如何解决解除投资组合的通用函数
我每天获得以下示例组合 p
的头寸 (signed_qty
):
p:([]date:(2013.07.01+1000#til 200);bb_code:((200#`TSLA),(200#`MSFT),(200#`GOOG),(200#`GS),(200#`AAPL));fill_px:1000?10e; signed_qty:(1000?(-1,1)) * 1000?10000);
我想估计不同平仓策略的每日交易和头寸。我想回测的一些不同策略如下:
- 持有 2 天,然后在接下来的 3 天内交易。
(n=2,m=3)
- 持有 1 天,然后在接下来的 4 天内交易。
(n=1,m=4)
- 在接下来的 4 天内平均交易。
(n=0,m=4)
...等等。
以下是为 n=3
、m=4
执行此操作的详细代码。有人可以建议一种更优雅和通用的方式来获取各种 n
和 m
的输出吗?
/Adding prev_bb_code column to facilitate computation
p:update prev_bb_code: prev bb_code from p;
/No Trading t+1.
p:update qty:prev signed_qty from p;
p:update signed_qty_p1:qty from p where bb_code = prev_bb_code;
p:update Traded_qty_p1:(signed_qty_p1 - qty) from p;
/No Trading t+2.
p:update qty:prev signed_qty_p1 from p;
p:update signed_qty_p2:qty from p where bb_code = prev_bb_code;
p:update Traded_qty_p2:(signed_qty_p2 - qty) from p;
/1st Trade (=1/4th position acquired on t) generated on t+3 with carry forward position = 3/4.
p:update qty:prev signed_qty_p2 from p;
p:update signed_qty_p3:"f"${$[x<0; ceiling((3%4.0)*x); floor((3%4.0)*x)]} each qty from p where bb_code = prev_bb_code;
p:update Traded_qty_p3:(signed_qty_p3 - qty) from p;
/2nd Trade (=1/4th position acquired on t) generated on t+4 with carry forward position = 2/4.
p:update qty:prev qty,prev_qty:prev signed_qty_p3 from p;
p:update signed_qty_p4:"f"${$[y=0n;0n;$[x<0; max((ceiling((2%4.0)*x)),y); min((floor((2%4.0)*x)),y)]]}'[qty; prev_qty] from p where bb_code=prev_bb_code;
p:update Traded_qty_p4:(signed_qty_p4 - prev_qty) from p;
/3rd Trade (=1/4th position acquired on t) generated on t+5 with carry forward position = 1/4.
p:update qty:prev qty,prev_qty:prev signed_qty_p4 from p;
p:update signed_qty_p5:"f"${$[y=0n;0n;$[x<0; max((ceiling((1%4.0)*x)),y); min((floor((1%4.0)*x)),y)]]}'[qty; prev_qty] from p where bb_code=prev_bb_code;
p:update Traded_qty_p5:(signed_qty_p5 - prev_qty) from p;
/4th Trade (=1/4th position acquired on t) generated on t+6 with carry forward position = 0. This abso
p:update qty:prev qty,prev_qty:prev signed_qty_p5 from p;
p:update signed_qty_p6:"f"${$[y=0n;0n;$[x<0; max((ceiling((0%4.0)*x)),y); min((floor((0%4.0)*x)),y)]]}'[qty; prev_qty] from p where bb_code=prev_bb_code;
p:update Traded_qty_p6:(signed_qty_p6 - prev_qty) from p;
/Aggregate Trades and positions.
p:update unwind_qty:((0^Traded_qty_p1) + (0^Traded_qty_p2) + (0^Traded_qty_p3) + (0^Traded_qty_p4) + (0^Traded_qty_p5) + (0^Traded_qty_p6)) from p;
p:update net_position:((0^signed_qty) + (0^signed_qty_p1) + (0^signed_qty_p2) + (0^signed_qty_p3) + (0^signed_qty_p4) + (0^signed_qty_p5) + (0^signed_qty_p6)) from p;
/ Finally only retain the columns of interest.
p: select date,bb_code,fill_px,signed_qty,unwind_qty,net_position from p;
解决方法
这可以通过 xprev
和函数形式来实现。
https://code.kx.com/q/ref/next/#xprev
https://code.kx.com/q/basics/funsql/
编辑:这实际上可以在没有函数形式的情况下实现:
g:{[n;m;x] sums[x]+sums sum each neg (flip xprev[;x] each n + til m)%m}
update net_position:g[3;4;signed_qty] by bb_code from t
原答案:
f:{[t;n;m]
//[table;where;by;cols]
![t;();(enlist `bb_code)!(enlist `bb_code);
(enlist `net_position)!enlist
// cumulative position change
(+;(sums;`signed_qty);
// cumulative unwind position change
// neg to invert the sign to unwind in opposite direction to original position
(sums;(each;sum;(neg;
// this part dynamically builds the number of xprev's required
// n being the start / hold period
// m for the number of unwind periods
(%;(flip;(enlist),(xprev),/:(n + til m),'`signed_qty);m)))))]
}
// No Comments
f:{[t;n;m]
![t;();(enlist `bb_code)!(enlist `bb_code);(enlist `net_position)!enlist (+;(sums;`signed_qty);(sums;(each;sum;(neg;(%;(flip;(enlist),'`signed_qty);m)))))]
};
q)f[p;3;4]
date bb_code fill_px signed_qty net_position
-----------------------------------------------------
2013.07.01 TSLA 4.695818 8159 8159
2013.07.02 TSLA 0.747672 2203 10362
2013.07.03 TSLA 8.014479 -566 9796
2013.07.04 TSLA 3.805866 -831 8965
2013.07.05 TSLA 2.884907 -7792 -866.75
2013.07.06 TSLA 1.303814 9188 5730.75
2013.07.07 TSLA 8.517136 2267 5548.75
2013.07.08 TSLA 5.645172 5352 8659.5
2013.07.09 TSLA 0.04426234 7867 18273
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