如何使用这些标记的联合来实现 zipPad？

Question

我正在尝试将以下 Haskell 代码转换为 Javascript：data These a b = This a | That b | These a bclass Align f where  align :: (These a b -> c) -> f a -> f b -> f cinstance Align [] where  align f []     []     = []  align f (x:xs) []     = f (This x)    : align f xs []  align f []     (y:ys) = f (That y)    : align f [] ys  align f (x:xs) (y:ys) = f (These x y) : align f xs ysliftAlign2 f a b = align t  where t (This l)    = f l b        t (That r)    = f a r        t (These l r) = f l rzipPad a b = liftAlign2 (,) a b我在翻译这个liftAlign2函数时遇到了麻烦。它需要一个元组构造函数和两个默认值，但我不知道从何t而来。这是我到目前为止的代码：const union = type => (tag, o) =>  (o.type = type.name || type, o.tag = tag.name || tag, o);const These_ = union("These");const This = _this => These_(This, {this: _this});const That = that => These_(That, {that});const These = _this => that => These_(These, {this: _this, that});const Pair = x => y => [x, y];const align = f => ([x, ...xs]) => ([y, ...ys]) =>  x === undefined && y === undefined ? []    : y === undefined ? [f(This(x)), ...align(f) (xs) ([])]    : x === undefined ? [f(That(y)), ...align(f) ([]) (ys)]    : [f(These(x) (y)), ...align(f) (xs) (ys)];const liftAlign2 = f => x => y => ?const zipPad = x => y =>  liftAlign2(Pair) (x) (y);const zipPad ("") (0) (["foo", "bar"]) ([2, 4, 6]); // [["foo", 2], ["bar", 4], ["", 6]]我知道 JS 代码效率很低，因为它不处理函数List类型，而是处理数组。效率对于这种翻译并不重要。

Answer 1

这就是我会做的。

/*

    data These a b

        = This  { fromThis :: a }

        | That  { fromThat :: b }

        | These { fromThis :: a, fromThat :: b }

*/

const This  = fromThis => ({ fromThis });

const That  = fromThat => ({ fromThat });

const These = fromThis => fromThat => ({ fromThis, fromThat });

// type Align f = forall a b c. (These a b -> c) -> f a -> f b -> f c

// alignArray :: Align []

const alignArray = f => xs => xs.reduceRight((next, x) => ([y, ...ys]) =>

        [f(y === undefined ? This(x) : These(x)(y)), ...next(ys)],

    ys => ys.map(y => f(That(y))));

// liftAlign2 :: Align f -> (a -> b -> c) -> a -> b -> f a -> f b -> f c

const liftAlign2 = align => f => x => y =>

    align(({ fromThis = x, fromThat = y }) =>

        f(fromThis)(fromThat));

// zipPad :: Align f -> a -> b -> f a -> f b -> f (a, b)

const zipPad = align => liftAlign2(align)(x => y => [x, y]);

// result :: [(String, Number)]

const result = zipPad(alignArray)("")(0)(["foo", "bar"])([2, 4, 6]);

// [["foo", 2], ["bar", 4], ["", 6]]

console.log(result);

注意liftAlign2和zipPad是多态的。它们不是专门用于数组的。

Answer 2

等效于的 javascriptliftAlign2是

const liftAlign2 = f => x => y => align (t => 

        f (t.this === undefined ? x : t.this) (t.that === undefined ? y : t.that)

    );

liftAlign2调用你的align函数。

它传递的参数是一个以 aThese作为参数的函数。

如果它们存在，它返回调用的结果，如果f它们These不存在则回退到。xy

通过这样做，它f从一个对两个普通值进行操作的函数“提升”到一个对两个列表进行操作的函数。

在zipPad fis 中Pair，因此这会从两个列表构造成对的值，直到最短列表的长度，然后重复回退值直到最长列表的长度。

值得注意的是，这并不完全等同于 Haskell 代码。这是因为 Haskell 代码更具多态性；它允许您调用liftAlign2（也ZipPad）作为类型类实例的任何Align类型。虽然这包括列表，但不仅限于列表。相比之下，javascript 代码专门用于 js 数组。