• kdb+

  Dynamically project function

  Posted by erichards on August 17, 2023 at 12:00 am

  Two questions on function projection

  Firstly, how can I project a function to one with no args, i.e. deferred execution.

  fn: {[x] x+1};

  Obviously fn[1] or fn @ 1 will invoke the function now. Is there a way without wrapping it in another fn?

   

  Secondly, how can I dynamically project a function, e.g.

  fn: {[x;y;z] x+y-z};
  args: (1; ::; 3);

  I’d like to achieve the equivalent of fn[1; ; 3], however fn . args invokes the function now with the second parameter as null rather than returning a projection.

   

  Thanks

  erichards replied 8 months, 2 weeks ago 2 Members · 1 Reply
• 1 Reply

• 

  gyorokpeter-kx

  Member

  August 17, 2023 at 12:00 am

  In reality there is no such thing as a “zero-argument function” in q. Even if you don’t have an argument list and you don’t use any of x, y and z in your function, or if you specify the argument list as [], the function still has one argument. If you call a function with “no arguments” [], that actually means calling it with :: as an argument.

  Projection for the purpose of a deferred function call is achieved by adding a dummy argument, e.g. instead of

  fn: {[x] x+1};

  you would have

  fn: {[x;u] x+1};

  then fn[1] creates a projection that will execute the calculation with any argument passed.

  For the second question, one possible way is by composing a projection of enlist with the dot-apply of your function:

  q)proj:(‘)[fn .;(1; ; 3)]; q)proj .[{[x;y;z] x+y-z}]enlist[1;;3] q)proj 2 0 q)proj 10 8

   

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