Parametric table name?

Community Support
1

Hi,

I wonder if there is a way to create parametric table (i.e. using the string store in a variable as the table name of a table).

For example in a master table I have something like this.

masterTable

instrutment_id | sym

1001 | `AAA

1002 | `BBB

1003 | `CCC 

Now for every stock in the master table, I would like to make a table to hold its order book. So I would like to make a few table, and I would like to have their names as

AAABook

BBBBook

CCCBook..

etc.

Any suggestions on how I can achieve this without interfacing with an external language?

Appreciate any helps.

Gary 

2

q)mas:(id:1001 1002 1003;sym:AAABBB`CCC)                                                                                                                                                                                                                                                       

q){@[.;$string,“Book”;:;(a:();b:())]} each exec sym from mas                                                                                                                                                                                                                                

..`.

q)\a                                                                                                                                                                                                                                                                                               

AAABookBBBBookCCCBookmas

q)AAABook                                                                                                                                                                                                                                                                                          

a b

3

Thanks Jonny,

I am new to kdb so wondering what does it mean by `. in the second statement.

Also is it a protected execution or a functional @ apply? I read the q for mortal and couldn’t find any clue…

Or are there other material mentioned this better?

Thanks

Gary

4

@ is amend and `. refers to the global context so you’re effectively amending the global text<o:p></o:p>

<o:p> </o:p>

Another to write it would be<o:p></o:p>

<o:p> </o:p>

q){(`$string,“Book”) set (a:();b:())} each exec sym from mas<o:p></o:p>

AAABookBBBBook`CCCBook<o:p></o:p>

<o:p> </o:p>

<o:p> </o:p>

<o:p> </o:p>

From: personal-kdbplus@googlegroups.com [mailto:personal-kdbplus@googlegroups.com] On Behalf Of Gary Chow
Sent: Friday, May 8, 2015 11:29 AM
To: personal-kdbplus@googlegroups.com
Subject: Re: [personal kdb+] Parametric table name?<o:p></o:p>

<o:p> </o:p>

Thanks Jonny,<o:p></o:p>

<o:p> </o:p>

I am new to kdb so wondering what does it mean by `. in the second statement.<o:p></o:p>

Also is it a protected execution or a functional @ apply? I read the q for mortal and couldn’t find any clue…<o:p></o:p>

<o:p> </o:p>

Or are there other material mentioned this better?<o:p></o:p>

<o:p> </o:p>

Thanks<o:p></o:p>

<o:p> </o:p>

Gary

5

It is an apply, the backtick means the root namespace.

6

On Saturday, May 9, 2015 at 10:19:57 AM UTC+8, Enoch Lam wrote:

It is an apply, the backtick means the root namespace.

Thanks Enoch and Jonny! 

7

Hi Enoch,

Thank you very much for your reply

So the bracket () in

(aaa) set value

sort of act as the dereferencing, and value will be assigned to the variable name stored as symbol in the variable name. Am I correct.

So now I have a further question. Is the bracket () in the statement specific to the verb “set”, or does it work in similar way in some other verb too? Or is it itself some sort of adverb or verb?

Thanks so much.

Gary

8

Gary,

not quite. (…) isn’t doing anything special beyond grouping elements in the expression to make sure that things evaluate correctly. If you have `somesym set …, this works fine, but if you wanted to build the name out of a string (which requires that you cast to symbol before setting) then you need the parenthesis there to make sure it casts to symbol first. Otherwise, you end up trying to set with your left-operand as string and casting the result to a symbol, which results in a type error. For example

q)`$“this” set 1

'type

q)(`$“this”) set 1

`this

Hope this helps