网格搜索分类的自定义评分功能

Posted 2023-03-12

【中文标题】网格搜索分类的自定义评分功能

【英文标题】：Custom scoring function for grid search classification

【发布时间】：2018-10-27 00:13:35

【问题描述】：

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。

只有在提供概率的情况下，评分函数才会起作用（例如，必须调用 rfc.predict_proba(...) 而不是 rfc.predict(...)）

如何指示 GridSearchCV 使用 predict_proba() 而不是 predict()？

from sklearn.model_selection import GridSearchCV
from sklearn.ensemble import RandomForestClassifier

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).max()
    return np.log(1 + diff)

param_grid = 'min_samples_leaf': [1, 2, 5, 10, 20, 50, 100], 'n_estimators': [100, 200, 300]
grid = GridSearchCV(RandomForestClassifier(), param_grid=param_grid,
                    scoring=my_custom_loss_func)

【参考方案1】：

参见文档here：可调用对象应具有参数（估计器、X、y）

然后你可以在你的定义中使用estimator.predict_proba(X)

或者，您可以将make_scorer 与needs_proba=True 一起使用

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简单

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，但我现在懒得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无需手动缩进所有内容？））

【讨论】：

使用sklearn==0.23.2，当使用记分器对象（scoring=custom_scorer）时，它会给出这个错误IndexError: too many indices for array: array is 1-dimensional, but 2 were indexed，它指向这一行diff = np.abs(ground_truth - predictions[:, 1]).max()。似乎暗示predictions 是一个一维数组，但它像二维数组一样被索引。这很令人困惑，因为.predict_proba() 应该返回一个二维数组，所以使用[:, 1] 对其进行索引应该没问题。关于可能导致此错误的任何线索？