Question

我使用GridSearchCV进行岭回归，但是使用matplotlib来显示模型性能与正则化器（alpha）无法

有人可以帮忙吗？

我的代码：

from sklearn.datasets import fetch_california_housing
cal=fetch_california_housing()
X = cal.data
y = cal.target 

from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=0)

param_grid = {'alpha': np.logspace(-3, 3, 13)}
print(param_grid)
grid = GridSearchCV(Ridge(normalize=True), param_grid, cv=10)
grid.fit(X_train, y_train)
print("Best cross-validation score: {:.2f}".format(grid.best_score_))
print("Best parameters: ", grid.best_params_)

import matplotlib.pyplot as plt
alphas = np.logspace(-3, 3, 13)
plt.semilogx(alphas, grid.fit(X_train, y_train), label='Train')
plt.semilogx(alphas, grid.fit(X_test, y_test), label='Test')

plt.legend(loc='lower left')
plt.ylim([0, 1.0])
plt.xlabel('alpha')
plt.ylabel('performance')

# the error code I got was "ValueError: x and y must have same first dimension"

基本上，我希望看到以下内容：

Answer 1

你应该绘制得分，而不是grid.fit()的结果。

首先使用return_train_score=True：

grid = GridSearchCV(Ridge(normalize=True), param_grid, cv=10, return_train_score=True)

然后在拟合模型图之后如下：

plt.semilogx(alphas, grid.cv_results_['mean_train_score'], label='Train')
plt.semilogx(alphas, grid.cv_results_['mean_test_score'], label='Test')
plt.legend()

结果：

Answer 2

在绘制使用GridSearch产生的模型选择性能时，通常可以绘制交叉验证折叠的测试和训练集的平均值和标准差。

还应注意确定在网格搜索中使用哪个评分标准来选择最佳模型。这通常是回归的R平方。

网格搜索返回一个字典（可通过.cv_results_访问），其中包含每个折叠列车/测试分数的分数以及训练/测试每个折叠所花费的时间。还使用均值和标准偏差包括该数据的摘要。 PS。在较新版本的熊猫中，您需要包含return_train_score=True PS.S.使用网格搜索时，模型选择不需要将数据拆分为训练/测试，因为网格搜索会自动拆分数据（cv = 10表示数据被拆分为10倍）

鉴于上述情况，我将代码修改为

import numpy as np
import matplotlib.pyplot as plt

from sklearn.linear_model import Ridge
from sklearn.model_selection import GridSearchCV
from sklearn.datasets import fetch_california_housing
cal = fetch_california_housing()
X = cal.data
y = cal.target


param_grid = {'alpha': np.logspace(-3, 3, 13)}
print(param_grid)
grid = GridSearchCV(Ridge(normalize=True), param_grid,
                    cv=10, return_train_score=True, scoring='r2')
grid.fit(X, y)
print("Best cross-validation score: {:.2f}".format(grid.best_score_))
print("Best parameters: ", grid.best_params_)


alphas = np.logspace(-3, 3, 13)

train_scores_mean = grid.cv_results_["mean_train_score"]
train_scores_std = grid.cv_results_["std_train_score"]
test_scores_mean = grid.cv_results_["mean_test_score"]
test_scores_std = grid.cv_results_["std_test_score"]

plt.figure()
plt.title('Model')
plt.xlabel('$\\alpha$ (alpha)')
plt.ylabel('Score')
# plot train scores
plt.semilogx(alphas, train_scores_mean, label='Mean Train score',
             color='navy')
# create a shaded area between [mean - std, mean + std]
plt.gca().fill_between(alphas,
                       train_scores_mean - train_scores_std,
                       train_scores_mean + train_scores_std,
                       alpha=0.2,
                       color='navy')
plt.semilogx(alphas, test_scores_mean,
             label='Mean Test score', color='darkorange')

# create a shaded area between [mean - std, mean + std]
plt.gca().fill_between(alphas,
                       test_scores_mean - test_scores_std,
                       test_scores_mean + test_scores_std,
                       alpha=0.2,
                       color='darkorange')

plt.legend(loc='best')
plt.show()

结果图如下所示

如何可视化模型性能的依赖性＆amp; alpha与matplotlib？

2 个答案: