KDB/Q关于解析字符串和连接的问题

发布于 2025-02-12 12:14:45 字数 3313 浏览 1 评论 0 原文

InstressID	OrderID	策略	填充
NG1	JHBUBUY-CLIENTNAME1-2022-07-01	VWAP	5000
NG2 NG2	NB8YVCE-CLIENTNAME2-2022-2022-07-01	POV	300
NG1 NG1	CEBCIUBE-CLIENTNAME 3-2022-2022-07-01	TWAP	1000

twap 1000与榜单相比。我是KDB/Q的初学者，所以请忍受我。任何指导都会有所帮助。

1.）我试图通过“ - ”来解析秩序，并根据我在Orderid

instransId	strundId	策略	上解析的内容创建一个新列“客户名称”，填充	客户名称
NG1	JHBUBUY-CLIENTNAME1-2022-07-01	VWAP	5000	CLUINTNAME 1 NG2 NB8YVCE-CLIENTNAME NAMENAME NAMENAME NAMENAME NAMENAME NAMENAME
VWAP	NAMENAME NAMENAME -2022-07-01	POV	300	ClientName124
NG1	CEBCIUBE-CLIENTNAMEABC-2022-07-01	TWAP	1000	CLIENTNAMEABC

2.）通过解析订单创建一个新列，然后将其与一个称为instanceID的单独列将其串联。它创建了一个instance_orderID

instance	instordID	策略	填充	instensId_orderID
ng1	jhbubuy-clientname1-2022-07-01	5000	ng1	-jhbubuy-
ng2	2022-222-22-22-22-22-22-2-22-22-22-22-22-22-22-22-22	vwap	因此，	NG2-NB8YVCE-CLIENTNAME124-2022-07-01
NG1	CEBCIUBE-CLIENTNAMEABC-2022-07-07-01	TWAP	1000	NG1-CLIENTNAMEABC-2022-2022-2022-07-07-01

原文

InstanceID	OrderID	Strategy	Fills
NG1	jhbubuy-ClientName1-2022-07-01	VWAP	5000
NG2	nb8yvce-ClientName2-2022-07-01	POV	300
NG1	cebciube-ClientName3-2022-07-01	TWAP	1000

Hi guys have the table above as an example. I am beginner in KDB/Q so please bear with me. Any guidance would help.

1.) I am trying to parse OrderID by "-" and create a new column "ClientName" based on what I parsed on OrderID

InstanceID	OrderID	Strategy	Fills	ClientName
NG1	jhbubuy-ClientName1-2022-07-01	VWAP	5000	ClientName1
NG2	nb8yvce-ClientName124-2022-07-01	POV	300	ClientName124
NG1	cebciube-ClientNameABC-2022-07-01	TWAP	1000	ClientNameABC

2.) Create a new column by parsing OrderID then concatenate it with a separate column called InstanceID. So it creates a InstanceID_OrderID

InstanceID	OrderID	Strategy	Fills	InstanceID_OrderID
NG1	jhbubuy-ClientName1-2022-07-01	VWAP	5000	NG1-jhbubuy-2022-07-01
NG2	nb8yvce-ClientName124-2022-07-01	POV	300	NG2-nb8yvce-ClientName124-2022-07-01
NG1	cebciube-ClientNameABC-2022-07-01	TWAP	1000	NG1-ClientNameABC-2022-07-01

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别想她 2025-02-19 12:14:45

要获得所需的表，您应该：

使用 要通过“ - ”拆分OrderID，
获取每个子名称的每个子列表的第二个元素
从每个sublist中删除第二个元素，请使用 sv instanceid_orderid的操作员。

如果表具有以下结构：

t: flip`InstanceID`OrderID`Strategy`Fills!
  (`NG1`NG2`NG3;
  ("hbubuy-ClientName1-2022-07-01";
     "nb8yvce-ClientName2-2022-07-01";
     "cebciube-ClientName3-2022-07-01");
  `VWAP`POV`TWAP;
  5000 300 1000);

下一个查询执行作业：

update
  ClientName: {("-" vs x) 1} each OrderID,
  InstanceID_OrderID: {x: "-" vs x; "-" sv enlist[y],(1#x),2_x}'[OrderID;string InstanceID]
from t

匿名函数

{（“ - ” vs x）1} splits OrderID by by “ - ” 并选择索引1
{x：“ - ” vs x; “ - ” sv enterist [y]，（1＃x），2_x} splits orderiD by “ - ” ，选择0和2+索引（那里） Q）中的索引操作不删除，预先启动InstanceID并将所有内容连接到“ - ”字符串中

To get desired table you should:

Use vs operator to split OrderID by "-"
Get second element of each sublist for ClientName
Remove second element from each sublist, prepend InstanceID and join new list using sv operator for InstanceID_OrderID.

If the table has following structure:

t: flip`InstanceID`OrderID`Strategy`Fills!
  (`NG1`NG2`NG3;
  ("hbubuy-ClientName1-2022-07-01";
     "nb8yvce-ClientName2-2022-07-01";
     "cebciube-ClientName3-2022-07-01");
  `VWAP`POV`TWAP;
  5000 300 1000);

The next query does the job:

update
  ClientName: {("-" vs x) 1} each OrderID,
  InstanceID_OrderID: {x: "-" vs x; "-" sv enlist[y],(1#x),2_x}'[OrderID;string InstanceID]
from t

where

anonymous function {("-" vs x) 1} splits OrderID by "-" and chooses index 1
{x: "-" vs x; "-" sv enlist[y],(1#x),2_x} splits OrderID by "-", chooses 0 and 2+ indices (there is no remove by index operation in Q), prepends InstanceID and joins everything into "-" delimited string