在 GridSearchCV 中使用精度作为评分时如何指定正标签

Question

model = sklearn.model_selection.GridSearchCV(
        estimator = est, 
        param_grid = param_grid,
        scoring = 'precision',
        verbose = 1,
        n_jobs = 1,
        iid = True,
        cv = 3)

在sklearn.metrics.precision_score(y, y_pred,pos_label=[0])中，我可以指定正标签，我如何在GridSearchCV中也指定这个？

如果无法指定，使用自定义评分时，如何定义？

我试过这个：

custom_score = make_scorer(precision_score(y, y_pred,pos_label=[0]),  
                          greater_is_better=True)

但我收到错误：

NameError: name 'y_pred' is not defined

Answer 1

读取 docs，您可以将任何 kwargs 传递给 make_scorer，它们将自动传递给 score_func 可调用对象。

from sklearn.metrics import precision_score, make_scorer
custom_scorer = make_scorer(precision_score, greater_is_better=True,  pos_label=0)

然后你把这个custom_scorer传给GridSearchCV:

gs = GridSearchCV(est, ..., scoring=custom_scorer)

在 GridSearchCV 中使用精度作为评分时如何指定正标签

How to specify positive label when use precision as scoring in GridSearchCV

scoring

scikit-learn

grid-search