如何将数组中的多个项目移动到数组中的不同索引？

Question

我有一个包含 7 个项目的数组。我想以与原始数组中相同的顺序将一些项目移动到不同的索引。到目前为止，我已将代码片段粘贴到我尝试过的任何内容上。let originalArray = ['a','b','c','d','e','f','g'];let itemsToBeMoved = ['c','f','e'];let newIndexToBeMoved = 4;//expected result is ['a','b','d','c','e','f','g'];let movableItemsIndex = [];movableItemsIndex.push(originalArray.indexOf('c'));movableItemsIndex.push(originalArray.indexOf('f'));movableItemsIndex.push(originalArray.indexOf('e'));//To be Moved items has to be sorted as in originalArraymovableItemsIndex.sort();let itemsToBeMovedSorted = [originalArray[movableItemsIndex[0]],originalArray[movableItemsIndex[1]],originalArray[movableItemsIndex[2]]];//Removing items before inserting to items to new positionwhile(movableItemsIndex.length) {    originalArray.splice(movableItemsIndex.pop(), 1);}let newUpdatedArray = [...originalArray],j=0;for(let i = newIndexToBeMoved ;i < originalArray.length ; i++){  newUpdatedArray[i] = itemsToBeMovedSorted[j];  j++;}console.log(newUpdatedArray);

Answer 1

假设所有元素都是唯一的：

let originalArray = ['a','b','c','d','e','f','g'];

let itemsToBeMoved = ['c','f','e'];

let newIndexToBeMoved = 4;

// find the value of the element the marks the insertion point

let insertBefore = originalArray[newIndexToBeMoved];

// in original sequence order, check for presence in the removal

// list, *and* remove them from the original array

let moved = [];

for (let i = 0; i < originalArray.length; ) {

    let value = originalArray[i];

    if (itemsToBeMoved.indexOf(value) >= 0) {

        moved.push(value);

        originalArray.splice(i, 1);

    } else {

        ++i;

    }

}

// find the new index of the insertion point

let insertionIndex = originalArray.indexOf(insertBefore);

if (insertionIndex < 0) {

    insertionIndex = originalArray.length;

}

// and add the elements back in

originalArray.splice(insertionIndex, 0, ...moved);

console.log(originalArray);

Answer 2

这可以通过多种方式解决，但这里简要介绍一下可以做什么：

1. 首先，您可以为要移动的项目创建一个“当前”地图（基于源数组中的位置）

2. 现在，过滤数组，删除项目，但跟踪它们是否存在

3. 将（排序的）当前过滤掉的值拼接到所需位置的更改数组中（基于源数组中的索引）

const move = (arr, items, index) => {

  const present = new Map(items.map(item => [item, arr.indexOf(item)]));

  const altered = arr.filter(item => !present.has(item));

  altered.splice(arr.indexOf(arr[index - 1]), 0, ...([...present.entries()]

    .filter(([k, v]) => v !== -1)

    .sort(([, k1], [, k2]) => k1 - k2)

    .map(([k, v]) => k)));

  return altered;

}

const moved = move(['a','b','c','d','e','f','g'], ['c','f','e'], 4);

console.log(moved);

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Answer 3

typescript这是一种使用and来实现的方法generics。它不会改变数组，速度非常快。

将物品移动到新位置时，总会有一系列物品受到移动的影响。受影响是指在该范围内，所有项目都必须移动。我计算范围的最小值/最大值，该范围之外的任何项目都将简单地从旧数组复制到新数组。

然后，根据项目是在数组中向上移动还是向下移动，我将每个项目从旧数组复制到新数组，同时考虑到移动的偏移量。

export const arrayMove = <ItemType>({

    arr,

    from,

    to,

    movedItemsCount = 1,

}: {

    arr: ItemType[]

    from: number

    to: number

    movedItemsCount?: number

}) => {

    if (from === to) return arr

    const minAffected = Math.min(from, to)

    const maxAffected = Math.max(to, from + movedItemsCount - 1)

    const pushedItemsCount = maxAffected - minAffected + 1 - movedItemsCount

    const newArr: ItemType[] = []

    newArr.length = arr.length

    for (let i = 0; i < arr.length; i++) {

        if (i < minAffected || i > maxAffected) {

            // When i is outside the affected range,

            // items are identical in new and old arrays

            newArr[i] = arr[i]

        } else {

            // Moving items down

            if (to > from) {

                if (i < to - movedItemsCount + 1) {

                    // Write pushed items

                    newArr[i] = arr[i + movedItemsCount]

                } else {

                    // Write moved items

                    newArr[i] = arr[i - pushedItemsCount]

                }

            } else {

                // Moving items up

                if (i < to + movedItemsCount) {

                    // Write moved items

                    newArr[i] = arr[i + pushedItemsCount]

                } else {

                    // Write pushed items

                    newArr[i] = arr[i - movedItemsCount]

                }

            }

        }

    }

    return newArr

}

Answer 4

尝试这个：

const originalArray = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];

const itemsToBeMoved = ['c', 'f', 'e'];

const newIndexToBeMoved = 4;

originalArray.splice(newIndexToBeMoved, 0, itemsToBeMoved)

console.log(originalArray.flat())