我有管道:
np.random.seed(42)
tf.random.set_seed(42)
pipeline = Pipeline([
('smote', SMOTE()),
('under',RandomUnderSampler()),
('cl', KerasClassifier(build_fn=create_model, verbose=0))
])
param_grid_pipeline = {
'smote__sampling_strategy':[.3],
'smote__random_state':[42],
'under__sampling_strategy':['auto'],
'under__random_state':[42],
'cl__batch_size':[128],
'cl__epochs':[20],
}
cv = StratifiedShuffleSplit(n_splits=40, test_size=0.2, random_state=42)
grid = GridSearchCV(estimator=pipeline, param_grid=param_grid_pipeline, cv=cv, scoring='f1', verbose=1, n_jobs=-1)
grid_result = grid.fit(X,y)
print("best_score_",grid_result.best_score_)best_score_是0.9981313067607172
但是,如果我将重新取样排除在管道之外并在外部执行:
np.random.seed(42)
tf.random.set_seed(42)
over = SMOTE(sampling_strategy=0.3,random_state=42)
under = RandomUnderSampler(sampling_strategy='auto',random_state=42)
X,y = over.fit_resample(X,y)
X,y = under.fit_resample(X,y)
pipeline = Pipeline([
('cl', KerasClassifier(build_fn=create_model, verbose=0))
])
param_grid_pipeline = {
'cl__batch_size':[128],
'cl__epochs':[20],
}
cv = StratifiedShuffleSplit(n_splits=40, test_size=0.2, random_state=42)
grid = GridSearchCV(estimator=pipeline, param_grid=param_grid_pipeline, cv=cv, scoring='f1', verbose=1, n_jobs=-1)
grid_result = grid.fit(X,y)
print("best_score_",grid_result.best_score_)我得到了(很多次)更好的结果: 0.9999888503305302
从管道外部再取样有什么区别?
发布于 2020-05-07 20:18:48
如果您在外部执行过采样和欠采样,则在整个数据集上执行它,然后将其拆分以进行交叉验证(这解释了较高的分数,因为在测试集中有关于您的培训集的“信息”)!如果将它包含在管道中,则只需要重新设置培训集(这实际上是正确的方法)。尝试手动进行交叉验证的混洗-拆分,然后只在训练集上重新采样,然后手动计算分数(因为在本例中实际上没有参数网格,这不是很多工作),那么您应该得到相同的结果。
https://stackoverflow.com/questions/61599901
复制相似问题