Related
I have an index where some data's has duplicate, all fields are similar except for latitude,longitude and id (field id is not realy ID, just generated row_number() OVER () AS id).
it's example:
mysql> select id,vacancy_id,prof_area_ids,latitude,longitude from jobVacancy;
+------+------------+---------------+----------+-----------+
| id | vacancy_id | prof_area_ids | latitude | longitude |
+------+------------+---------------+----------+-----------+
| 1 | 917 | 11,199,202 | 0.973178 | 0.743566 |
| 2 | 916 | 17,283,288 | 0.973178 | 0.743566 |
| 3 | 915 | 17,288 | 0.973178 | 0.743566 |
| 4 | 914 | 30,482 | 0.973178 | 0.743566 |
| 5 | 919 | 15,243 | 0.825153 | 0.692837 |
| 6 | 919 | 15,243 | 0.825162 | 0.692828 |
| 7 | 918 | 8,154 | 0.825153 | 0.692837 |
| 8 | 918 | 8,154 | 0.825162 | 0.692828 |
| 9 | 920 | 17,283,288 | 0.958914 | 1.282161 |
| 10 | 920 | 17,283,288 | 0.958915 | 1.282215 |
| 11 | 924 | 12,208 | 0.97333 | 0.658246 |
| 12 | 924 | 12,208 | 0.973336 | 0.658237 |
| 13 | 923 | 21,365 | 0.97333 | 0.658246 |
| 14 | 923 | 21,365 | 0.973336 | 0.658237 |
| 15 | 922 | 20,359 | 0.97333 | 0.658246 |
| 16 | 922 | 20,359 | 0.973336 | 0.658237 |
| 17 | 921 | 19,346 | 0.97333 | 0.658246 |
| 18 | 921 | 19,346 | 0.973336 | 0.658237 |
| 19 | 926 | 12,17,208,292 | 0.88396 | 2.389868 |
| 20 | 925 | 12,208 | 0.88396 | 2.389868 |
+------+------------+---------------+----------+-----------+
20 rows in set (0.00 sec)
Now I want to group data by vacancy_id
mysql> select id,vacancy_id,prof_area_ids,latitude,longitude from jobVacancy group by vacancy_id;
+------+------------+---------------+----------+-----------+
| id | vacancy_id | prof_area_ids | latitude | longitude |
+------+------------+---------------+----------+-----------+
| 1 | 917 | 11,199,202 | 0.973178 | 0.743566 |
| 2 | 916 | 17,283,288 | 0.973178 | 0.743566 |
| 3 | 915 | 17,288 | 0.973178 | 0.743566 |
| 4 | 914 | 30,482 | 0.973178 | 0.743566 |
| 5 | 919 | 15,243 | 0.825153 | 0.692837 |
| 7 | 918 | 8,154 | 0.825153 | 0.692837 |
| 9 | 920 | 17,283,288 | 0.958914 | 1.282161 |
| 11 | 924 | 12,208 | 0.97333 | 0.658246 |
| 13 | 923 | 21,365 | 0.97333 | 0.658246 |
| 15 | 922 | 20,359 | 0.97333 | 0.658246 |
| 17 | 921 | 19,346 | 0.97333 | 0.658246 |
| 19 | 926 | 12,17,208,292 | 0.88396 | 2.389868 |
| 20 | 925 | 12,208 | 0.88396 | 2.389868 |
| 21 | 961 | 4,105 | 0.959217 | 1.280721 |
| 23 | 960 | 8,155 | 0.959217 | 1.280721 |
| 25 | 959 | 12,208 | 0.959217 | 1.280721 |
| 27 | 928 | 1,60 | 0.963734 | 1.070297 |
| 29 | 927 | 32,513 | 0.963734 | 1.070297 |
| 31 | 929 | 6,140 | 0.786553 | 0.678649 |
| 33 | 932 | 1,40,46 | 0.824627 | 0.694182 |
+------+------------+---------------+----------+-----------+
20 rows in set (0.00 sec)
Result is awesome! But problem begins when I want to get all grouped data with faceted
mysql> select id,vacancy_id,prof_area_ids,latitude,longitude from jobVacancy where prof_area_ids=199 group by vacancy_id facet prof_area_ids;
+------+------------+-----------------+----------+-----------+
| id | vacancy_id | prof_area_ids | latitude | longitude |
+------+------------+-----------------+----------+-----------+
| 1 | 917 | 11,199,202 | 0.973178 | 0.743566 |
| 191 | 1004 | 11,196,199 | 0.925335 | 2.768874 |
| 313 | 1072 | 1,11,60,197,199 | 0.963968 | 1.070624 |
| 318 | 1136 | 11,196,199 | 0.96071 | 1.448998 |
| 374 | 1097 | 11,199 | 0.785255 | 0.678504 |
+------+------------+-----------------+----------+-----------+
5 rows in set (0.00 sec)
+---------------+----------+
| prof_area_ids | count(*) |
+---------------+----------+
| 202 | 1 |
| 199 | 12 |
| 11 | 12 |
| 196 | 5 |
| 197 | 3 |
| 60 | 3 |
| 1 | 3 |
+---------------+----------+
7 rows in set (0.02 sec)
Faceted result is incorrect. Because in fact data's count where prof_area_ids=199 must be 5 and not 12. So how I can group field for faceted?
Additionaly
I fount here http://sphinxsearch.com/blog/2013/06/21/faceted-search-with-sphinx/ but just written "If you have a MVA facet, you need to use the GROUPBY() function which returns the actual value on which the grouping was made." and without examle.
mysql> select id,vacancy_id,prof_area_ids,latitude,longitude,GROUPBY() as selected,COUNT(*) from jobVacancy where prof_area_ids=199 group by vacancy_id facet prof_area_ids;
+------+------------+-----------------+----------+-----------+----------+----------+
| id | vacancy_id | prof_area_ids | latitude | longitude | selected | count(*) |
+------+------------+-----------------+----------+-----------+----------+----------+
| 1 | 917 | 11,199,202 | 0.973178 | 0.743566 | 917 | 1 |
| 191 | 1004 | 11,196,199 | 0.925335 | 2.768874 | 1004 | 2 |
| 313 | 1072 | 1,11,60,197,199 | 0.963968 | 1.070624 | 1072 | 3 |
| 318 | 1136 | 11,196,199 | 0.96071 | 1.448998 | 1136 | 3 |
| 374 | 1097 | 11,199 | 0.785255 | 0.678504 | 1097 | 3 |
+------+------------+-----------------+----------+-----------+----------+----------+
5 rows in set (0.00 sec)
+---------------+----------+
| prof_area_ids | count(*) |
+---------------+----------+
| 202 | 1 |
| 199 | 12 |
| 11 | 12 |
| 196 | 5 |
| 197 | 3 |
| 60 | 3 |
| 1 | 3 |
+---------------+----------+
7 rows in set (0.02 sec)
Also faceted result is wrong.
Seems, wanting effectively COUNT(DISTINCT vacancy_id) on the FACET rather than the default COUNT(*), but alas it turns out
... FACET prof_area_ids,COUNT(DISTINCT vacancy_id) AS vacancies BY prof_area_ids
doesnt work. The bit before BY only supports attributes, not custom functions.
... will just have to write it out the long way, with full queries...
select id,vacancy_id,prof_area_ids,latitude,longitude from jobVacancy
where prof_area_ids=199 group by vacancy_id;
SELECT GROUPBY() AS prof_area_id, COUNT(DISTINCT vacancy_id) FROM jobVacancy
WHERE prof_area_ids=199 GROUP BY prof_area_id;
Same results, just slightly more verbose. ie rather than using FACET shorthand, write it
out in full, as multiple seperate queries.
Faceted result is incorrect. Because in fact data's count where prof_area_ids=199 must be 5 and not 12. So how I can group field for faceted?
It looks like you misunderstand how FACET works. It seems to me, that you think it takes as a base the main query's result, but it actually just does another grouping. E.g. here:
mysql> select g, t from idx_mva where t = 11 group by g facet t;
+------+----------+
| g | t |
+------+----------+
| 1 | 11,12 |
| 2 | 11,13,15 |
| 3 | 9,11 |
| 5 | 11,12,15 |
+------+----------+
4 rows in set (0.00 sec)
+------+----------+
| t | count(*) |
+------+----------+
| 12 | 2 |
| 11 | 6 |
| 15 | 4 |
| 13 | 1 |
| 9 | 1 |
| 3 | 1 |
+------+----------+
6 rows in set (0.00 sec)
for t=11 you can see that as in your case it's found 3 times in the 1st query's result, but the count for that is 6 in the FACET's query result. This is because it actually occurs 6 times in the index:
mysql> select * from idx_mva where t = 11;
+------+------+----------+
| id | g | t |
+------+------+----------+
| 2 | 1 | 11,12 |
| 3 | 1 | 11,15 |
| 3 | 2 | 11,13,15 |
| 6 | 3 | 9,11 |
| 8 | 5 | 11,12,15 |
| 11 | 2 | 3,11,15 |
+------+------+----------+
6 rows in set, 1 warning (0.00 sec)
and it happens 3 times in the 1st case only because the t's value is returned only once for each of the groups. You can use group_concat() to see more values from the same group:
mysql> select g, group_concat(to_string(t)) from idx_mva where t = 11 group by g facet t;
+------+----------------------------+
| g | group_concat(to_string(t)) |
+------+----------------------------+
| 1 | 11,12,11,15 |
| 2 | 11,13,15,3,11,15 |
| 3 | 9,11 |
| 5 | 11,12,15 |
+------+----------------------------+
4 rows in set (0.00 sec)
+------+----------+
| t | count(*) |
+------+----------+
| 12 | 2 |
| 11 | 6 |
| 15 | 4 |
| 13 | 1 |
| 9 | 1 |
| 3 | 1 |
+------+----------+
6 rows in set (0.00 sec)
If you want to learn more about faceting here's an interactive course about that - https://play.manticoresearch.com/faceting/
I'm trying to generate a serial number based on a few conditions.
My dataset:
+--------+------------+------------+---------+--------+
| Client | Start_Date | End_date | Product | Ser_No |
+--------+------------+------------+---------+--------+
| 44 | 22-01-2018 | 31-12-2018 | A | |
+--------+------------+------------+---------+--------+
| 44 | 24-02-2018 | 01-01-2019 | B | |
+--------+------------+------------+---------+--------+
| 44 | 12-03-2018 | 01-01-2019 | C | |
+--------+------------+------------+---------+--------+
| 100 | 24-01-2018 | 30-11-2018 | A | |
+--------+------------+------------+---------+--------+
| 100 | 26-01-2018 | 15-12-2018 | D | |
+--------+------------+------------+---------+--------+
| 100 | 26-01-2018 | 01-02-2019 | E | |
+--------+------------+------------+---------+--------+
| 100 | 01-03-2018 | 31-01-2019 | F | |
+--------+------------+------------+---------+--------+
What I did to configure my serial number:
RANK() OVER(PARTITION BY Client ORDER BY Client, Start_date ASC)
So now it generates a serial number for my which looks like this:
+--------+------------+------------+---------+--------+
| Client | Start_Date | End_date | Product | Ser_No |
+--------+------------+------------+---------+--------+
| 44 | 22-01-2018 | 31-12-2018 | A | 1 |
+--------+------------+------------+---------+--------+
| 44 | 24-02-2018 | 01-01-2019 | B | 2 |
+--------+------------+------------+---------+--------+
| 44 | 12-03-2018 | 01-01-2019 | C | 3 |
+--------+------------+------------+---------+--------+
| 100 | 24-01-2018 | 30-11-2018 | A | 1 |
+--------+------------+------------+---------+--------+
| 100 | 26-01-2018 | 15-12-2018 | D | 2 |
+--------+------------+------------+---------+--------+
| 100 | 26-01-2018 | 01-02-2019 | E | 2 |
+--------+------------+------------+---------+--------+
| 100 | 01-03-2018 | 31-01-2019 | F | 4 |
+--------+------------+------------+---------+--------+
What goes wrong for my analysis is the last line, it generates the serial number. What it has to be is 3.
Can anayone help me to generate it in this order?
Thanks in advance!
Extra
In addition to my question from yesterday, there is something extra that I need to do. Because the Ser_No has to be the same when my Start_Date is the same, but the Ser_No has also be the same when my folowing records is the same product (also when it has a different Start_Date)
So what I I expect and what I get right now:
+--------+------------+------------+---------+--------+------------+
| Client | Start_Date | End_date | Product | Ser_No | Ser_No New |
+--------+------------+------------+---------+--------+------------+
| 44 | 22-01-2018 | 31-12-2018 | A | 1 | 1 |
+--------+------------+------------+---------+--------+------------+
| 44 | 24-02-2018 | 01-01-2019 | B | 2 | 2 |
+--------+------------+------------+---------+--------+------------+
| 44 | 12-03-2018 | 01-01-2019 | C | 2 | 2 |
+--------+------------+------------+---------+--------+------------+
| 100 | 24-01-2018 | 30-11-2018 | A | 1 | 1 |
+--------+------------+------------+---------+--------+------------+
| 100 | 26-01-2018 | 15-12-2018 | D | 2 | 2 |
+--------+------------+------------+---------+--------+------------+
| 100 | 26-01-2018 | 01-02-2019 | E | 2 | 2 |
+--------+------------+------------+---------+--------+------------+
| 100 | 01-03-2018 | 31-01-2019 | F | 3 | 3 |
+--------+------------+------------+---------+--------+------------+
| 100 | 11-04-2018 | 31-03-2019 | F | 4 | 3 |
+--------+------------+------------+---------+--------+------------+
| 100 | 20-04-2018 | 31-01-2019 | G | 5 | 4 |
+--------+------------+------------+---------+--------+------------+
| 100 | 21-04-2018 | 31-01-2019 | A | 6 | 5 |
+--------+------------+------------+---------+--------+------------+
| 100 | 21-04-2018 | 31-01-2019 | B | 6 | 5 |
+--------+------------+------------+---------+--------+------------+
| 100 | 01-05-2018 | 31-01-2019 | B | 7 | 5 |
+--------+------------+------------+---------+--------+------------+
Any idea on how to achieve this, because I won't get it
You need to use DENSE_RANK instead:
This function returns the rank of each row within a result set partition, with no gaps in the ranking values.
DENSE_RANK() OVER(PARTITION BY Client ORDER BY Start_date) AS Ser_no
Additionaly the Client in ORDER BY has no effect because it has the same value per partition.
I have the following table:
SELECT * FROM trips_motion_xtics
+------------+---------+-------------+------------+-------------+------------+-----+------------------+------------------+-------+--------------+-------+-------------+
| session_id | trip_id | lat_start | lat_end | lon_start | lon_end | alt | distance | segments_length | speed | acceleration | track | travel_mode |
+------------+---------+-------------+------------+-------------+------------+-----+------------------+------------------+-------+--------------+-------+-------------+
| 652 | 303633 | 41.1523521 | 41.1524966 | -8.6097233 | -8.6096833 | 0 | 42.7424443438547 | 28.0353622436523 | 0 | 74.208 | 0 | foot |
| 652 | 303633 | 41.1523521 | 41.1524966 | -8.6097233 | -8.6096833 | 0 | 42.7424443438547 | 28.0353622436523 | 0 | 74.154 | 0 | foot |
| 652 | 303633 | 41.1523521 | 41.1524966 | -8.6097233 | -8.6096833 | 0 | 42.7424443438547 | 28.0353622436523 | 0 | 68.226 | 0 | foot |
| 656 | 303637 | 41.14454009 | 41.1631127 | -8.56292593 | -8.5870161 | 0 | 5921.07030809987 | 2785.6088546142 | 0 | 99.028 | 0 | car |
| 656 | 303637 | 41.14454009 | 41.1631127 | -8.56292593 | -8.5870161 | 0 | 5921.07030809987 | 2785.6088546142 | 0 | 109.992 | 0 | car |
+------------+---------+-------------+------------+-------------+------------+-----+------------------+------------------+-------+--------------+-------+-------------+
Now would like to compute the average value for columns alt, distance, speed ... for unique value of session_id, trip_id, lat_start,...
Query:
SELECT DISTINCT(session_id, trip_id, lat_start, lat_end, lon_start, lon_end, travel_mode), AVG(alt) AS avg_alt, AVG(distance) AS avg_disntance, AVG(speed) AS avg_speed, AVG(acceleration) AS avg_acc FROM akil.trips_motion_xtics;
ERROR: column "trips_motion_xtics.session_id" must appear in the GROUP BY clause or be used in an aggregate function
LINE 1: SELECT DISTINCT(session_id, trip_id, lat_start, lat_end, lon...
Required result:
+------------+---------+-------------+------------+-------------+------------+-----+------------------+------------------+-------+--------------+-------+-------------+
| session_id | trip_id | lat_start | lat_end | lon_start | lon_end | alt | distance | segments_length | speed | acceleration | track | travel_mode |
+------------+---------+-------------+------------+-------------+------------+-----+------------------+------------------+-------+--------------+-------+-------------+
| 652 | 303633 | 41.1523521 | 41.1524966 | -8.6097233 | -8.6096833 | 0 | 42.7424443438547 | 28.0353622436523 | 0 | 72.196 | 0 | foot |
| 656 | 303637 | 41.14454009 | 41.1631127 | -8.56292593 | -8.5870161 | 0 | 5921.07030809987 | 2785.6088546142 | 0 | 104.51 | 0 | car |
+------------+---------+-------------+------------+-------------+------------+-----+------------------+------------------+-------+--------------+-------+-------------+
You want aggregation. You will get a unique record for each combination of the column listed in the GROUP BY clause, and you can apply aggregate functions (such as AVG()) on other columns:
SELECT
session_id,
trip_id,
lat_start,
lat_end,
lon_start,
lon_end,
travel_mode,
AVG(alt) AS avg_alt,
AVG(distance) AS avg_disntance,
AVG(speed) AS avg_speed,
AVG(acceleration) AS avg_acc
FROM akil.trips_motion_xtics
GROUP BY
session_id,
trip_id,
lat_start,
lat_end,
lon_start,
lon_end,
travel_mode
Suppose such a spreadsheet in org table
|------------+-------+------------+--------+--------+------------|
| Date | Items | Unit Price | Amount | Amount | Categories |
|------------+-------+------------+--------+--------+------------|
| 2019/09/17 | A | 2.64 | 1 | 2.64 | materials |
| | B | 52.67 | 2 | 105.34 | diagnosis |
| | C | 3.08 | 1 | 3.08 | materials |
| | D | 3.85 | 2 | 7.7 | materials |
| | E | 33.66 | 2 | 67.32 | materials |
| | F | 40 | 1 | 40 | treatments |
| | G | 16.5 | 1 | 16.5 | materials |
| | H | 4 | 3 | 12 | treatments |
| | I | 40 | 1 | 40 | bed |
| x | M | 6 | 13 | 78 | treatments |
|------------+-------+------------+--------+--------+------------|
#+TBLFM: $5=$3*$4
I want to sum up the material fees.
Is it possible to calculate it by grouping like vsum(where Categories == materials)?
One way to do this with an elisp expression will be:
|------------+-------+------------+--------+--------+------------|
| Date | Items | Unit Price | Amount | Amount | Categories |
|------------+-------+------------+--------+--------+------------|
| 2019/09/17 | A | 2.64 | 1 | 2.64 | materials |
| | B | 52.67 | 2 | 105.34 | diagnosis |
| | C | 3.08 | 1 | 3.08 | materials |
| | D | 3.85 | 2 | 7.7 | materials |
| | E | 33.66 | 2 | 67.32 | materials |
| | F | 40 | 1 | 40 | treatments |
| | G | 16.5 | 1 | 16.5 | materials |
| | H | 4 | 3 | 12 | treatments |
| | I | 40 | 1 | 40 | bed |
| x | M | 6 | 13 | 78 | treatments |
|------------+-------+------------+--------+--------+------------|
| TOTAL: | | | | 97.24 | |
|------------+-------+------------+--------+--------+------------|
#+TBLFM: $5=$3*$4
#+TBLFM: #12$5='(apply #'+ (cl-mapcar (lambda (num category) (if (eq category 'materials) num 0)) '(#II$5..#III$5) '(#II$6..#III$6)));L
cl-mapcar applies + to cell #12$5 by comparing the list which is column 6 to symbol'materials.
This solution and a `calc solution in emacsSE
My question is about forming Postgres SQL query for below use case
Approach#1
I have a table like below where I generate the same uuid across different types(a,b,c,d) like mapping different types.
+----+------+-------------+
| id | type | master_guid |
+----+------+-------------+
| 1 | a | uuid-1 |
| 2 | a | uuid-2 |
| 3 | a | uuid-3 |
| 4 | a | uuid-4 |
| 5 | a | uuid-5 |
| 6 | b | uuid-1 |
| 7 | b | uuid-2 |
| 8 | b | uuid-3 |
| 9 | b | uuid-6 |
| 10 | c | uuid-1 |
| 11 | c | uuid-2 |
| 12 | c | uuid-3 |
| 13 | c | uuid-6 |
| 14 | c | uuid-7 |
| 15 | d | uuid-6 |
| 16 | d | uuid-2 |
+----+------+-------------+
Approach#2
I have a created two tables for id to type and then id to master_guid, like below
table1:
+----+------+
| id | type |
+----+------+
| 1 | a |
| 2 | a |
| 3 | a |
| 4 | a |
| 5 | a |
| 6 | b |
| 7 | b |
| 8 | b |
| 9 | b |
| 10 | c |
| 11 | c |
| 12 | c |
| 13 | c |
| 14 | c |
| 15 | d |
| 16 | d |
+----+------+
table2
+----+-------------+
| id | master_guid |
+----+-------------+
| 1 | uuid-1 |
| 2 | uuid-2 |
| 3 | uuid-3 |
| 4 | uuid-4 |
| 5 | uuid-5 |
| 6 | uuid-1 |
| 7 | uuid-2 |
| 8 | uuid-3 |
| 9 | uuid-6 |
| 10 | uuid-1 |
| 11 | uuid-2 |
| 12 | uuid-3 |
| 13 | uuid-6 |
| 14 | uuid-7 |
| 15 | uuid-6 |
| 16 | uuid-2 |
+----+-------------+
I want to get output like below with both approaches:
+----+------+--------+------------+
| id | type | uuid | mapped_ids |
+----+------+--------+------------+
| 1 | a | uuid-1 | [6,10] |
| 2 | a | uuid-2 | [7,11] |
| 3 | a | uuid-3 | [8,12] |
| 4 | a | uuid-4 | null |
| 5 | a | uuid-5 | null |
+----+------+--------+------------+
I have tried self-joins with array_agg on ids and grouping based on uuid but not able to get the desired output.
Use below query to populate data:
Approach#1
insert into table1 values
(1,'a','uuid-1'),
(2,'a','uuid-2'),
(3,'a','uuid-3'),
(4,'a','uuid-4'),
(5,'a','uuid-5'),
(6,'b','uuid-1'),
(7,'b','uuid-2'),
(8,'b','uuid-3'),
(9,'b','uuid-6'),
(10,'c','uuid-1'),
(11,'c','uuid-2'),
(12,'c','uuid-3'),
(13,'c','uuid-6'),
(14,'c','uuid-7'),
(15,'d','uuid-6'),
(16,'d','uuid-2')
Approach#2
insert into table1 values
(1,'a'),
(2,'a'),
(3,'a'),
(4,'a'),
(5,'a'),
(6,'b'),
(7,'b'),
(8,'b'),
(9,'b'),
(10,'c'),
(11,'c'),
(12,'c'),
(13,'c'),
(14,'c'),
(15,'d'),
(16,'d')
insert into table2 values
(1,'uuid-1'),
(2,'uuid-2'),
(3,'uuid-3'),
(4,'uuid-4'),
(5,'uuid-5'),
(6,'uuid-1'),
(7,'uuid-2'),
(8,'uuid-3'),
(9,'uuid-6'),
(10,'uuid-1'),
(11,'uuid-2'),
(12,'uuid-3'),
(13,'uuid-6'),
(14,'uuid-7'),
(15,'uuid-6'),
(16,'uuid-2')
demo: db<>fiddle
Using window function ARRAY_AGG allows you to aggregate your ids per groups (in your case the groups are your uuids)
SELECT
id, type, master_guid as uuid,
array_agg(id) OVER (PARTITION BY master_guid) as mapped_ids
FROM table1
ORDER BY id
Result:
| id | type | uuid | mapped_ids |
|----|------|--------|------------|
| 1 | a | uuid-1 | 10,6,1 |
| 2 | a | uuid-2 | 16,2,7,11 |
| 3 | a | uuid-3 | 8,3,12 |
| 4 | a | uuid-4 | 4 |
| 5 | a | uuid-5 | 5 |
| 6 | b | uuid-1 | 10,6,1 |
| 7 | b | uuid-2 | 16,2,7,11 |
| 8 | b | uuid-3 | 8,3,12 |
| 9 | b | uuid-6 | 15,13,9 |
| 10 | c | uuid-1 | 10,6,1 |
| 11 | c | uuid-2 | 16,2,7,11 |
| 12 | c | uuid-3 | 8,3,12 |
| 13 | c | uuid-6 | 15,13,9 |
| 14 | c | uuid-7 | 14 |
| 15 | d | uuid-6 | 15,13,9 |
| 16 | d | uuid-2 | 16,2,7,11 |
These arrays currently contain also the id of the current row (mapped_ids of id = 1 contains the 1). This can be corrected by remove this element with array_remove:
SELECT
id, type, master_guid as uuid,
array_remove(array_agg(id) OVER (PARTITION BY master_guid), id) as mapped_ids
FROM table1
ORDER BY id
Result:
| id | type | uuid | mapped_ids |
|----|------|--------|------------|
| 1 | a | uuid-1 | 10,6 |
| 2 | a | uuid-2 | 16,7,11 |
| 3 | a | uuid-3 | 8,12 |
| 4 | a | uuid-4 | |
| 5 | a | uuid-5 | |
| 6 | b | uuid-1 | 10,1 |
| 7 | b | uuid-2 | 16,2,11 |
| 8 | b | uuid-3 | 3,12 |
| 9 | b | uuid-6 | 15,13 |
| 10 | c | uuid-1 | 6,1 |
| 11 | c | uuid-2 | 16,2,7 |
| 12 | c | uuid-3 | 8,3 |
| 13 | c | uuid-6 | 15,9 |
| 14 | c | uuid-7 | |
| 15 | d | uuid-6 | 13,9 |
| 16 | d | uuid-2 | 2,7,11 |
Now for example id=4 contains an empty array instead of a NULL value. This can be achieved by using the NULLIF function. This gives NULL if both parameters are equal, else it gives out the first parameter.
SELECT
id, type, master_guid as uuid,
NULLIF(
array_remove(array_agg(id) OVER (PARTITION BY master_guid), id),
'{}'::int[]
) as mapped_ids
FROM table1
ORDER BY id
Result:
| id | type | uuid | mapped_ids |
|----|------|--------|------------|
| 1 | a | uuid-1 | 10,6 |
| 2 | a | uuid-2 | 16,7,11 |
| 3 | a | uuid-3 | 8,12 |
| 4 | a | uuid-4 | (null) |
| 5 | a | uuid-5 | (null) |
| 6 | b | uuid-1 | 10,1 |
| 7 | b | uuid-2 | 16,2,11 |
| 8 | b | uuid-3 | 3,12 |
| 9 | b | uuid-6 | 15,13 |
| 10 | c | uuid-1 | 6,1 |
| 11 | c | uuid-2 | 16,2,7 |
| 12 | c | uuid-3 | 8,3 |
| 13 | c | uuid-6 | 15,9 |
| 14 | c | uuid-7 | (null) |
| 15 | d | uuid-6 | 13,9 |
| 16 | d | uuid-2 | 2,7,11 |
Try this:
select
t1.id, t1.type, t1.master_guid, array_agg (distinct t2.id)
from
table1 t1
left join table1 t2 on
t1.master_guid = t2.master_guid and
t1.id != t2.id
group by
t1.id, t1.type, t1.master_guid
I don't come up with exactly the same results you listed, but I thought it was close enought that maybe there was a mistaken expectation on your side or only a small error on mine... either way, a potential starting point.
-- EDIT --
For approach #2, I think you just need to add an inner join to Table2 to get the GUID:
select
t1.id, t1.type, t2.master_guid,
array_agg (t2a.id)
from
table1 t1
join table2 t2 on t1.id = t2.id
left join table2 t2a on
t2.master_guid = t2a.master_guid and
t2a.id != t1.id
where
t1.type = 'a'
group by
t1.id, t1.type, t2.master_guid