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/
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 |
| | M | 6 | 13 | 78 | treatments |
|------------+-------+------------+--------+--------+------------|
#+TBLFM: $5=$3*$4
How could copy the date 2019/09.17 to the bottom of data column?
The link that #manandearth posted in the comments describes how to duplicate (perhaps with slight modifications) the entries in a column. Briefly, pressing S-RET in a cell duplicates its contents from the cell above (if it is not empty) - if the cell is full and the next cell is empty then it duplicates the full cell to the empty cell. If the contents are numeric, then the "duplication" involves a slight modification: it increases the value by 1. The same happens with a date: it increases the date to next day (but the date has to be in a format that Org mode recognizes: either an active date <YYYY-MM-DD> or an inactive data [YYYY-MM-DD]). The increment by default is 1 in these cases, but can be set to something else by setting the variable org-table-copy-increment to a different value. That's the "interactive" case I mention in my comment.
The other way to fill a column in a table is by using a formula. For example here's a formula to fill the first column with a copy of the first entry in the column:
#+TBLFM: #3$1..#>$1 = #2$1
This says: Set all rows from row 3 (#3) to the last row (#>) of column 1 ($1) to the value of the cell in row 2 (#2), column 1 ($1). Note that row 1 is the header. Press C-c C-c on the table formula line above and ... wait, what happened?
|------------+-------+------------+--------+--------+------------|
| Date | Items | Unit Price | Amount | Amount | Categories |
|------------+-------+------------+--------+--------+------------|
| 2019/09/17 | A | 2.64 | 1 | 2.64 | materials |
| 13.196078 | B | 52.67 | 2 | 105.34 | diagnosis |
| 13.196078 | C | 3.08 | 1 | 3.08 | materials |
| 13.196078 | D | 3.85 | 2 | 7.7 | materials |
| 13.196078 | E | 33.66 | 2 | 67.32 | materials |
| 13.196078 | F | 40 | 1 | 40 | treatments |
| 13.196078 | G | 16.5 | 1 | 16.5 | materials |
| 13.196078 | H | 4 | 3 | 12 | treatments |
| 13.196078 | I | 40 | 1 | 40 | bed |
| 13.196078 | M | 6 | 13 | 78 | treatments |
|------------+-------+------------+--------+--------+------------|
#+TBLFM: #3$1..#>$1 = #2$1
It does not quite work in this case for a technical reason: Org mode uses Calc in table formula calculations and Calc looks at 2019/09/17 and says: "Aha, I have to divide 2019 by 9 and then divide the result by 17", and fills the rest of the column with the result of the divisions: 13.196078. You may have meant 2019/09/17 to be a date, but Org mode does not know that: it gives it to Calc which interprets it as an arithmetic expression. The solution here is the same as in the linked answer: make Org mode aware that it's a date by making it either an active date: <2019-09-17> or an inactive date: [2019-09-17]:
|------------------+-------+------------+--------+--------+------------|
| Date | Items | Unit Price | Amount | Amount | Categories |
|------------------+-------+------------+--------+--------+------------|
| [2019-09-17] | A | 2.64 | 1 | 2.64 | materials |
| [2019-09-17 Tue] | B | 52.67 | 2 | 105.34 | diagnosis |
| [2019-09-17 Tue] | C | 3.08 | 1 | 3.08 | materials |
| [2019-09-17 Tue] | D | 3.85 | 2 | 7.7 | materials |
| [2019-09-17 Tue] | E | 33.66 | 2 | 67.32 | materials |
| [2019-09-17 Tue] | F | 40 | 1 | 40 | treatments |
| [2019-09-17 Tue] | G | 16.5 | 1 | 16.5 | materials |
| [2019-09-17 Tue] | H | 4 | 3 | 12 | treatments |
| [2019-09-17 Tue] | I | 40 | 1 | 40 | bed |
| [2019-09-17 Tue] | M | 6 | 13 | 78 | treatments |
|------------------+-------+------------+--------+--------+------------|
#+TBLFM: #3$1..#>$1 = #2$1
This does not do automatic incrementation but if that's what you want, it's easy to accomplish: Calc can do calculations on dates, so we can increment daily by adding to the date in each row the row number minus 2 (e.g. row 3 would get an increment of 3 - 2 = 1, row 4 would get 4 - 2 = 2, etc). To accomplish this, you have to get the row number of the current row: the idiom is ##. Then the formula becomes:
#+TBLFM: #3$1..#>$1 = #2$1 + ## - 2
and the table becomes:
|------------------+-------+------------+--------+--------+------------|
| Date | Items | Unit Price | Amount | Amount | Categories |
|------------------+-------+------------+--------+--------+------------|
| [2019-09-17] | A | 2.64 | 1 | 2.64 | materials |
| [2019-09-18 Wed] | B | 52.67 | 2 | 105.34 | diagnosis |
| [2019-09-19 Thu] | C | 3.08 | 1 | 3.08 | materials |
| [2019-09-20 Fri] | D | 3.85 | 2 | 7.7 | materials |
| [2019-09-21 Sat] | E | 33.66 | 2 | 67.32 | materials |
| [2019-09-22 Sun] | F | 40 | 1 | 40 | treatments |
| [2019-09-23 Mon] | G | 16.5 | 1 | 16.5 | materials |
| [2019-09-24 Tue] | H | 4 | 3 | 12 | treatments |
| [2019-09-25 Wed] | I | 40 | 1 | 40 | bed |
| [2019-09-26 Thu] | M | 6 | 13 | 78 | treatments |
|------------------+-------+------------+--------+--------+------------|
#+TBLFM: #3$1..#>$1 = #2$1+ ## - 2
The various anomalies of the display of dates (do we include the day of the week? do we include the time?) might be worked around using org-time-stamp-custom-formats but that gets us into waters that I have not explored.
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
I looked through similar questions like this one, but they seem to have a definite number of columns. I would like to input a table that I do not know the number of columns.
Question:
How to calculate aggregate functions (e.g. avg() or sum() ) for each row across several columns if number of columns is not known in advance?
I have put the input table panel_stats_rnd csv and a DLL to create it here.
I would like to calculate for each row the rnd_avg_parcelcount as average of all columns c_1_avg_parcelcount, c_2_avg_parcelcount, ... where I can have input tables with any number (say 100) columns of _avg_parcelcount. And for columns rnd_sum_parcelcount I would like to calculate sum() of all columns that start with c_ and end with _sum_parcelcount.
The table looks like this:
SELECT * FROM panel_stats_rnd;
gid | d | dist_from | dist_to | distlabel | rnd_avg_parcelcount | rnd_sum_parcelcount | rnd_avg_callcount | rnd_sum_callcount | rnd_avg_perccalled | called_avg_parcelcount | called_sum_parcelcount | called_avg_callcount | called_sum_callcount | called_avg_perccalled | c_1_avg_parcelcount | c_1_sum_parcelcount | c_1_avg_callcount | c_1_sum_callcount | c_1_avg_perccalled | c_2_avg_parcelcount | c_2_sum_parcelcount | c_2_avg_callcount | c_2_sum_callcount | c_2_avg_perccalled
-----+----+-----------+---------+-----------+---------------------+---------------------+-------------------+-------------------+--------------------+------------------------+------------------------+----------------------+----------------------+-----------------------+---------------------+---------------------+-------------------+-------------------+----------------------+---------------------+---------------------+-------------------+-------------------+----------------------
1 | 0 | 0 | 100 | 0-100 | | | | | | 119045 | 119045 | 119045 | 23 | 0.000193204250493511 | 119045 | 119045 | 119045 | 16 | 0.000134402956865051 | 119045 | 119045 | 119045 | 16 | 0.000134402956865051
2 | 1 | 100 | 200 | 100-200 | | | | | | 163140 | 163140 | 163140 | 22 | 0.000134853500061297 | 163140 | 163140 | 163140 | 17 | 0.000104204977320093 | 163140 | 163140 | 163140 | 18 | 0.000110334681868334
3 | 2 | 200 | 300 | 200-300 | | | | | | 135934 | 135934 | 135934 | 10 | 7.3565112481057e-05 | 135934 | 135934 | 135934 | 18 | 0.000132417202465903 | 135934 | 135934 | 135934 | 15 | 0.000110347668721585
4 | 3 | 300 | 400 | 300-400 | | | | | | 116874 | 116874 | 116874 | 13 | 0.000111230898232284 | 116874 | 116874 | 116874 | 11 | 9.41184523503944e-05 | 116874 | 116874 | 116874 | 18 | 0.000154012012937009
5 | 4 | 400 | 500 | 400-500 | | | | | | 93216 | 93216 | 93216 | 12 | 0.000128733264675592 | 93216 | 93216 | 93216 | 10 | 0.000107277720562993 | 93216 | 93216 | 93216 | 12 | 0.000128733264675592
6 | 5 | 500 | 600 | 500-600 | | | | | | 69992 | 69992 | 69992 | 7 | 0.0001000114298777 | 69992 | 69992 | 69992 | 10 | 0.000142873471253858 | 69992 | 69992 | 69992 | 7 | 0.0001000114298777
7 | 6 | 600 | 700 | 600-700 | | | | | | 50816 | 50816 | 50816 | 10 | 0.000196788413098237 | 50816 | 50816 | 50816 | 6 | 0.000118073047858942 | 50816 | 50816 | 50816 | 0 | 0
8 | 7 | 700 | 800 | 700-800 | | | | | | 34814 | 34814 | 34814 | 0 | 0 | 34814 | 34814 | 34814 | 6 | 0.000172344459125639 | 34814 | 34814 | 34814 | 4 | 0.000114896306083759
9 | 8 | 800 | 900 | 800-900 | | | | | | 23023 | 23023 | 23023 | 1 | 4.34348260435217e-05 | 23023 | 23023 | 23023 | 4 | 0.000173739304174087 | 23023 | 23023 | 23023 | 1 | 4.34348260435217e-05
10 | 9 | 900 | 1000 | 900-1000 | | | | | | 14215 | 14215 | 14215 | 1 | 7.03482237073514e-05 | 14215 | 14215 | 14215 | 1 | 7.03482237073514e-05 | 14215 | 14215 | 14215 | 5 | 0.000351741118536757
11 | 10 | 1000 | 5000 | 1000-5000 | | | | | | 23527 | 23527 | 23527 | 0 | 0 | 23527 | 23527 | 23527 | 0 | 0 | 23527 | 23527 | 23527 | 3 | 0.000127513070089684
(11 rows)
I tried the following for 2 columns (works but I'd rather not write it 5 times for 100 columns, besides the number of columns has to be a parameter):
SELECT d,c_1_avg_parcelcount,c_2_avg_parcelcount,
(SELECT avg(c) FROM (VALUES (c_1_avg_parcelcount) , (c_2_avg_parcelcount) ) T (c)) AS Avg_,
(SELECT sum(c) FROM (VALUES (c_1_avg_parcelcount) , (c_2_avg_parcelcount) ) T (c)) AS sum_
FROM panel_stats_rnd;
I also tried the following but doesn't work.
WITH cols AS (
select value(column_name) from information_schema.columns
where table_name = 'panel_stats_rnd'
AND column_name SIMILAR TO 'c_%avg_parcelcount'
AND column_name != 'called_avg_parcelcount'
)
SELECT *, (SELECT avg(Col) FROM cols V(Col) ) AS col_average
FROM panel_stats_rnd;
I am almost there but something is missing...
select
*,
(select avg(v::numeric)
from json_each_text(row_to_json(panel_stats_rnd.*)) as j(k,v)
where k like 'c\_%\_avg\_parcelcount') as rnd_avg_parcelcount,
(select sum(v::numeric)
from json_each_text(row_to_json(panel_stats_rnd.*)) as j(k,v)
where k like 'c\_%\_sum\_parcelcount') as rnd_sum_parcelcount
from
panel_stats_rnd;
Look at the documentation about functions involved.
There are escapes for underlying characters (\_) because for like operator it is meaning any single character, for example select 'a' like '_'; is true.