https://learninghub.kx.com/forums/topic/dynamically-project-function
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
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