Combining arrays of properties in two objects

Question

Combining arrays of properties in two objects

Below are the JavaScript objects I am working with:

Object A:

0300 : ["295", "293", "298"],
0800 : ["293", "298"],
0930 : ["295"],
1130 : ["297", "293", "298"],
1230 : ["295"]

Object B:

0300 : ["297"],
0800 : ["297"],
0930 : ["293"],

I aim to merge these two objects to create a final object that looks like this:

0300 : ["295", "293", "298", "297"],
0800 : ["293", "298", "297"],
0930 : ["295", "298", "293"],
1130 : ["297", "293", "298"],
1230 : ["295", "298"]

I attempted using new_object['0300'].push within a loop, but it did not yield the desired outcome.

What is the best way to merge JavaScript objects efficiently? Any recommended practices?

javascript object

Answer 1

Answer №1

Give this a shot:

for (let property in fresh_object){
    if(stale_object.has(property)){
        fresh_object[property] = fresh_object[property].combine(stale_object[property]);
    } 
}

Answer 2

Give this a shot:

for (let property in fresh_object){
    if(stale_object.has(property)){
        fresh_object[property] = fresh_object[property].combine(stale_object[property]);
    } 
}

Answer 3

Answer №2

Another option is utilizing jquery's $.merge() function.

var combinedArray = $.merge(existingArray, newArray);

Click here

Answer 4

Another option is utilizing jquery's $.merge() function.

var combinedArray = $.merge(existingArray, newArray);

Click here

Answer 5

Answer №3

I believe the correct syntax should be new_object[0300].push, not new_object['0300'].push. Give it a try with new_object[0300].push and observe the results, although they may seem unusual.

var newObject = {0300 : ["295", "293", "298"],0800 : ["293", "298"],0930 : ["295"],1130 : ["297", "293", "298"],1230 : ["295"]}
var oldObject = {0300 : ["297"],0800 : ["297"],0930 : ["293"]}
newObject[0300].push(oldObject[0300])
//console outputs: ["295", "293", "298", ["297"]]

To resolve this issue, I recommend using the concat method.

var newObject = {0300 : ["295", "293", "298"],0800 : ["293", "298"],0930 : ["295"],1130 : ["297", "293", "298"],1230 : ["295"]}
var oldObject = {0300 : ["297"],0800 : ["297"],0930 : ["293"]}

newObject[0300].concat(oldObject[0300])
//console outputs: ["295", "293", "298", "297"]

To merge two Json objects, follow these steps:

For further information on how to dynamically merge properties of two JavaScript objects, please visit How can I merge properties of two JavaScript objects dynamically?

for (var attrname in obj2) { obj1[attrname] = obj2[attrname]; }

Answer 6

I believe the correct syntax should be new_object[0300].push, not new_object['0300'].push. Give it a try with new_object[0300].push and observe the results, although they may seem unusual.

var newObject = {0300 : ["295", "293", "298"],0800 : ["293", "298"],0930 : ["295"],1130 : ["297", "293", "298"],1230 : ["295"]}
var oldObject = {0300 : ["297"],0800 : ["297"],0930 : ["293"]}
newObject[0300].push(oldObject[0300])
//console outputs: ["295", "293", "298", ["297"]]

To resolve this issue, I recommend using the concat method.

var newObject = {0300 : ["295", "293", "298"],0800 : ["293", "298"],0930 : ["295"],1130 : ["297", "293", "298"],1230 : ["295"]}
var oldObject = {0300 : ["297"],0800 : ["297"],0930 : ["293"]}

newObject[0300].concat(oldObject[0300])
//console outputs: ["295", "293", "298", "297"]

To merge two Json objects, follow these steps:

For further information on how to dynamically merge properties of two JavaScript objects, please visit How can I merge properties of two JavaScript objects dynamically?

for (var attrname in obj2) { obj1[attrname] = obj2[attrname]; }

Answer 7

Answer №4

In order to combine properties with the same key in two objects, a utility function must be written that takes a merge strategy as an argument.

The process involves iterating over the keys of the second object and determining whether to merge the value into the first object (if there is already a property with the same key) or simply copy it over. The current implementation mutates the first object (o1):

function merge_keys(o1, o2, merge) {
  var keys2 = Object.keys(o2), key;

  for (var i = 0; i < keys2.length; i++) {
    key = keys2[i];
    o1[key] = o1[key] ? merge(o1[key], o2[key]) : o2[key];
  }
}

The merging strategy can then be defined, such as:

function concat(a, b) { return a.concat(b); }

Finally, the function can be called using the specified merge strategy:

merge_keys(new_object, old_object, concat);

To achieve this more concisely in ES6 using for...of:

function merge_keys(o1, o2, merge) {
  for (var k of Object.keys(o2)) {
    o1[k] = o1[k] ? merge(o1[k], o2[k]) : o2[k];
  }
}

Alternatively, if you prefer using Array#forEach:

function merge_keys(o1, o2, merge) {
  function maybe_merge(k) { o1[k] = o1[k] ? merge(o1[k], o2[k]) : o2[k]; }

  Object.keys(o2).forEach(maybe_merge);
}

Answer 8

In order to combine properties with the same key in two objects, a utility function must be written that takes a merge strategy as an argument.

The process involves iterating over the keys of the second object and determining whether to merge the value into the first object (if there is already a property with the same key) or simply copy it over. The current implementation mutates the first object (o1):

function merge_keys(o1, o2, merge) {
  var keys2 = Object.keys(o2), key;

  for (var i = 0; i < keys2.length; i++) {
    key = keys2[i];
    o1[key] = o1[key] ? merge(o1[key], o2[key]) : o2[key];
  }
}

The merging strategy can then be defined, such as:

function concat(a, b) { return a.concat(b); }

Finally, the function can be called using the specified merge strategy:

merge_keys(new_object, old_object, concat);

To achieve this more concisely in ES6 using for...of:

function merge_keys(o1, o2, merge) {
  for (var k of Object.keys(o2)) {
    o1[k] = o1[k] ? merge(o1[k], o2[k]) : o2[k];
  }
}

Alternatively, if you prefer using Array#forEach:

function merge_keys(o1, o2, merge) {
  function maybe_merge(k) { o1[k] = o1[k] ? merge(o1[k], o2[k]) : o2[k]; }

  Object.keys(o2).forEach(maybe_merge);
}

Answer 9

Answer №5

Finally cracked the code...

const updated_object = {};

const new_props = Object.getOwnPropertyNames(new_data);
const old_props = Object.getOwnPropertyNames(old_data);
let new_props_length = new_props.length;
let old_props_length = old_props.length;

if (new_props_length > old_props_length) {
    while (--new_props_length) {
        const prop_name = new_props[new_props_length];
        
        if (old_data[prop_name]) {
            updated_object[prop_name] = new_data[prop_name].concat(old_data[prop_name]);
        } else {
            updated_object[prop_name] = new_props[new_props_length];
        }
    }

} else {
    while (--old_props_length) {
        const prop_name = old_props[old_props_length];
        
        if (new_data[prop_name]) {
            updated_object[prop_name] = old_data[prop_name].concat(new_data[prop_name]);
        } else {
            updated_object[prop_name] = old_props[old_props_length];
        }
    }
}

Answer 10

Finally cracked the code...

const updated_object = {};

const new_props = Object.getOwnPropertyNames(new_data);
const old_props = Object.getOwnPropertyNames(old_data);
let new_props_length = new_props.length;
let old_props_length = old_props.length;

if (new_props_length > old_props_length) {
    while (--new_props_length) {
        const prop_name = new_props[new_props_length];
        
        if (old_data[prop_name]) {
            updated_object[prop_name] = new_data[prop_name].concat(old_data[prop_name]);
        } else {
            updated_object[prop_name] = new_props[new_props_length];
        }
    }

} else {
    while (--old_props_length) {
        const prop_name = old_props[old_props_length];
        
        if (new_data[prop_name]) {
            updated_object[prop_name] = old_data[prop_name].concat(new_data[prop_name]);
        } else {
            updated_object[prop_name] = old_props[old_props_length];
        }
    }
}

Combining arrays of properties in two objects

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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