I-SQL HowTo: ukubhala i-loop yesikhashana ngqo embuzweni, noma "Izinyathelo ezintathu zokuqala"

Ngezikhathi ezithile, umsebenzi wokusesha idatha ehlobene usebenzisa isethi yokhiye uyavela. size sithole inani elidingekayo lamarekhodi.

Isibonelo "sempilo yangempela" kakhulu esokuboniswa 20 izinkinga ezindala kakhulu, ohlwini ohlwini lwabasebenzi (isibonelo, ngaphakathi kwesigaba esisodwa). Kumadeshibhodi okuphatha ahlukahlukene anezifinyezo ezimfushane zezindawo zokusebenza, isihloko esifanayo siyadingeka kaningi.

Kulesi sihloko sizobheka ukuqaliswa ku-PostgreSQL kwesisombululo "esingenangqondo" senkinga enjalo, i-algorithm "ehlakaniphile" neyinkimbinkimbi kakhulu. "iluphu" ku-SQL enesimo sokuphuma kudatha etholiwe, okungaba usizo kokubili ekuthuthukisweni okujwayelekile nasekusetshenzisweni kwezinye izimo ezifanayo.

Ake sithathe isethi yedatha yokuhlola isihloko esandulele. Ukuvimbela amarekhodi abonisiwe ukuthi "agxume" ngezikhathi ezithile lapho amanani ahlungiwe ehlangana, nwebisa inkomba yesihloko ngokwengeza ukhiye oyinhloko. Ngasikhathi sinye, lokhu kuzokunikeza ngokushesha okuhlukile futhi kusiqinisekise ukuthi uhlelo lokuhlunga alunambithisi:

CREATE INDEX ON task(owner_id, task_date, id);
-- а старый - удалим
DROP INDEX task_owner_id_task_date_idx;

Njengoba lizwakala, kanjalo kulotshiwe

Okokuqala, ake sidwebe inguqulo elula yesicelo, sidlulise omazisi babadlali amalungu afanayo njengepharamitha yokufaka:

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;

[buka kokuthi explain.tensor.ru]

Okudabukisayo - si-ode amarekhodi angama-20 kuphela, kodwa i-Index Scan isibuyisele wona 960 imigqa, okwabe sekudingeka futhi kuhlungwe... Ake sizame ukufunda kancane.

unnest + ARRAY

Ukucatshangelwa kokuqala okungasisiza yikuthi sidinga 20 kuphela ahlungiwe amarekhodi, bese ufunda nje angabi ngaphezu kwama-20 ahlelwe ngendlela efanayo ngayinye ukhiye. Kuhle, inkomba efanele (i-id_yomnikazi, usuku_lomsebenzi, i-id) sinayo.

Masisebenzise indlela efanayo yokukhipha kanye “nokusabalalisa kumakholomu” irekhodi lethebula elibalulekile, njengaku isihloko sokugcina. Singasebenzisa futhi ukugoqa ohlwini sisebenzisa umsebenzi 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

[buka kokuthi explain.tensor.ru]

O, sekungcono kakhulu! 40% ngokushesha futhi 4.5 izikhathi ngaphansi idatha Kwadingeka ngiyifunde.

Ukwenziwa kwezinto zamarekhodi ethebula nge-CTEAke ngidonse ukunaka kwakho eqinisweni lokuthi kwezinye izimo Umzamo wokusebenza ngokushesha nezinkambu zerekhodi ngemva kokulicinga kumbuzo omncane, ngaphandle “kokulisonga” ku-CTE, kungaholela ekutheni "phindaphindeka" InitPlan ngokulingana nenani lalezi zinkambu ezifanayo:

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

Irekhodi elifanayo "labhekwa phezulu" izikhathi ezingu-4 ... Kuze kube yi-PostgreSQL 11, lokhu kuziphatha kwenzeka njalo, futhi isisombululo "ukusonga" ku-CTE, okuwumkhawulo ophelele we-optimizer kulezi zinguqulo.

I-accumulator ephindaphindayo

Enguqulweni yangaphambili, sisonke sifundile 200 imigqa ngenxa yalokho okudingekayo 20. Hhayi 960, kodwa ngisho ngaphansi - kungenzeka?

Ake sizame ukusebenzisa ulwazi esiludingayo isiyonke 20 amarekhodi. Okusho ukuthi, sizophindaphinda ukufundwa kwedatha kuphela size sifinyelele inani esilidingayo.

Isinyathelo 1: Uhlu Lokuqala

Ngokusobala, uhlu lwethu "oluqondisiwe" lwamarekhodi angu-20 kufanele luqale ngamarekhodi "okuqala" okhiye bethu bobunikazi_bobunikazi. Ngakho-ke, okokuqala sizothola abanjalo “kokuqala kakhulu” kukhiye ngamunye futhi siyengeze ohlwini, siyihlele ngendlela esiyifunayo - (task_date, id).

Isinyathelo sesi-2: Thola okufakiwe "okulandelayo".

Manje uma sithatha ukungena kokuqala ohlwini lwethu bese siqala “sinyathelo” ngokuqhubekayo eduze kwenkomba ukugcina ukhiye ongumnikazi_id, khona-ke wonke amarekhodi atholakele yilawa alandelayo ekukhethweni okuwumphumela. Yebo, kuphela size siwele ukhiye wezinqe ukungena kwesibili ohlwini.

Uma kuvela ukuthi "siwele" irekhodi lesibili, ke okokugcina okufundiwe kufanele kwengezwe ohlwini esikhundleni sokuqala (nge-id efanayo), ngemva kwalokho sihlunga kabusha uhlu futhi.

Okusho ukuthi, sihlala sithola ukuthi uhlu alunakho okungaphezulu kokukodwa kokhiye ngamunye (uma okufakiwe kuphelelwa yisikhathi futhi "singaweli", khona-ke okufakiwe kokuqala okuvela ohlwini kuzomane kunyamalale futhi akukho lutho oluzongezwa. ), futhi njalo ihlelwa ngohlelo olukhuphukayo lokhiye wohlelo lokusebenza (idethi_yomsebenzi, i-id).

Isinyathelo sesi-3: hlunga futhi "unwebe" amarekhodi

Kweminye yemigqa yokukhetha kwethu okuphindaphindayo, amanye amarekhodi rv ziyimpinda - okokuqala sithola njengokuthi "weqa umngcele wokufakwa kwesi-2 ohlwini", bese sifaka esikhundleni soku-1 ohlwini. Ngakho isenzakalo sokuqala sidinga ukuhlungwa.

Umbuzo wokugcina owesabekayo

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

[buka kokuthi explain.tensor.ru]

Ngakho, thina kuhwetshwe u-50% wedatha efundwa u-20% wesikhathi sokwenza. Okusho ukuthi, uma unezizathu zokukholelwa ukuthi ukufunda kungathatha isikhathi eside (isibonelo, idatha ngokuvamile ayikho kunqolobane, futhi kufanele uye kudiski kuyo), ngakho-ke ngale ndlela ungancika kancane ekufundeni. .

Kunoma yikuphi, isikhathi sokwenza sibonakale singcono kunenketho yokuqala "engenangqondo". Kodwa ikuphi kulezi zinketho ezi-3 ongazisebenzisa kukuwe.

Source: www.habr.com