加速計數不同

Question

我的查詢返回數據不爲空的每個字段的音量。

SELECT COUNT(field1) AS field1, COUNT(field2) AS field2, COUNT(field3) AS field3 
FROM (
    SELECT field1, field2, field3 
    FROM table1, table2 
    WHERE table1.id=table2.idt1 
    ORDER BY table1.id ASC 
    LIMIT 10000 
) AS rq 

table1.id是table1的主鍵，table2.idt1是table2的輔助鍵。 此查詢工作得很好，但如果我需要返回每個字段的不同的卷，這樣

SELECT COUNT(DISTINCT(field1)) AS field1, COUNT(DISTINCT(field2)) AS field2, COUNT(DISTINCT(field3)) AS field3 
FROM (
    SELECT field1, field2, field3 
    FROM table1, table2 
    WHERE table1.id=table2.idt1 
    ORDER BY table1.id ASC 
    LIMIT 10000 
) AS rq 

問題開始...查詢正在做的工作，但表演的當然比沒有DISTINCT子句要慢很多。

Table 1和表2是指標與B樹

CREATE INDEX field1_index ON table1 USING btree (field1) 
CREATE INDEX field2_index ON table1 USING btree (field2) 
CREATE INDEX field3_index ON table2 USING btree (field3) 

我如何可以加快這種重複計數的每個字段？也許有更好的索引？

感謝您的幫助

Answer 1

Postgres沒有優化COUNT(DISTINCT)很好。你有多個這樣的表達式，這使得它更難一點。我打算利用窗口函數和條件聚集建議：

SELECT SUM(CASE WHEN seqnum_1 = 1 THEN 1 ELSE 0 END) as field1, 
     SUM(CASE WHEN seqnum_2 = 1 THEN 1 ELSE 0 END) as field2, 
     SUM(CASE WHEN seqnum_3 = 1 THEN 1 ELSE 0 END) as field3 
FROM (SELECT field1, field2, field3, 
      ROW_NUMBER() OVER (PARTITION BY field1 ORDER BY field1) as seqnum_1, 
      ROW_NUMBER() OVER (PARTITION BY field2 ORDER BY field2) as seqnum_2, 
      ROW_NUMBER() OVER (PARTITION BY field3 ORDER BY field3) as seqnum_3 
     FROM table1 JOIN 
      table2 
      ON table1.id=table2.idt1 
     ORDER BY table1.id ASC 
     LIMIT 10000 
    ) rq 

編輯：

認爲row_number()可能的limit前處理發生在我身上。試試這個版本：

SELECT SUM(CASE WHEN seqnum_1 = 1 THEN 1 ELSE 0 END) as field1, 
     SUM(CASE WHEN seqnum_2 = 1 THEN 1 ELSE 0 END) as field2, 
     SUM(CASE WHEN seqnum_3 = 1 THEN 1 ELSE 0 END) as field3 
FROM (SELECT field1, field2, field3, 
      ROW_NUMBER() OVER (PARTITION BY field1 ORDER BY field1) as seqnum_1, 
      ROW_NUMBER() OVER (PARTITION BY field2 ORDER BY field2) as seqnum_2, 
      ROW_NUMBER() OVER (PARTITION BY field3 ORDER BY field3) as seqnum_3 
     FROM (SELECT field1, field2, field3 
      FROM table1 JOIN 
       table2 
       ON table1.id = table2.idt1 
      ORDER BY table1.id ASC 
      LIMIT 10000 
      ) t 
    ) rq 

Answer 2

我試過類似的東西在一張大桌子裏。 （12百萬行）

沒有DISTINCT需要10秒鐘。

隨着DISTINCT喜歡你的代碼它需要19秒。

Puting子查詢裏面的DISTINCT需要11秒

SELECT COUNT(field1) AS field1, COUNT(field2) AS field2, COUNT(field3) AS field3 
FROM (
    SELECT DISTINCT(field1) AS field1, DISTINCT(field2) AS field2, DISTINCT(field3) AS field3 
    FROM table1, table2 
    WHERE table1.id=table2.idt1 
    ORDER BY table1.id ASC 
    LIMIT 10000 
) AS rq 

其他的事情，如果你只想要過濾NULL數據，可以作出這樣的where子句而不是使用不同英寸