如何计算 Leave One Out Cross Validation 的敏感性分数？

Question

我尝试对分类模型实施保留一组交叉验证。到目前为止，我用这段代码来做简历。

from sklearn.model_selection import LeaveOneGroupOut

X = X
y = np.array(df.loc[:, df.columns == 'label'])

scores=[]
groups = df["cow_id"].values
logo = LeaveOneGroupOut()

logo.get_n_splits(X, y, groups)

cv=logo.split(X, y, groups)

for train_index, test_index in cv:
    print("Train Index: ", train_index, "\n")
    print("Test Index: ", test_index)
    X_train, X_test, y_train, y_test = X[train_index], X[test_index], y[train_index], y[test_index]
    model.fit(X_train, y_train.ravel())
    scores.append(model.score(X_test, y_test.ravel()))

从这段代码中，我得到每个折叠的准确度分数。例如，如果我有 35 个组，我将获得 35 的准确度分数。我的问题：如何获得每个折叠的敏感度分数？

Answer 1

你只需要从sklearn.metrics导入recall_score并使用它，如下所示：

from sklearn.model_selection import LeaveOneGroupOut
from sklearn.metrics import recall_score

X = X
y = np.array(df.loc[:, df.columns == 'label'])

scores=[]
senstivities = []
groups = df["cow_id"].values
logo = LeaveOneGroupOut()

logo.get_n_splits(X, y, groups)

cv=logo.split(X, y, groups)

for train_index, test_index in cv:
    print("Train Index: ", train_index, "\n")
    print("Test Index: ", test_index)
    X_train, X_test, y_train, y_test = X[train_index], X[test_index], y[train_index], y[test_index]
    model.fit(X_train, y_train.ravel())
    y_pred = model.predict(X_test)
    scores.append(model.score(X_test, y_test.ravel()))
    senstivities.append(recall_score(y_test.ravel(), y_pred))

希望这会有所帮助！