删除多个数组之间的公共元素

Question

我有 3 个数组（或更多/更少，不是必须为 3，我只是举了一个例子），我想删除它们之间的所有公共元素。例如，在第一个 2 之间，公共元素是 x and z，在第二个和第三个数组之间，公共元素是 t。在第一个和第三个之间的共同元素是k。基本上我想删除在多个数组中出现超过 1 次的任何元素。

！！第一个数组可以和第三个数组有共同的元素！！

这是我迄今为止尝试过的，但它无法正常工作。

let y = [{
    id: 'a',
    elems: ['x', 'y', 'z', 'k']
  },
  {
    id: 'b',
    elems: ['x', 't', 'u', 'i', 'z']
  },
  {
    id: 'c',
    elems: ['m', 'n', 'k', 'o', 't']
  },
]

// x, z, t

for (let i = 0; i < y.length - 1; i++) {
  let current = y[i].elems
  let current2 = y[i + 1].elems

  if (current[i] == current2[i]) {
    const index = current.indexOf(current[i]);
    if (index > -1) {
      current.splice(index, 1);
      current2.splice(index, 1);
    }
  }
}

console.log(y)

想要的结果是

[
  {

    "id": "a",
    "elems": [
      "y"
    ]
  },
  {
    "id": "b",
    "elems": [
      "u",
      "i"
    ]
  },
  {
    "id": "c",
    "elems": [
      "m",
      "n",
      "o"
    ]
  }
]

这将是一个正确和最佳的解决方案？我还尝试连接 3 个数组并删除重复项，但后来我不知道如何重新创建 3 个数组。谢谢！

Answer 1

我会首先遍历所有元素并计算已经看到的次数。之后，我会再次循环并过滤掉多次看到的任何内容。

const myData = [{
    id: 'a',
    elems: ['x', 'y', 'z']
  },
  {
    id: 'b',
    elems: ['x', 't', 'u', 'i', 'z']
  },
  {
    id: 'c',
    elems: ['m', 'n', 'o', 't']
  },
]

// count up every elem so we know which ones are duplicated
const allElems = myData.reduce((acc, item) => {
  item.elems.forEach( key => { 
    acc[key] = acc[key] || 0;
    acc[key]++;
  });
  return acc;
}, {})

// loop over all the elems and select only the elems that we have seen once
myData.forEach(item => {
  item.elems = item.elems.filter(key => allElems[key] === 1);
})

console.log(myData)

Answer 2

let x = ['a', 'b']
let y = [{
    id: 'a',
    elems: ['x', 'y', 'z', 'k']
  },
  {
    id: 'b',
    elems: ['x', 't', 'u', 'i', 'z']
  },
  {
    id: 'c',
    elems: ['m', 'n', 'k', 'o', 't']
  },
]


// x, z, t

for (let i = 0; i < y.length - 1; i++) {
  for (let j = 1; j < y.length; j++) {
    let current = y[i].elems
    let current2 = y[j].elems

    current2.forEach((item,index)=>{
      if(current.includes(item)){
        current.splice(current.indexOf(item),1)
        current2.splice(index,1)
      }
    })
  }
}

console.log(y)

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Answer 3

const y = [
  { id: 'a', elems: ['x', 'y', 'z'] },
  { id: 'b', elems: ['x', 't', 'u', 'i', 'z'] },
  { id: 'c', elems: ['m', 'n', 'o', 't'] },
];

// get number of occurences for each elem
const elems 
  = y.flatMap(e => e.elems).reduce((acc,elem) => { 
      acc[elem] = acc[elem] ? acc[elem]+1 : 1; 
      return acc;
    }, {});
      
// get unique elems
const unique = Object.keys(elems).filter(elem => elems[elem]===1);

// remove non-unique elems from each item
const res = y.map(item => 
  ({ ...item, elems: item.elems.filter(e => unique.includes(e)) })
);

console.log(res);

Answer 4

在一个循环后使用 Map 跟踪计数，然后在过滤器中使用该 Map 的计数以获得最终结果

let x = ['a', 'b']
let y = [{
    id: 'a',
    elems: ['x', 'y', 'z']
  },
  {
    id: 'b',
    elems: ['x', 't', 'u', 'i', 'z']
  },
  {
    id: 'c',
    elems: ['m', 'n', 'o', 't']
  },
]

const counts = new Map()
// first iteration to count values
y.forEach(({ elems }) => elems.forEach(v => counts.set(v, (counts.get(v) || 0) + 1)));
// second iteration to filter out dups 
y.forEach(e => e.elems = e.elems.filter(v => counts.get(v) === 1))


console.log(y)

Answer 5

让我知道这是否适合你。

let y = [
  {
    id: "a",
    elems: ["x", "y", "z", "k"],
  },
  {
    id: "b",
    elems: ["x", "t", "u", "i", "z"],
  },
  {
    id: "c",
    elems: ["m", "n", "x", "z", "t"],
  },
];

// For every element in first array
for (let el of y[0].elems) {

  //If we find that every other array includes it
  if (y.every((obj) => obj.elems.includes(el))) {

    //Remove it from all arrays
    for (let obj of y) {
      obj.elems = obj.elems.filter((x) => x !== el);
    }
  }
}

Answer 6

let y = [{
    id: 'a',
    elems: ['x', 'y', 'z']
},
{
    id: 'b',
    elems: ['x', 't', 'u', 'i', 'z']
},
{
    id: 'c',
    elems: ['m', 'n', 'o', 't']
},
];
//

const notExist = (x, arr) => !arr.find(el => el == x);
const restToArrays = (i, arr) => arr.reduce((a, b, index) => index == i ? a : [...a, ...b.elems], []);
const result = y.map((ligne, index, arr) => ({
    id: ligne.id,
    elems: ligne.elems.filter(v => notExist(v, restToArrays(index, arr)))
}))

console.log(result);