javascript - 使用Ramda从多个阵列中轮流获取前X个项目

我有一个数组数组，并且想编写一个函数，该函数通过按顺序从每个数组中获取项目来返回最上面的x项数。

这是我所追求的一个例子:

    const input = [
      ["1a", "2a", "3a", "4a", "5a"],
      ["1b", "2b", "3b", "4b", "5b"],
      ["1c", "2c", "3c", "4c", "5c"],
      ["1d", "2d", "3d", "4d", "5d"]
    ];

    const takeRoundRobin = count => arr => {
      // implementation here
    };

    const actual = takeRoundRobin(5)(input);

    const expected = [
      "1a", "1b", "1c", "1d", "2a"
    ];

我看到一个有关Scala问题的建议，该问题使用zip解决了这一问题，但在Ramda中，您只能将2个列表传递给zip。

最佳答案

在这里，Ramda的 transpose 可以成为您的基础。添加一团 unnest ，一点点 take ，您会得到:

const {take, unnest, transpose} = R

const takeRoundRobin = (n) => (vals) => take(n, unnest(transpose(vals)))

const input = [
  ['1a', '2a', '3a', '4a', '5a'],
  ['1b', '2b', '3b', '4b', '5b'],
  ['1c', '2c', '3c', '4c', '5c'],
  ['1d', '2d', '3d', '4d', '5d']
]

console.log(takeRoundRobin(5)(input))

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

还要注意，这可以处理不同长度的数组:

如果您希望能够重新开始并继续使用值，则可以将take替换为recursiveTake，如下所示:

const {take, unnest, transpose, concat } = R

//recursive take
const recursiveTake = (n) => (vals) => {
  const recur = (n,vals,result) =>
    (n<=0)
      ? result
      : recur(n-vals.length,vals,result.concat(take(n,vals)))
  return recur(n,vals,[]);
};

const takeRoundRobin = (n) => (vals) =>
  recursiveTake(n)(unnest(transpose(vals)));

const input = [
  ['1a', '2a', '3a', '4a'],
  ['1b'],
  ['1c', '2c', '3c', '4c', '5c'],
  ['1d', '2d']
]

console.log(takeRoundRobin(14)(input))

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

没有显式递归的该函数的另一个版本如下所示:

const takeCyclic = (n) => (vals) => take(
  n,
  unnest(times(always(vals), Math.ceil(n / (vals.length || 1))))
)

关于javascript - 使用Ramda从多个阵列中轮流获取前X个项目，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/52875280/