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我在使用 postgresql 9.5 的“LEFT JOIN LATERAL”函数时遇到了一些困难。

在我的表中，有“ID”、“DATE”、“CODE”三列。一个人 (ID) 有多个行，如下所示。ID 数为 362，总行数约为 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    /  ...      / ...

我需要总结药物 A 在药物 B 的第一天和最后一天之间使用的信息。

在上述情况下，ID（1）[在20020103~20020106之间；用药期B]．

1    /  20020104 / drugA
1    /  20020105 / drugA

为了完成这项工作，我使用“LEFT LATERAL JOIN”编写 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，但这个 postgresql 代码大约需要 2 分钟。

它太慢了。因此，我尝试了另一个使用子查询的 SQL 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);

这段代码并不简单，但非常快（只需 1.6 秒）。

当我比较两个代码的解释时，第二个代码使用哈希连接，但第一个代码没有。

我可以得到一些提示，以更有效地改进具有“左横向连接”功能的第一个代码吗？

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1 回答 1

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为什么不只使用joinand 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，也许更有效的是窗口函数：

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;
于 2016-02-15T15:46:54.570 回答