How to translate following python expression to postgesql?
>>> ','.join([30 for x in range(3)])
30,30,30
I have table with colums:
id | entry | type | size
1 10 0 10
2 20 0 10
3 30 1 10
4 30 2 15
I want to query it like this:
SELECT id,
CASE WHEN type = 1
THEN
--For entry 30 and size 10 (300) - get 150,90,60
WHEN type = 2
THEN
--For entry 10 and size 15 (150) - get 30,30,30,30,30
ELSE
entry*size
END
FROM table;
UPD Expected result:
id | prize_pool
1 | 100
2 | 200
3 | 150,90,60
4 | 30,30,30,30,30
UPD2 Equivalent function in python:
def prize_pool(entry,type,size):
prize = entry*size
if type == 0:
return [prize]
if type == 1:
return [prize * x for x in [0.5,0.3,0.2]]
if type == 2:
return [prize/int(size/3) for x in range(int(size/3))]
Assuming your starting table is named plop
SELECT
plop.id,
CASE
WHEN plop.type = 1 THEN (SELECT array_agg(plop.entry * plop.size * val.x) FROM (VALUES (0.5), (0.3), (0.2)) val (x))::int4[]
WHEN plop.type = 2 THEN (SELECT array_agg(3 * plop.entry * x/x ) FROM generate_series(1, plop.size / 3) x)::int4[]
ELSE ARRAY[plop.entry * plop.size]::int4[]
END AS prize_pool
FROM plop
;
That returns:
┌────┬──────────────────┐
│ id │ prize_pool │
├────┼──────────────────┤
│ 1 │ {100} │
│ 2 │ {200} │
│ 3 │ {150,90,60} │
│ 4 │ {90,90,90,90,90} │
└────┴──────────────────┘
Because entry x size / ( size / 3 ) = 3 x entry
Note the x/x
is always equal to 1 and is needed to indicate to Postgres on which set it must aggregate the results as an array.
Hope it helps.
The two functions to look at are array_agg() which allows you to take an aggregate of elements as an array, and string_agg which allows you to append these in a string.
For example:
SELECT type, array_agg(size) from mytable group by type;
would show
type | size
-----------------
0 | {10,10}
1 | {10}
2 | {15}
(3 rows)
I don't know about Python's drivers but it might be one that will parse it into Python structures for you.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.