python-LASSO回归模型

article/2025/9/28 22:02:07

https://zhuanlan.zhihu.com/p/76055830

这里写图片描述

LASSO回归模型的应用

①可视化方法确定 $\lambda$ 值

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib
from sklearn import model_selection
from sklearn.linear_model import Lasso, LassoCV
font = {'family': 'FangSong','weight': 'bold','size': 12
}
matplotlib.rc("font", **font)
# 读取糖尿病数据集
diabetes = pd.read_excel('./data/diabetes.xlsx')
# 构造自变量(剔除患者性别、年龄和因变量)
predictors = diabetes.iloc[:, 2: -1]
# 将数据集拆分成训练集和测试集
x_train, x_test, y_train, y_test = model_selection.train_test_split(predictors, diabetes['Y'], test_size=0.2, random_state=1234)
'''
可视化方法确定λ的值
'''
# 构造不同的Lambda值
Lambdas = np.logspace(-5, 2, 200)
# 构造空列表，用于存储模型的偏回归系数
lasso_cofficients = []
for Lambda in Lambdas:lasso = Lasso(alpha=Lambda, normalize=True, max_iter=10000)lasso.fit(x_train, y_train)lasso_cofficients.append(lasso.coef_)
'''
可视化方法确定λ的值
'''
# 绘制Lambda与回归线的折线图
plt.plot(Lambdas, lasso_cofficients)
# 对x轴做对数变换
plt.xscale('log')
# 设置折线图x轴和y轴标签
plt.xlabel('Lambda')
plt.ylabel('Cofficients')
# 显示图形
plt.show()

②交叉验证法确定λ的值

# LASSO回归模型的交叉验证
lasso_cv = LassoCV(alphas=Lambdas, normalize=True, cv=10, max_iter=10000)
lasso_cv.fit(x_train, y_train)
# 输出最佳的lambda值
lasso_best_alpha = lasso_cv.alpha_  # 0.06294988990221888
print(lasso_best_alpha)

③模型的预测


# 基于最佳的lambda值建模
lasso = Lasso(alpha=lasso_best_alpha, normalize=True, max_iter=10000)
# 对"类"加以数据实体，执行回归系数的运算
lasso.fit(x_train, y_train)
# 返回LASSO回归的系数
res = pd.Series(index=['Intercept'] + x_train.columns.tolist(), data=[lasso.intercept_] + lasso.coef_.tolist())
'''
Intercept   -278.560358
BMI            6.188602
BP             0.860826
S1            -0.127627
S2            -0.000000
S3            -0.488408
S4             0.000000
S5            44.487738
S6             0.324076
系数中含有两个0，分别是S2和S4，说明这两个变量对糖尿病指数Y没有显著意义
'''
print(res)
# 模型预测
lasso_predict = lasso.predict(x_test)
# 验证预测效果
from sklearn.metrics import mean_squared_error
RMSE = np.sqrt(mean_squared_error(y_test, lasso_predict))  # 53.061437258225745
print(RMSE)

用线性回归做比较

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib
from sklearn import model_selection
from statsmodels import api as sms
from sklearn.metrics import mean_squared_error
font = {'family': 'FangSong','weight': 'bold','size': 12
}
matplotlib.rc("font", **font)
# 读取糖尿病数据集
diabetes = pd.read_excel('./data/diabetes.xlsx')
# 构造自变量(剔除患者性别、年龄和因变量)
predictors = diabetes.iloc[:, 2: -1]
# 将数据集拆分成训练集和测试集
x_train, x_test, y_train, y_test = model_selection.train_test_split(predictors, diabetes['Y'], test_size=0.2, random_state=1234)
# 为自变量x添加常数列1，用于拟合截距项
x_train2 = sms.add_constant(x_train)
x_test2 = sms.add_constant(x_test)
# 构建多元线性回归模型
liner = sms.formula.OLS(y_train, x_train2).fit()
# 返回线性回归模型的系数
params = liner.params
'''
const   -406.699716
BMI        6.217649
BP         0.948245
S1        -1.264772
S2         0.901368
S3         0.962373
S4         6.694215
S5        71.614661
S6         0.376004
dtype: float64
'''
print(params)
# 模型预测
linear_predict = liner.predict(x_test2)
# 预测效果验证
RMSE = np.sqrt(mean_squared_error(y_test, linear_predict))  # 53.42623939722992
print(RMSE)