我创建了一个 5 折交叉验证模型并使用 cross_val_score 函数计算交叉验证模型的精度和召回率，如下所示：

def print_accuracy_report(classifier, X, y, num_validations=5):
    precision = cross_validation.cross_val_score(classifier, 
            X, y, scoring='precision', cv=num_validations)
    print "Precision: " + str(round(100*precision.mean(), 2)) + "%"


    recall = cross_validation.cross_val_score(classifier, 
            X, y, scoring='recall', cv=num_validations)
    print "Recall: " + str(round(100*recall.mean(), 2)) + "%" 

我想知道是否允许我做这些行：

    print "Precision: " + str(round(100*precision.mean(), 2)) + "%"
    print "Recall: " + str(round(100*recall.mean(), 2)) + "%" 

我的意思是这是否代表了整个模型的精度和召回率precision.mean()？recall.mean()

只是为了比较起见，在scikit-learn 的 文档中，我看到模型的准确性计算为：

from sklearn.model_selection import cross_val_score
clf = svm.SVC(kernel='linear', C=1)
scores = cross_val_score(clf, iris.data, iris.target, cv=5)


print(scores)                                          

数组([0.96..., 1. ..., 0.96..., 0.96..., 1. ])

    print("Accuracy: %0.2f (+/- %0.2f)" % (scores.mean(), scores.std() * 2))

准确度：0.98 (+/- 0.03)