MySQL 8 - 表达式中的用户变量已弃用（复杂计算）

Question

// 编辑 //

下面是原始问题，但是我的问题真的可以用一个问题来简化...我怎样才能在不使用用户变量的情况下得到下面的输出？！

SELECT ID, @cumu_return:=IF(id = 1, 1, @cumu_return + (@cumu_return * ret)) AS cumulative_return 
FROM (
    SELECT 1 AS ID, 1 AS num, 0.1 AS ret UNION ALL
    SELECT 2 AS ID, 1 AS num, 0.1 AS ret UNION ALL
    SELECT 3 AS ID, 1 AS num, 0.1 AS ret UNION ALL
    SELECT 4 AS ID, 1 AS num, 0.1 AS ret UNION ALL
    SELECT 5 AS ID, 1 AS num, 0.1 AS ret
) t

// 结束编辑 //

我有一个table如下...

CREATE TABLE `daily_return` (
  `id` int(11) NOT NULL,
  `list_id` int(11) NOT NULL,
  `last_updated` datetime NOT NULL DEFAULT '2000-01-01 00:00:00',
  `daily_return` float NOT NULL,
  `last_return` float NOT NULL, KEY (`id`)
)

'daily return' 是 list_id 的价格百分比变化，每天计算，last_return 是 first/starting return 数字。

我需要使用以下逻辑计算累积 return 指标：

'Previous Day Return' + ('Previous Day Return' * 'Daily Return')

因此，我有以下查询...

INSERT INTO cmc_cumulative_return (list_id, last_updated, cumulative_return)
SELECT list_id, last_updated, cumulative_return FROM (
    SELECT id, list_id, last_updated, daily_return, last_return,
        @cumu_return:=IF(id = 1, last_return, @cumu_return + (@cumu_return * daily_return)) AS cumulative_return 
    FROM daily_return c
) t WHERE id <> 1;

当我在程序中运行 this 时，它会抛出警告：

Setting user variables within expressions is deprecated and will be removed in a future release. Please set variables in separate statements instead.

我在Whosebug上看到过一些其他类似的问题，但都是简单的增量计算，可以用ROW_NUMBER() OVER或SUM() OVER代替，但我无法弄清楚如何删除上面查询中的变量。

最初，我使用了 3-4 个变量，但现在我已经减少到一个，请求您帮助删除最后一个。

编辑 2

查询实际数据：（包括 Gordon 的计算）

SELECT id, list_id, last_updated, daily_return, last_return,
    @cumu_return:=IF(id = 1, last_return, @cumu_return + (@cumu_return * daily_return)) AS cumulative_return,
    (
        EXP(SUM(LN(1 + daily_return)) OVER (ORDER BY id)) / (1 + daily_return)
    ) as cumulative_return2
FROM (
    SELECT 1 AS id, 2 AS list_id, '2019-02-20' AS last_updated, 0 AS daily_return, 1.15 AS last_return UNION ALL
    SELECT 2 AS id, 2 AS list_id, '2019-02-21' AS last_updated, 0.0145999858 AS daily_return, 1.15 AS last_return UNION ALL
    SELECT 3 AS id, 2 AS list_id, '2019-02-22' AS last_updated, -0.0503679203 AS daily_return, 1.15 AS last_return UNION ALL
    SELECT 4 AS id, 2 AS list_id, '2019-02-23' AS last_updated, 0.0111594238 AS daily_return, 1.15 AS last_return
) t

Answer 1

您可以使用自然对数和求幂来计算累积积。

INSERT INTO cmc_cumulative_return (list_id, last_updated, cumulative_return)
    SELECT list_id, last_updated, 
           (MAX(CASE WHEN id = 1 THEN last_return END) OVER () *
            EXP(SUM(LN(1 + daily_return)) OVER (ORDER BY id)
               ) / (1 + daily_return)
           ) as cumulative_return
    FROM daily_return c;

如果您真的想排除 id = 1 的位置，则需要额外的子查询级别。

Here 是一个 db<>fiddle.

MySQL 8 - 表达式中的用户变量已弃用（复杂计算）

MySQL 8 - user variables within expressions is deprecated (complex calculation)

mysql

sql

user-variables

deprecation-warning