在对象数组数组中查找交叉点

Question

假设我有以下包含对象数组（歌曲）的数组：

[
  [
    {
      name: 'I Want You Back',
      id: 1
    },
    {
      name: 'ABC',
      id: 2
    }
  ],
  [
    {
      name: 'I Want You Back',
      id: 1
    },
    {
      name: 'Dont Stop Me Now',
      id: 3
    }
  ],
  [
    {
      name: 'I Want You Back',
      id: 1
    },
    {
      name: 'ABC',
      id: 2
    }
  ],
]

我希望浏览我的数据并返回每个数组中找到的对象，因此在这种情况下，返回值将是：

{
   name: 'I Want You Back',
   id: 1
}

到目前为止，我只设法为字符串阵列而不是对象这样做，例如：

const arrays = [
  ['I Want You Back', 'ABC'],
  ['I Want You Back', 'Dont stop me now'],
  ['I Want You Back', 'ABC']
];
const result = arrays.shift().filter((v) => {
  return arrays.every((a) => {
    return a.indexOf(v) !== -1;
  });
});
// result = ["I Want You Back"]

我一直在尝试，但是我无法设法使用我的对象（例如它们的名称或ID）应用相同的逻辑。如果你们中有些人向我展示您将如何做。

Answer 1

您只能按唯一的 id计数对象，然后浏览与数组长度相同的对象：

const songs = [[{name: 'I Want You Back',id: 1},{name: 'ABC',id: 2}],[{name: 'I Want You Back',id: 1},{name: 'Dont Stop Me Now',id: 3}],[{name: 'I Want You Back',id: 1},{name: 'ABC',id: 2}],]
// count of each unique id
let counts  = songs.reduce((count, arr) => {
  arr.forEach(obj => {
    if (!count[obj.id]) count[obj.id] = {obj, count: 0}
    count[obj.id].count += 1  
  })
  return count
}, {})
// now filter out all whose count isn't the length
// and map to retreive just the object
let intersect = Object.values(counts)
                .filter(o => o.count == songs.length)
                .map(o => o.obj)
console.log(intersect)

当然，如果对象平等按值进行测试，这将容易得多，但是由于不是您需要跳过一些额外的篮球。