Hive: SemanticException [Error 10002]: Line 3:21 Invalid column reference 'name' - hadoop

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.

Related

How to groupBy on one column in laravel?

I have a section table and class Table
class table is designed in this way
(id,class_name,section_id)
one class has many sections like
--------------------------------------------
| SN | ClassName | Section_id |
--------------------------------------------
| 1 | ClassOne | 1 |
| 2 | ClassOne | 2 |
| 3 | ClassOne | 3 |
| 4 | ClassOne | 4 |
--------------------------------------------
Now i want to groupBy Only ClassName and display all the sections of that class
$data['classes'] = SectionClass::groupBy('class_name')->paginate(10);
i have groupby like this but it only gives me one section id
Try this way...
$things = SectionClass::paginate(10);
$data['classes']= $things->groupBy('class_name');
You are getting just one row because that is what GROUP BY does, groups a set of rows into a set of summary rows and returns one row for each group. In standard SQL, a query that includes a GROUP BY clause cannot refer to nonaggregated columns in the select list that are not named in the GROUP BY clause. For example, in SQL Server if you try the next clause
SELECT * FROM [Class] GROUP BY [ClassName]
You'll get the next error
"Column 'SN' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause"
Think about it, you are grouping by ClassName, and following your sample data, this will return just one row. Your SELECT clause includes column ClassName, which is easy to get because is the same in every single row, but when you are selecting another, which one should be return if only one has to be selected?
Now, things change a little bit in MySQL. MySQL extends the standard SQL use of GROUP BY so that the select list can refer to nonaggregated columns not named in the GROUP BY clause. This means that the preceding query is legal in MySQL. However, this is useful primarily when all values in each nonaggregated column not named in the GROUP BY are the same for each group. The server is free to choose any value from each group, so unless they are the same, the values chosen are nondeterministic. You can find a complete explanation about this topic here https://dev.mysql.com/doc/refman/5.6/en/group-by-handling.html
If you are expecting a result in one row, you can use GROUP_CONCAT() function to get something like
--------------------------------
| ClassName | Sections |
--------------------------------
| ClassOne | 1,2,3,4 |
--------------------------------
Your query must be something like:
select `ClassName`, group_concat(Section_id) from `class` group by `ClassName`
You can get this with a raw query in laravel or its up to you to find a way to get the same result using query builder ;)

In hiveql, what is the most elegant/performatic way of calculating an average value if some of the data is implicitly not present?

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.

How to compare two columns in the same table

First this is how my table looks like:
tbl
------------------------------------------
| USERID | requestID
|test1#gmail.com | sunsun#gmail.com
|sunsun#gmail.com | test1#gmail.com
|test2#gmail.com | kittyhsk#gmail.com
|sunsun#gmail.com | test2#gmail.com
|test#gmail.com | sunsun#gmail.com
|sunsun#gmail.com | test3#gmail.com
I named my columns wrong but
userIds are the ids that are following requestIds,
and requestIds are the ids that are being followed.
What I want to do is to find the cases that the ids are following each other.
Like for example, I log in with the id sunsun#gmail.com(this is not real address), then I find ids that I'm following and also the ids that follows me, but under the ids that are following each other, I want to print out some text saying that they are following each other. (So under test1 and test2, I should have that text.)
I found this but this does not really apply to my situation as I have to get the results under one logged in ID.
I was trying to do this by myself but I'm all out of ideas. Please help me out. Thanks in advance.
You will have to join the table with itself and compare. So something like
SELECT *
FROM table as t1
JOIN table as t2
ON t1.requestid = t2.useriD and t1.userid = t2.requestid

cassandra query on map in select clause

i am new to cassandra and i am trying to read a row from database which contains values
siteId | country | someMap
1 | US | {a:b, x:z}
2 | PR | {a:b, x:z}
I have also created an index on table using create index on columnfamily(keys(someMap));
but still when i query as select * from table where siteId=1 and someMap contains key 'a'
it returns an entiremap as
1 | US | {a:b, x:z}
Can somebody help me on what should i do to get the value as
1 | US | {a:b}
You can not: even if internally each entry of a Map|List|Set is stored as a column you can only retrieve the whole collection but not part of it. You are not asking cassandra give me the entry of the map containing X, but the row whom map contains X.
HTH,
Carlo

How display two fields sums in the same query in HIve

I have a Hive table with the following fields:
id STRING , x STRING
where x can have values such as 'c'.
I need a query that display number of rows where column x contains a value 'c' and the number of rows where x has values are other than 'c'.
id | count(x='c') | count(x<>'c')
---|--------------|--------------
1 | 3 | 7
I don't know if it's possible.
You can try :
SELECT sum(if(x='c',1,0)), sum(if(x!='c',1,0)) FROM table_name;
This will print two columns. I didn't understand the id field in your sample output.

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