Question

我有一个对象数组，需要遍历并检查对象的键是否具有不同的值。如果是这样，请返回此密钥。

const arrayOfObjects = [
  { item: { first_name: "Joe", last_name: "Dow", age: 15 } },
  { item: { first_name: "Joe", last_name: "d", age: 15 } },
  { item: { first_name: "Joe", last_name: "Dow", age: 20 } }
];

预期结果应为

const result = {
  last_name: true,
  age: true
}

PS：每个对象始终具有相同数量的键

Answer 1

您可以遍历arrayOfObjects。检查当前item的键值是否与先前对象的item的键值不同。如果是，请在true

中将该密钥设置为result

const arrayOfObjects = [{ item: { first_name: "Joe", last_name: "Dow", age: 15 } }, { item: { first_name: "Joe", last_name: "d", age: 15 } }, { item: { first_name: "Joe", last_name: "Dow", age: 20 } }];
const result = {};

arrayOfObjects.forEach(({ item }, i, arr) => {
  if (i === 0) return; // skip the first item
  
  const prev = arr[i - 1].item; // get the previous item
  for (const key in item) {
    if (item[key] !== prev[key])
      result[key] = true
  }
})

console.log(result)

Answer 2

只需使用find。更清楚

const data = [
    { item: { first_name: "Joe", last_name: "Dow", age: 15 } },
    { item: { first_name: "Joe", last_name: "d", age: 15 } },
    { item: { first_name: "Joe", last_name: "Dow", age: 20 } }
];

let result = {};

for (var key in data[0].item) {
  if (data.find((el) => el.item[key] !== data[0].item[key])) {
    result[key] = true;
  }
}
console.log(result);

Answer 3

您可以检查一下嵌套对象。

function check(source, target, result) {
    Object.entries(source).forEach(([k, v]) => {
        if (v && typeof v === 'object') return check(v, target[k], result);
        if (result[k] === undefined) result[k] = true;
        result[k] = result[k] && v === target[k];
    });
    return result;
}

var arrayOfObjects = [{ item: { first_name: "Joe", last_name: "Dow", age: 15 } }, { item: { first_name: "Joe", last_name: "d", age: 15 } }, { item: { first_name: "Joe", last_name: "Dow", age: 20 } }],
    result = arrayOfObjects.reduce((r, o, _, [q]) => check(q, o, r), {});

console.log(result);

Answer 4

这是一个保留唯一性标志，过滤出任何唯一项，将它们映射为所需格式并创建结果条目对象的解决方案。时间复杂度：O（n * length_of_item_with_most_keys），空间复杂度：O（n）。

const arrayOfObjects = [
  { item: { first_name: "Joe", last_name: "Dow", age: 15 } },
  { item: { first_name: "Joe", last_name: "d", age: 15 } },
  { item: { first_name: "Joe", last_name: "Dow", age: 20 } }
];

const result = Object.fromEntries(
  Object.entries(arrayOfObjects.reduce((a, e) => {
    for (const k in e.item) {
      if (!a[k]) {
        a[k] = {val: e.item[k], unique: true};
      }
      else if (a[k].unique) {
        a[k].unique = a[k].val === e.item[k];
      }
    }

    return a;
  }, {})
).filter(e => !e[1].unique).map(e => [e[0], true]));

console.log(result);

Answer 5

您还可以使用Array.reduce和Set来解决此问题，您可以在其中分组键并继续添加值。然后根据集合的大小进行过滤...如果大于1，则说明您有多个值：

const data = [ { item: { first_name: "Joe", last_name: "Dow", age: 15 } }, { item: { first_name: "Joe", last_name: "d", age: 15 } }, { item: { first_name: "Joe", last_name: "Dow", age: 20 } } ];

let result = data.reduce((r,{item},i,a) => {
 Object.keys(item).forEach(k => r[k] = (r[k] || new Set()).add(item[k]))	
 return i == (a.length-1) 
 ? Object.keys(r).filter(k => r[k].size > 1).reduce((r,k) => (r[k] = true, r), {})
 : r
}, {})

console.log(result)

Javascript循环对象的槽数组，并返回一个仅包含其值已更改的键的对象

5 个答案: