Postgresql

Question

我在 postgresql 9.5 中使用 'LEFT JOIN LATERAL' 函数时遇到一些困难。

在我的table中，有'ID'、'DATE'、'CODE'三列。一个人（ID）有多个行，如下所示。 ID个数362，总行数约2500000

ID   /  DATE     / CODE
1    /  20020101 / drugA
1    /  20020102 / drugA
1    /  20020103 / drugB
1    /  20020104 / drugA
1    /  20020105 / drugA
1    /  20020106 / drugB
1    /  20020107 / drugA
2    /  ...      / ...

我需要汇总药物A从第一天到最后一天使用药物B的信息。

在上面的例子中，ID(1)只需要保留两行[介于20020103 ~ 20020106之间；药物B的时期。

1    /  20020104 / drugA
1    /  20020105 / drugA

为了接受这份工作，我使用 'LEFT LATERAL JOIN' 编写了 SQL 代码，如下所示。

SELECT * FROM (SELECT ID, min(DATE) as start_date, max(DATE) as end_date from MAIN_TABLE WHERE CODE = 'drugA' GROUP BY ID) AA
LEFT JOIN LATERAL (SELECT ID, COUNT(ID) as no_tx, min(DATE) as fday_tx, max(DATE) lday_tx from MAIN_TABLE WHERE CODE = 'drugB' AND DATE > AA.start_date AND DATE < AA.end_date GROUP BY ID) as BB USING(ID);

只有362个人ID，但是这个postgresql代码大概需要2分钟。

太慢了。因此，我尝试了另一个 SQL 使用子查询的代码。

SELECT * FROM (SELECT ID, min(DATE) as start_date, max(DATE) as end_date from MAIN_TABLE WHERE CODE ='drugA' GROUP BY ID) AA
LEFT JOIN (
       SELECT ID, COUNT(ID) as no_tx, min(DATE) as fday_tx, max(DATE) lday_tx FROM (SELECT ID, DATE, CODE FROM MAIN_TABLE) BB
            LEFT JOIN (SELECT ID, min(DATE) as start_date, max(DATE) as end_date from MAIN_TABLE WHERE CODE ='drugA' GROUP BY ID) CC USING (ID)
            WHERE CODE = 'drugB' and DATE > start_date and DATE < end_date GROUP BY ID
            ) DD USING (ID);

这段代码不简单但非常快（仅需 1.6 秒）。

当我比较两个代码的解释时，第二个代码使用散列连接，但第一个代码没有。

我能得到一些提示来更有效地改进第一个代码 'LEFT LATERAL JOIN' 函数吗？

Answer 1

为什么不直接使用 join 和 group by？

SELECT AA.ID, COUNT(B.ID) as no_tx, min(B.DATE) as fday_tx, max(B.DATE) as lday_tx,
       AA.start_date, AA.end_date
FROM (SELECT ID, min(DATE) as start_date, max(DATE) as end_date 
      FROM MAIN_TABLE
      WHERE CODE = 'drugA'
      GROUP BY ID
     ) AA LEFT JOIN
     MAIN_TABLE b
     ON b.CODE = 'drugB' AND b.DATE > AA.start_date AND b.DATE < AA.end_date
GROUP BY AA.ID,  AA.start_date, AA.end_date;

或者，也许更有效，window 函数：

SELECT ID, SUM(CASE WHEN code = 'drugB' THEN 1 ELSE 0 END) as no_tx,
       MIN(CASE WHEN code = 'drugB' THEN DATE END) as fday_tx,
       MIN(CASE WHEN code = 'drugB' THEN DATE END) as lday_tx,
       start_date, end_date
FROM (SELECT t.*,
             MIN(CASE WHEN code = 'drugA' THEN date END) as start_date,
             MAX(CASE WHEN code = 'drugB' THEN date END) as end_date
      FROM MAIN_TABLE t
     ) t
WHERE code in ('drugA', 'drugB') AND
      date between start_date and end_date
GROUP BY t.id;

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