python - 将 sklearn 管道 + 嵌套交叉验证放在一起进行 KNN 回归

Question

我正在尝试弄清楚如何为 sklearn.neighbors.KNeighborsRegressor 构建一个工作流程，其中包括:

• 归一化特征
• 特征选择(20 个数字特征的最佳子集，没有具体总数)
• 在 1 到 20 的范围内交叉验证超参数 K
• 交叉验证模型
• 使用 RMSE 作为误差指标

scikit-learn 中有太多不同的选项，以至于我在决定我需要哪些类时有点不知所措。

除了sklearn.neighbors.KNeighborsRegressor，我想我还需要:

sklearn.pipeline.Pipeline  
sklearn.preprocessing.Normalizer
sklearn.model_selection.GridSearchCV
sklearn.model_selection.cross_val_score

sklearn.feature_selection.selectKBest
OR
sklearn.feature_selection.SelectFromModel

谁能告诉我这个管道/工作流程的定义是什么样的？我认为应该是这样的:

import numpy as np
from sklearn.pipeline import Pipeline
from sklearn.preprocessing import Normalizer
from sklearn.feature_selection import SelectKBest, f_classif
from sklearn.neighbors import KNeighborsRegressor
from sklearn.model_selection import cross_val_score, GridSearchCV

# build regression pipeline
pipeline = Pipeline([('normalize', Normalizer()),
                     ('kbest', SelectKBest(f_classif)),
                     ('regressor', KNeighborsRegressor())])

# try knn__n_neighbors from 1 to 20, and feature count from 1 to len(features)
parameters = {'kbest__k':  list(range(1, X.shape[1]+1)),
              'regressor__n_neighbors': list(range(1,21))}

# outer cross-validation on model, inner cross-validation on hyperparameters
scores = cross_val_score(GridSearchCV(pipeline, parameters, scoring="neg_mean_squared_error", cv=10), 
                         X, y, cv=10, scoring="neg_mean_squared_error", verbose=2)

rmses = np.abs(scores)**(1/2)
avg_rmse = np.mean(rmses)
print(avg_rmse)

它似乎没有出错，但我的一些担忧是:

• 我是否正确执行了嵌套交叉验证以使我的 RMSE 没有偏差？
• 如果我希望根据最佳 RMSE 选择最终模型，我是否应该对 cross_val_score 和 GridSearchCV 使用 scoring="neg_mean_squared_error" ？
• SelectKBest, f_classif 是用于为 KNeighborsRegressor 模型选择特征的最佳选择吗？
• 我怎样才能看到:
  • 哪个特征子集被选为最佳
  • 哪个 K 被选为最佳

非常感谢任何帮助!

Answer 1

您的代码似乎没问题。

对于 cross_val_score 和 GridSearchCV 的 scoring="neg_mean_squared_error"，我会做同样的事情来确保一切正常，但是对此进行测试的唯一方法是删除两者之一，然后查看结果是否发生变化。

SelectKBest 是一个很好的方法，但您也可以使用 SelectFromModel 或者您可以找到的其他方法 here

最后，为了获得最佳参数和特征得分，我对您的代码进行了如下修改:

import ...


pipeline = Pipeline([('normalize', Normalizer()),
                     ('kbest', SelectKBest(f_classif)),
                     ('regressor', KNeighborsRegressor())])

# try knn__n_neighbors from 1 to 20, and feature count from 1 to len(features)
parameters = {'kbest__k':  list(range(1, X.shape[1]+1)),
              'regressor__n_neighbors': list(range(1,21))}

# changes here

grid = GridSearchCV(pipeline, parameters, cv=10, scoring="neg_mean_squared_error")

grid.fit(X, y)

# get the best parameters and the best estimator
print("the best estimator is \n {} ".format(grid.best_estimator_))
print("the best parameters are \n {}".format(grid.best_params_))

# get the features scores rounded in 2 decimals
pip_steps = grid.best_estimator_.named_steps['kbest']

features_scores = ['%.2f' % elem for elem in pip_steps.scores_ ]
print("the features scores are \n {}".format(features_scores))

feature_scores_pvalues = ['%.3f' % elem for elem in pip_steps.pvalues_]
print("the feature_pvalues is \n {} ".format(feature_scores_pvalues))

# create a tuple of feature names, scores and pvalues, name it "features_selected_tuple"

featurelist = ['age', 'weight']

features_selected_tuple=[(featurelist[i], features_scores[i], 
feature_scores_pvalues[i]) for i in pip_steps.get_support(indices=True)]

# Sort the tuple by score, in reverse order

features_selected_tuple = sorted(features_selected_tuple, key=lambda 
feature: float(feature[1]) , reverse=True)

# Print
print 'Selected Features, Scores, P-Values'
print features_selected_tuple

使用我的数据的结果:

the best estimator is
Pipeline(steps=[('normalize', Normalizer(copy=True, norm='l2')), ('kbest', SelectKBest(k=2, score_func=<function f_classif at 0x0000000004ABC898>)), ('regressor', KNeighborsRegressor(algorithm='auto', leaf_size=30, metric='minkowski',
         metric_params=None, n_jobs=1, n_neighbors=18, p=2,
         weights='uniform'))])

the best parameters are
{'kbest__k': 2, 'regressor__n_neighbors': 18}

the features scores are
['8.98', '8.80']

the feature_pvalues is
['0.000', '0.000']

Selected Features, Scores, P-Values
[('correlation', '8.98', '0.000'), ('gene', '8.80', '0.000')]