From ebffd12d0ff006c2eba1bfefa9c42c7221c3ac34 Mon Sep 17 00:00:00 2001
From: Your Name <you@example.com>
Date: Fri, 5 Jul 2024 21:43:00 +0800
Subject: [PATCH] liyuan

---
 ly.py | 38 ++++++++++++++++++++++++++++++++++++++
 1 file changed, 38 insertions(+)
 create mode 100644 ly.py

diff --git a/ly.py b/ly.py
new file mode 100644
index 0000000..b62417c
--- /dev/null
+++ b/ly.py
@@ -0,0 +1,38 @@
+import numpy as np
+
+# 假设数据集：humidity为湿度，temperature为温度
+# 这些数据应该是从实际观测中获取的
+data = np.array([
+    [20, 25],  # 湿度20%，温度25°C
+    [30, 23],  # 湿度30%，温度23°C
+    [40, 22],  # 湿度40%，温度22°C
+    [50, 20],  # 湿度50%，温度20°C
+    # 更多数据...
+])
+
+# 分割数据集：前四分之三用于训练，后四分之一用于测试
+split_idx = int(len(data) * 0.75)
+train_data, test_data = data[:split_idx], data[split_idx:]
+
+# 训练过程：计算斜率和截距
+X_train, y_train = train_data[:, 0], train_data[:, 1]  # 分别为湿度和温度
+mean_x = np.mean(X_train)
+mean_y = np.mean(y_train)
+numerator = np.sum((X_train - mean_x) * (y_train - mean_y))
+denominator = np.sum((X_train - mean_x) ** 2)
+slope = numerator / denominator
+intercept = mean_y - slope * mean_x
+
+# 预测函数
+def predict(humidity):
+    return slope * humidity + intercept
+
+# 测试模型：对测试数据集进行预测，并计算预测误差的均方差作为评估
+X_test, y_test = test_data[:, 0], test_data[:, 1]
+predictions = [predict(hum) for hum in X_test]
+mse = np.mean((predictions - y_test) ** 2)
+print(f"模型的均方误差(MSE)为: {mse}")
+
+# 示例：预测湿度为60%时的温度
+predicted_temp = predict(60)
+print(f"预测湿度为60%时的温度为: {predicted_temp}°C")
\ No newline at end of file
-- 
Gitee