auc计算 代码

# 　coding=utf-8# 　auc值的大小可以理解为: 随机抽一个正样本和一个负样本，正样本预测值比负样本大的概率# 根据这个定义，我们可以自己实现计算aucimport randomimport timedef timeit(func):    """    装饰器，计算函数执行时间    """    def wrapper(*args, **kwargs):        time_start = time.time()        result = func(*args, **kwargs)        time_end = time.time()        exec_time = time_end - time_start        print "{function} exec time: {time}s".format(function=func.__name__, time=exec_time)        return result    return wrapperdef gen_label_pred(n_sample):    """    随机生成n个样本的标签和预测值    """    labels = [random.randint(0, 1) for _ in range(n_sample)]    preds = [random.random() for _ in range(n_sample)]    return labels, preds@timeitdef naive_auc(labels, preds):    """    最简单粗暴的方法　　　先排序，然后统计有多少正负样本对满足：正样本预测值>负样本预测值, 再除以总的正负样本对个数     复杂度 O(NlogN), N为样本数    """    n_pos = sum(labels)    n_neg = len(labels) - n_pos    total_pair = n_pos * n_neg    labels_preds = zip(labels, preds)    labels_preds = sorted(labels_preds, key=lambda x: x[1])    accumulated_neg = 0    satisfied_pair = 0    for i in range(len(labels_preds)):        if labels_preds[i][0] == 1:            satisfied_pair += accumulated_neg        else:            accumulated_neg += 1    return satisfied_pair / float(total_pair)@timeitdef approximate_auc(labels, preds, n_bins=100):    """    近似方法，将预测值分桶(n_bins)，对正负样本分别构建直方图，再统计满足条件的正负样本对    复杂度 O(N)    这种方法有什么缺点？怎么分桶？    """    n_pos = sum(labels)    n_neg = len(labels) - n_pos    total_pair = n_pos * n_neg    pos_histogram = [0 for _ in range(n_bins)]    neg_histogram = [0 for _ in range(n_bins)]    bin_width = 1.0 / n_bins    for i in range(len(labels)):        nth_bin = int(preds[i] / bin_width)        if labels[i] == 1:            pos_histogram[nth_bin] += 1        else:            neg_histogram[nth_bin] += 1    accumulated_neg = 0    satisfied_pair = 0    for i in range(n_bins):        satisfied_pair += (pos_histogram[i] * accumulated_neg + pos_histogram[i] * neg_histogram[i] * 0.5)        accumulated_neg += neg_histogram[i]    return satisfied_pair / float(total_pair)# 思考：mapreduce版本的auc该怎么写if __name__ == "__main__":    labels, preds = gen_label_pred(10000000)    naive_auc_rst = naive_auc(labels, preds)    approximate_auc_rst = approximate_auc(labels, preds)    print "naive auc result:{},approximate auc result:{}".format(naive_auc_rst, approximate_auc_rst)    """    naive_auc exec time: 31.7306630611s    approximate_auc exec time: 2.32403683662s    naive auc result:0.500267265728,approximate auc result:0.50026516844    """