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我试图弄清楚如何为此构建一个工作流程sklearn.neighbors.KNeighborsRegressor，包括：

  • 规范化特征
  • 特征选择（20 个数字特征的最佳子集，没有具体的总数）
  • 在 1 到 20 范围内交叉验证超参数 K
  • 交叉验证模型
  • 使用 RMSE 作为误差度量

scikit-learn 中有很多不同的选项，以至于我在决定我需要哪些课程时有点不知所措。

此外sklearn.neighbors.KNeighborsRegressor，我想我需要：

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

有人可以告诉我定义这个管道/工作流的样子吗？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 没有偏见？
  • 如果我想根据最佳 RMSE 选择最终模型，我应该同时使用scoring="neg_mean_squared_error"和cross_val_score吗GridSearchCV？
  • SelectKBest, f_classif用于选择模型特征的最佳选项是什么KNeighborsRegressor？
  • 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

任何帮助是极大的赞赏！

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1 回答 1

6

您的代码似乎还可以。

对于scoring="neg_mean_squared_error"两者cross_val_score和GridSearchCV，我会做同样的事情以确保一切正常，但测试这一点的唯一方法是删除两者之一，看看结果是否改变。

SelectKBest是一种很好的方法，但您也可以使用SelectFromModel甚至可以在此处找到的其他方法

最后，为了获得最佳参数和特征分数，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')]
于 2017-07-17T19:09:32.373 回答