each-both creating a projection

[davidcrossey]

Your usage of square brackets has created the projection on the right-hand side, which is expecting an x argument:

q).my.func:{[x;y;z] (x;y;z)} q)tab col1 col2 ——— a 1 b 2 c 3 / projection q){.my.func[x;y;`field]}'[flip value exec col1,col2 from tab] {.my.func[x;y;`field]}'[((`a;1);(`b;2);(`c;3))] q)type {.my.func[x;y;`field]}'[flip value exec col1,col2 from tab] 104h / right-hand side projection; note .’ as type (‘[x]) will return 106 q)type .'[flip value exec col1,col2 from tab] 104h / using apply-each without square brackets q)type {.my.func[x;y;`field]}.’flip value exec col1,col2 from tab 0h q){.my.func[x;y;`field]}.’flip value exec col1,col2 from tab `a 1 `field `b 2 `field `c 3 `field

Note the usage of apply-each .’

Here is a simplier example of your projection syntax:

/ missing x arg q)@'[1 2] @'[1 2] q)type @'[1 2] 104h / providing x arg q)@'[10+;1 2] 11 12 q)type @'[10+;1 2] 7h

Hope this helps

 

[cillianreilly]

The exec statement creates a list of lists – enclosing these in square brackets means kdb+ treats this as a singular item passed to your function.  In order to treat the list as the first and second arguments respectively, you need apply:

q)tab:([]col1:`a`b`c;col2:1 2 3) q).my.func:{0N!(x;y;z);} q){.my.func[x;y;`field]}./:flip value exec col1,col2 from tab; (`a;1;`field) (`b;2;`field) (`c;3;`field)

 

In this case though, you can keep the entire operation contained in the qsql statement. kdb+ will extend the atomic symbol to match the length of the table columns:

q)exec .my.func'[col1;col2;`field]from tab; (`a;1;`field) (`b;2;`field) (`c;3;`field)

 

 

[kdb_newbie]

thank you so much for all the help! super clear explanation!