转换IsolationForest决策分数为概率算法

我希望创建一个通用函数,将sklearn的IsolationForest的输出decision_scores转换为真正的概率[0.0, 1.0]

我已经阅读并了解了原始论文,我明白从数学角度来看,该函数的输出不是概率,而是每个基础估计器构建的路径长度的平均值,用于隔离异常值。

问题

我想将该输出转换为概率形式的tuple (x,y),其中x=P(anomaly)y=1-x

当前方法

def convert_probabilities(predictions, scores):    from sklearn.preprocessing import MinMaxScaler    new_scores = [(1,1) for _ in range(len(scores))]    anomalous_idxs = [i for i in (range(len(predictions))) if predictions[i] == -1]    regular_idxs = [i for i in (range(len(predictions))) if predictions[i] == 1]    anomalous_scores = np.asarray(np.abs([scores[i] for i in anomalous_idxs]))    regular_scores = np.asarray(np.abs([scores[i] for i in regular_idxs]))    scaler = MinMaxScaler()    anomalous_scores_scaled = scaler.fit_transform(anomalous_scores.reshape(-1,1))    regular_scores_scaled = scaler.fit_transform(regular_scores.reshape(-1,1))    for i, j in zip(anomalous_idxs, range(len(anomalous_scores_scaled))):        new_scores[i] = (anomalous_scores_scaled[j][0], 1-anomalous_scores_scaled[j][0])        for i, j in zip(regular_idxs, range(len(regular_scores_scaled))):        new_scores[i] = (1-regular_scores_scaled[j][0], regular_scores_scaled[j][0])    return new_scoresmodified_scores = convert_probabilities(model_predictions, model_decisions)

最小可重现示例

import pandas as pdfrom sklearn.datasets import make_classification, load_irisfrom sklearn.ensemble import IsolationForestfrom sklearn.decomposition import PCAfrom sklearn.model_selection import train_test_split# Get dataX, y = load_iris(return_X_y=True, as_frame=True)anomalies, anomalies_classes = make_classification(n_samples=int(X.shape[0]*0.05), n_features=X.shape[1], hypercube=False, random_state=60, shuffle=True)anomalies_df = pd.DataFrame(data=anomalies, columns=X.columns)# Split into train/testX_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.15, random_state=60)# Combine testing dataX_test['anomaly'] = 1anomalies_df['anomaly'] = -1X_test = X_test.append(anomalies_df, ignore_index=True)y_test = X_test['anomaly']X_test.drop('anomaly', inplace=True, axis=1)# Build a modelmodel = IsolationForest(n_jobs=1, bootstrap=False, random_state=60)# Fit itmodel.fit(X_train)# Test itmodel_predictions = model.predict(X_test)model_decisions = model.decision_function(X_test)# Print resultsfor a,b,c in zip(y_test, model_predictions, model_decisions):    print_str = """    Class: {} | Model Prediction: {} | Model Decision Score: {}    """.format(a,b,c)    print(print_str)

问题

modified_scores = convert_probabilities(model_predictions, model_decisions)# Print resultsfor a,b in zip(model_predictions, modified_scores):    ans = False    if a==-1:        if b[0] > b[1]:            ans = True        else:            ans = False    elif a==1:        if b[1] > b[0]:            ans=True        else:            ans=False    print_str = """    Model Prediction: {} | Model Decision Score: {} | Correct: {}    """.format(a,b, str(ans))    print(print_str)

显示了一些奇怪的结果,例如:

Model Prediction: 1 | Model Decision Score: (0.17604259932311161, 0.8239574006768884) | Correct: TrueModel Prediction: 1 | Model Decision Score: (0.7120367886017022, 0.28796321139829784) | Correct: FalseModel Prediction: 1 | Model Decision Score: (0.7251531538304419, 0.27484684616955807) | Correct: FalseModel Prediction: -1 | Model Decision Score: (0.16776449326185877, 0.8322355067381413) | Correct: FalseModel Prediction: 1 | Model Decision Score: (0.8395087028516501, 0.1604912971483499) | Correct: False

Model Prediction: 1 | Model Decision Score: (0.0, 1.0) | Correct: True

如何可能预测为-1 (异常),但概率仅为37%?或者预测为1 (正常),但概率为26%?

请注意,玩具数据集是有标签的,但显然无监督的异常检测算法假设没有标签。


回答:

尽管几个月后,这个问题有了答案。

2011年发表了一篇论文,试图展示关于将异常分数统一为概率的研究。

事实上,pyod库有一个通用的predict_proba方法,该方法提供了使用这种统一方法的选项。

这是受他们的源代码启发的代码实现:

def convert_probabilities(data, model):    decision_scores = model.decision_function(data)    probs = np.zeros([data.shape[0], int(model.classes)])    pre_erf_score = ( decision_scores - np.mean(decision_scores) ) / ( np.std(decision_scores) * np.sqrt(2) )    erf_score = erf(pre_erf_score)    probs[:, 1] = erf_score.clip(0, 1).ravel()    probs[:, 0] = 1 - probs[:, 1]    return probs

(作为参考,pyod确实有一个Isolation Forest实现

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