循环直到...与 Ramda

Question

我正在尝试使用 Ramda 重构几段代码，我想知道，在 Ramda/Functional 编程中有什么好的方法可以解决以下代码：

let arrayOfSomething = initArray();

for(let i = 0; SOME_INDEX_CONDITION(i)|| SOME_CONDITION(arrayOfSomething); i++) {
    const value = operation(arrayOfSomething);
    const nextValue = anotherOperation(value);

   arrayOfSomething = clone(nextValue)
}

所以基本上我想在 arrayOfSomething 上迭代并应用相同的 pipe/composition 操作，直到满足其中一个条件。重要的是我得到最后一个值 (nextValue) 作为对 forLoop 组合的反馈。

Answer 1

我不知道这是否符合您的要求，但 Ramda 的 until 可能是您需要的：

const operation = ({val, ctr}) => ({val: val % 2 ? (3 * val + 1) : (val / 2), ctr: ctr + 1})

const indexCondition = ({ctr}) => ctr > 100
const valCondition = ({val}) =>  val === 1
const condition = R.either(indexCondition, valCondition)

const check = R.until(condition, operation)

const collatz = n => check({ctr: 0, val: n})

console.log(collatz(12)) 
// 12 -> 6 -> 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 //=> {"ctr": 9, "val": 1}
console.log(collatz(5)) 
// 5 -> 16 -> 8 -> 4 -> 2 -> 1 //=> {"ctr": 5, "val": 1}
console.log(collatz(27)) 
//27 -> 82 -> 41 -> 124 -> 62 -> .... //=> {"ctr": 101, "val": 160}

<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.js"></script>

Answer 2

您似乎在寻找反向弃牌，或 unfold。

大多数人都熟悉 reduce：它需要一组值，然后 将其减少 为单个值 – unfold 恰恰相反：它取一个值，展开它到一个值集合

如果库中已经存在类似的函数，其他更熟悉 Ramda 的人可以发表评论

const unfold = (f, init) =>
  f ( (x, next) => [ x, ...unfold (f, next) ]
    , () => []
    , init
    )

const nextLetter = c =>
  String.fromCharCode (c.charCodeAt (0) + 1)

const alphabet =
  unfold
    ( (next, done, c) =>
        c > 'z'
          ? done ()
          : next (c, nextLetter (c))
    , 'a'
    )

console.log (alphabet)
// [ a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z ]

unfold很厉害

const fib = (n = 0) =>
  unfold
    ( (next, done, [ n, a, b ]) =>
        n < 0
          ? done ()
          : next (a, [ n - 1, b, a + b ])
    , [ n, 0, 1 ]
    )

console.log (fib (20))
// [ 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765 ]

我们可以使用 unfold

实现您的 iterateUntil

const unfold = (f, init) =>
  f ( (x, acc) => [ x, ...unfold (f, acc) ]
    , () => []
    , init
    )
    
const iterateUntil = (f, init) =>
  unfold
    ( (next, done, [ arr, i ]) =>
        i >= arr.length || f (arr [i], i, arr)
          ? done ()
          : next (arr [i], [ arr, i + 1 ])
    , [ init, 0 ]
    )
  
const data =
  [ 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i' ]
  
console.log (iterateUntil ((x, i) => i > 3, data))
// [ 'a', 'b', 'c', 'd' ]

console.log (iterateUntil ((x, i) => x === 'd', data))
// [ 'a', 'b', 'c', 'd' ]

我们可以通过 async 和 await 使其轻松支持异步。下面我们使用 asyncUnfold 执行递归数据库查找，从单个节点 ID 开始，0

db.getChildren 接受一个节点 id 和 returns 只有节点的直接子节点
traverse 接受一个节点 id 并递归地获取所有后代子节点（按 depth-first 顺序）

const asyncUnfold = async (f, init) =>
  f ( async (x, acc) => [ x, ...await asyncUnfold (f, acc) ]
    , async () => []
    , init
    )

// demo async function
const Db =
  { getChildren : (id) =>
      new Promise (r => setTimeout (r, 100, data [id] || []))
  }

const Empty =
  Symbol ()

const traverse = (id) =>
  asyncUnfold
    ( async (next, done, [ id = Empty, ...rest ]) =>
        id === Empty
          ? done ()
          : next (id, [ ...await Db.getChildren (id), ...rest ])
    , [ id ]
    )
    
const data =
  { 0 : [ 1, 2, 3 ]
  , 1 : [ 11, 12, 13 ]
  , 2 : [ 21, 22, 23 ]
  , 3 : [ 31, 32, 33 ]
  , 11 : [ 111, 112, 113 ]
  , 33 : [ 333 ]
  , 333 : [ 3333 ]
  }

traverse (0) .then (console.log, console.error)
// => Promise
// ~2 seconds later
// [ 0, 1, 11, 111, 112, 113, 12, 13, 2, 21, 22, 23, 3, 31, 32, 33, 333, 3333 ]

其他适合 unfold

的程序

"starting with a page URL /, crawl all descendant pages"
"starting with search "foo" and page 1, collect results from all pages"
"starting with user Alice, show me her friends, and all of her friends' friends"

循环直到...与 Ramda

Loop until... with Ramda

javascript

functional-programming

ramda.js