字节安全风控数开实习一面凉经

针对简历的项目提问,
介绍一下mapreduce的过程,
spark的优势,
数据倾斜的解决方案,
数仓分层的好处,
介绍一下flink,
如何实现精准一次
sql题:现在有一张表存的是这两个字段,student_id, score 数据格式如下,lihua01,['math':98 english':87, chinese':99].
写一个hivesql实现查询每个学科排名第三名的同学

sql题:hivesql实现查询今日订单量是过去7日订单量均值的2倍的商家id

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校友啊,数开看看我们这里呀,我可以帮你tui
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发布于 04-16 17:14 天津
SELECT subject, student_id, score FROM ( SELECT student_id, subject, score, ROW_NUMBER() OVER (PARTITION BY subject ORDER BY score DESC) as rn FROM ( SELECT student_id, kv.key as subject, kv.value as score FROM student_score LATERAL VIEW explode(score) kv AS key, value ) t1 ) t2 WHERE rn = 3;
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发布于 07-07 15:57 北京
SELECT shop_id FROM ( SELECT shop_id, SUM(CASE WHEN order_date = CURRENT_DATE THEN 1 ELSE 0 END) AS today_cnt, SUM(CASE WHEN order_date >= date_sub(CURRENT_DATE, 7) AND order_date < CURRENT_DATE THEN 1 ELSE 0 END) / 7.0 AS avg_7_cnt FROM orders WHERE order_date >= date_sub(CURRENT_DATE, 7) -- 近7天及今天 AND order_date <= CURRENT_DATE GROUP BY shop_id ) t WHERE today_cnt >= 2 * avg_7_cnt ;
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发布于 07-07 15:57 北京
WITH -- 1. 统计今日订单量 today_orders AS ( SELECT shop_id, COUNT(*) AS today_cnt FROM orders WHERE order_date = CURRENT_DATE GROUP BY shop_id ), -- 2. 统计过去7天(不含今天)日均订单量 last7_orders AS ( SELECT shop_id, COUNT(*) / 7.0 AS avg_7_cnt FROM orders WHERE order_date >= date_sub(CURRENT_DATE, 7) AND order_date < CURRENT_DATE GROUP BY shop_id ) -- 3. 筛选今日订单量为过去7天均值2倍的商家 SELECT t.shop_id, t.today_cnt, l.avg_7_cnt FROM today_orders t JOIN last7_orders l ON t.shop_id = l.shop_id WHERE t.today_cnt >= 2 * l.avg_7_cnt ;
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发布于 07-07 15:56 北京

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