如何使用van Laarhoven镜头删除对象的聚焦属性？

Question

使用我的简单镜头实现，我可以执行常见的修改、设置、获取和删除操作：

​

// 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);

​

然而，delete并不能令人满意，因为我想删除整个属性，而不是只将值替换为空。

我如何才能做到这一点？

*请注意，我通常会使用适当的Option类型来表示没有值。

Answer 1

这就是我要做的：

​

// 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 }

​

可以看到，像foo这样的属性镜头的类型签名是Lens Object (Maybe Value)。这是有道理的，因为如果您尝试使用get(foo)({ bar: 20 })，您应该什么也得不到。del函数适用于任何聚焦于可能值的镜头，并将其值设置为空(即null)。

归功于Bergi的showing me，它可以在计算属性上进行模式匹配。

Answer 2

一种可能的方法是定义一个表示删除的特殊类型：

​

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);

​

然而，_const(Id(Deleter))并不是特别直观，但看起来有点老土。希望有更好的方法。