我对结果感到困惑，可能我没有得到交叉验证和GridSearch的概念。我遵循了这篇文章背后的逻辑:https://randomforests.wordpress.com/2014/02/02/basics-of-k-fold-cross-validation-and-gridsearchcv-in-scikit-learn/
argd = CommandLineParser(argv)
folder,fname=argd['dir'],argd['fname']
df = pd.read_csv('../../'+folder+'/Results/'+fname, sep=";")
explanatory_variable_columns = set(df.columns.values)
response_variable_column = df['A']
explanatory_variable_columns.remove('A')
y = np.array([1 if e else 0 for e in response_variable_column])
X =df[list(explanatory_variable_columns)].as_matrix()
kf_total = KFold(len(X), n_folds=5, indices=True, shuffle=True, random_state=4)
dt=DecisionTreeClassifier(criterion='entropy')
min_samples_split_range=[x for x in range(1,20)]
dtgs=GridSearchCV(estimator=dt, param_grid=dict(min_samples_split=min_samples_split_range), n_jobs=1)
scores=[dtgs.fit(X[train],y[train]).score(X[test],y[test]) for train, test in kf_total]
# SAME AS DOING: cross_validation.cross_val_score(dtgs, X, y, cv=kf_total, n_jobs = 1)
print scores
print np.mean(scores)
print dtgs.best_score_
# score [0.81818181818181823, 0.78181818181818186, 0.7592592592592593, 0.7592592592592593, 0.72222222222222221]
# mean score 0.768
# .best_score_ 0.683486238532
我使用解释变量的另一个组合(只使用其中的一些)运行它，我得到了反问题。现在.bestrongcore_比交叉验证数组中的所有值都高。
# score [0.74545454545454548, 0.70909090909090911, 0.79629629629629628, 0.7407407407407407, 0.64814814814814814]
# mean score 0.728
# .best_score_ 0.802752293578
这段代码混淆了几个地方。dtgs.fit(X[train_],y[train_])对来自param_grid的每个参数组合进行内部3-fold交叉验证，产生包含20个结果的网格，您可以通过调用dtgs.grid_scores_打开该网格。
[dtgs.fit(X[train_],y[train_]).score(X[test],y[test]) for train_, test in kf_total]因此，这一行适合网格搜索五次，然后使用5-Fold交叉验证取其分数。结果是5-Fold验证的分数数组。
当你调用dtgs.best_score_时，你在最后 fit (of 5)的超参数的3倍验证结果的网格中得到最好的分数。