力扣之1369.获取最近第二次的活动

• 题目：

• 

• sql建表语句

• Create table If Not Exists UserActivity (username varchar(30), activity varchar(30), startDate date, endDate date);
  Truncate table UserActivity;
  insert into UserActivity (username, activity, startDate, endDate) values ('Alice', 'Travel', '2020-02-12', '2020-02-20');
  insert into UserActivity (username, activity, startDate, endDate) values ('Alice', 'Dancing', '2020-02-21', '2020-02-23');
  insert into UserActivity (username, activity, startDate, endDate) values ('Alice', 'Travel', '2020-02-24', '2020-02-28');
  insert into UserActivity (username, activity, startDate, endDate) values ('Bob', 'Travel', '2020-02-11', '2020-02-18');

• 分析：首先，我们先按照姓名分组，然后按照开始时间排序，找出每个人的活动次序，然后我们再算出每个人的最大次数，然后筛选出最大值不为1，而且排序问最大值-1的信息，或者最大值和排序都为1的信息，下面是图解分析：

• 

• sql实现：

  with t1 as (select username,activity,startDate,endDate,row_number() over (partition by username order by startDate) rn from UserActivity        -- 首先按照姓名分组，然后按照开始时间排序，找出每个人的活动次序
  ),t2 as (select username,activity,startDate,endDate,rn,max(rn) over(partition by username) max_rn from t1  -- 然后我们再算出每个人的最大次数)
  select username,activity,startDate,endDate from t2  where (rn=max_rn-1 and max_rn!=1) or (rn=1 and max_rn=1) --筛选出最大值不为1，而且排序问最大值-1的信息，或者最大值和排序都为1的信息

• pandas例子：

• data = [['Alice', 'Travel', '2020-02-12', '2020-02-20'], ['Alice', 'Dancing', '2020-02-21', '2020-02-23'], ['Alice', 'Travel', '2020-02-24', '2020-02-28'], ['Bob', 'Travel', '2020-02-11', '2020-02-18']]
  user_activity = pd.DataFrame(data, columns=['username', 'activity', 'startDate', 'endDate']).astype({'username':'object', 'activity':'object', 'startDate':'datetime64[ns]', 'endDate':'datetime64[ns]'})
  

• Pandas 分析与sql分析一样：

• pandas实现：

  import pandas as pddef second_most_recent(user_activity: pd.DataFrame) -> pd.DataFrame:user_activity['rn']=user_activity.groupby('username')['startDate'].rank()  --按照姓名分组，然后按照开始时间排序user_activity['max_rn']=user_activity.groupby('username')['rn'].transform('max') -- 算出每个人的最大次数user_activity=(user_activity[((user_activity['rn']==1) & (user_activity['max_rn']==1)) |((user_activity['rn']==user_activity['max_rn']-1) & (user_activity['max_rn']!=1))].reset_index(drop=True)) -- 筛选出最大值不为1，而且排序问最大值-1的信息，或者最大值和排序都为1的信息user_activity=user_activity[['username','activity','startDate','endDate']] -- 取出所需要的列return user_activity