Lokaci-lokaci, aikin neman bayanai masu alaƙa ta amfani da saitin maɓalli yana tasowa. har sai mun sami adadin adadin da ake buƙata.
Misalin “rayuwa na gaske” shine a nuna 20 mafi tsufa matsaloli, jera a cikin jerin ma'aikata (misali, a cikin rukuni ɗaya). Don “dashboards” gudanarwa daban-daban tare da taƙaitaccen taƙaitaccen wuraren aiki, ana buƙatar irin wannan batu sau da yawa.
A cikin wannan labarin za mu dubi aiwatarwa a cikin PostgreSQL na "naive" maganin irin wannan matsala, "mafi wayo" da kuma hadaddun algorithm. "Madauki" a cikin SQL tare da yanayin fita daga bayanan da aka samo, wanda zai iya zama da amfani duka don ci gaban gaba ɗaya da kuma amfani da su a wasu lokuta masu kama.
Bari mu ɗauki saitin bayanan gwaji daga . Don hana bayanan da aka nuna daga "tsalle" daga lokaci zuwa lokaci lokacin da ƙimar da aka jera suka zo daidai, faɗaɗa jigon jigo ta ƙara maɓalli na farko. A lokaci guda, wannan zai ba shi keɓantacce nan da nan kuma ya ba mu garantin cewa tsarin rarraba ba shi da tabbas:
CREATE INDEX ON task(owner_id, task_date, id);
-- а старый - удалим
DROP INDEX task_owner_id_task_date_idx;Kamar yadda aka ji, haka nan aka rubuta
Da farko, bari mu zana mafi sauƙi sigar buƙatun, wucewa da ID na masu yin :
SELECT
*
FROM
task
WHERE
owner_id = ANY('{1,2,4,8,16,32,64,128,256,512}'::integer[])
ORDER BY
task_date, id
LIMIT 20;
Abin baƙin ciki kaɗan - mun ba da umarnin rikodin 20 kawai, amma Index Scan ya mayar mana da shi 960 layi, wanda kuma dole ne a jera su... Bari mu yi ƙoƙari mu karanta ƙasa.
unnest + ARRAY
Tattaunawa ta farko da za ta taimake mu ita ce idan muna bukata 20 kawai aka jera records, to kawai karanta ba za a jera sama da 20 a cikin tsari iri ɗaya ga kowane ba key. Da kyau, dace index (owner_id, task_date, id) muna da.
Bari mu yi amfani da wannan inji don cirewa da "watsawa cikin ginshiƙai" rikodin tebur mai mahimmanci, kamar a cikin . Hakanan zamu iya amfani da nadawa cikin tsararru ta amfani da aikin ARRAY():
WITH T AS (
SELECT
unnest(ARRAY(
SELECT
t
FROM
task t
WHERE
owner_id = unnest
ORDER BY
task_date, id
LIMIT 20 -- ограничиваем тут...
)) r
FROM
unnest('{1,2,4,8,16,32,64,128,256,512}'::integer[])
)
SELECT
(r).*
FROM
T
ORDER BY
(r).task_date, (r).id
LIMIT 20; -- ... и тут - тоже
Oh, mafi kyau riga! 40% sauri da 4.5 sau ƙasa da bayanai Dole ne in karanta shi.
Kayan aiki na bayanan tebur ta hanyar CTEBari in jawo hankalin ku ga gaskiyar cewa a wasu lokuta Ƙoƙarin yin aiki nan da nan tare da filayen rikodin bayan neman shi a cikin wani yanki, ba tare da "nannade" a cikin CTE ba, zai iya haifar da " ninka" InitPlan daidai da adadin waɗannan filayen guda ɗaya:
SELECT
((
SELECT
t
FROM
task t
WHERE
owner_id = 1
ORDER BY
task_date, id
LIMIT 1
).*);Result (cost=4.77..4.78 rows=1 width=16) (actual time=0.063..0.063 rows=1 loops=1)
Buffers: shared hit=16
InitPlan 1 (returns $0)
-> Limit (cost=0.42..1.19 rows=1 width=48) (actual time=0.031..0.032 rows=1 loops=1)
Buffers: shared hit=4
-> Index Scan using task_owner_id_task_date_id_idx on task t (cost=0.42..387.57 rows=500 width=48) (actual time=0.030..0.030 rows=1 loops=1)
Index Cond: (owner_id = 1)
Buffers: shared hit=4
InitPlan 2 (returns $1)
-> Limit (cost=0.42..1.19 rows=1 width=48) (actual time=0.008..0.009 rows=1 loops=1)
Buffers: shared hit=4
-> Index Scan using task_owner_id_task_date_id_idx on task t_1 (cost=0.42..387.57 rows=500 width=48) (actual time=0.008..0.008 rows=1 loops=1)
Index Cond: (owner_id = 1)
Buffers: shared hit=4
InitPlan 3 (returns $2)
-> Limit (cost=0.42..1.19 rows=1 width=48) (actual time=0.008..0.008 rows=1 loops=1)
Buffers: shared hit=4
-> Index Scan using task_owner_id_task_date_id_idx on task t_2 (cost=0.42..387.57 rows=500 width=48) (actual time=0.008..0.008 rows=1 loops=1)
Index Cond: (owner_id = 1)
Buffers: shared hit=4"
InitPlan 4 (returns $3)
-> Limit (cost=0.42..1.19 rows=1 width=48) (actual time=0.009..0.009 rows=1 loops=1)
Buffers: shared hit=4
-> Index Scan using task_owner_id_task_date_id_idx on task t_3 (cost=0.42..387.57 rows=500 width=48) (actual time=0.009..0.009 rows=1 loops=1)
Index Cond: (owner_id = 1)
Buffers: shared hit=4
Haka rikodin an "duba sama" sau 4 ... Har zuwa PostgreSQL 11, wannan hali yana faruwa akai-akai, kuma mafita shine "nannade" shi a cikin CTE, wanda shine cikakken iyaka ga ingantawa a cikin waɗannan nau'o'in.
Recursive tarawa
A cikin sigar da ta gabata, duka mun karanta 200 layi don kare kanka da ake buƙata 20. Ba 960 ba, amma ko da ƙasa - yana yiwuwa?
Mu yi ƙoƙari mu yi amfani da ilimin da muke bukata jimla 20 rubuce-rubuce. Wato, za mu sake maimaita karatun bayanai kawai har sai mun kai adadin da muke bukata.
Mataki 1: Farawa
Babu shakka, jerin “manufa” na bayananmu 20 yakamata su fara da bayanan “farko” na ɗaya daga cikin maɓallai_id ɗin mu. Saboda haka, da farko za mu sami irin wannan "sosai na farko" ga kowane makullin kuma ƙara shi zuwa lissafin, rarraba shi cikin tsari da muke so - (aiki_date, id).
Mataki 2: Nemo shigarwar "na gaba".
Yanzu idan muka ɗauki farkon shigarwa daga jerinmu kuma mu fara "mataki" gaba tare da index adana maɓallin owner_id, sannan duk bayanan da aka samo su ne ainihin na gaba a cikin sakamakon zaɓin. Tabbas, kawai sai mun haye butt key shigarwa na biyu a cikin jerin.
Idan ya bayyana cewa mun "ketare" rikodin na biyu, to ya kamata a ƙara shigarwar ƙarshe zuwa jerin maimakon na farko (tare da mai shi_id iri ɗaya), bayan haka mun sake tsara lissafin.
Wato, koyaushe muna samun cewa jerin ba su da shigarwa fiye da ɗaya ga kowane maɓalli (idan shigarwar ta ƙare kuma ba mu “ƙetare” ba, shigar da farko daga jerin za ta ɓace kawai kuma ba za a ƙara komai ba. ), kuma su kullum ana jerawa a cikin tsari mai hawa na maɓallin aikace-aikacen (aiki_date, id).
Mataki na 3: tace da kuma "fadada" records
A cikin wasu layuka na zaɓinmu na maimaitawa, wasu bayanai rv ana kwafi - da farko muna samun irin su "Ketare iyakar shiga ta 2 na lissafin", sannan mu musanya shi a matsayin na 1 daga jerin. Don haka abin da ya faru na farko yana buƙatar tace.
Tambayar ƙarshe mai ban tsoro
WITH RECURSIVE T AS (
-- #1 : заносим в список "первые" записи по каждому из ключей набора
WITH wrap AS ( -- "материализуем" record'ы, чтобы обращение к полям не вызывало умножения InitPlan/SubPlan
WITH T AS (
SELECT
(
SELECT
r
FROM
task r
WHERE
owner_id = unnest
ORDER BY
task_date, id
LIMIT 1
) r
FROM
unnest('{1,2,4,8,16,32,64,128,256,512}'::integer[])
)
SELECT
array_agg(r ORDER BY (r).task_date, (r).id) list -- сортируем список в нужном порядке
FROM
T
)
SELECT
list
, list[1] rv
, FALSE not_cross
, 0 size
FROM
wrap
UNION ALL
-- #2 : вычитываем записи 1-го по порядку ключа, пока не перешагнем через запись 2-го
SELECT
CASE
-- если ничего не найдено для ключа 1-й записи
WHEN X._r IS NOT DISTINCT FROM NULL THEN
T.list[2:] -- убираем ее из списка
-- если мы НЕ пересекли прикладной ключ 2-й записи
WHEN X.not_cross THEN
T.list -- просто протягиваем тот же список без модификаций
-- если в списке уже нет 2-й записи
WHEN T.list[2] IS NULL THEN
-- просто возвращаем пустой список
'{}'
-- пересортировываем словарь, убирая 1-ю запись и добавляя последнюю из найденных
ELSE (
SELECT
coalesce(T.list[2] || array_agg(r ORDER BY (r).task_date, (r).id), '{}')
FROM
unnest(T.list[3:] || X._r) r
)
END
, X._r
, X.not_cross
, T.size + X.not_cross::integer
FROM
T
, LATERAL(
WITH wrap AS ( -- "материализуем" record
SELECT
CASE
-- если все-таки "перешагнули" через 2-ю запись
WHEN NOT T.not_cross
-- то нужная запись - первая из спписка
THEN T.list[1]
ELSE ( -- если не пересекли, то ключ остался как в предыдущей записи - отталкиваемся от нее
SELECT
_r
FROM
task _r
WHERE
owner_id = (rv).owner_id AND
(task_date, id) > ((rv).task_date, (rv).id)
ORDER BY
task_date, id
LIMIT 1
)
END _r
)
SELECT
_r
, CASE
-- если 2-й записи уже нет в списке, но мы хоть что-то нашли
WHEN list[2] IS NULL AND _r IS DISTINCT FROM NULL THEN
TRUE
ELSE -- ничего не нашли или "перешагнули"
coalesce(((_r).task_date, (_r).id) < ((list[2]).task_date, (list[2]).id), FALSE)
END not_cross
FROM
wrap
) X
WHERE
T.size < 20 AND -- ограничиваем тут количество
T.list IS DISTINCT FROM '{}' -- или пока список не кончился
)
-- #3 : "разворачиваем" записи - порядок гарантирован по построению
SELECT
(rv).*
FROM
T
WHERE
not_cross; -- берем только "непересекающие" записи
Don haka, mu yayi ciniki 50% na bayanan da aka karanta don 20% na lokacin aiwatarwa. Wato, idan kuna da dalilan da za ku yarda cewa karatun na iya ɗaukar lokaci mai tsawo (misali, yawancin bayanai ba sa cikin cache, kuma dole ne ku je faifai don yin su), to ta wannan hanyar za ku iya dogara da karatu kaɗan. .
A kowane hali, lokacin kisa ya juya ya zama mafi kyau fiye da na farko na "naive". Amma wanne daga cikin waɗannan zaɓuɓɓuka 3 da za ku yi amfani da su ya rage na ku.
source: www.habr.com