做任务赚q币的网站最大的搜索网站排名

本题用到知识点：row_number(), union, date_sub(), to_timedelta()……

目录

思路

pandas

Mysql

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思路

链接：报告系统状态的连续日期

思路：

判断连续性常用的一个方法，增量相同的两个列的差值是固定的。

让日期与行号 * 天数相减，连续的区域就会的得到相同的结果。

不同的连续区域得到不同结果（因为来连续到不连续，中间出现了增量不为1的情况。但是行号总是增量为1，所以下一个连续区域与行号的差值与上一个连续区域不同。）

如：1~2 连续，7~8连续，但是与行号的差值tag并不相同。

根据相同的tag分组后，找到分组种的最大最小时间分别作为开始时间和结束时间，

两个表都根据上述操作后之际纵向拼接后按照开始时间升序输出。

pandas

import pandas as pddef report_contiguous_dates(failed: pd.DataFrame, succeeded: pd.DataFrame) -> pd.DataFrame:f_date_vaild = (failed['fail_date'] >= '2019-01-01') & (failed['fail_date'] <= '2019-12-31')failed = failed[f_date_vaild]failed['tag'] = failed['fail_date'] - pd.to_timedelta(failed.index,unit='d')failed = failed.groupby(by='tag')['fail_date'].agg(start_date='min',end_date='max').reset_index()failed['tag'] = 'failed's_date_vaild = (succeeded['success_date'] >= '2019-01-01') & (succeeded['success_date'] <= '2019-12-31')succeeded = succeeded[s_date_vaild]succeeded['tag'] = succeeded['success_date'] - pd.to_timedelta(succeeded.index, unit='d')succeeded = succeeded.groupby(by='tag')['success_date'].agg(start_date='min', end_date='max').reset_index()succeeded['tag'] = 'succeeded'return pd.concat([succeeded,failed],axis=0).rename(columns={'tag':'period_state'}).sort_values(by='start_date',ascending=True)

中间结果示例：failed表获取tag后

Mysql

pandas直接使用index作为行号，mysql则使用排名row_number()来获得固定增量的行号。

with f as 
(
select fail_date,
row_number() over(order by fail_date asc) as rn from failed 
where fail_date between '2019-01-01' and '2019-12-31')
,fg as 
(
select 'failed' as period_state,
min(fail_date) as start_date,
max(fail_date) as end_date
from 
(
select 
fail_date,
date_sub(fail_date,interval rn  day) as tag
from f 
) t
group by tag
)
,
s as 
(
select success_date,
row_number() over(order by success_date asc) as rn from succeeded
where success_date between '2019-01-01' and '2019-12-31'
)
,
sg as 
(
select 'succeeded' as period_state,
min(success_date) as start_date,
max(success_date) as end_date
from 
(
select 
success_date,
date_sub(success_date,interval rn  day) as tag
from s 
) t
group by tag
)select * from 
(
(select * from fg)union all (select * from sg)
) t
order by start_date asc