获取GridSearchCV的标准差 [英] Get standard deviation for a GridSearchCV
问题描述
在scikit学习0.20之前,我们可以使用 result.grid_scores_ [result.best_index _]
来获取标准偏差。 (例如:平均值:0.76172,标准:0.05225,参数:{'n_neighbors':21}
)
Before scikit-learn 0.20 we could use result.grid_scores_[result.best_index_]
to get the standard deviation. (It returned for exemple: mean: 0.76172, std: 0.05225, params: {'n_neighbors': 21}
)
在scikit-learn 0.20中获得最佳分数的标准偏差的最佳方法是什么?
What's the best way in scikit-learn 0.20 to get the standard deviation of the best score ?
推荐答案
在较新的版本中, grid_scores _
重命名为 cv_results _
。遵循文档,您需要以下内容:
In newer versions, the grid_scores_
is renamed as cv_results_
. Following the documentation, you need this:
best_index_ : int
The index (of the cv_results_ arrays) which corresponds to the best >
candidate parameter setting.
The dict at search.cv_results_['params'][search.best_index_] gives the >
parameter setting for the best model, that gives the highest mean
score (search.best_score_).
因此,在您的情况下,您需要
So in your case, you need
- 最佳参数:-
result.cv_results _ ['params'] [result.best_index _]
或result.best_params _
-
最高平均得分:-
result.cv_results _ ['mean_test_score'] [result.best_index_ ]
或result.best_score _
最佳std:- result.cv_results _ ['std_test_score'] [result.best_index _]
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