Python实现ridge和lasso

# -*- coding: utf-8 -*-

"""

Created on Mon Nov 12 17:07:16 2018

@author: wp:lasso|ridge

"""

#经典鸢尾花数据集

from sklearn.datasets import load_iris

iris = load_iris()

data_x = iris.data

data_y = iris.target

#带入需要的包、库

import pandas as pd

import numpy as np

import matplotlib.pyplot as plt

from sklearn.cross_validation import train_test_split

from sklearn.linear_model import Ridge,RidgeCV

from sklearn.linear_model import Lasso,LassoCV

from sklearn.metrics import mean_squared_error

x_tr,x_te,y_tr,y_te = train_test_split(data_x,data_y,train_size = 0.7,random_state =22)

######################ridge########################################

#通过不同的alpha值 生成不同的ridge模型

alphas = 10**np.linspace(-10,10,100)

ridge_cofficients = []

for alpha in alphas:

ridge = Ridge(alpha = alpha, normalize=True)

ridge.fit(x_tr, y_tr)

ridge_cofficients.append(ridge.coef_)

#画出alpha和回归系数的关系

plt.rcParams['font.sans-serif'] = ['Microsoft YaHei']

plt.rcParams['axes.unicode_minus'] = False

# 设置绘图风格

plt.style.use('ggplot')

plt.plot(alphas, ridge_cofficients)

plt.xscale('log')

plt.axis('tight')

plt.title(r'alpha系数与岭回归系数的关系')

plt.xlabel('Log Alpha')

plt.ylabel('Cofficients')

plt.show()

#ridge交叉验证

ridge_cv = RidgeCV(alphas = alphas, normalize=True, scoring='mean_squared_error', cv = 10)

ridge_cv.fit(x_tr, y_tr)

# 取出最佳的lambda值ridge_best_alpha = ridge_cv.alpha_

ridge_best_alpha = ridge_cv.alpha_ #得到最佳lambda值

#基于最佳lambda值建模

ridge = Ridge(alpha = ridge_best_alpha,normalize = True)

ridge.fit(x_tr,y_tr)

ridge_predict = ridge.predict(x_te)

rmse = np.sqrt(mean_squared_error(y_te,ridge_predict))

######################lasso##################################

# LASSO回归模型的交叉验证

lasso_cv = LassoCV(alphas = alphas, normalize=True, cv = 10, max_iter=10000)

lasso_cv.fit(x_tr,y_tr)

# 取出最佳的lambda值

lasso_best_alpha = lasso_cv.alpha_

lasso_best_alpha

#基于最佳lambda值建模

lasso = Lasso(alpha = lasso_best_alpha, normalize=True, max_iter=10000)

lasso.fit(x_tr, y_tr)

lasso_predict = lasso.predict(x_te) #预测

RMSE = np.sqrt(mean_squared_error(y_te,lasso_predict))