大厂面试题（一）

问题1答案：

思路：先找用户最近一次未签到日期，再用今天减那个日期

create table t_user_consecutive_days as 
select fuser_id
,datediff(‘20200322‘,fdate_max) fconsecutive_days
from
    (select fuser_id
    ,max(fdate) fdate_max
    from t_user_attendence
    where fis_sign_in = 0
    group by fuser_id
    ) t1
;

问题1修改：

思路：我觉得还得限制用户是今天的。

create table t_user_consecutive_days as 
select fuser_id
,datediff(‘20200322‘,fdate_max) fconsecutive_days
from
    (select fuser_id
    ,max(fdate) fdate_max
    from t_user_attendence
    where fis_sign_in = 0
    group by fuser_id
    ) t1
where fuser_id in ( select fuser_id
    from t_user_attendence
    where fdate = ‘20200322‘）
;

 1 问题2答案：把用户所有签到记录转化成一条0-1字符串序列，用0做split切割，计算切出来的1序列组中的最大长度
 2 
 3 create table t_user_max_days as
 4 select fuser_id
 5 ,max(length(cut_fsign_record)) as fmax_days
 6 (select fuser_id
 7 ,fsign_record
 8 ,cut_fsign_record
 9 from
10     (select fuser_id
11     ,wm_concat(fis_sign_in) fsign_record
12     from t_user_attendence
13     group by fuser_id
14     ) t1
15 lateral view explode(split(fsign_record,‘0‘)) t as cut_fsign_record
16 ) t2
17 where cut_fsign_record<>‘‘
18 group by fuser_id
19 ;

 1 问题2答案：没看懂，修改
 2 
 3 create table t_user_max_days as
 4 select fuser_id
 5 ,max(sign_in) 
 6 from
 7     (select fuser_id
 8     ,fdate-lag(fate) over (partition by fuser_id order by fate desc )  as sign_in from 
 9     (select fuser_id,fdate 
10     from t_user_attendence
11     where fis_sign_in =0
12     order by fdate desc)a
13  ) t1
14 group by fuser_id
15 ;

 1 最大回撤率：输入参数都是按照日期降序排列的净值序列
 2 基础实现版本：
 3 
 4 def max_drawdown(accnavArr):
 5 mdd = 0
 6 for i in range(0, len(accnavArr)):
 7 for j in range(i + 1, len(accnavArr)):
 8 drawdown = accnavArr[i] / accnavArr[j] - 1
 9 if drawdown < mdd:
10 mdd = drawdown
11 return mdd
12 空间换时间实现版本：
13 
14 把每个时间点计算的最大值都存到一个列表结构中，最大回撤的计算只需要再依赖这个列表进行多一次循环计算。
15 
16 def maxDrawdownGainCal(accnavArr):      # 默认accnavArr按日期降序排列
17 maxDrawdown = 10000
18 maxGain =0
19 arr_len = len(accnavArr)
20 maxList = [0.0] * arr_len
21 minList = [0.0] * arr_len
22 maxList[arr_len-1] = accnavArr[arr_len-1]
23 minList[arr_len-1] = accnavArr[arr_len-1]
24 for i in range(arr_len-2,-1,-1):
25 if accnavArr[i] > maxList[i+1]:
26 maxList[i] = accnavArr[i]
27 else:
28 maxList[i] = maxList[i+1]
29 if accnavArr[i] < minList[i+1]:
30 minList[i] = accnavArr[i]
31 else:
32 minList[i] = minList[i+1]
33 for i in range(0,arr_len):
34 mdd = (accnavArr[i]/maxList[i]-1)
35 mg = (accnavArr[i]/minList[i]-1)
36 if mdd < maxDrawdown : maxDrawdown = mdd
37 if mg > maxGain : maxGain = mg
38 return maxDrawdown,maxGain
39 当前最优版本：
40 
41 每个时间点同时更新最大值和最大回撤，两个指标，不需要额外空间，且只做一次列表循环计算。
42 
43 def maxDrawdownGainCal(accnavArr):      # 默认accnavArr按日期降序排列
44     maxDrawdown = 10000
45     maxGain =0
46     arrLen = len(accnavArr)
47     startMdd = accnavArr[arrLen-1]
48     startGain = accnavArr[arrLen-1]
49     for i in range(arrLen-2,-1,-1):
50         if accnavArr[i] > startMdd:
51             startMdd = accnavArr[i]
52         mdd = (accnavArr[i]/startMdd-1)
53         if accnavArr[i] < startGain:
54             startGain = accnavArr[i]
55         mg = (accnavArr[i]/startGain-1)
56         if mdd < maxDrawdown : maxDrawdown = mdd
57         if mg > maxGain : maxGain = mg
58     return maxDrawdown,maxGain