ML——AUC和GAUC

本文介绍AUC和GAUC

参考链接

编程实现

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def calculate_auc(ground_truth, predictions):
    # 将预测结果和真实标签按照预测结果从大到小的顺序进行排序
    sorted_predictions = [p for _, p in sorted(zip(predictions, ground_truth), reverse=True)]
    print(sorted_predictions)
    # 统计正样本和负样本的数量
    positive_count = sum(ground_truth)
    negative_count = len(ground_truth) - positive_count

    neg_found_count = 0
    pos_gt_neg_count = 0
    # 计算正样本大于负样本的数量之和
    for label in sorted_predictions:
        if label == 1:
            pos_gt_neg_count += negative_count - neg_found_count
        else:
            neg_found_count += 1

    # 计算AUC
    auc = 1.0 * pos_gt_neg_count / (positive_count * negative_count)

    return auc

# 真实标签
ground_truth = [1, 0, 1, 0, 1, 1]
# 预测结果
predictions = [0.5, 0.3, 0.1, 0.2, 0.8, 0.9]

# 计算AUC
auc = calculate_auc(ground_truth, predictions)

print("AUC:", auc)

SQL实现

  • 详情见：深入理解AUC

  • 推导思路：

    • 统计每个正样本大于负样本的概率（排在该正样本后面的负样本数/总的负样本数）
    • 对所有正样本的概率求均值

  • SQL实现

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    select
        (ry - 0.5*n1*(n1+1))/n0/n1 as auc
    from(
        select
            sum(if(y=0, 1, 0)) as n0,
            sum(if(y=1, 1, 0)) as n1,
            sum(if(y=1, r, 0)) as ry
        from(
            select y, row_number() over(order by score asc) as r
            from(
                select y, score
                from some.table
            )A
        )B
    )C

  • SQL实现（分场景+pcoc实现）

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    select 
        scene,
        (ry - 0.5*n1*(n1+1))/n0/n1 as auc,
        n1/(n1+n0) as ctr,
        pctr,
        pctr/(n1/(n1+n0)) as pcoc,
    from(
        select 
            scene,
            sum(if(y=0, 1, 0)) as n0,
            sum(if(y=1, 1, 0)) as n1,
            sum(if(y=1, r, 0)) as ry,
            avg(score) as pctr
        from(
            select scene, score, y, row_number() over(partition by scene order by score asc) as r
            from(
                select scene, y, score
                from some.table
            )A
        )B
    )C