如何用ChatGPT优化SQL实现的协同过滤算法

-- 初始实现：计算物品相似度矩阵
SELECT 
    a.item_id as item1,
    b.item_id as item2,
    COUNT(DISTINCT a.user_id) as common_users,
    COUNT(DISTINCT a.user_id) / SQRT(COUNT(DISTINCT b.user_id) * COUNT(DISTINCT c.user_id)) as similarity
FROM user_behavior a
JOIN user_behavior b ON a.user_id = b.user_id AND a.item_id != b.item_id
JOIN (SELECT item_id, COUNT(DISTINCT user_id) as total_users FROM user_behavior GROUP BY item_id) c 
    ON a.item_id = c.item_id
JOIN (SELECT item_id, COUNT(DISTINCT user_id) as total_users FROM user_behavior GROUP BY item_id) d 
    ON b.item_id = d.item_id
GROUP BY a.item_id, b.item_id;

WITH item_popularity AS (
    SELECT 
        item_id, 
        COUNT(DISTINCT user_id) as total_users
    FROM user_behavior 
    GROUP BY item_id
),
user_item_interactions AS (
    SELECT
        user_id,
        item_id,
        COUNT(*) as interaction_count
    FROM user_behavior
    GROUP BY user_id, item_id
)

-- 使用HLL近似计算替代精确COUNT DISTINCT
SELECT 
    a.item_id as item1,
    b.item_id as item2,
    APPROX_COUNT_DISTINCT(a.user_id) as common_users,
    APPROX_COUNT_DISTINCT(a.user_id) / 
        SQRT(APPROX_COUNT_DISTINCT(b.user_id) * APPROX_COUNT_DISTINCT(c.user_id)) as similarity
FROM user_behavior a
JOIN user_behavior b ON a.user_id = b.user_id AND a.item_id != b.item_id
-- ... 其余部分类似

-- 按时间分区计算，每天只计算新增数据
WITH recent_interactions AS (
    SELECT *
    FROM user_behavior 
    WHERE event_date >= CURRENT_DATE - INTERVAL '7 days'
)
-- 只计算热门物品的相似度，减少计算量
, popular_items AS (
    SELECT item_id
    FROM item_popularity
    WHERE total_users > 1000
    ORDER BY total_users DESC
    LIMIT 1000
)