在不平衡数据中，如何以中等精度和召回率解释高AUC-ROC？

Question

我有一些机器学习的结果，我正试图弄明白。任务是预测/标记“爱尔兰人”与“非爱尔兰人”。Python 2.7的输出：

1= ir
0= non-ir
Class count:
0    4090942
1     940852
Name: ethnicity_scan, dtype: int64
Accuracy: 0.874921350119
Classification report:
             precision    recall  f1-score   support

          0       0.89      0.96      0.93   2045610
          1       0.74      0.51      0.60    470287

avg / total       0.87      0.87      0.87   2515897

Confusion matrix:
[[1961422   84188]
 [ 230497  239790]]
AUC-ir= 0.901238104773

如您所见，准确率和召回率一般，但AUC-ROC较高（~0.90）。我试图找出原因，我怀疑这是由于数据不平衡（大约1:5）。基于混淆矩阵，并使用Irish作为目标（），我计算了TPR=0.51和FPR=0.04。如果我将非爱尔兰人视为（），那么TPR=0.96，FPR=0.49。那么，当TPR在FPR=0.04时仅为0.5时，如何获得0.9 AUC？

代码：

try:
    for i in mass[k]:
        df = df_temp # reset df before each loop
        #$$
        #$$ 
        if 1==1:
        ###if i == singleEthnic:
            count+=1
            ethnicity_tar = str(i) # fr, en, ir, sc, others, ab, rus, ch, it, jp
            # fn, metis, inuit; algonquian, iroquoian, athapaskan, wakashan, siouan, salish, tsimshian, kootenay
            ############################################
            ############################################

            def ethnicity_target(row):
                try:
                    if row[ethnicity_var] == ethnicity_tar:
                        return 1
                    else:
                        return 0
                except: return None
            df['ethnicity_scan'] = df.apply(ethnicity_target, axis=1)
            print '1=', ethnicity_tar
            print '0=', 'non-'+ethnicity_tar

            # Random sampling a smaller dataframe for debugging
            rows = df.sample(n=subsample_size, random_state=seed) # Seed gives fixed randomness
            df = DataFrame(rows)
            print 'Class count:'
            print df['ethnicity_scan'].value_counts()

            # Assign X and y variables
            X = df.raw_name.values
            X2 = df.name.values
            X3 = df.gender.values
            X4 = df.location.values
            y = df.ethnicity_scan.values

            # Feature extraction functions
            def feature_full_name(nameString):
                try:
                    full_name = nameString
                    if len(full_name) > 1: # not accept name with only 1 character
                        return full_name
                    else: return '?'
                except: return '?'

            def feature_full_last_name(nameString):
                try:
                    last_name = nameString.rsplit(None, 1)[-1]
                    if len(last_name) > 1: # not accept name with only 1 character
                        return last_name
                    else: return '?'
                except: return '?'

            def feature_full_first_name(nameString):
                try:
                    first_name = nameString.rsplit(' ', 1)[0]
                    if len(first_name) > 1: # not accept name with only 1 character
                        return first_name
                    else: return '?'
                except: return '?'

            # Transform format of X variables, and spit out a numpy array for all features
            my_dict = [{'last-name': feature_full_last_name(i)} for i in X]
            my_dict5 = [{'first-name': feature_full_first_name(i)} for i in X]

            all_dict = []
            for i in range(0, len(my_dict)):
                temp_dict = dict(
                    my_dict[i].items() + my_dict5[i].items()
                    )
                all_dict.append(temp_dict)

            newX = dv.fit_transform(all_dict)

            # Separate the training and testing data sets
            X_train, X_test, y_train, y_test = cross_validation.train_test_split(newX, y, test_size=testTrainSplit)

            # Fitting X and y into model, using training data
            classifierUsed2.fit(X_train, y_train)

            # Making predictions using trained data
            y_train_predictions = classifierUsed2.predict(X_train)
            y_test_predictions = classifierUsed2.predict(X_test)

插入用于重新采样的代码：

try:
    for i in mass[k]:
        df = df_temp # reset df before each loop
        #$$
        #$$ 
        if 1==1:
        ###if i == singleEthnic:
            count+=1
            ethnicity_tar = str(i) # fr, en, ir, sc, others, ab, rus, ch, it, jp
            # fn, metis, inuit; algonquian, iroquoian, athapaskan, wakashan, siouan, salish, tsimshian, kootenay
            ############################################
            ############################################

            def ethnicity_target(row):
                try:
                    if row[ethnicity_var] == ethnicity_tar:
                        return 1
                    else:
                        return 0
                except: return None
            df['ethnicity_scan'] = df.apply(ethnicity_target, axis=1)
            print '1=', ethnicity_tar
            print '0=', 'non-'+ethnicity_tar

            # Resampled
            df_resampled = df.append(df[df.ethnicity_scan==0].sample(len(df)*5, replace=True))

            # Random sampling a smaller dataframe for debugging
            rows = df_resampled.sample(n=subsample_size, random_state=seed) # Seed gives fixed randomness
            df = DataFrame(rows)
            print 'Class count:'
            print df['ethnicity_scan'].value_counts()

            # Assign X and y variables
            X = df.raw_name.values
            X2 = df.name.values
            X3 = df.gender.values
            X4 = df.location.values
            y = df.ethnicity_scan.values

            # Feature extraction functions
            def feature_full_name(nameString):
                try:
                    full_name = nameString
                    if len(full_name) > 1: # not accept name with only 1 character
                        return full_name
                    else: return '?'
                except: return '?'

            def feature_full_last_name(nameString):
                try:
                    last_name = nameString.rsplit(None, 1)[-1]
                    if len(last_name) > 1: # not accept name with only 1 character
                        return last_name
                    else: return '?'
                except: return '?'

            def feature_full_first_name(nameString):
                try:
                    first_name = nameString.rsplit(' ', 1)[0]
                    if len(first_name) > 1: # not accept name with only 1 character
                        return first_name
                    else: return '?'
                except: return '?'

            # Transform format of X variables, and spit out a numpy array for all features
            my_dict = [{'last-name': feature_full_last_name(i)} for i in X]
            my_dict5 = [{'first-name': feature_full_first_name(i)} for i in X]

            all_dict = []
            for i in range(0, len(my_dict)):
                temp_dict = dict(
                    my_dict[i].items() + my_dict5[i].items()
                    )
                all_dict.append(temp_dict)

            newX = dv.fit_transform(all_dict)

            # Separate the training and testing data sets
            X_train, X_test, y_train, y_test = cross_validation.train_test_split(newX, y, test_size=testTrainSplit)

            # Fitting X and y into model, using training data
            classifierUsed2.fit(X_train, y_train)

            # Making predictions using trained data
            y_train_predictions = classifierUsed2.predict(X_train)
            y_test_predictions = classifierUsed2.predict(X_test)

Answer 1

您的模型为其得分的测试集中的每一行输出概率P（介于0和1之间）。摘要统计数据（精度、召回率等）针对单个值P作为预测阈值，可能P=0.5，除非您在代码中对此进行了更改。然而，ROC包含更多信息，其想法是您可能不希望使用此默认值作为预测阈值，因此通过计算0和1之间每个预测阈值的真阳性与假阳性的比率来绘制ROC。

如果你在数据中对非爱尔兰人的抽样不足，那么你认为AUC和精确度会被高估是正确的；如果数据集只有5000行，那么在更大的训练集上运行模型不会有问题；只需重新平衡您的数据集（通过引导抽样来增加您的非爱尔兰人），直到您的数据集准确反映您的样本人口。