【发布时间】:2017-12-13 10:26:22
【问题描述】:
我在 param_grid 中尝试了来自 SelectKBest 的 k 和 PCA 的 n_components 的参数组合。我可以使用下面的代码打印 k 值 和 n_components。我将发布整个代码,以便您了解功能是从哪个列表中获取的
#THE FIRST FEATURE HAS TO BE THE LABEL
featurelist = ['poi', 'exercised_stock_options', 'expenses', 'from_messages',
'from_poi_to_this_person', 'from_this_person_to_poi', 'other',
'restricted_stock', 'salary', 'shared_receipt_with_poi',
'to_messages', 'total_payments', 'total_stock_value',
'ratio_from_poi', 'ratio_to_poi']
enronml = pd.DataFrame(enron[['poi', 'exercised_stock_options', 'expenses', 'from_messages',
'from_poi_to_this_person', 'from_this_person_to_poi', 'other',
'restricted_stock', 'salary', 'shared_receipt_with_poi',
'to_messages', 'total_payments', 'total_stock_value',
'ratio_from_poi', 'ratio_to_poi']].copy())
enronml = enronml.to_dict(orient="index")
dataset = enronml
#featureFormat, takes the dictionary as the dataset, converts the first
feature in featurelist into label
data = featureFormat(dataset, featurelist, sort_keys = True)
labels, features = targetFeatureSplit(data)
from sklearn.cross_validation import train_test_split
from sklearn.naive_bayes import GaussianNB
from sklearn.cross_validation import train_test_split
X_train, X_test, y_train, y_test = train_test_split(features, labels,
test_size=0.20, random_state=0)
pca = PCA()
gnba = GaussianNB()
steps = [('scaler', MinMaxScaler()),
('best', SelectKBest()),
('pca', pca),
('gnba', gnba)]
pipeline = Pipeline(steps)
parameters = [
{
'best__k':[3],
'pca__n_components': [1,2]
},
{
'best__k':[4],
'pca__n_components': [1,2,3]
},
{
'best__k':[5],
'pca__n_components': [1,2,3,4]
},
]
cv = StratifiedShuffleSplit(test_size=0.2, random_state=42)
gnbawithpca = GridSearchCV(pipeline, param_grid = parameters, cv=cv,
scoring="f1")
gnbawithpca.fit(X_train,y_train)
means = gnbawithpca.cv_results_['mean_test_score']
stds = gnbawithpca.cv_results_['std_test_score']
for mean, std, params in zip(means, stds,
gnbawithpca.cv_results_['params']):
print("%0.3f (+/-%0.03f) for %r"
% (mean, std * 2, params))
我可以得到这样的结果
0.480 (+/-0.510) for {'best__k': 3, 'pca__n_components': 1}
0.534 (+/-0.409) for {'best__k': 3, 'pca__n_components': 2}
0.480 (+/-0.510) for {'best__k': 4, 'pca__n_components': 1}
0.534 (+/-0.409) for {'best__k': 4, 'pca__n_components': 2}
0.565 (+/-0.342) for {'best__k': 4, 'pca__n_components': 3}
0.480 (+/-0.510) for {'best__k': 5, 'pca__n_components': 1}
0.513 (+/-0.404) for {'best__k': 5, 'pca__n_components': 2}
0.473 (+/-0.382) for {'best__k': 5, 'pca__n_components': 3}
0.448 (+/-0.353) for {'best__k': 5, 'pca__n_components': 4}
我想知道选择了哪些特征,例如当best_k = 5时,我想知道这5个特征的名称。
【问题讨论】:
-
你能澄清你的问题吗?您只是说明了您在做什么,而不是问题所在。
-
我很抱歉。我忘了最后一行。
标签: python machine-learning scikit-learn grid-search