主要来自两个方面的提升：

1. 更快的AUC函数：https://www.kaggle.com/c/riiid-test-answer-prediction/discussion/208031
2. 并行处理多个行为

经过 用户数量 = [500, 1000, 2000, 5000, 10000, 20000]，每个用户行为数 = [50, 100, 200, 500, 1000] 的随机生成数据的测试，大约比优化前的实现快 5-8 倍。

import numpy as np
import pandas as pd
from numba import njit
from scipy.stats import rankdata
from joblib import Parallel, delayed

[@njit](/user/njit)
def _auc(actual, pred_ranks):
    actual = np.asarray(actual)
    pred_ranks = np.asarray(pred_ranks)
    n_pos = np.sum(actual)
    n_neg = len(actual) - n_pos
    return (np.sum(pred_ranks[actual == 1]) - n_pos*(n_pos+1)/2) / (n_pos*n_neg)

def auc(actual, predicted):
    pred_ranks = rankdata(predicted)
    return _auc(actual, pred_ranks)

def uAUC(y_true, y_pred, userids):
    num_labels = y_pred.shape[1]

    def uAUC_infunc(i):
        uauc_df = pd.DataFrame()
        uauc_df['userid'] = userids
        uauc_df['y_true'] = y_true[:, i]
        uauc_df['y_pred'] = y_pred[:, i]

        label_nunique = uauc_df.groupby(by='userid')['y_true'].transform('nunique')
        uauc_df = uauc_df[label_nunique == 2]
        
        aucs = uauc_df.groupby(by='userid').apply(
            lambda x: auc(x['y_true'].values, x['y_pred'].values))
        return np.mean(aucs)

    uauc = Parallel(n_jobs=4)(delayed(uAUC_infunc)(i) for i in range(num_labels))
    return np.average(uauc, weights=[4, 3, 2, 1]), uauc