0

我有这个 Bigquery 数据框，其中 long_entry 或 short_entry 中的 1 表示当时进入交易，对应的多头/空头头寸。long_exit 或 short_exit 中的 1 表示退出交易。我想有 2 个新列，一个称为 long_pnl 将单个多头交易生成的 PnL 制成表格，另一个称为 short_pnl 将单个空头交易生成的 PnL 制成表格。

此回测在任何时间点最多只能有 1 笔交易/头寸。

下面是我的数据框。如我们所见，多头交易于 2019 年 2 月 26 日进入并于 2019 年 1 月 3 日结束，盈亏为 64.45 美元，而空头交易于 2019 年 4 月 3 日进入并于 2019 年 5 月 3 日结束盈亏为 -$119.11（亏损）。

        date    price       long_entry  long_exit   short_entry short_exit
0   24/2/2019   4124.25           0          0           0              0
1   25/2/2019   4130.67           0          0           0              0
2   26/2/2019   4145.67           1          0           0              0
3   27/2/2019   4180.10           0          0           0              0
4   28/2/2019   4200.05           0          0           0              0
5   1/3/2019    4210.12           0          1           0              0
6   2/3/2019    4198.10           0          0           0              0
7   3/3/2019    4210.34           0          0           0              0
8   4/3/2019    4100.12           0          0           1              0
9   5/3/2019    4219.23           0          0           0              1

我希望有这样的输出，另外一列是 short_pnl：

        date    price       long_entry  long_exit   short_entry short_exit  long_pnl         
0   24/2/2019   4124.25           0          0           0             0    NaN  
1   25/2/2019   4130.67           0          0           0             0    NaN
2   26/2/2019   4145.67           1          0           0             0  64.45
3   27/2/2019   4180.10           0          0           0             0    NaN
4   28/2/2019   4200.05           0          0           0             0    NaN
5   1/3/2019    4210.12           0          1           0             0    NaN
6   2/3/2019    4198.10           0          0           0             0    NaN
7   3/3/2019    4210.34           0          0           0             0    NaN
8   4/3/2019    4100.12           0          0           1             0    NaN
9   5/3/2019    4219.23           0          0           0             1    NaN
4

1 回答 1

1

以下是 BigQuery 标准 SQL 

#standardSQL
WITH temp1 AS (
  SELECT PARSE_DATE('%d/%m/%Y', dt) dt, CAST(price AS numeric) price, long_entry, long_exit, short_entry, short_exit
  FROM `project.dataset.table`
), temp2 AS (
  SELECT dt, price, long_entry, long_exit, short_entry, short_exit,
    SUM(long_entry) OVER(ORDER BY dt) + SUM(long_exit) OVER(ORDER BY dt ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING) long_grp,
    SUM(short_entry) OVER(ORDER BY dt) + SUM(short_exit) OVER(ORDER BY dt ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING) short_grp
  FROM temp1
)
SELECT dt, price, long_entry, long_exit, short_entry, short_exit,
  IF(long_entry = 0, NULL, 
    FIRST_VALUE(price) OVER(PARTITION BY long_grp ORDER BY dt DESC) - 
    LAST_VALUE(price) OVER(PARTITION BY long_grp ORDER BY dt DESC)
  ) long_pnl,
  IF(short_entry = 0, NULL, 
    LAST_VALUE(price) OVER(PARTITION BY short_grp ORDER BY dt DESC) - 
    FIRST_VALUE(price) OVER(PARTITION BY short_grp ORDER BY dt DESC)
  ) short_pnl
FROM temp2

如果将上述应用到您问题中的示例数据 

#standardSQL
WITH `project.dataset.table` AS (
  SELECT '24/2/2019' dt, 4124.25 price, 0 long_entry, 0 long_exit, 0 short_entry, 0 short_exit UNION ALL
  SELECT '25/2/2019', 4130.67, 0, 0, 0, 0 UNION ALL
  SELECT '26/2/2019', 4145.67, 1, 0, 0, 0 UNION ALL
  SELECT '27/2/2019', 4180.10, 0, 0, 0, 0 UNION ALL
  SELECT '28/2/2019', 4200.05, 0, 0, 0, 0 UNION ALL
  SELECT '1/3/2019', 4210.12, 0, 1, 0, 0 UNION ALL
  SELECT '2/3/2019', 4198.10, 0, 0, 0, 0 UNION ALL
  SELECT '3/3/2019', 4210.34, 0, 0, 0, 0 UNION ALL
  SELECT '4/3/2019', 4100.12, 0, 0, 1, 0 UNION ALL
  SELECT '5/3/2019', 4219.23, 0, 0, 0, 1 
), temp1 AS (
  SELECT PARSE_DATE('%d/%m/%Y', dt) dt, CAST(price AS numeric) price, long_entry, long_exit, short_entry, short_exit
  FROM `project.dataset.table`
), temp2 AS (
  SELECT dt, price, long_entry, long_exit, short_entry, short_exit,
    SUM(long_entry) OVER(ORDER BY dt) + SUM(long_exit) OVER(ORDER BY dt ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING) long_grp,
    SUM(short_entry) OVER(ORDER BY dt) + SUM(short_exit) OVER(ORDER BY dt ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING) short_grp
  FROM temp1
)
SELECT dt, price, long_entry, long_exit, short_entry, short_exit,
  IF(long_entry = 0, NULL, 
    FIRST_VALUE(price) OVER(PARTITION BY long_grp ORDER BY dt DESC) - 
    LAST_VALUE(price) OVER(PARTITION BY long_grp ORDER BY dt DESC)
  ) long_pnl,
  IF(short_entry = 0, NULL, 
    LAST_VALUE(price) OVER(PARTITION BY short_grp ORDER BY dt DESC) - 
    FIRST_VALUE(price) OVER(PARTITION BY short_grp ORDER BY dt DESC)
  ) short_pnl
FROM temp2
-- ORDER BY dt

结果将是 

Row dt          price   long_entry  long_exit   short_entry short_exit  long_pnl    short_pnl    
1   2019-02-24  4124.25 0           0           0           0           null        null     
2   2019-02-25  4130.67 0           0           0           0           null        null     
3   2019-02-26  4145.67 1           0           0           0           64.45       null     
4   2019-02-27  4180.1  0           0           0           0           null        null     
5   2019-02-28  4200.05 0           0           0           0           null        null     
6   2019-03-01  4210.12 0           1           0           0           null        null     
7   2019-03-02  4198.1  0           0           0           0           null        null     
8   2019-03-03  4210.34 0           0           0           0           null        null     
9   2019-03-04  4100.12 0           0           1           0           null        -119.11  
10  2019-03-05  4219.23 0           0           0           1           null        null

我觉得应该有一个“更短”的解决方案 - 但我认为上面仍然足够好使用

于 2019-05-19T18:09:29.463 回答