Redshift PostgreSQL Distinct ON 运算符

Question

我有一个数据集，我想对其进行解析以查看多点触控归因。该数据集由响应营销活动的潜在客户及其营销来源组成。

每个潜在客户都可以响应多个营销活动，我想在同一个 table 中获得他们的第一个营销来源和最后一个营销来源。

我想我可以创建两个 tables 并使用来自两者的 select 语句。第一个 table 将尝试创建一个 table，其中包含每个人的最新营销来源（使用电子邮件作为他们的唯一 ID）。

create table temp.multitouch1 as (
select distinct on (email) email, date, market_source as last_source 
from sf.campaignmember
where date >= '1/1/2016' ORDER BY DATE DESC);

然后我会创建一个 table，其中包含删除重复的电子邮件，但这次是针对第一个来源。

create table temp.multitouch2 as (
select distinct on (email) email, date, market_source as first_source 
from sf.campaignmember
where date >= '1/1/2016' ORDER BY DATE ASC);

最后我想简单地 select 电子邮件并将第一个和最后一个市场资源添加到它们自己的列中。

select a.email, a.last_source, b.first_source, a.date 
from temp.multitouch1 a
left join temp.multitouch b on b.email = a.email

由于 distinct on 在 redshift 的 postgresql 版本上不起作用，我希望有人有想法以另一种方式解决这个问题。

编辑 2/22：为了获得更多背景信息，我正在处理他们已经回应的人和活动。每条记录都是 "campaign response"，每个人都可以有多个来源的多个活动响应。我正在尝试制作一个 select 声明，该声明将按人进行重复数据删除，然后分别为他们响应的第一个 campaign/marketing 源和他们响应的最后一个 campaign/marketing 源提供列.

编辑 2/24：理想的输出是 table，有 4 列：电子邮件、last_source、first_source、日期。

对于只有 1 个活动成员记录的人来说，第一个和最后一个源列是相同的，对于拥有超过 1 个活动成员记录的每个人来说都是不同的。

Answer 1

您可以使用旧的左连接分组最大值。

SELECT DISTINCT c1.email, c1.date, c1.market_source
FROM sf.campaignmember c1
  LEFT JOIN sf.campaignmember c2 
    ON c1.email = c2.email AND c1.date > c2.date AND c1.id > c2.id
  LEFT JOIN sf.campaignmember c3
    ON c1.email = c3.email AND c1.date < c3.date AND c1.id > c3.id
WHERE c1.date >= '1/1/2016' AND c2.date >= '1/1/2016'
      AND (c2.email IS NULL OR c3.email IS NULL)

这假设您有一个唯一的 ID 列，如果（日期，电子邮件）是唯一的 ID 则不需要。

Answer 2

我相信你可以像这样在 case 表达式中使用 row_number()：

SELECT
      email
    , MIN(first_source) AS first_source
    , MIN(date) first_date
    , MAX(last_source) AS last_source
    , MAX(date) AS last_date
FROM (
      SELECT
            email
          , date
          , CASE
                  WHEN ROW_NUMBER() OVER (PARTITION BY email ORDER BY date ASC) = 1 THEN market_source
                  ELSE NULL
            END AS first_source
          , CASE
                  WHEN ROW_NUMBER() OVER (PARTITION BY email ORDER BY date DESC) = 1 THEN market_source
                  ELSE NULL
            END AS last_source
      FROM sf.campaignmember
      WHERE date >= '2016-01-01'
      ) s
WHERE first_source IS NOT NULL
      OR last_source IS NOT NULL
GROUP BY
      email

在此处测试：SQL Fiddle

PostgreSQL 9.3 架构设置:

CREATE TABLE campaignmember
    (email varchar(3), date timestamp, market_source varchar(1))
;

INSERT INTO campaignmember
    (email, date, market_source)
VALUES
    ('a@a', '2016-01-02 00:00:00', 'x'),
    ('a@a', '2016-01-03 00:00:00', 'y'),
    ('a@a', '2016-01-04 00:00:00', 'z'),
    ('b@b', '2016-01-02 00:00:00', 'x')
;

查询 1:

SELECT
      email
    , MIN(first_source) AS first_source
    , MIN(date) first_date
    , MAX(last_source) AS last_source
    , MAX(date) AS last_date
FROM (
      SELECT
            email
          , date
          , CASE
                  WHEN ROW_NUMBER() OVER (PARTITION BY email ORDER BY date ASC) = 1 THEN market_source
                  ELSE NULL
            END AS first_source
          , CASE
                  WHEN ROW_NUMBER() OVER (PARTITION BY email ORDER BY date DESC) = 1 THEN market_source
                  ELSE NULL
            END AS last_source
      FROM campaignmember
      WHERE date >= '2016-01-01'
      ) s
WHERE first_source IS NOT NULL
      OR last_source IS NOT NULL
GROUP BY
      email

Results:

| email | first_source |                first_date | last_source |                 last_date |
|-------|--------------|---------------------------|-------------|---------------------------|
|   a@a |            x | January, 02 2016 00:00:00 |           z | January, 04 2016 00:00:00 |
|   b@b |            x | January, 02 2016 00:00:00 |           x | January, 02 2016 00:00:00 |

&请求的小扩展，统计联系点数

SELECT
      email
    , MIN(first_source) AS first_source
    , MIN(date) first_date
    , MAX(last_source) AS last_source
    , MAX(date) AS last_date
    , MAX(numof) AS Numberof_Contacts 
FROM (
      SELECT
            email
          , date
          , CASE
                  WHEN ROW_NUMBER() OVER (PARTITION BY email ORDER BY date ASC) = 1 THEN market_source
                  ELSE NULL
            END AS first_source
          , CASE
                  WHEN ROW_NUMBER() OVER (PARTITION BY email ORDER BY date DESC) = 1 THEN market_source
                  ELSE NULL
            END AS last_source
          , COUNT(*) OVER (PARTITION BY email) as numof
      FROM campaignmember
      WHERE date >= '2016-01-01'
      ) s
WHERE first_source IS NOT NULL
      OR last_source IS NOT NULL
GROUP BY
      email

Redshift PostgreSQL Distinct ON 运算符

Redshift PostgreSQL Distinct ON Operator

postgresql

distinct

distinct-on

amazon-redshift

postgresql-8.0