如何在sklearn的分类报告中获得我们唯一需要的分数？

Question

我正在使用 sklearn cross_val_score 得到如下结果。

import numpy as np
my_results = cross_val_score(clf, X, y, cv=10, scoring = 'accuracy')
print(np.mean(my_results))

因此，这使我的交叉验证更加准确。要获得我的其他分数，我必须再次执行分类才能获得结果。

precision = cross_val_score(clf, X, y, cv=k_fold, scoring = 'precision_weighted')
#print(precision.mean())
recall = cross_val_score(clf, X, y, cv=k_fold, scoring = 'recall_weighted')
#print(recall.mean())
f1 = cross_val_score(clf, X, y, cv=k_fold, scoring = 'f1_weighted')
#print(f1.mean())

我只对accuracy, weighted average precision, weighted average recall, weighted avareage f-measure and auc 感兴趣，想通过只执行一次cross_val_score 来获得这些分数。

我遇到了sklearn的classification_report：https://scikit-learn.org/stable/modules/generated/sklearn.metrics.classification_report.html。

但是，它不支持cross_val_score。我想知道在 sklearn 中是否有办法做到这一点。

如果需要，我很乐意提供更多详细信息。

编辑：

我尝试按照@t_e_o 的建议执行此操作。但是，它产生的结果是错误的。例如，考虑 iris 数据集。

from sklearn import datasets
iris = datasets.load_iris()
X = iris.data[:, :2]  # we only take the first two features.
y = iris.target

from sklearn.ensemble import RandomForestClassifier
clf=RandomForestClassifier(random_state = 0, class_weight="balanced")

from sklearn.model_selection cross_val_score, cross_validate
cross_val_score(clf, X, y, cv=10, scoring = 'accuracy')
cross_validate(clf, X, y, cv=10, scoring = ('accuracy'))

我得到的结果如下。

#cross_val_score
array([0.53333333, 0.73333333, 0.66666667, 0.8       , 0.73333333,
       0.6       , 0.86666667, 0.86666667, 0.6       , 0.8       ])
#cross_validate
{'fit_time': array([0.01695228, 0.01396275, 0.01296329, 0.01496124, 0.0149579 ,
        0.01097107, 0.01395988, 0.01495957, 0.01396394, 0.0149579 ]),
 'score_time': array([0.00199723, 0.00199652, 0.00199461, 0.00199294, 0.00199652,
        0.00199389, 0.00199485, 0.00199366, 0.00099611, 0.00099897]),
 'test_score': array([0.53333333, 0.73333333, 0.66666667, 0.8       , 0.73333333,
        0.6       , 0.86666667, 0.86666667, 0.6       , 0.8       ]),
 'train_score': array([0.94074074, 0.91851852, 0.92592593, 0.93333333, 0.91851852,
        0.91851852, 0.91111111, 0.8962963 , 0.91851852, 0.91851852])}

Answer 1

你听说过cross_validate 函数吗？它可能是您的正确选择，因此您可以计算多个值。示例：

>>> scores = cross_validate(lasso, X, y, cv=3,
     ...                         scoring=('r2', 'neg_mean_squared_error'),
      ...                         return_train_score=True)
>>> print(scores['test_neg_mean_squared_error'])
    [-3635.5... -3573.3... -6114.7...]
>>> print(scores['train_r2'])
    [0.28010158 0.39088426 0.22784852]

https://scikit-learn.org/stable/modules/generated/sklearn.model_selection.cross_validate.html#sklearn.model_selection.cross_validate

希望对你有一个好的解决方案！

Answer 2

我遇到了类似的问题，但我找不到最佳解决方案，所以我只是使用字典来解决这个问题。在您的情况下，我认为您应该查看以下内容：

metrics = {'precision_weighted':[],'recall_weighted':[],'f1_weighted':[],'accuracy':[]}
for i in metrics:
   metrics[i].append(cross_val_score(clf, X, y, cv=k_fold, scoring = i)