基于概率的 GridSearchCV 超参数调整随机森林分类器

Question

刚开始为随机森林二元分类调整超参数，我想知道是否有人 knew/could 建议如何将评分设置为基于预测概率而不是预测分类。理想情况下，我想要一些可以考虑 roc_auc 的概率（即 [0.2,0.6,0.7,0.1,0.0]）而不是分类（即 [0,1,1,0,0]).

from sklearn.metrics import roc_auc_score
from sklearn.ensemble import RandomForestClassifier as rfc
from sklearn.grid_search import GridSearchCV

rfbase = rfc(n_jobs = 3, max_features = 'auto', n_estimators = 100, bootstrap=False)

param_grid = {
    'n_estimators': [200,500],
    'max_features': [.5,.7],
    'bootstrap': [False, True],
    'max_depth':[3,6]
}

rf_fit = GridSearchCV(estimator=rfbase, param_grid=param_grid
      , scoring = 'roc_auc')

我认为目前 roc_auc 正在脱离实际分类。想在开始创建自定义评分函数之前检查是否有更有效的方法，在此先感谢您的帮助！

Answer 1

最终使用 Jarad 提供的参考求解：

from sklearn.metrics import roc_auc_score
from sklearn.ensemble import RandomForestClassifier as rfc
from sklearn.grid_search import GridSearchCV

rfbase = rfc(n_jobs = 3, max_features = 'auto', n_estimators = 100, bootstrap=False)

param_grid = {
    'n_estimators': [200,500],
    'max_features': [.5,.7],
    'bootstrap': [False, True],
    'max_depth':[3,6]
}

def roc_auc_scorer(y_true, y_pred):
    return roc_auc_score(y_true, y_pred[:, 1])
scorer = make_scorer(roc_auc_scorer, needs_proba=True)

rf_fit = GridSearchCV(estimator=rfbase, param_grid=param_grid
      , scoring = scorer)

基于概率的 GridSearchCV 超参数调整随机森林分类器

Hyperparameter tuning Random Forest Classifier with GridSearchCV based on probability

python

random-forest

hyperparameters