Python 网格搜索分类的自定义评分函数
我想在scikit learn中为Python 网格搜索分类的自定义评分函数,python,scikit-learn,classification,grid-search,Python,Scikit Learn,Classification,Grid Search,我想在scikit learn中为RandomForestClassifier执行GridSearchCV,并且我有一个我想使用的自定义评分函数 评分函数仅在提供概率的情况下有效(例如,必须调用rfc.predict\u probabila(…),而不是rfc.predict(…)) 我如何指示GridSearchCV使用predict\u proba()而不是predict() 请参阅文档:可调用项应具有参数(估计器,X,y) 然后你可以在定义中使用,estimator.predict\u p
RandomForestClassifier
执行GridSearchCV
,并且我有一个我想使用的自定义评分函数
评分函数仅在提供概率的情况下有效(例如,必须调用rfc.predict\u probabila(…)
,而不是rfc.predict(…)
)
我如何指示GridSearchCV使用predict\u proba()
而不是predict()
请参阅文档:可调用项应具有参数(估计器,X,y)
然后你可以在定义中使用,estimator.predict\u proba(X)
或者,您可以使用withneeds\u proba=True
完整的代码示例:
from sklearn.datasets import make_classification
from sklearn.model_selection import GridSearchCV
from sklearn.ensemble import RandomForestClassifier
from sklearn.metrics import make_scorer
import pandas as pd
import numpy as np
X, y = make_classification()
def my_custom_loss_func_est(estimator, X, y):
# predictions must be probabilities - e.g. model.predict_proba()
# example code here:
diff = np.abs(y - estimator.predict_proba(X)[:, 1]).max()
return -np.log(1 + diff)
def my_custom_loss_func(ground_truth, predictions):
# predictions must be probabilities - e.g. model.predict_proba()
# example code here:
diff = np.abs(ground_truth - predictions[:, 1]).max()
return np.log(1 + diff)
custom_scorer = make_scorer(my_custom_loss_func,
greater_is_better=False,
needs_proba=True)
使用记分器对象:
param_grid = {'min_samples_leaf': [10, 50], 'n_estimators': [100, 200]}
grid = GridSearchCV(RandomForestClassifier(), param_grid=param_grid,
scoring=custom_scorer, return_train_score=True)
grid.fit(X, y)
pd.DataFrame(grid.cv_results_)[['mean_test_score',
'mean_train_score',
'param_min_samples_leaf',
'param_n_estimators']]
mean_test_score mean_train_score param_min_samples_leaf param_n_estimators
0 -0.505201 -0.495011 10 100
1 -0.509190 -0.498283 10 200
2 -0.406279 -0.406292 50 100
3 -0.406826 -0.406862 50 200
直接使用损失函数也很容易
grid = GridSearchCV(RandomForestClassifier(), param_grid=param_grid,
scoring=my_custom_loss_func_est, return_train_score=True)
grid.fit(X, y)
pd.DataFrame(grid.cv_results_)[['mean_test_score',
'mean_train_score',
'param_min_samples_leaf',
'param_n_estimators']]
mean_test_score mean_train_score param_min_samples_leaf param_n_estimators
0 -0.509098 -0.491462 10 100
1 -0.497693 -0.490936 10 200
2 -0.409025 -0.408957 50 100
3 -0.409525 -0.409500 50 200
结果因cv折叠不同而不同(我假设,但我现在太懒了,无法设置种子并再次编辑(或者有没有更好的方法来粘贴代码而不需要手动缩进?)使用scorer对象(
scoring=custom\u scorer
)时,使用sklearn==0.23.2
),它给出了这个错误索引器:数组的索引太多:数组是一维的,但有2个索引了,它指向这一行diff=np.abs(ground\u truth-predictions[:,1]).max()。这似乎表明predictions
是一个1D数组,但它被索引为一个2D数组。这是令人困惑的,因为.predict_proba()
应该返回一个2D数组,所以使用[:,1]
对其进行索引应该是可以的。有没有线索说明是什么导致了这个错误?
grid = GridSearchCV(RandomForestClassifier(), param_grid=param_grid,
scoring=my_custom_loss_func_est, return_train_score=True)
grid.fit(X, y)
pd.DataFrame(grid.cv_results_)[['mean_test_score',
'mean_train_score',
'param_min_samples_leaf',
'param_n_estimators']]
mean_test_score mean_train_score param_min_samples_leaf param_n_estimators
0 -0.509098 -0.491462 10 100
1 -0.497693 -0.490936 10 200
2 -0.409025 -0.408957 50 100
3 -0.409525 -0.409500 50 200