Question

我试图对分类模型实施一组交叉验证。到目前为止，我已经使用此代码进行了简历。

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()))

从此代码中，我获得每一折的准确性得分。例如，如果我有35个小组，我将获得35个准确性得分。我的问题：如何获得每折的灵敏度得分？

Answer 1

您只需要从recall_score导入sklearn.metrics即可使用它，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！

如何计算留一法交叉验证的敏感性得分？

1 个答案: