如何解决使用 scikit-learn 计算 AUC 的正确方法是什么?
我注意到下面两个代码的结果是不同的。
isabelle build -bv -d <path to afp>/thys -o timeout_scale=3 Jordan_normal_Form
那么,哪种方法是正确的?
#1
metrics.plot_roc_curve(classifier,X_test,y_test,ax=plt.gca())
#2
metrics.plot_roc_curve(classifier,ax=plt.gca(),label=clsname + ' (AUC = %.2f)' % roc_auc_score(y_test,y_predicted))
输出(#1):
from sklearn.metrics import roc_auc_score
from sklearn import metrics
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.svm import SVC
from sklearn.datasets import load_breast_cancer
data = load_breast_cancer()
X = data.data
y = data.target
X_train,y_train,y_test = train_test_split(X,y,test_size=0.20,random_state=12)
svclassifier = SVC(kernel='rbf')
svclassifier.fit(X_train,y_train)
y_predicted = svclassifier.predict(X_test)
print('AUC = %.2f' % roc_auc_score(y_test,y_predicted)) #1
metrics.plot_roc_curve(svclassifier,ax=plt.gca()) #2
plt.show()
虽然(#2):
解决方法
这里的区别可能是sklearn在内部使用predict_proba()
来获取每个类的概率,并从中找到auc
示例,当您使用 classifier.predict()
import matplotlib.pyplot as plt
from sklearn import datasets,metrics,model_selection,svm
X,y = datasets.make_classification(random_state=0)
X_train,X_test,y_train,y_test = model_selection.train_test_split(X,y,random_state=0)
clf = svm.SVC(random_state=0,probability=False)
clf.fit(X_train,y_train)
clf.predict(X_test)
>> array([1,1,0])
# calculate auc
metrics.roc_auc_score(y_test,clf.predict(X_test))
>>>0.8782051282051283 # ~0.88
如果您使用 classifier.predict_proba()
X,random_state=0)
# set probability=True
clf = svm.SVC(random_state=0,probability=True)
clf.fit(X_train,y_train)
clf.predict_proba(X_test)
>> array([[0.13625954,0.86374046],[0.90517034,0.09482966],[0.19754525,0.80245475],[0.96741274,0.03258726],[0.80850602,0.19149398],......................,[0.31927198,0.68072802],[0.8454472,0.1545528 ],[0.75919018,0.24080982]])
# calculate auc
# when computing the roc auc metrics,by default,estimators.classes_[1] is
# considered as the positive class here 'clf.predict_proba(X_test)[:,1]'
metrics.roc_auc_score(y_test,clf.predict_proba(X_test)[:,1])
>> 0.9102564102564102
因此对于您的问题 metrics.plot_roc_curve(classifier,y_test,ax=plt.gca())
可能使用默认 predict_proba()
来预测 auc
,而对于 metrics.plot_roc_curve(classifier,ax=plt.gca(),label=clsname + ' (AUC = %.2f)' % roc_auc_score(y_test,y_predicted))
,您正在计算 roc_auc_score
并通过分数作为标签。
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