How to make functions callable with multiple parameters, so instead of f(x)(y) you can do f(x,y)

Question

given the following data structure

var categories = [
  {_id: 1, order: 0, ancestors: []},
    {_id: 2, order: 0, ancestors: [1]},
    {_id: 3, order: 0, ancestors: [1]},
  {_id: 4, order: 0, ancestors: []},
  {_id: 5, order: 0, ancestors: [4]},
]

and the functions

var byLevel = R.curry((level) => R.filter(o => o.ancestors.length == level))
var levelZero = byLevel(0);

I'd like to be able to call both of these like so :-

console.log(byLevel(0,categories));
console.log(levelZero(categories));

and get the same result (an array of two categories with ids 1 and 4)

however byLevel returns a function. I can call it like

console.log(byLevel(0)(categories));

and it does what I want. But if I write byLevel like :-

var byLevel = R.curry((level, l) => R.filter(o => o.ancestors.length == level, l))

then

console.log(byLevel(0,categories));

works fine and curries fine. However, it seems not as "clean".

Question

Is there a "Ramda" way to make it so I can define the function so that I can do f(x,y) instead of f(x)(y)

you pass to `curry` function with _one_ argument, so `curry` can't make from it function with _two_ arguments. So this variant seems normal `R.curry((level, l) => R.filter(o => o.ancestors.length == level, l))` — Grundy, Oct 28 '15 at 04:24

score 1 · Accepted Answer · answered Oct 28 '15 at 03:48

I think this is all it would take:

var byLevel = R.curry((level, categories) => 
    R.filter(o => o.ancestors.length == level, categories)
);

If you want to curry a function to allow yourself to supply the parameters that way, you have to have them listed.

How to make functions callable with multiple parameters, so instead of f(x)(y) you can do f(x,y)

1 Answers1