如何将数组中指定元素从原位置移动至数组开头？

要实现将elementeSelected中的元素移到allElements数组开头，同时保留两类元素的指定顺序（选中元素按elementeSelected顺序，剩余元素按原数组顺序），可以用以下几种方法：

方法一：过滤拼接（直观简洁） #

通过两次过滤分别提取选中元素和剩余元素，再合并结果：

const elementeSelected = ["a", "b"];
const allElements = ["m", "s", "e", "r", "q", "d", "a", "b", "c"];

// 按elementeSelected的顺序提取存在于allElements中的元素
const selectedItems = elementeSelected.filter(item => allElements.includes(item));
// 提取allElements中未被选中的元素，保留原顺序
const remainingItems = allElements.filter(item => !elementeSelected.includes(item));

const result = [...selectedItems, ...remainingItems];
console.log(result); // 输出: ["a", "b", "m", "s", "e", "r", "q", "d", "c"]

方法二：Set优化查找（适合大数据量） #

当数组元素较多时，用Set优化查找性能，避免多次数组includes的低效遍历：

const elementeSelected = ["a", "b"];
const allElements = ["m", "s", "e", "r", "q", "d", "a", "b", "c"];

const selectedSet = new Set(elementeSelected);
// 按指定顺序提取选中元素
const selectedItems = elementeSelected.filter(item => selectedSet.has(item) && allElements.includes(item));
// 提取剩余元素
const remainingItems = allElements.filter(item => !selectedSet.has(item));

const result = [...selectedItems, ...remainingItems];
console.log(result);

方法三：Reduce一次遍历（处理重复元素场景） #

如果原数组存在重复元素，且需要确保选中元素仅出现一次，可通过reduce一次遍历分类，再调整顺序：

const elementeSelected = ["a", "b"];
const allElements = ["m", "s", "e", "r", "q", "d", "a", "b", "c"];

const selectedSet = new Set(elementeSelected);
const { selected, remaining } = allElements.reduce((acc, item) => {
  if (selectedSet.has(item)) {
    // 避免重复添加同一元素
    if (!acc.selected.includes(item)) {
      acc.selected.push(item);
    }
  } else {
    acc.remaining.push(item);
  }
  return acc;
}, { selected: [], remaining: [] });

// 确保选中元素顺序与elementeSelected一致
const orderedSelected = elementeSelected.filter(item => selectedSet.has(item));
const result = [...orderedSelected, ...remaining];
console.log(result);

内容的提问来源于stack exchange，提问作者DenisPaul