使用Bigquery进行回测的制表损益表

时间:2019-05-19 17:10:43

标签: google-bigquery back-testing

我有这个Bigquery数据框,其中long_entry或short_entry中的1代表当时对应的多头/空头头寸进入交易。 long_exit或short_exit中的1表示退出交易。我想添加2个新列,一列称为long_pnl,它列出由各个多头交易产生的PnL,另一列称为short_pnl,它对从各个空头交易中产生的PnL进行列表。

此回测在任何时间点最多只能进行1个交易/头寸。

下面是我的数据框。如我们所见,在26/2/2019输入多头交易并在1/3/2019关闭,Pnl将为$ 64.45,而在4/3/2019输入空头交易并在2019/5/3关闭PNL为-119.11美元(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,并为short_pnl提供另一列:

        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

1 个答案:

答案 0 :(得分:1)

以下是用于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

如果将以上内容应用于问题中的样本数据

#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-但以上内容仍然足够我认为可以使用