在Hiveql中,当数据中存在“间隙”且之间存在隐式重复值时,计算平均值的最优雅,最有效的方法是什么?即,考虑包含以下数据的表:
+----------+----------+----------+
| Employee | Date | Balance |
+----------+----------+----------+
| John | 20181029 | 1800.2 |
| John | 20181105 | 2937.74 |
| John | 20181106 | 3000 |
| John | 20181110 | 1500 |
| John | 20181119 | -755.5 |
| John | 20181120 | -800 |
| John | 20181121 | 1200 |
| John | 20181122 | -400 |
| John | 20181123 | -900 |
| John | 20181202 | -1300 |
+----------+----------+----------+
如果我尝试计算11月行的简单平均值,它将返回〜722.78,但平均值应考虑未显示的天数与上一个寄存器具有相同余额的情况。在上述数据中,例如,约翰在20181101和20181104之间的比率为1800.2。
假设该表对于每个日期/余额始终仅具有一行,并且鉴于我无法更改此数据的存储方式(并且可能不应该这样做,因为在余额保持不变的情况下连续几天写入行会浪费存储空间),我一直在设法从查询的月份中所有日期的子查询的选择中获取平均值,在缺席的日子中返回NULL,然后使用case以相反的顺序从上一个可用日期获取余额。所有这些只是为了避免编写临时表。
答案 0 :(得分:1)
第1步:原始数据
第一步是使用原始数据重新创建表。假设原始表格称为daily_employee_balance
。
daily_employee_balance
use default;
drop table if exists daily_employee_balance;
create table if not exists daily_employee_balance (
employee_id string,
employee string,
iso_date date,
balance double
);
将示例数据插入原始表daily_employee_balance
insert into table daily_employee_balance values
('103','John','2018-10-25',1800.2),
('103','John','2018-10-29',1125.7),
('103','John','2018-11-05',2937.74),
('103','John','2018-11-06',3000),
('103','John','2018-11-10',1500),
('103','John','2018-11-19',-755.5),
('103','John','2018-11-20',-800),
('103','John','2018-11-21',1200),
('103','John','2018-11-22',-400),
('103','John','2018-11-23',-900),
('103','John','2018-12-02',-1300);
第2步:维度表
您将需要一个维度表,其中将有一个日历(包含所有可能日期的表),将其命名为dimension_date
。这是具有日历表的正常行业标准,您可能可以通过Internet下载此示例数据。
use default;
drop table if exists dimension_date;
create external table dimension_date(
date_id int,
iso_date string,
year string,
month string,
month_desc string,
end_of_month_flg string
);
插入2018年11月整个月的一些示例数据:
insert into table dimension_date values
(6880,'2018-11-01','2018','2018-11','November','N'),
(6881,'2018-11-02','2018','2018-11','November','N'),
(6882,'2018-11-03','2018','2018-11','November','N'),
(6883,'2018-11-04','2018','2018-11','November','N'),
(6884,'2018-11-05','2018','2018-11','November','N'),
(6885,'2018-11-06','2018','2018-11','November','N'),
(6886,'2018-11-07','2018','2018-11','November','N'),
(6887,'2018-11-08','2018','2018-11','November','N'),
(6888,'2018-11-09','2018','2018-11','November','N'),
(6889,'2018-11-10','2018','2018-11','November','N'),
(6890,'2018-11-11','2018','2018-11','November','N'),
(6891,'2018-11-12','2018','2018-11','November','N'),
(6892,'2018-11-13','2018','2018-11','November','N'),
(6893,'2018-11-14','2018','2018-11','November','N'),
(6894,'2018-11-15','2018','2018-11','November','N'),
(6895,'2018-11-16','2018','2018-11','November','N'),
(6896,'2018-11-17','2018','2018-11','November','N'),
(6897,'2018-11-18','2018','2018-11','November','N'),
(6898,'2018-11-19','2018','2018-11','November','N'),
(6899,'2018-11-20','2018','2018-11','November','N'),
(6900,'2018-11-21','2018','2018-11','November','N'),
(6901,'2018-11-22','2018','2018-11','November','N'),
(6902,'2018-11-23','2018','2018-11','November','N'),
(6903,'2018-11-24','2018','2018-11','November','N'),
(6904,'2018-11-25','2018','2018-11','November','N'),
(6905,'2018-11-26','2018','2018-11','November','N'),
(6906,'2018-11-27','2018','2018-11','November','N'),
(6907,'2018-11-28','2018','2018-11','November','N'),
(6908,'2018-11-29','2018','2018-11','November','N'),
(6909,'2018-11-30','2018','2018-11','November','Y');
第3步:事实表
从原始表创建事实表。在通常的实践中,您将数据提取到hdfs / hive中,然后处理原始数据并创建一个包含历史数据的表,并在其中以增量方式不断插入。您可以深入研究数据仓库以获得正确的定义,但我将其称为事实表-f_employee_balance
。
这将重新创建缺少日期的原始表,并使用早期已知余额填充缺少的余额。
--inner query to get all the possible dates
--outer self join query will populate the missing dates and balance
drop table if exists f_employee_balance;
create table f_employee_balance
stored as orc tblproperties ("orc.compress"="SNAPPY") as
select q1.employee_id, q1.iso_date,
nvl(last_value(r.balance, true) --initial dates to be populated with 0 balance
over (partition by q1.employee_id order by q1.iso_date rows between unbounded preceding and current row),0) as balance,
month, year from (
select distinct
r.employee_id,
d.iso_date as iso_date,
d.month, d.year
from daily_employee_balance r, dimension_date d )q1
left outer join daily_employee_balance r on
(q1.employee_id = r.employee_id) and (q1.iso_date = r.iso_date);
第4步:分析
以下查询将为您提供每月的真实平均值:
select employee_id, monthly_avg, month, year from (
select employee_id,
row_number() over (partition by employee_id,year,month) as row_num,
avg(balance) over (partition by employee_id,year,month) as monthly_avg, month, year from
f_employee_balance)q1
where row_num = 1
order by year, month;
第5步:结论
您可以将步骤3和4组合在一起;这样可以避免创建额外的表格。在大数据世界中,您不必担心浪费额外的磁盘空间或开发时间。您可以轻松添加另一个磁盘或节点,并使用工作流自动执行该过程。有关更多信息,请查看数据仓库概念和配置单元分析查询。