How can I delete a focused property of an object with a van Laarhoven Lens?

user_5536315 0
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With my simple lens implementation i can perform the usual modify, set, get and delete operations:

// Lens type

const Lens = f => ({runLens: f, [Symbol.toStringTag]: "Lens"});

const objLens = map => k =>
  Lens(f => o =>
    map(x => Object.assign({}, o, {[k]: x})) (f(o[k]))); // object lens

// Id type

const Id = x => ({runId: x, [Symbol.toStringTag]: "Id"});
const idMap = f => tx => Id(f(tx.runId)); // functor

// Const type

const Const = x => ({runConst: x, [Symbol.toStringTag]: "Const"});
const constMap = f => tx => Const(tx.runConst); // functor

// auxiliary function

const _const = x => y => x;

// MAIN

const o = {foo: "abc", bar: 123};

const get = objLens(constMap) ("foo").runLens(x => Const(x)) (o),
  set = objLens(idMap) ("bat").runLens(_const(Id(true))) (o),
  mod = objLens(idMap) ("foo").runLens(s => Id(s.toUpperCase())) (o),
  del = objLens(idMap) ("foo").runLens(_const(Id(null))) (o); //*

console.log("get", get.runConst);
console.log("set", set.runId);
console.log("mod", mod.runId);
console.log("del", del.runId);

However, delete is unsatisfying as I want to delete the whole property rather than merely replacing the value with nothing.

How can I achieve this?

*Please note that i usually would have used a proper Option type to indicate no values.

lenses functional-programming javascript
2 Answers

This is what I would do:

// type Lens a b = forall f. Functor f => (b -> f b) -> a -> f a

// newtype Const b a = Const { getConst :: b } deriving Functor
const Const = getConst => ({ getConst, map: _ => Const(getConst) });

// newtype Identity a = Identity { runIdentity :: a } deriving Functor
const Identity = runIdentity => ({ runIdentity, map: f => Identity(f(runIdentity)) });

// remove :: String -> Object -> Object
const remove = key => ({ [key]: _, ...rest }) => rest;

// prop :: String -> Lens Object (Maybe Value)
const prop = key => fun => obj =>
    fun(obj.hasOwnProperty(key) ? { fromJust: obj[key] } : null)
        .map(data => Object.assign(remove(key)(obj), data && { [key]: data.fromJust }));

// get :: Lens a b -> a -> b
const get = lens => data => lens(Const)(data).getConst;

// modify :: Lens a b -> (b -> b) -> a -> a
const modify = lens => fun => data => lens(x => Identity(fun(x)))(data).runIdentity;

// set :: Lens a b -> b -> a -> a
const set = lens => value => modify(lens)(_ => value);

// del :: Lens a (Maybe b) -> a -> a
const del = lens => set(lens)(null);

// foo :: Lens Object (Maybe Value)
const foo = prop("foo");

console.log(get(foo)({ foo: 10, bar: 20 })); // { fromJust: 10 }
console.log(del(foo)({ foo: 10, bar: 20 })); // { bar: 20 }

As seen, the type signature of a property lens like foo is Lens Object (Maybe Value). This makes sense because if you try get(foo)({ bar: 20 }) you should get nothing. The del function works on any lens which focuses on a maybe value, and sets its value to nothing (i.e. null).

Credit goes to Bergi for showing me that it's possible to pattern match on computed properties.

user_7837430

One possible way would be to define a special type that denotes deletion:

const Lens = f => ({runLens: f, [Symbol.toStringTag]: "Lens"});

const objLens = map => k =>
  Lens(f => o => map(x => {
    if (x[Symbol.toStringTag] === "Deleter") {
      const p = Object.assign({}, o);
      delete p[k];
      return p;
    }

    else
      return Object.assign({}, o, {[k]: x});
    }) (f(o[k])));

const Id = x => ({runId: x, [Symbol.toStringTag]: "Id"});
const idMap = f => tx => Id(f(tx.runId)); // functor

const _const = x => y => x;

// deletion type

const Deleter =
  ({get runDeleter() {return Deleter}, [Symbol.toStringTag]: "Deleter"});

// MAIN

const o = {foo: "abc", bar: 123};

const del = objLens(idMap) ("foo").runLens(_const(Id(Deleter))) (o);

console.log("del", del.runId);

However, _const(Id(Deleter)) isn't particularly intuitive, but looks hackish. Hopefully there is a better approach.

user_55363150
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