In Hiveql, what is the most elegant and performatic way of calculating an average value when there are 'gaps' in the data, with implicit repeated values between them? i.e. Considering a table with the following data:
+----------+----------+----------+
| 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 |
+----------+----------+----------+
If I try to calculate a simple average of the november rows, it will return ~722.78, but the average should take into account the days that are not shown have the same balance as the previous register. In the above data, John had 1800.2 between 20181101 and 20181104, for example.
Assuming that the table always have exactly one row for each date/balance and given that I cannot change how this data is stored (and probably shouldn't since it would be a waste of storage to write rows for days with unchanged balances), I've been tinkering with getting the average from a select with subqueries for all the days in the queried month, returning a NULL for the absent days, and then using case to get the balance from the previous available date in reverse order. All of this just to avoid writing temporary tables.
Step 1: Original Data
The 1st step is to recreate a table with the original data. Let's say the original table is called 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
);
Insert Sample Data in original table 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);
Step 2: Dimension Table
You will need a dimension table where you will have a calendar (table with all the possible dates), call it dimension_date. This is a normal industry standard to have a calendar table, you could probably download this sample data over the internet.
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
);
Insert some sample data for entire month of Nov 2018:
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');
Step 3: Fact Table
Create a fact table from the original table. In normal practice, you ingest the data to hdfs/hive then process the raw data and create a table with historical data where you keep inserting in increment manner. You can look more into data warehousing to get the proper definition but I call this a fact table - f_employee_balance.
This will re-create the original table with missing dates and populate the missing balance with earlier known balance.
--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);
Step 4: Analytics
The query below will give you the true average for by month:
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;
Step 5: Conclusion
You could have just combined step 3 and 4 together; this would save you from creating extra table. When you are in the big data world, you don't worry much about wasting extra disk space or development time. You can easily add another disk or node and automate the process using workflows. For more information, please look into data warehousing concept and hive analytical queries.
I am using the following hive query script for the version 0.13.0
DROP TABLE IF EXISTS movies.movierating;
DROP TABLE IF EXISTS movies.list;
DROP TABLE IF EXISTS movies.rating;
DROP DATABASE IF EXISTS movies;
ADD JAR /usr/local/hadoop/hive/hive/lib/RegexLoader.jar;
CREATE DATABASE IF NOT EXISTS movies;
CREATE EXTERNAL TABLE IF NOT EXISTS movies.list (id STRING, name STRING, genre STRING)
ROW FORMAT SERDE 'com.cisco.hadoop.loaders.RegexSerDe'with SERDEPROPERTIES(
"input.regex"="^(.*)\\:\\:(.*)\\:\\:(.*)$",
"output.format.string"="%1$s %2$s %3$s");
CREATE EXTERNAL TABLE IF NOT EXISTS movies.rating (id STRING, userid STRING, rating STRING, timestamp STRING)
ROW FORMAT SERDE 'com.cisco.hadoop.loaders.RegexSerDe'
with SERDEPROPERTIES(
"input.regex"="^(.*)\\:\\:(.*)\\:\\:(.*)\\:\\:(.*)$",
"output.format.string"="%1$s %2$s %3$s %4$s");
LOAD DATA LOCAL INPATH 'ml-10M100K/movies.dat' into TABLE movies.list;
LOAD DATA LOCAL INPATH 'ml-10M100K/ratings.dat' into TABLE movies.rating;
CREATE TABLE movies.movierating(id STRING, name STRING, genre STRING, rating STRING);
INSERT OVERWRITE TABLE movies.movierating
SELECT list.id, list.name, list.genre, rating.rating from movies.list list LEFT JOIN movies.rating rating ON (list.id=rating.id) GROUP BY list.id;
The issue is when I execute the script without the "GROUP BY" clause it works fine.
But when I execute it with the "GROUP BY" clause, I get the following error
FAILED: SemanticException [Error 10002]: Line 4:21 Invalid column reference 'name'
Any ideas what is happening here?
Appreciate your help
Thanks!
If you group by a column, your select statement can only select a) that column, b) columns derived only from that column, or c) a UDAF applied to other columns.
In this case, you're only grouping by list.id, so when you try to select list.name, that's invalid. Think about it this way: what if your list table contained the following two entries:
id|name |genre
--+-----+------
01|name1|comedy
01|name2|horror
What would you expect this query to return:
select list.id, list.name, list.genre from list group by list.id;
In this case it's nonsensical. I'm guessing that id in reality is a primary key, but note that hive does not know this, so the above data set is perfectly valid.
With all that in mind, it's not clear to me how to fix it because I don't know the desired output. For example, let's say without the group by (just the join), you have as output:
id|name |genre |rating
--+-----+------+-------
01|name1|comedy|'pretty good'
01|name1|comedy|'bad'
02|name2|horror|'9/10'
03|name3|action|NULL
What would you want the output to be with the group by? What are you trying to accomplish by doing the group by?
OK let me see if I can ask this in a better way.
Here are my two tables
Movies list table - Consists of movies information
ID | Movie Name | Genre
1 | Movie 1 | comedy
2 | movie 2 | action
3 | movie 3 | thriller
And I have ratings table
MOVIE_ID | USER ID | RATING on 5 | TIMESTAMP
1 | xyz | 5 | 12345612
1 | abc | 4 | 23232312
2 | zvc | 1 | 12321123
2 | zyx | 2 | 12312312
What I would like to do is get the output in the following way:
Movie ID | Movie Name | Genre | Rating Average
1 | Movie 1 | comedy | 4.5
2 | Movie 2 | action | 1.5
I am not a db expert but I understand this, when you group the data together you need to convert the multiple values to the scalar values or all the values, if string should be same right?
For example in my previous case, I was grouping them together as a string. So which is okay for list.id, list.name and list.genre, but the list.rating, well that is always going to give some problem here (I just learnt PIG along with hive, so grouping works differently there)
So to tackle the problem, I casted the rating and averaged it out and stored it in the float table. Have a look at my code below:
CREATE TABLE movies.movierating(id STRING, name STRING, genre STRING, rating FLOAT);
INSERT OVERWRITE TABLE movies.movierating
SELECT list.id, list.name, list.genre, AVG(cast(rating.rating as FLOAT)) from movies.list list LEFT JOIN movies.rating rating ON (list.id=rating.id) GROUP BY list.id, list.name,list.genre order by list.id DESC;
Thank you for your explanation. I might save the following question for the next thread but here is my observation:
The performance of the Overall job is reduced when performing Grouping and Joining together than to do it in two separate queries. For the same job, I had changed the code a bit to perform the grouping first and then joining the data and the over all time was reduced by 40 seconds. Earlier it was taking 140 seconds and now it is taking 100 seconds. Any reasons to that?
Once again thank you for your explanation.
I came across same issue:
org.apache.hadoop.hive.ql.parse.SemanticException: Invalid column reference "charge_province"
After I put the "charge_province" in the group by, the issue is gone. I don't know why.
I am using Oracle 10g Enterprise edition.
A table in our Oracle database stores the soundex value representation of another text column. We are using a custom soundex implementation in which the soundex values are longer than are generated by traditional soundex algorithms (such as the one Oracle uses). That's really beside the point.
Basically I have a varchar2 column that has values containing a single character followed by a dynamic number of numeric values (e.g. 'A12345', 'S382771', etc). The table is partitioned by another column, but I'd like to add a partitioned index to the soundex column since it is often searched. When trying to add a range partitioned index using the first character of the soundex column it worked great:
create index IDX_NAMES_SOUNDEX on NAMES_SOUNDEX (soundex)
global partition by range (soundex) (
partition IDX_NAMES_SOUNDEX_PART_A values less than ('B'), -- 'A%'
partition IDX_NAMES_SOUNDEX_PART_B values less than ('C'), -- 'B%'
...
);
However, I in order to more evenly distribute the size of the partitions, I want to define some partitions by the first two chars, like so:
create index IDX_NAMES_SOUNDEX on NAMES_SOUNDEX (soundex)
global partition by range (soundex) (
partition IDX_NAMES_SOUNDEX_PART_A5 values less than ('A5'), -- 'A0% - A4%'
partition IDX_NAMES_SOUNDEX_PART_A values less than ('B'), -- 'A4% - A9%'
partition IDX_NAMES_SOUNDEX_PART_B values less than ('C'), -- 'B%'
...
);
I'm not sure how to properly range partition using varchar2 columns. I'm sure this is a less than ideal choice, so perhaps someone can recommend a better solution. Here's a distribution of the soundex data in my table:
-----------------------------------
| SUBSTR(SOUNDEX,1,1) | COUNT |
-----------------------------------
| A | 6476349 |
| B | 854880 |
| D | 520676 |
| F | 1200045 |
| G | 280647 |
| H | 3048637 |
| J | 711031 |
| K | 1336522 |
| L | 348743 |
| M | 3259464 |
| N | 1510070 |
| Q | 276769 |
| R | 1263008 |
| S | 3396223 |
| V | 533844 |
| W | 555007 |
| Y | 348504 |
| Z | 1079179 |
-----------------------------------
As you can see, the distribution is not evenly spread, which is why I want to define range partitions using the first two characters instead of just the first character.
Suggestions?
Thanks!
What exactly is your question?
Don't you know how you can split your table in n equal parts to avoid skew?
You can do that with analytic function percentile_disc().
Here an SQL PLUS example with n=100, I admit that it isn't very sophisticated but it will do the job.
set pages 0
set lines 200
drop table random_strings;
create table random_strings
as
select upper(dbms_random.string('A', 12)) rndmstr
from dual
connect by level < 1000;
spool parts
select 'select '||level||'/100,percentile_disc('||level||
'/100) within group (order by RNDMSTR) from random_strings;'
sql_statement
from dual
connect by level <= 100
/
spool off
This will output in file parts.lst:
select 1/100,percentile_disc(1/100) within group (order by RNDMSTR) from random_strings;
select 2/100,percentile_disc(2/100) within group (order by RNDMSTR) from random_strings;
select 3/100,percentile_disc(3/100) within group (order by RNDMSTR) from random_strings;
...
select 100/100,percentile_disc(100/100) within group (order by RNDMSTR) from random_strings;
Now you can run script parts.lst to get the partition values. Each partition will contain 1% of the data initially.
Script parts.lst will output:
,01 AJUDRRSPGMNP
,02 AOMJZQPZASQZ
,03 AWDQXVGLLUSJ
,04 BIEPUHAEMELR
....
,99 ZTMHDWTXUJAR
1 ZYVJLNATVLOY
Is the table is being searched by the partitioning key in addition to the SOUNDEX value? Or is it being searched just by the SOUNDEX column?
If you are just trying to achieve an even distribution of data among partitions, have you considered using hash partitions rather than range partitions? Assuming you choose a power of 2 for the number of partitions, that should give you a pretty even distribution of data between partitions.
Talk to me!
Can you tell me what your reason is for partitioning this table? It sounds like it is an OLTP table and may not need to be partition. We don’t want to partition just to say we are partitioned. Tell me what you are trying to accomplish by partitioning this table and I can help you pick a correct partitioning scheme. Partitioning does not equal faster queries. It actually can cause your queries to be slower in some cases.
I see some of your additional thoughts above and I don’t believe you need to partition your table. If your queries are going to be doing aggregates on entire partitions then you may want to partition. If you are going to have hundreds of millions of rows of data you may want to partition to help with DBA maintenance. If you just want you queries to run fast then the primary key index will suffice. Please let me know
Just create a global index on your desired columns.