按id合并数组，并考虑具有重复关联索引的唯一值

function mergeArrays(arr) {
  const groups = {};
  
  // 分组
  arr.forEach(item => {
    if (!groups[item.id]) {
      groups[item.id] = [];
    }
    groups[item.id].push(item);
  });
  
  const mergedArray = [];
  
  // 合并数组
  for (const id in groups) {
    const group = groups[id];
    const mergedItem = { id };
    
    for (const item of group) {
      for (const key in item) {
        if (key !== 'id') {
          if (!mergedItem[key]) {
            mergedItem[key] = new Set();
          }
          mergedItem[key].add(item[key]);
        }
      }
    }
    
    // 将Set转换为数组
    for (const key in mergedItem) {
      if (key !== 'id') {
        mergedItem[key] = Array.from(mergedItem[key]);
      }
    }
    
    mergedArray.push(mergedItem);
  }
  
  return mergedArray;
}

// 示例数据
const arr = [
  { id: 1, name: 'A', value: 'X' },
  { id: 2, name: 'B', value: 'Y' },
  { id: 1, name: 'C', value: 'Z' },
  { id: 2, name: 'D', value: 'W' }
];

const merged = mergeArrays(arr);
console.log(merged);

[
  { id: 1, name: ['A', 'C'], value: ['X', 'Z'] },
  { id: 2, name: ['B', 'D'], value: ['Y', 'W'] }
]