How to merge two objects containing same keys but different values

Question

I'm wanting to merge two array that carry objects with the same keys, but one object may have a value, but the other object, carrying the same key, may have a null value. I'm wanting to merge the two to try and replace all the null values and have one final object.

I've written the following, but at the moment, I'm only getting undefined as the final result.

const objectOne = [
  {
    one: null,
    two: "Two",
    three: null
  }
];

const objectTwo = [
  {
    one: "One",
    two: null,
    three: "Three"
  }
];

const compareObjects = (searchKey, searchValue) =>
  objectTwo.map((retrieve) =>
    Object.entries(retrieve).forEach(([retrieveKey, retrieveValue]) => {
      if (!searchValue) {
        objectOne.searchKey = retrieveValue;
        console.log(objectOne);
      }
    })
  );

const newObject = objectOne.map((search) =>
  Object.entries(search).forEach(([searchKey, searchValue]) =>
    compareObjects(searchKey, searchValue)
  )
);

console.log(newObject);

Interesting names choice objectOne and objectTwo are actually Arrays ;) — Roko C. Buljan
– Roko C. Buljan, Commented Aug 6, 2021 at 18:31
I thought I explained that in the description, but the end result is to have one a single array carrying objects with less null values. — pingeyeg
– pingeyeg, Commented Aug 6, 2021 at 18:34
What to do when neither object has a null value for a given key? — Gershom Maes
– Gershom Maes, Commented Aug 6, 2021 at 18:37

Charlie Bamford · Accepted Answer · 2021-08-06 18:38:23Z

2

Reduce seems more useful in this case than forEach. Reduce will build a new thing from the input and return that.

const objectOne = [
  {
    one: null,
    two: "Two",
    three: null
  }
];

const objectTwo = [
  {
    one: "One",
    two: null,
    three: "Three"
  }
];

const merged = objectOne.map((item, idx) => {
  return Object.entries(item).reduce((acc, [key, value]) => {
    acc[key] = value === null && objectTwo.length > idx ? objectTwo[idx][key] : value;
    return acc;
  }, {})
});
console.log(merged);

edited Aug 6, 2021 at 18:38

answered Aug 6, 2021 at 18:34

Charlie Bamford

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1 Comment

pingeyeg Over a year ago

Very nice, I forgot about the reduce method. This makes it much cleaner.

Kelvin Schoofs · Accepted Answer · 2021-08-06 18:33:51Z

Your mapping of keys/entries is pretty close. It might be easiest to do something like this though:

const objectOnes = [
  {
    one: null,
    two: "Two",
    three: null
  },
  {
     a: 123,
     b: undefined,
  },
];

const objectTwos = [
  {
    one: "One",
    two: null,
    three: "Three"
  },
  {
      b: 456,
  }
];

function merge(a, b) {
    const result = { ...a }; // copy a
    for (const key in result) {
        // Let's say we want to pick the non-null/undefined value
        if (result[key] !== null && result[key] !== undefined) continue;
        result[key] = b[key]; // copy from b
    }
    return result;
}

const merged = objectOnes.map((obj, i) => merge(obj, objectTwos[i]));
console.log(merged);

Aniketh Malyala · Accepted Answer · 2021-08-06 18:55:36Z

One method you can use is to create a function that combines/merges the two dictionaries. First, we create our newObject:

const newObject = [

]

Then, we create our function using an O(n^2) approach and call it:

combine(objectOne,objectTwo)
console.log(newObject)

function combine(objectOne, objectTwo){
    for (let [key1,value1] of Object.entries(objectOne[0])){
        for (let [key2,value2] of Object.entries(objectTwo[0])){
            if(key1 == key2){
                if(value1 == null){
                    newObject.push({
                        key: key1,
                        value: value2
                    })
                }
                else{
                    newObject.push({
                        key: key1,
                        value: value1
                    })
                }
            }
        }
    }
}

This is the following output:

[
  { key: 'one', value: 'One' },
  { key: 'two', value: 'Two' },
  { key: 'three', value: 'Three' }
]

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