大数据各大厂面试真题答案及代码解析_小红书面试题:如何分析用户行为-优快云博客

该博客包含Soul、微信、京东、滴滴等多平台面试真题答案及代码解析。涉及用户活跃、留存和粘性分析，红包流向探索，电商购物漏斗，订单呼叫完答率分析等内容，还包括连续登录专题，如美团、小鹏汽车、微保的相关分析。

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面试真题 答案及代码解析

一、Soul 用户活跃、留存和粘性分析.....................................................................................................................................................................................................................6

1、2020 年 6 月的活跃用户数为？.......................................................................................................................................................................................................................6

2、７月份工作日期间，各时间段的月活分布，通勤（7：00-9：00、18：00-20：00），午休（11：00-13：00）,临睡（22：00-1：00），哪

段时间的活跃用户数最高？....................................................................................................................................................................................................................................7

3、单日登录次数大于等于 5 次的用户数？......................................................................................................................................................................................................8

4、6 月 12 日的 T+1 日留存、6 月 15 日的 T+3 日留存、6 月 20 日的 T+7 日留存分别为................................................................................................................8

5、6 月份连续 7 天登录的用户数为.................................................................................................................................................................................................................. 11

二、微信-情人节红包流向探索分析 ..................................................................................................................................................................................................................... 13

1、红包发送方用户的基本信息缺失率有多高？（即有多少红包发送方用户无法在用户基本信息表中匹配？）.............................................................. 14

2、哪一组红包金额的拒收率最高？................................................................................................................................................................................................................. 15

3、最受二线城市欢迎的红包金额为？（即发出次数最多）................................................................................................................................................................... 16

4、北上广深 4 大城市中，哪座城市的男性用户发出的 520 红包比例最低？................................................................................................................................... 17

5、将用户划分为两大群体，都市丽人（年龄 25-35 岁，性别女，一线城市）和时尚大妈（年龄 45-55 岁，性别女，三四线城市）收到的红包

平均金额分别是？................................................................................................................................................................................................................................................... 18

三、京东电商购物漏斗............................................................................................................................................................................................................................................... 19

1、从展示到浏览、浏览到加购、加购到购买的转化率分别为？(按照用户数而非点击量算)(1 分) ...................................................................................... 20

2、以下哪个商品的加购率最高？ ..................................................................................................................................................................................................................... 21

3、购买哪个商品的用户的平均年龄最高？................................................................................................................................................................................................... 224、以下哪组价格区间的购买人数最多？........................................................................................................................................................................................................ 23

5.以下说法正确的是？.......................................................................................................................................................................................................................................... 24

四、滴滴面试真题-订单呼叫完答率分析............................................................................................................................................................................................................ 26

1、订单应答率为...................................................................................................................................................................................................................................................... 27

2、订单完单率为...................................................................................................................................................................................................................................................... 28

3、呼叫量最高的是哪一个小时？ ..................................................................................................................................................................................................................... 28

4、第二天继续呼叫的比例为？.......................................................................................................................................................................................................................... 29

5、哪个小时的呼叫应答时间最短？................................................................................................................................................................................................................. 29

五、货拉拉面试题........................................................................................................................................................................................................................................................ 31

1、用车方和司机被禁止(banned=1)的比率分别为？(保留两位小数)................................................................................................................................................. 33

2、2020 年 1 月 25 日的订单完成率为？......................................................................................................................................................................................................... 33

3、用车至少两次，且主动取消过至少 1 次的用车方有多少名？.......................................................................................................................................................... 33

4、北京、上海的非禁止用户的用车取消率分别为？（要求输出结果保留两位小数）................................................................................................................. 34

5、长沙、北京被用车方取消率排第一的司机编号为？............................................................................................................................................................................ 35

六、哔哩哔哩面试真题-观看偏好分析................................................................................................................................................................................................................. 36

1、2020 年 1 月 4 日、1 月 6 日的新增会员分别为？................................................................................................................................................................................. 38

2、用户观看高峰期为？........................................................................................................................................................................................................................................ 38

3、鬼畜区用户里，有多少用户看过汽车，番剧区用户里，有多少用户看过放映厅？................................................................................................................. 38

4、哪一类用户的观看视频个数最多？（以每个用户观看的视频个数平均数衡量）..................................................................................................................... 405、当天最受欢迎的放映厅、番剧分别是？................................................................................................................................................................................................... 41

七、滴滴热门目的地................................................................................................................................................................................................................................................... 42

1、以下哪个地址的用车人数最多？................................................................................................................................................................................................................. 43

2、以前海湾休闲会所为目的地的订单高峰期是几点？............................................................................................................................................................................ 43

3、用车人次最高的住宅、用车人次第一的酒吧分别是？........................................................................................................................................................................ 44

4、从机场到酒店，单量最高的车型为？........................................................................................................................................................................................................ 44

5、以下哪种说法错误？........................................................................................................................................................................................................................................ 45

八、小红书面试真题-用户行为分析 ..................................................................................................................................................................................................................... 46

1、被收藏次数最多的商品为？.......................................................................................................................................................................................................................... 48

2、购买人数最多的商品类目为？ ..................................................................................................................................................................................................................... 48

3、以下哪个商品只被收藏，却未被购买？................................................................................................................................................................................................... 49

4、以下哪个商品只被购买，却从未被收藏？............................................................................................................................................................................................... 49

5、以下哪个商品，既被同一个用户购买，又被同一个用户收藏，且购买人数最多？................................................................................................................. 50

九、快手直播-直播间观看人数峰值分析............................................................................................................................................................................................................ 50

1、进入直播间的高峰期为？(以进入用户数衡量) ..................................................................................................................................................................................... 51

2、晚上 11 点，哪个直播间的进入人数最多？............................................................................................................................................................................................ 52

3、20：00-23：00，娱乐类、搞笑类，进入人数最多直播间分别是？ .............................................................................................................................................. 53

4、娱乐类、搞笑类，人均在线时长（退出时间-进入时间）最长的直播间分别是？................................................................................................................... 54

5、关于同时在线人数，以下哪个说法错误？............................................................................................................................................................................................... 55十、哔哩哔哩面试真题-大会员收入均摊折算................................................................................................................................................................................................... 56

十一、连续登录专题................................................................................................................................................................................................................................................... 57

1、美团连续登录...................................................................................................................................................................................................................................................... 57

2、小鹏汽车连续快充 ............................................................................................................................................................................................................................................ 59

3、微保连续点击...................................................................................................................................................................................................................................................... 60

一、Soul 用户活跃、留存和粘性分析

1、2020 年 6 月的活跃用户数为？

select

substr(load_dt, 1, 7) load_month --题干要求 6 月份，但是实际场景中，都是取更长周期，比如近 6 个月、近 1 年等

, count(distinct usr_id) cst_cnt --活跃用户数是去重用户数，初学者容易写成 count(1)

from

td_load_rcd

group by

substr(load_dt, 1, 7)

;2、７月份工作日期间，各时间段的月活分布，通勤（7：00-9：00、18：00-20：00），午休（11：00-13：00）,临睡（22：00-1：00），

哪段时间的活跃用户数最高？

select

case when hour(load_tm) between 7 and 8 or hour(load_tm) between 18 and 19 then 'commute'

when hour(load_tm) between 11 and 12 then 'lunch'

when hour(load_tm) in (22, 23, 0) then 'before_sleep'

end as time_prd

,count(distinct usr_id) as cst_dt

from

td_load_rcd where load_dt in ('2020-07-01', '2020-07-02',

'2020-07-03', '2020-07-06', '2020-07-07', --题干要求工作日，这里通过枚举

'2020-07-08', '2020-07-09', '2020-07-10') --大家想想有没有其他方式呢？提示（mysql 提取星期函数，大家可自行百度）

group by

case when hour(load_tm) between 7 and 8 or hour(load_tm) between 18 and 19 then 'commute'

when hour(load_tm) between 11 and 12 then 'lunch'

when hour(load_tm) in (22, 23, 0) then 'before_sleep'

end --注意两处 case when 的区别，此处没有 as time_prd;

3、单日登录次数大于等于 5 次的用户数？

select count(distinct usr_id) as usr_cnt

from

(

select usr_id, load_dt, count(1) as load_times --本题的思路是子查询，保存每个用户在每天的登录次数

from td_load_rcd

group by usr_id, load_dt

having count(1) >=5 --把 count(1) 改成 load_times 行不行？大家可以试一下

;

4、6 月 12 日的 T+1 日留存、6 月 15 日的 T+3 日留存、6 月 20 日的 T+7 日留存分别为

--经典题，T+N 留存率查询，同学们需要反复琢磨

--首先明确留存率的定义：T 日新增用户中，在第 n 日（即 T+n 日）再次活跃的用户，占 T 日新增用户的比例。

--谷歌的官方说法更简洁，叫：Percentage of new users who return each day--总之，一定得是新用户

create view td_distinct_load_rcd_min as

select usr_id,min(load_dt) load_dt

from td_load_rcd

group by usr_id --既然是新用户，首先得找到每天新增用户

;

create view td_distinct_load_rcd as

select load_dt, usr_id

from td_load_rcd

group by load_dt, usr_id --这是 sql 代码优化的一种思路。建中间表，减少代码量，提升查询速度。中间表保存了用户在每天

的去重登录情况

;

select

t0.load_dt

, count(t0.usr_id) as cst_dt_0

, count(t1.usr_id) as cst_dt_1

, count(t1.usr_id)/count(t0.usr_id) as cst_dt_pct_1, count(t2.usr_id) as cst_dt_2

, count(t2.usr_id)/count(t0.usr_id) as cst_dt_pct_2

, count(t3.usr_id) as cst_dt_3

, count(t3.usr_id)/count(t0.usr_id) as cst_dt_pct_7

from

td_distinct_load_rcd_min t0

left join

td_distinct_load_rcd t1

on t0.usr_id=t1.usr_id and t0.load_dt=date_sub(t1.load_dt, interval 1 day) --修改本处的 1，3，7 即可得到任意一天的

任意 N 日留存率

left join

td_distinct_load_rcd t2

on t0.usr_id=t2.usr_id and t0.load_dt=date_sub(t2.load_dt, interval 3 day)

left join

td_distinct_load_rcd t3

on t0.usr_id=t3.usr_id and b(t3.load_dt, interval 7 day)

group by t0.load_dt

order by

t0.load_dt

;

写法２

select a.*, b.v_d as vd2, datediff(b.v_d, a.v_d) d_diff

from

(select usr_id, min(load_dt) v_d from td_load_rcd group by usr_id)a

left join

(select usr_id, load_dt v_d from td_load_rcd group by usr_id, load_dt)b

on a.usr_id = b.usr_id

5、6 月份连续 7 天登录的用户数为

--本题的复杂程度，已经超过文字所能描述的范围。

select count(distinct usr_id)

from(

select usr_id, load_dt2, count(1) load_days

from

(

select usr_id, load_dt, rnk, date_sub(load_dt, interval rnk day) as load_dt2

from

(

select

a.usr_id

, a.load_dt

, row_number()

over(partition by a.usr_id order by a.load_dt) rnk

from

(

select

usr_id

, load_dt from td_load_rcd where substr(load_dt, 1, 7)='2020-06'

group by

usr_id, load_dt

group by usr_id,load_dt2

having load_days >= 7

;

二、微信-情人节红包流向探索分析

本场景主要考察多表连接，凡是涉及到多表关联，建议用画图的方式理解

1、红包发送方用户的基本信息缺失率有多高？（即有多少红包发送方用户无法在用户基本信息表中匹配？）

select count(1), count(b.usr_id), 1-count(b.usr_id)/ count(1)from

(select distinct snd_usr_id from tx_red_pkt_rcd) a --以有红包记录的用户为左表

left join

(select distinct usr_id from tx_usr_bas_inf) b --以用户记录表为右表

on a.snd_usr_id = b.usr_id

;

2、哪一组红包金额的拒收率最高？

select

case when pkt_amt between 0 and 50 then 'bin1'

when pkt_amt between 50 and 200 then 'bin2'

else 'bin3' end as pkt_amt_bin, sum(if_ref)

, count(1)

,sum(if_ref)/count(1) c1

from

(select *, case when date(rcv_datetime)='1900-01-01' --关键考点，如何识别未接收红包。通常数据开发人员在设计物理模

型时，会以特殊日期标注then 1 else 0 end as if_ref from tx_red_pkt_rcd)t

group by

case when pkt_amt between 0 and 50 then 'bin1'

when pkt_amt between 50 and 200 then 'bin2'

else 'bin3' end

order by c1 desc

；

3、最受二线城市欢迎的红包金额为？（即发出次数最多）

本题没有标准答案，只是表达了一种数据处理的思路，4、5 题同理。原因是用户基本信息表模拟了错误的场景，一个用户对应了多个

信息，需要想办法强制唯一匹配

select pkt_amt, count(1) c1

from

tx_red_pkt_rcd a

inner join

(select usr_id, max(cty) cty --从第 3 题开始，涉及到 tx_usr_bas_inf 表，这张表是有问题的（为什么会有问题？），一个用

户对应了多条用户信息。需要强制对应唯一信息from tx_usr_bas_inf group by usr_id) b

on a.snd_usr_id = b.usr_id

inner join

tx_cty_map c on

b.cty = c.cty

and c.cty_cls like '%二线%'

group by pkt_amt order by c1 desc;

4、北上广深 4 大城市中，哪座城市的男性用户发出的 520 红包比例最低？

--本题没有标准答案，只是表达了一种数据处理的思路

select cty, count(1), sum(if_520), sum(if_520)/count(1)

from

(

select *,

case when a.pkt_amt=520 then 1 else 0 end as if_520 --给符合条件的红包打上 1 的标签

from

tx_red_pkt_rcd ainner join --不要用 in 子查询来圈定北上广男性用户，in 查询的效率比不上 inner join，大家切记！

(select usr_id

, max(cty) cty

, max(gdr) gdr --如上，仍需要唯一对应，也可以是 min

from tx_usr_bas_inf group by usr_id)b

on a.snd_usr_id = b.usr_id

where b.cty in ('北京市', '上海市', '广州市')

and b.gdr='M')a

group by cty

5、将用户划分为两大群体，都市丽人（年龄 25-35 岁，性别女，一线城市）和时尚大妈（年龄 45-55 岁，性别女，三四线城市）收到

的红包平均金额分别是？

select cls, avg(pkt_amt)

from

(

select a.*, case when c.cty_cls='一线' and b.gdr='F' and age between 25 and 35 then 'dslr'

when c.cty_cls in ('三线', '四线') and b.gdr='F' and age between 45 and 55 then 'ssdm' end as clsfrom

(select * from tx_red_pkt_rcd

where year(rcv_datetime)<>1900 --细节，不要把这个条件漏了，得是接收成功的红包

) a

inner join

(select usr_id, max(cty) cty, max(gdr) gdr, datediff(now(), max(bth_dt))/365.25 --细节，求年龄除以 365.25

as age from tx_usr_bas_inf group by usr_id) b

on a.snd_usr_id = b.usr_id

inner join

tx_cty_map c

on b.cty = c.cty

) t

group by cls

;

三、京东电商购物漏斗 1、从展示到浏览、浏览到加购、加购到购买的转化率分别为？(按照用户数而非点击量算)(1 分)

经典场景，漏斗转化率的求法,左连接

select count(a.cust_uid)

, count(b.cust_uid)

, count(c.cust_uid)

, count(d.cust_uid)

from(select distinct cust_uid, prd_id from tb_clk_rcd where if_snd=1)a --step1:触达

left join

(select distinct cust_uid, prd_id from tb_clk_rcd where if_vw=1)b --step2:浏览

on a.cust_uid=b.cust_uid and a.prd_id=b.prd_id --细节，关联条件必须为 cust_uid & prd_id，两个都要写。很容易漏掉 prd_id

left join

(select distinct cust_uid, prd_id from tb_clk_rcd where if_cart=1)c --step3:加购

on b.cust_uid=c.cust_uid and b.prd_id=c.prd_id

left join

(select distinct cust_uid, prd_id from tb_clk_rcd where if_buy=1)d --step4:购买(付款)，实际电商漏斗比这个长，之后至少

还有两步，付款成功、签收成功

on c.cust_uid = d.cust_uid and c.prd_id=d.prd_id

;

2、以下哪个商品的加购率最高？

select t2.prd_nm, t1.*

from

(select a.prd_id, count(b.cust_uid), count(a.cust_uid),

count(b.cust_uid)/count(a.cust_uid) pct --求出每个商品的加购率 from

(select distinct cust_uid, prd_id from tb_clk_rcd where if_vw=1)a--从浏览

left join

(select distinct cust_uid,prd_id from tb_clk_rcd where if_cart=1)b--到加购，是加购率

on a.cust_uid=b.cust_uid and a.prd_id = b.prd_id

group by prd_id)t1

inner join

tb_prd_map t2

on t1.prd_id = t2.prd_id

order by pct desc;

3、购买哪个商品的用户的平均年龄最高？

select c.prd_nm, avg(age)

from

(select cust_uid,age from tb_cst_bas_inf)a

inner join

(select cust_uid, prd_id from tb_clk_rcd where if_buy=1 group by cust_uid, prd_id)bon a.cust_uid = b.cust_uid

inner join tb_prd_map c

on b.prd_id = c.prd_id

group by c.prd_nm order by 2 desc

4、以下哪组价格区间的购买人数最多？

select case when price<= 100 then 'bin1'

when price >100 and price <=500 then 'bin2'

else 'bin3'

end as price_bin, count(distinct cust_uid)

from

(select a.*, b.price

from

tb_clk_rcd a

inner join

tb_prd_map b on a.prd_id=b.prd_id

where a.if_buy=1)tgroup by

case when price<= 100 then 'bin1'

when price >100 and price <=500 then 'bin2'

else 'bin3'

end

5.以下说法正确的是？

select prd_nm,count(distinct cust_uid)

from

(select a.*, b.age,b.gdr,c.prd_nm from tb_clk_rcd a

inner join tb_cst_bas_inf b

on a.cust_uid =b.cust_uid

inner join tb_prd_map c

on a.prd_id = c.prd_id

where b.gdr='M' and a.if_vw=1 and b.age between 20 and 35) t --A 组用户每个产品的浏览量

group by prd_nm

;

select prd_nm,count(distinct cust_uid)from

(select a.*, b.age,b.gdr,c.prd_nm from tb_clk_rcd a

inner join tb_cst_bas_inf b

on a.cust_uid =b.cust_uid

inner join tb_prd_map c

on a.prd_id = c.prd_id

where b.gdr='F' and a.if_vw=1 and b.age between 45 and 55) t --B 组用户每个产品的浏览量

group by prd_nm

;

select count(distinct a.cust_uid) from tb_cst_bas_inf a

inner join

(select distinct cust_uid from tb_clk_rcd where if_vw=1) b --A 组用户总浏览量

on a.cust_uid = b.cust_uid

where a.age between 20 and 35 and a.gdr='M'

group by gdr

;select count(distinct a.cust_uid) from tb_cst_bas_inf a

inner join

(select distinct cust_uid from tb_clk_rcd where if_vw=1) b --B 组用户总浏览量

on a.cust_uid = b.cust_uid

where a.age between 45 and 55 and a.gdr='F'

group by gdr

四、滴滴面试真题-订单呼叫完答率分析

https://blog.youkuaiyun.com/SeizeeveryDay/article/details/112914590

1、订单应答率为

select sum(if_grab)/count(1)

from(select *, case when year(grab_time)=1970 --类似与微信红包场景，通过特殊日期识别取消用户

then 0 else 1 end as if_grab

from didi_order_rcd)t

；

2、订单完单率为

select sum(if_finish)/count(1)

from

(select *, case when year(finish_time)<>1970 then 1 else 0 end as if_finish

from didi_order_rcd)t

；

3、呼叫量最高的是哪一个小时？

select hour(call_time), count(1) c1

from didi_order_rcd

group by hour(call_time)

order by 2 desc； -- 规范的写法是 order by c1,甚至 c1 也是不规范的，群主图快，你可不要学哦4、第二天继续呼叫的比例为？

select count(b.cust_uid)/count(a.cust_uid)

from

(select distinct cust_uid from

didi_order_rcd where

substr(call_time, 1, 10) = '2021-05-02')a --类似与留存率的写法，这里又偷懒了，因为只有两天数据

left join

(select distinct cust_uid from

didi_order_rcd where substr(call_time, 1, 10) = '2021-05-03')b

on a.cust_uid = b.cust_uid

；

5、哪个小时的呼叫应答时间最短？

select hour(call_time),

sum(TIMESTAMPDIFF(second,call_time, grab_time))/count(1)/60 --日期，时间相减函数需要掌握

from didi_order_rcdwhere year(grab_time)<>1970

group by hour(call_time)

order by 2

；

五、货拉拉面试题 1、用车方和司机被禁止(banned=1)的比率分别为？(保留两位小数)

select role, count(1), sum(banned),

round(sum(banned)/count(1), 4) --百分比保留两位小数，就是小数保留 4 位小数。同学们也可以研究直接生成百分比形式

from hll_t2 group by role

；

2、2020 年 1 月 25 日的订单完成率为？

Select

order_dt, --顺带求出每天的完成率

sum(if(status='completed',1,0))/count(1) '完成率'

from hll_t1

group by order_dt

;

3、用车至少两次，且主动取消过至少 1 次的用车方有多少名？

--错误写法 13 名

select count(a.usr_id) from

(select usr_id, count(1) from hll_t1 group by usr_id having count(1) >=2) ainner join

(select usr_id, count(1) from hll_t1 where status= 'cancel' group by usr_id having count(1)>=1)b

on a.usr_id =b.usr_id

--正确写法 9 名

select count(a.usr_id) from

(select usr_id, count(1) from hll_t1 group by usr_id having count(1) >=2) a --限制用车至少两次

inner join

(select usr_id, count(1) from hll_t1

where status= 'cancel_by_usr' --限制主动取消至少 1 次，不要写成 canel，必须是 cancel_by_usr

group by usr_id having count(1)>=1)b

on a.usr_id =b.usr_id

4、北京、上海的非禁止用户的用车取消率分别为？（要求输出结果保留两位小数）

select cty

, count(1),

sum(if(status<>'completed',1, 0)) --if 函数的用法，可节省代码量，请大家比较 case when 的写法

,sum(if(status<>'completed',1, 0))/count(1) from(

select a.* from hll_t1 a

inner join

hll_t2 b

on a.usr_id = b.usr_id --1.先找到被禁止的用车方

where b.banned=0

union --3.二者联合，union all 和 union 的区别，请大家自行百度

select a.* from hll_t1 a

inner join

hll_t2 b

on a.driver_id = b.usr_id --2.再找到被禁止的司机方

where b.banned=0)t

group by cty

5、长沙、北京被用车方取消率排第一的司机编号为？

select * from

(select cty,driver_id,cancel_rate, dense_rank() --有 3 种 rank，请大家思考这里应该用什么 rank？百度排序窗口函数

over(partition by cty order by cancel_rate desc) rnk

from

(

select cty, driver_id, sum(if(status='cancel_by_usr', 1, 0))/count(1) as cancel_rate

from hll_t1

group by cty, driver_id)t

where rnk=1

六、哔哩哔哩面试真题-观看偏好分析 1、2020 年 1 月 4 日、1 月 6 日的新增会员分别为？

select v_date, count(1) from

(

select usr_id,

min(v_date) v_date --新增会员的写法，就是看这个用户最早登录的那天。看答案才能想到的话，你已经落后了

from bilibili_t1 where m_flg=1 group by usr_id ) t

group by v_date

；

2、用户观看高峰期为？

select hour(v_time), count(1) from bilibili_t2 group by hour(v_time)

order by 2 desc

；

3、鬼畜区用户里，有多少用户看过汽车，番剧区用户里，有多少用户看过放映厅？

create view bilibili_view_2 as

select a.*, b.v_typ v_typ2 from (select usr_id, v_typ --这题很难，难到我不想说话。from (select a.*, b.v_typ --啤酒尿布 4 个字看起来简单，背后需要很长的代码实现

from bilibili_t2 a --我之后会开专题讲这类题的写法及应用

inner join bilibili_t3 b

on a.v_id = b.v_id)t

group by usr_id, v_typ) a

left join

(select usr_id, v_typ from (select a.*, b.v_typ

from bilibili_t2 a

inner join bilibili_t3 b

on a.v_id = b.v_id)t group by usr_id, v_typ) b

on a.usr_id = b.usr_id

order by usr_id

;

select a.*,b.c2,c1/c2

from

(select v_typ, v_typ2, count(distinct usr_id) c1from bilibili_view_2

group by v_typ, v_typ2

order by v_typ, v_typ2) a

inner join

(select v_typ, count(distinct usr_id) c2

from bilibili_view_2

group by v_typ) b

on a.v_typ= b.v_typ

；

4、哪一类用户的观看视频个数最多？（以每个用户观看的视频个数平均数衡量）

select

v_typ,avg(c1)

from

(select b.v_typ, a.usr_id, count(1) c1

from bilibili_t2 a

inner joinbilibili_t3 b

on a.v_id = b.v_id

group by b.v_typ, a.usr_id)t

group by v_typ

order by 2 desc

；

5、当天最受欢迎的放映厅、番剧分别是？

select * from

(

select t.*, dense_rank() over(partition by v_typ order by c1 desc) as rnk --不要偷懒，使用窗口函数找出每一类最受欢迎

的视频

from

(select v_typ, v_nm, count(1) c1

from bilibili_t2 a

inner join

bilibili_t3 b

on a.v_id = b.v_idgroup by v_typ, v_nm) t

order by v_typ, rnk

) t

having rnk=1

;

七、滴滴热门目的地 1、以下哪个地址的用车人数最多？

select

start_loc,count(distinct cust_uid)

from

didi_sht_rcd

group by

start_loc

order by 2 desc

；

2、以前海湾休闲会所为目的地的订单高峰期是几点？

select

hour(start_tm),

count(1)

from didi_sht_rcd

where end_loc like '前海湾休闲%'

group by hour(start_tm) order by 2 desc

;

3、用车人次最高的住宅、用车人次第一的酒吧分别是？

select b.loc_ctg, a.start_loc, c1, dense_rank()over(partition by loc_ctg order by c1 desc)rnk

from

(select start_loc, count(1) c1

from didi_sht_rcd group by start_loc) a

inner join

( select loc_nm, loc_ctg from loc_nm_ctg group by loc_nm, loc_ctg)b

on a.start_loc = b.loc_nm

where loc_ctg in ('住宅', '酒吧')

；

4、从机场到酒店，单量最高的车型为？

select

r.car_cls,

count(distinct cust_uid) 'cnt'

from didi_sht_rcd r

inner join loc_nm_ctg s on r.start_loc=s.loc_nm

inner join loc_nm_ctg e on r.end_loc=e.loc_nmwhere s.loc_ctg = '机场'

and e.loc_ctg = '酒店'

group by r.car_cls

order by cnt desc

;

5、以下哪种说法错误？

select * from

(

select b.loc_ctg as start_ctg,

a.start_loc,c.loc_ctg as end_ctg,

a.end_loc, count(1) c1,

dense_rank()over(partition by start_ctg, end_ctg order by c1 desc ) rnk

from didi_sht_rcd a

inner join loc_nm_ctg b

on a.start_loc = b.loc_nm

inner join loc_nm_ctg c

on a.end_loc = c.loc_nmgroup by b.loc_ctg , a.start_loc,c.loc_ctg, a.end_loc

order by b.loc_ctg)t

where start_ctg in ('酒店', '住宅', '写字楼') and rnk=1

；

八、小红书面试真题-用户行为分析 1、被收藏次数最多的商品为？

select

i.gd_nm,

count(1) 'cnt'

from xhs_fav_rcd r

join gd_inf i on r.mch_id = i.gd_id

group by i.gd_nm

order by cnt desc

;

2、购买人数最多的商品类目为？

select

i.gd_typ,

count(distinct r.cust_uid) 'cnt'

from xhs_pchs_rcd r

join gd_inf i on r.mch_id = i.gd_id

group by i.gd_typ

order by cnt desc;

3、以下哪个商品只被收藏，却未被购买？

select

i.gd_nm

from (select distinct mch_id from xhs_fav_rcd )f

left join (select distinct mch_id from xhs_pchs_rcd )p

on f.mch_id = p.mch_id

join gd_inf i on f.mch_id = i.gd_id

where p.mch_id is null

;

4、以下哪个商品只被购买，却从未被收藏？

select

i.gd_nm

from (select distinct mch_id from xhs_fav_rcd )f

right join (select distinct mch_id from xhs_pchs_rcd )p

on f.mch_id = p.mch_idjoin gd_inf i on p.mch_id = i.gd_id

where f.mch_id is null

;

5、以下哪个商品，既被同一个用户购买，又被同一个用户收藏，且购买人数最多？

select

i.gd_nm,

count(1) 'cnt'

from xhs_fav_rcd f

join xhs_pchs_rcd p on f.cust_uid =p.cust_uid and f.mch_id=p.mch_id

join gd_inf i on f.mch_id = i.gd_id

group by i.gd_nm

order by cnt desc

九、快手直播-直播间观看人数峰值分析 1、进入直播间的高峰期为？(以进入用户数衡量)

select hour(enter_time), count(distinct usr_id)

from ks_live_t1group by hour(enter_time)

order by 2 desc

；

2、晚上 11 点，哪个直播间的进入人数最多？

select b.live_nm, usr_cnt

from

(select live_id, count(distinct usr_id) usr_cnt

from ks_live_t1

where hour(enter_time)=23

group by live_id)a

inner join

ks_live_t2 b

on a.live_id = b.live_id

order by usr_cnt desc

；3、20：00-23：00，娱乐类、搞笑类，进入人数最多直播间分别是？

select * from

(

select live_type, live_nm, dense_rank()over(partition by live_type order by cst_cnt desc) rnk

from

(select live_type, live_nm, count(distinct usr_id) cst_cnt

from

(

select a.*, b.live_nm, b.live_type

from ks_live_t1 a

inner join

ks_live_t2 b

on a.live_id = b.live_id)t

where hour(enter_time) in (20, 21, 22)

group by live_type, live_nm)t

where rnk=1；

4、娱乐类、搞笑类，人均在线时长（退出时间-进入时间）最长的直播间分别是？

select * from

(

select *,dense_rank()over(partition by live_type order by retain_time desc) rnk

from

(select live_type, live_nm, avg(retain_time) retain_time

from

(

select live_type, live_nm, usr_id, (leave_time - enter_time) as retain_time

from ks_live_t1 a

inner join

ks_live_t2 b

on a.live_id = b.live_id)t

group by live_type, live_nm)t

where rnk=1；

5、关于同时在线人数，以下哪个说法错误？

select b.live_nm, a.*

from

(select live_id, max(num) max_num

from(

select live_id,tms, sum(tag)over(partition by live_id order by tms) as num

from(

select usr_id, live_id,enter_time tms, 1 as tag from ks_live_t1

union all

select usr_id, live_id,leave_time tms, -1 as tag from ks_live_t1)t

group by live_id)a

inner join ks_live_t2 b

on a.live_id = b.live_id

；

十、哔哩哔哩面试真题-大会员收入均摊折算

select y_m, sum(avg_day_amt) from

(

select *, pay_amount/datediff(end_date, begin_date) as avg_day_amt

from (select * from bilibili_m2 where m_date between '2021-01-01' and '2021-05-31' ) a

left join

(select * from bilibili_m1) b

on 1 –无实意，本质是笛卡尔连接

where user_id like '%1014%' –只是为了验证，实际查数时需要去掉

and m_date>= begin_date and m_date <=end_date

group by y_m

十一、连续登录专题

1、美团连续登录

select count(distinct usr_id)

from

(select usr_id, load_dt2, count(1) load_days

from

(

select usr_id, load_dt, rnk, date_sub(load_dt, interval rnk day) as load_dt2

from

(

select

a.usr_id

, a.load_dt

, row_number()

over(partition by a.usr_id order by a.load_dt) rnk

from

(

select

usr_id

, load_date load_dt

from mt_t1)a

group by usr_id,load_dt2

having load_days >= 2 --大于等于 2

;

2、小鹏汽车连续快充

select usr_id,max(times)

from

(

select usr_id, rnk3, count(1) times

from

(select

, row_number() over(partition by usr_id order by charge_time) rnk1

, sum(charge_type) over(partition by usr_id order by charge_time) rnk2 ,row_number() over(partition by usr_id order by charge_time)- sum(charge_type) over(partition by usr_id order by

charge_time) rnk3

from xp_t1)t

group by usr_id, rnk3)t

group by usr_id

having max(times)>=13

3、微保连续点击

select distinct usr_id

from

( select *, rank_1- rank_2 as diff

from

(select *,

row_number() over(order by click_time) as rank_1,