ROC_AUC百分比计算给NAN

发布于 2025-01-26 05:10:54 字数 413 浏览 4 评论 0原文

我正在尝试计算数据的ROC分数，但导致NAN。

代码：

scoring = 'roc_auc'
kfold= KFold(n_splits=10, random_state=42, shuffle=True)
model = LinearDiscriminantAnalysis()
results = cross_val_score(model, df_n, y, cv=kfold, scoring=scoring)
print("AUC: %.3f (%.3f)" % (results.mean(), results.std()))

df_n是来自归一化值的数组，我还尝试使用数据集的x数据值尝试。 Y是二进制值的数组。

DF_N形状：（ 150，4） Y形：（150，）

我很难过，应该有效！

原文

I am trying to calculate the ROC score for my data but it is resulting in nan.

The code:

scoring = 'roc_auc'
kfold= KFold(n_splits=10, random_state=42, shuffle=True)
model = LinearDiscriminantAnalysis()
results = cross_val_score(model, df_n, y, cv=kfold, scoring=scoring)
print("AUC: %.3f (%.3f)" % (results.mean(), results.std()))

df_n is an array from the normalised values, I also tried it just with the X data value from the dataset.
y is an array of binary values.

df_n shape: (150, 4)
y shape: (150,)

I am stumped, it should work!

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只为守护你 2025-02-02 05:10:54

问题在于roc_auc_score期望在多类分类的情况下，而不是预测。但是，使用该代码，得分将获得预测的输出。

使用新得分手：

from sklearn.metrics import roc_auc_score, make_scorer

multi_roc_scorer = make_scorer(lambda y_in, y_p_in: roc_auc_score(y_in, y_p_in, multi_class='ovr'), needs_proba=True)
scores = cross_validate(model, X_s, y_s, scoring=multi_roc_scorer, cv=cv, error_score="raise")

The problem is that roc_auc_score expects the probabilities and not the predictions in the case of multi-class classification. However, with that code the score is getting the output of predict instead.

Use a new scorer:

from sklearn.metrics import roc_auc_score, make_scorer

multi_roc_scorer = make_scorer(lambda y_in, y_p_in: roc_auc_score(y_in, y_p_in, multi_class='ovr'), needs_proba=True)
scores = cross_validate(model, X_s, y_s, scoring=multi_roc_scorer, cv=cv, error_score="raise")

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