如何在cross_validate sklearn函数中集成G-均值

Question

from sklearn.model_selection import cross_validate
scores = cross_validate(LogisticRegression(class_weight='balanced',max_iter=100000),
X,y, cv=5, scoring=('roc_auc', 'average_precision','f1','recall','balanced_accuracy'))
scores['test_roc_auc'].mean(), scores['test_average_precision'].mean(),scores['test_f1'].mean(),scores['test_recall'].mean(),scores['test_balanced_accuracy'].mean()

在上述交叉验证评分参数下，我如何计算以下G-means：

from imblearn.metrics import geometric_mean_score
print('The geometric mean is {}'.format(geometric_mean_score(y_test, y_test_pred)))

或

from sklearn.metrics import accuracy_score
g_mean = 1.0
#
for label in np.unique(y_test):
idx = (y_test == label)
g_mean *= accuracy_score(y_test[idx], y_test_pred[idx])
#
g_mean = np.sqrt(g_mean)
score = g_mean
print(score)

Answer 1

只需将其作为自定义得分手传递即可

from sklearn.metrics import make_scorer
from imblearn.metrics import geometric_mean_score
gm_scorer = make_scorer(geometric_mean_score, greater_is_better=True, average='binary')

将CCD_ 1设置为最佳值更接近于1。geometrics_mean_score的附加参数可以直接传递给make_scorer

完整示例

from sklearn.model_selection import cross_validate
from sklearn.metrics import make_scorer
from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression
from imblearn.metrics import geometric_mean_score
X, y = load_breast_cancer(return_X_y=True)
gm_scorer = make_scorer(geometric_mean_score, greater_is_better=True)
scores = cross_validate(
LogisticRegression(class_weight='balanced',max_iter=100000),
X,y, 
cv=5, 
scoring=gm_scorer
)
scores
>>>
{'fit_time': array([0.76488066, 0.69808364, 1.22158527, 0.94157672, 1.01577377]),
'score_time': array([0.00103951, 0.00100923, 0.00065804, 0.00071168, 0.00068736]),
'test_score': array([0.91499142, 0.93884403, 0.9860133 , 0.92439026, 0.9525989 ])}

编辑

要指定多个度量，请将dict传递给scoring参数

scores = cross_validate(
LogisticRegression(class_weight='balanced',max_iter=100000),
X,y, 
cv=5, 
scoring={'gm_scorer': gm_scorer, 'AUC': 'roc_auc', 'Avg_Precision': 'average_precision'}
)
scores
>>>
{'fit_time': array([1.03509665, 0.96399784, 1.49760461, 1.13874388, 1.32006526]),
'score_time': array([0.00560617, 0.00357151, 0.0057447 , 0.00566769, 0.00549698]),
'test_gm_scorer': array([0.91499142, 0.93884403, 0.9860133 , 0.92439026, 0.9525989 ]),
'test_AUC': array([0.99443171, 0.99344907, 0.99801587, 0.97949735, 0.99765258]),
'test_Avg_Precision': array([0.99670544, 0.99623085, 0.99893162, 0.98640759, 0.99861043])}

Answer 2

您需要创建一个自定义记分器，下面是一个示例：https://stackoverflow.com/a/53850851/12384070然后，如果它是你想要的唯一得分手，你可以这样做：

scores = cross_validate(LogisticRegression(class_weight='balanced',max_iter=100000),
X,y, cv=5, scoring=your_custom_function)

我认为你可以使用另一个得分手，正如文档中所解释的：

If scoring reprents multiple scores, one can use:
a list or tuple of unique strings;
a callable returning a dictionary where the keys are the metric names and the values are the metric scores;
a dictionary with metric names as keys and callables a values.