Scikit learn 在GridSearchCV中将精度用作评分时如何指定正标签
在Scikit learn 在GridSearchCV中将精度用作评分时如何指定正标签,scikit-learn,grid-search,scoring,Scikit Learn,Grid Search,Scoring,在sklearn.metrics.precision\u score(y,y\u pred,pos\u label=[0])中,我可以指定正标签,如何在GridSearchCV中也指定它 如果在使用自定义评分时无法指定,如何定义 我试过这个: model = sklearn.model_selection.GridSearchCV( estimator = est, param_grid = param_grid, scoring = 'prec
sklearn.metrics.precision\u score(y,y\u pred,pos\u label=[0])
中,我可以指定正标签,如何在GridSearchCV中也指定它
如果在使用自定义评分时无法指定,如何定义
我试过这个:
model = sklearn.model_selection.GridSearchCV(
estimator = est,
param_grid = param_grid,
scoring = 'precision',
verbose = 1,
n_jobs = 1,
iid = True,
cv = 3)
但我有一个错误:
custom_score = make_scorer(precision_score(y, y_pred,pos_label=[0]),
greater_is_better=True)
阅读,您可以将任何kwargs
传递到make_scorer
中,它们将自动传递到score_func
可调用中
NameError: name 'y_pred' is not defined
然后将此自定义记分器
传递给GridSearchCV
:
from sklearn.metrics import precision_score, make_scorer
custom_scorer = make_scorer(precision_score, greater_is_better=True, pos_label=0)
gs = GridSearchCV(est, ..., scoring=custom_scorer)