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

Question

I have an array of objects I need to loop through and check if the key of the object has different value. If so return this key.

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 } }
];

expected result should be

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

PS: each object has always the same number of keys

Answer 1

You can loop through the arrayOfObjects . Check if current item 's keys have different value compared to previous object's item . If yes, set that key to true in 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

Just use find . It's more clear

 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

You could take a check which looks into the nested object.

 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

Here's a solution which keeps a flag for uniqueness, filters out any unique items, maps them to your desired format and creates an object of the resulting entries. Time complexity: O(n * length_of_item_with_most_keys), space complexity: 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

You can also solve this with Array.reduce and Set where you would group on the key and keep adding the values. Then filter on the size of the set ... if larger than 1 you had more than one value:

 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)