2020大厂数据分析师SQL笔试真题

1. 提要

笔者作为一名互联网商业数据分析师，SQL是日常工作中最常用的数据提取&简单预处理语言。因为其使用的广泛性和易学程度也被其他岗位比如产品经理、研发广泛学习使用，本篇文章主要结合经典面试题，给出通过数据分析师面试的SQL方法论。
以下提均来与笔者经历&网上分享的中高难度SQL题

2. 解题思路

简单——会考察一些group by & limit之类的用法，或者平时用的不多的函数比如rand()类；会涉及到一些表之间的关联；
中等——会考察一些窗口函数的基本用法；会有表之间的关联，相对tricky的地方在于会有一些自关联的使用；
困难——会有中位数或者更加复杂的取数概念，可能要求按照某特定要求生成列；一般这种题建中间表会解得清晰些；

3. SQL真题

3.1

order订单表，字段为： goods_id， amount ；
pv 浏览表，字段为： goods_id，uid；
goods按照总销售金额排序，分成top10,top10~top20,其他三组
求每组商品的浏览用户数（同组内同一用户只能算一次）

create table if not exists test.nil_goods_category as 
select goods_id
,case when nn<= 10 then 'top10'
      when nn<= 20 then 'top10~top20'
      else 'other' end as goods_group
from
(
    select goods_id
    ,row_number() over(partition by goods_id order by sale_sum desc) as nn
    from
    (
        select goods_id,sum(amount) as sale_sum
        from order 
        group by 1
    ) aa
) bb;
select b.goods_group,count(distinct a.uid) as num
from pv a 
left join test.nil_goods_category b 
on a.goods_id = b.goods_id
group by 1;

3.2

商品活动表 goods_event，g_id（有可能重复），t1（开始时间）,t2（结束时间）
给定时间段（t3,t4)，求在时间段内做活动的商品数

1.
select count(distinct g_id) as event_goods_num
from goods_event
where (t1<=t4 and t1>=t3) 
or (t2>=t3 and t2<=t4)

2.
select count(distinct g_id) as event_goods_num
from goods_event
where (t1<=t4 and t1>=t3) 
union all

3.3

商品活动流水表，表名为event，字段：goods_id， time；
求参加活动次数最多的商品的最近一次参加活动的时间

select a.goods_id,a.time
from event a 
inner join
(
    select goods_id,count(*)
    from event
    group by gooods_id
    order by count(*) desc
    limit 1
) b
on a.goods_id = b.goods_id
order by a.goods_id,a.time desc

3.4

用户登录的log数据，划定session，同一个用户一个小时之内的登录算一个session；
生成session列

drop table if exists koo.nil_temp0222_a2;
create table if not exists koo.nil_temp0222_a2 as
select *
    ,row_number() over(partition by userid order by inserttime) as nn1
from 
(
    select a.*
    ,b.inserttime as inserttime_aftr
    ,datediff(b.inserttime,a.inserttime) as session_diff
  from
  (
    select userid,inserttime
      ,row_number() over(partition by userid order by inserttime asc) nn
    from koo.nil_temp0222 
    where userid = 1900000169
  ) a   
  left join 
  (
     select userid,inserttime
      ,row_number() over(partition by userid order by inserttime asc) nn
    from koo.nil_temp0222 
    where userid = 1900000169 
  ) b
  on a.userid =  b.userid and a.nn = b.nn-1
) aa
where session_diff >10 or nn = 1
order by userid,inserttime;

drop table if exists koo.nil_temp0222_a2_1;
create table if not exists koo.nil_temp0222_a2_1 as
select a.*
,case when b.nn is null then a.nn+3 else b.nn end as nn_end
from koo.nil_temp0222_a2 a 
left join koo.nil_temp0222_a2 b 
on a.userid = b.userid 
and a.nn1 = b.nn1 - 1;

select a.*,b.nn1 as session_id
from
(
  select userid,inserttime
    ,row_number() over(partition by userid order by inserttime asc) nn
  from koo.nil_temp0222 
  where userid = 1900000169
) a
left join koo.nil_temp0222_a2_1 b 
on a.userid = b.userid
and a.nn>=b.nn
and a.nn

3.5

订单表，字段有订单编号和时间；
取每月最后一天的最后三笔订单

select *
from 
(
  select *
  ,rank() over(partition by mm order by dd desc) as nn1
  ,row_number() over(partition by mm,dd order by inserttime desc) as nn2
  from
  (select cast(right(to_date(inserttime),2) as int) as dd,month(inserttime) as mm,userid,inserttime
  from koo.nil_temp0222) aa 
) bb 
where nn1 = 1 and nn2<=3;

3.6

数据库表Tourists，记录了某个景点7月份每天来访游客的数量如下：
id date visits 1 2017-07-01 100 …… 非常巧，id字段刚好等于日期里面的几号。
现在请筛选出连续三天都有大于100天的日期。 上面例子的输出为： date 2017-07-01 ……

select a.*,b.num as num2,c.num as num3
from table  a 
left join table b
on a.userid = b.userid
and a.dt = date_add(b.dt,-1)
left join table c
on a.userid = c.userid
and a.dt = date_add(c.dt,-2)
where b.num>100
and a.num>100
and c.num>100

3.7

现有A表，有21个列，第一列id，剩余列为特征字段，列名从d1-d20，共10W条数据！
另外一个表B称为模式表，和A表结构一样，共5W条数据
请找到A表中的特征符合B表中模式的数据，并记录下相对应的id
有两种情况满足要求：
1 每个特征列都完全匹配的情况下。
2 最多有一个特征列不匹配，其他19个特征列都完全匹配，但哪个列不匹配未知

1.
select aa.*
from 
(
  select *,concat(d1,d2,d3……d20) as mmd
  from table
) aa 
left join 
(
  select id,concat(d1,d2,d3……d20) as mmd
  from table
) bb 
on aa.id = bb.id
and aa.mmd = bb.mmd

2.
select a.*,sum(d1_jp,d2_jp……,d20_jp) as same_judge
from 
(
  select a.*
  ,case when a.d1 = b.d1 then 1 else 0 end as d1_jp
  ,case when a.d2 = b.d2 then 1 else 0 end as d2_jp
  ,case when a.d3 = b.d3 then 1 else 0 end as d3_jp
  ,case when a.d4 = b.d4 then 1 else 0 end as d4_jp
  ,case when a.d5 = b.d5 then 1 else 0 end as d5_jp
  ,case when a.d6 = b.d6 then 1 else 0 end as d6_jp
  ,case when a.d7 = b.d7 then 1 else 0 end as d7_jp
  ,case when a.d8 = b.d8 then 1 else 0 end as d8_jp
  ,case when a.d9 = b.d9 then 1 else 0 end as d9_jp
  ,case when a.d10 = b.d10 then 1 else 0 end as d10_jp
  ,case when a.d20 = b.d20 then 1 else 0 end as d20_jp
  ,case when a.d11 = b.d11 then 1 else 0 end as d11_jp
  ,case when a.d12 = b.d12 then 1 else 0 end as d12_jp
  ,case when a.d13 = b.d13 then 1 else 0 end as d13_jp
  ,case when a.d14 = b.d14 then 1 else 0 end as d14_jp
  ,case when a.d15 = b.d15 then 1 else 0 end as d15_jp
  ,case when a.d16 = b.d16 then 1 else 0 end as d16_jp
  ,case when a.d17 = b.d17 then 1 else 0 end as d17_jp
  ,case when a.d18 = b.d18 then 1 else 0 end as d18_jp
  ,case when a.d19 = b.d19 then 1 else 0 end as d19_jp
  from table a 
  left join table b 
  on a.id = b.id 
) aa
where sum(d1_jp,d2_jp……,d20_jp) = 19

3.8

我们把用户对商品的评分用稀疏向量表示，保存在数据库表t里面：
t的字段有：uid，goods_id，star。 uid是用户id；
goodsid是商品id；
star是用户对该商品的评分，值为1-5。
现在我们想要计算向量两两之间的内积，内积在这里的语义为：
对于两个不同的用户，如果他们都对同样的一批商品打了分，那么对于这里面的每个人的分数乘起来，并对这些乘积求和。
例子，数据库表里有以下的数据：
U0 g0 2
U0 g1 4
U1 g0 3
U1 g1 1
计算后的结果为：
U0 U1 23+41=10 ……

select aa.uid1,aa.uid2
,sum(star_multi) as result
from 
(
  select a.uid as uid1
  ,b.uid as uid2
  ,a.goods_id
  ,a.star * b.star as star_multi
  from t a 
  left join t b 
  on a.goods_id = b.goods_id
  and a.udi<>b.uid  
) aa 
group by 1,2

select uid1,uid2,sum(multiply) as result
from
(select t.uid as uid1, t.uid as uid2, goods_id,a.star*star as multiply
from a left join b 
on a.goods_id = goods_id
and a.uid<>uid) aa
group by goods 

3.9

给出一堆数和频数的表格，统计这一堆数中位数

select a.*
,b.s_mid_n
,c.l_mid_n
,avg(b.s_mid_n,c.l_mid_n)
from 
(
  select 
  case when mod(count(*),2) = 0 then count(*)/2 else (count(*)+1)/2 end as s_mid
  ,case when mod(count(*),2) = 0 then count(*)/2+1 else (count(*)+1)/2 end as l_mid  
  from table 
) a 
left join 
(
  select id,num,row_number() over(partition by id order by num asc) nn
  from table
) b 
on a.s_mid = b.nn
left join 
(
  select id,num,row_number() over(partition by id order by num asc) nn
  from table
) c  
on a.l_mid = c.nn

3.10

表order有三个字段，店铺ID，订单时间，订单金额
查询一个月内每周都有销量的店铺

select distinct credit_level
from 
(
  select credit_level,count(distinct nn) as number
  from
  (
    select userid,credit_level,inserttime,month(inserttime) as mm
    ,weekofyear(inserttime) as week
    ,dense_rank() over(partition by credit_level,month(inserttime) order by weekofyear(inserttime) asc) as nn
    from koo.nil_temp0222 
    where substring(inserttime,1,7) = '2019-12'
    order by credit_level ,inserttime  
  ) aa 
  group by 1  
) bb
where number = (select count(distinct weekofyear(inserttime))
from koo.nil_temp0222 
where substring(inserttime,1,7) = '2019-12')