1 - 1e-10)
+        
         gradient = np.dot(X.T, (predictions - y)) / y.size
         weights -= learning_rate * gradient
-        
-        # 可选：输出损失，帮助调试
+
+        # 输出损失帮助调试
         if i % 100 == 0:
             loss = -np.mean(y * np.log(predictions) + (1 - y) * np.log(1 - predictions))
             print(f"迭代 {i}，损失: {loss}")
 
     return weights
 
-# 对率回归预测函数
-def predict_logreg(inX, weights):
-    """
-    对率回归预测函数。
-    Args:
-    - inX (numpy array): 输入待分类的样本。
-    - weights (numpy array): 训练得到的权重。
-    Returns:
-    - int: 预测的分类标签。
-    """
-    probability = 1 / (1 + np.exp(-np.dot(inX, weights)))
-    return 1 if probability >= 0.5 else 0
-
 # 数据加载函数
 def load_data(filepath):
     """