特征选择和预测

Question

from sklearn.feature_selection import RFECV
from sklearn.metrics import accuracy_score
from sklearn.model_selection import cross_val_predict, KFold
from sklearn.preprocessing import StandardScaler
from sklearn.ensemble import RandomForestClassifier
from sklearn.pipeline import Pipeline
from sklearn.datasets import load_iris

我有 X 和 Y 数据。

data = load_iris()    
X = data.data
Y = data.target 

我想使用 k-fold 验证方法来实现 RFECV 特征选择和预测。

从答案@https://stackoverflow.com/users/3374996/vivek-kumar 纠正的代码

clf = RandomForestClassifier()

kf = KFold(n_splits=2, shuffle=True, random_state=0)  

estimators = [('standardize' , StandardScaler()),
              ('clf', clf)]

class Mypipeline(Pipeline):
    @property
    def coef_(self):
        return self._final_estimator.coef_
    @property
    def feature_importances_(self):
        return self._final_estimator.feature_importances_ 

pipeline = Mypipeline(estimators)

rfecv = RFECV(estimator=pipeline, cv=kf, scoring='accuracy', verbose=10)
rfecv_data = rfecv.fit(X, Y)

print ('no. of selected features =', rfecv_data.n_features_) 

编辑（少量剩余）：

X_new = rfecv.transform(X)
print X_new.shape

y_predicts = cross_val_predict(clf, X_new, Y, cv=kf)
accuracy = accuracy_score(Y, y_predicts)
print ('accuracy =', accuracy)

Answer 1

不要将 StandardScaler 和 RFECV 包装在同一管道中，而是对 StandardScaler 和 RandomForestClassifier 执行此操作，并将该管道作为估计器传递给 RFECV。在此不会泄露任何 traininf 信息。

estimators = [('standardize' , StandardScaler()),
              ('clf', RandomForestClassifier())]

pipeline = Pipeline(estimators)


rfecv = RFECV(estimator=pipeline, scoring='accuracy')
rfecv_data = rfecv.fit(X, Y)

更新：关于错误'RuntimeError: The classifier does not expose "coef_" or "feature_importances_" attributes'

是的，这是 scikit-learn 管道中的一个已知问题。您可以查看我的其他 answer here for 更多详细信息并使用我在那里创建的新管道。

像这样定义一个自定义管道：

class Mypipeline(Pipeline):
    @property
    def coef_(self):
        return self._final_estimator.coef_
    @property
    def feature_importances_(self):
        return self._final_estimator.feature_importances_ 

然后使用它：

pipeline = Mypipeline(estimators)

rfecv = RFECV(estimator=pipeline, scoring='accuracy')
rfecv_data = rfecv.fit(X, Y)

更新 2：

@brute，对于您的数据和代码，算法会在一分钟内在我的 PC 上完成。这是我使用的完整代码：

import numpy as np
import glob
from sklearn.utils import resample
files = glob.glob('/home/Downloads/Untitled Folder/*') 
outs = [] 
for fi in files: 
    data = np.genfromtxt(fi, delimiter='|', dtype=float) 
    data = data[~np.isnan(data).any(axis=1)] 
    data = resample(data, replace=False, n_samples=1800, random_state=0) 
    outs.append(data) 

X = np.vstack(outs) 
print X.shape 
Y = np.repeat([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 1800) 
print Y.shape

#from sklearn.utils import shuffle
#X, Y = shuffle(X, Y, random_state=0)

from sklearn.feature_selection import RFECV
from sklearn.model_selection import KFold
from sklearn.preprocessing import StandardScaler
from sklearn.ensemble import RandomForestClassifier
from sklearn.pipeline import Pipeline

clf = RandomForestClassifier()

kf = KFold(n_splits=10, shuffle=True, random_state=0)  

estimators = [('standardize' , StandardScaler()),
              ('clf', RandomForestClassifier())]

class Mypipeline(Pipeline):
    @property
    def coef_(self):
        return self._final_estimator.coef_
    @property
    def feature_importances_(self):
        return self._final_estimator.feature_importances_ 

pipeline = Mypipeline(estimators)

rfecv = RFECV(estimator=pipeline, scoring='accuracy', verbose=10)
rfecv_data = rfecv.fit(X, Y)

print ('no. of selected features =', rfecv_data.n_features_) 

更新 3：对于 cross_val_predict

X_new = rfecv.transform(X)
print X_new.shape

# Here change clf to pipeline, 
# because RFECV has found features according to scaled data,
# which is not present when you pass clf 
y_predicts = cross_val_predict(pipeline, X_new, Y, cv=kf)
accuracy = accuracy_score(Y, y_predicts)
print ('accuracy =', accuracy)

Answer 2

我们将这样做：

适合训练集

from sklearn.feature_selection import RFECV
from sklearn.metrics import accuracy_score
from sklearn.model_selection import cross_val_predict, KFold
from sklearn.preprocessing import StandardScaler
from sklearn.ensemble import RandomForestClassifier
from sklearn.pipeline import Pipeline
from sklearn.datasets import load_iris
from sklearn.model_selection import train_test_split

data = load_iris()    
X = data.data, Y = data.target

# split into train and test sets
X_train, X_test, y_train, y_test = train_test_split(X, Y, shuffle=True)

# create model
clf = RandomForestClassifier()    
# instantiate K-Fold
kf = KFold(n_splits=10, shuffle=True, random_state=0)

# pipeline estimators
estimators = [('standardize' , StandardScaler()),
             ('rfecv', RFECV(estimator=clf, cv=kf, scoring='accuracy'))]

# instantiate pipeline
pipeline = Pipeline(estimators)    
# fit rfecv to train model
rfecv_model = rfecv_model = pipeline.fit(X_train, y_train)

# print number of selected features
print ('no. of selected features =', pipeline.named_steps['rfecv'].n_features_)
# print feature ranking
print ('ranking =', pipeline.named_steps['rfecv'].ranking_)

'Output':
no. of selected features = 3
ranking = [1 2 1 1]

在测试集上预测

# make predictions on the test set
predictions = rfecv_model.predict(X_test)

# evaluate the model performance using accuracy metric
print("Accuracy on test set: ", accuracy_score(y_test, predictions))

'Output':
Accuracy:  0.9736842105263158