Question

如何使用RamdaJS通过partition实现index？

/*
 * @param {number} index
 * @param {[]} list

 * @returns {[found, rest[]]} - array whose index 0 has the found element 
 * * and index 1 has the rest of the given list
*/
const partitionByIndex = (index, list) => {};

// this is what I got so far, but I really think it is too verbose

export const partitionByIndex = R.curry((i, cards) => R.pipe(
  R.partition(R.equals(R.nth(i, cards))),
  ([found, rest]) => [R.head(found), rest],
)(cards));

const list = [1, 2, 3, 4, 5, 6, 7];
const index = 1;

const [found, rest] = partitionByIndex(index, list);

console.log({ found, rest });

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

Answer 1

另一种解决方法是在R.take元素周围R.drop和R.nth，例如：

R.converge(R.pair, [R.nth, (n, xs) => R.concat(R.take(n, xs), R.drop(n + 1, xs))])

或者没有Ramda：

(n, xs) => [xs[n], xs.slice(0, n).concat(xs.slice(n + 1))]

Answer 2

一个相当简单的无点解决方案是

const partitionByIndex = converge(pair, [compose(of, nth), flip(remove)(1)])

partitionByIndex(2, ['a', 'b', 'c', 'd', 'e']) //=> [['c'], ['a', 'b', 'd', 'e']]

flip使我意识到remove的参数可能是错误排列的。

更新

Yogi指出期望的响应可能是['c', ['a', 'b', 'd', 'e']]，而不是上面的结果：[['c'], ['a', 'b', 'd', 'e']]。如果是这样，此代码可以变得更简单：

const partitionByIndex = converge(pair, [nth, flip(remove)(1)])

partitionByIndex(2, ['a', 'b', 'c', 'd', 'e']) //=> ['c', ['a', 'b', 'd', 'e']]

Ramda-按索引划分

2 个答案:

更新