Replacing objects in array
JavascriptLodashJavascript Problem Overview
I have this javascript object:
var arr1 = [{id:'124',name:'qqq'},
{id:'589',name:'www'},
{id:'45',name:'eee'},
{id:'567',name:'rrr'}]
var arr2 = [{id:'124',name:'ttt'},
{id:'45',name:'yyy'}]
I need to replace objects in arr1 with items from arr2 with same id.
So here is the result I want to get:
var arr1 = [{id:'124',name:'ttt'},
{id:'589',name:'www'},
{id:'45',name:'yyy'},
{id:'567',name:'rrr'}]
How can I implement it using javascript?
Javascript Solutions
Solution 1 - Javascript
You can use Array#map with Array#find.
arr1.map(obj => arr2.find(o => o.id === obj.id) || obj);
var arr1 = [{
id: '124',
name: 'qqq'
}, {
id: '589',
name: 'www'
}, {
id: '45',
name: 'eee'
}, {
id: '567',
name: 'rrr'
}];
var arr2 = [{
id: '124',
name: 'ttt'
}, {
id: '45',
name: 'yyy'
}];
var res = arr1.map(obj => arr2.find(o => o.id === obj.id) || obj);
console.log(res);
Here, arr2.find(o => o.id === obj.id) will return the element i.e. object from arr2 if the id is found in the arr2. If not, then the same element in arr1 i.e. obj is returned.
Solution 2 - Javascript
What's wrong with Object.assign(target, source) ?
Arrays are still type object in Javascript, so using assign should still reassign any matching keys parsed by the operator as long as matching keys are found, right?
Solution 3 - Javascript
Since you're using Lodash you could use _.map and _.find to make sure major browsers are supported.
In the end I would go with something like:
function mergeById(arr) {
return {
with: function(arr2) {
return _.map(arr, item => {
return _.find(arr2, obj => obj.id === item.id) || item
})
}
}
}
var result = mergeById([{id:'124',name:'qqq'},
{id:'589',name:'www'},
{id:'45',name:'eee'},
{id:'567',name:'rrr'}])
.with([{id:'124',name:'ttt'}, {id:'45',name:'yyy'}])
console.log(result);
<script src="https://raw.githubusercontent.com/lodash/lodash/4.13.1/dist/lodash.js"></script>
Solution 4 - Javascript
Thanks to ES6 we can made it with easy way -> for example on util.js module ;))).
-
Merge 2 array of entity
export const mergeArrays = (arr1, arr2) => arr1 && arr1.map(obj => arr2 && arr2.find(p => p.id === obj.id) || obj);
> gets 2 array and merges it.. Arr1 is main array which is priority is > high on merge process
-
Merge array with same type of entity
export const mergeArrayWithObject = (arr, obj) => arr && arr.map(t => t.id === obj.id ? obj : t);
> it merges the same kind of array of type with some kind of type for
example: array of person ->
[{id:1, name:"Bir"},{id:2, name: "Iki"},{id:3, name:"Uc"}]
second param Person {id:3, name: "Name changed"}
result is
[{id:1, name:"Bir"},{id:2, name: "Iki"},{id:3, name:"Name changed"}]
Solution 5 - Javascript
There is always going to be a good debate on time vs space, however these days I've found using space is better for the long run.. Mathematics aside let look at a one practical approach to the problem using hashmaps, dictionaries, or associative array's whatever you feel like labeling the simple data structure..
var marr2 = new Map(arr2.map(e => [e.id, e]));
arr1.map(obj => marr2.has(obj.id) ? marr2.get(obj.id) : obj);
I like this approach because though you could argue with an array with low numbers you are wasting space because an inline approach like @Tushar approach performs indistinguishably close to this method. However I ran some tests and the graph shows how performant in ms both methods perform from n 0 - 1000. You can decide which method works best for you, for your situation but in my experience users don't care to much about small space but they do care about small speed.
Here is my performance test I ran for source of data
var n = 1000;
var graph = new Array();
for( var x = 0; x < n; x++){
var arr1s = [...Array(x).keys()];
var arr2s = arr1s.filter( e => Math.random() > .5);
var arr1 = arr1s.map(e => {return {id: e, name: 'bill'}});
var arr2 = arr2s.map(e => {return {id: e, name: 'larry'}});
// Map 1
performance.mark('p1s');
var marr2 = new Map(arr2.map(e => [e.id, e]));
arr1.map(obj => marr2.has(obj.id) ? marr2.get(obj.id) : obj);
performance.mark('p1e');
// Map 2
performance.mark('p2s');
arr1.map(obj => arr2.find(o => o.id === obj.id) || obj);
performance.mark('p2e');
graph.push({ x: x, r1: performance.measure('HashMap Method', 'p1s', 'p1e').duration, r2: performance.measure('Inner Find', 'p2s','p2e').duration});
}
Solution 6 - Javascript
I like to go through arr2 with foreach() and use findIndex() for checking for occurrence in arr1:
var arr1 = [{id:'124',name:'qqq'},
{id:'589',name:'www'},
{id:'45',name:'eee'},
{id:'567',name:'rrr'}]
var arr2 = [{id:'124',name:'ttt'},
{id:'45',name:'yyy'}]
arr2.forEach(element => {
const itemIndex = arr1.findIndex(o => o.id === element.id);
if(itemIndex > -1) {
arr1[itemIndex] = element;
} else {
arr1 = arr1.push(element);
}
});
console.log(arr1)
Solution 7 - Javascript
Considering that the accepted answer is probably inefficient for large arrays, O(nm), I usually prefer this approach, O(2n + 2m):
function mergeArrays(arr1 = [], arr2 = []){
//Creates an object map of id to object in arr1
const arr1Map = arr1.reduce((acc, o) => {
acc[o.id] = o;
return acc;
}, {});
//Updates the object with corresponding id in arr1Map from arr2,
//creates a new object if none exists (upsert)
arr2.forEach(o => {
arr1Map[o.id] = o;
});
//Return the merged values in arr1Map as an array
return Object.values(arr1Map);
}
Unit test:
it('Merges two arrays using id as the key', () => {
var arr1 = [{id:'124',name:'qqq'}, {id:'589',name:'www'}, {id:'45',name:'eee'}, {id:'567',name:'rrr'}];
var arr2 = [{id:'124',name:'ttt'}, {id:'45',name:'yyy'}];
const actual = mergeArrays(arr1, arr2);
const expected = [{id:'124',name:'ttt'}, {id:'589',name:'www'}, {id:'45',name:'yyy'}, {id:'567',name:'rrr'}];
expect(actual.sort((a, b) => (a.id < b.id)? -1: 1)).toEqual(expected.sort((a, b) => (a.id < b.id)? -1: 1));
})
Solution 8 - Javascript
// here find all the items that are not it the arr1
const temp = arr1.filter(obj1 => !arr2.some(obj2 => obj1.id === obj2.id))
// then just concat it
arr1 = [...temp, ...arr2]
Solution 9 - Javascript
I'd like to suggest another solution:
const objectToReplace = this.array.find(arrayItem => arrayItem.id === requiredItem.id);
Object.assign(objectToReplace, newObject);
Solution 10 - Javascript
If you don't care about the order of the array, then you may want to get the difference between arr1 and arr2 by id using differenceBy() and then simply use concat() to append all the updated objects.
var result = _(arr1).differenceBy(arr2, 'id').concat(arr2).value();
var arr1 = [{
id: '124',
name: 'qqq'
}, {
id: '589',
name: 'www'
}, {
id: '45',
name: 'eee'
}, {
id: '567',
name: 'rrr'
}]
var arr2 = [{
id: '124',
name: 'ttt'
}, {
id: '45',
name: 'yyy'
}];
var result = _(arr1).differenceBy(arr2, 'id').concat(arr2).value();
console.log(result);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.13.1/lodash.js"></script>
Solution 11 - Javascript
I am only submitting this answer because people expressed concerns over browsers and maintaining the order of objects. I recognize that it is not the most efficient way to accomplish the goal.
Having said this, I broke the problem down into two functions for readability.
// The following function is used for each itertion in the function updateObjectsInArr
const newObjInInitialArr = function(initialArr, newObject) {
let id = newObject.id;
let newArr = [];
for (let i = 0; i < initialArr.length; i++) {
if (id === initialArr[i].id) {
newArr.push(newObject);
} else {
newArr.push(initialArr[i]);
}
}
return newArr;
};
const updateObjectsInArr = function(initialArr, newArr) {
let finalUpdatedArr = initialArr;
for (let i = 0; i < newArr.length; i++) {
finalUpdatedArr = newObjInInitialArr(finalUpdatedArr, newArr[i]);
}
return finalUpdatedArr
}
const revisedArr = updateObjectsInArr(arr1, arr2);
Solution 12 - Javascript
Here a more transparent approach. I find the oneliners harder to read and harder to debug.
export class List {
static replace = (object, list) => {
let newList = [];
list.forEach(function (item) {
if (item.id === object.id) {
newList.push(object);
} else {
newList.push(item);
}
});
return newList;
}
}
Solution 13 - Javascript
function getMatch(elem) {
function action(ele, val) {
if(ele === val){
elem = arr2[i];
}
}
for (var i = 0; i < arr2.length; i++) {
action(elem.id, Object.values(arr2[i])[0]);
}
return elem;
}
var modified = arr1.map(getMatch);
Solution 14 - Javascript
I went with this, because it makes sense to me. Comments added for readers!
masterData = [{id: 1, name: "aaaaaaaaaaa"},
{id: 2, name: "Bill"},
{id: 3, name: "ccccccccc"}];
updatedData = [{id: 3, name: "Cat"},
{id: 1, name: "Apple"}];
updatedData.forEach(updatedObj=> {
// For every updatedData object (dataObj), find the array index in masterData where the IDs match.
let indexInMasterData = masterData.map(masterDataObj => masterDataObj.id).indexOf(updatedObj.id); // First make an array of IDs, to use indexOf().
// If there is a matching ID (and thus an index), replace the existing object in masterData with the updatedData's object.
if (indexInMasterData !== undefined) masterData.splice(indexInMasterData, 1, updatedObj);
});
/* masterData becomes [{id: 1, name: "Apple"},
{id: 2, name: "Bill"},
{id: 3, name: "Cat"}]; as you want.`*/
Solution 15 - Javascript
The accepted answer using array.map is correct but you have to remember to assign it to another variable since array.map doesnt change original array, it actually creates a new array.
//newArr contains the mapped array from arr2 to arr1.
//arr1 still contains original value
var newArr = arr1.map(obj => arr2.find(o => o.id === obj.id) || obj);
Solution 16 - Javascript
Array.prototype.update = function(...args) {
return this.map(x=>args.find((c)=>{return c.id===x.id}) || x)
}
const result =
[
{id:'1',name:'test1'},
{id:'2',name:'test2'},
{id:'3',name:'test3'},
{id:'4',name:'test4'}
]
.update({id:'1',name:'test1.1'}, {id:'3',name:'test3.3'})
console.log(result)
Solution 17 - Javascript
This is how I do it in TypeScript:
const index = this.array.indexOf(this.objectToReplace);
this.array[index] = newObject;