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Compare two array of objects based on their id's

I am trying to compare two arrays of objects based on their IDs and return a boolean. I have filtered out items from the both arrays based on is_selected === 1. If IDs of both arrays are same will return isModified false or it will return true.
Both of these are an array of checkboxes. The initial array is what I am getting from the backend and the current is modified by the user. The current array length can be changed by changing checkboxes.
With my current approach for loop is not executing from the second time.
function cancel() {
const initialArr = [
{id: 8, name: "Celery", is_selected: 1},
{id: 9, name: "Crustaceans", is_selected: 1},
{id: 2, name: "Eggs", is_selected: 1},
{id: 6, name: "Fish", is_selected: 1},
];
const currentArr = [
{id: 8, name: "Celery", is_selected: 1},
{id: 4, name: "Mustard", is_selected: 1},
{id: 2, name: "Eggs", is_selected: 1},
{id: 6, name: "Fish", is_selected: 1},
];
let isModified!: boolean;
for (let index = 0; index < currentArr.length; index++) {
const id = currentArr[index].id;
isModified = initialArr.some((o2) => o2.id !== id);
if (isModified) break;
}
if(isModified){
alert('are you sure?');
} else {
console.log('exit the page');
}
}

There's no issue with the for loop in your code. There are few changes that you could do to improve it
Use a map instead of an array for faster lookups. Running an iteration on array everytime is slow, and it will only increase as the size of arr grows.
Check-in each iteration if the value is changed in the currentArr.
You also need to establish what will happen when there is a new item in the currentArr(i.e with new id), will it be counted as modified or as not modified?
Assuming the answer to the above question is yes, the below code should work
function cancel() {
const initialArr = [
{id: 8, name: "Celery", is_selected: 1},
{id: 9, name: "Crustaceans", is_selected: 1},
{id: 2, name: "Eggs", is_selected: 1},
{id: 6, name: "Fish", is_selected: 1},
];
const initalArrMap = initialArr.reduce((prev, current) => {
return {...prev, [current.id]: current}
}, {});
const currentArr = [
{id: 8, name: "Celery", is_selected: 1},
{id: 9, name: "Mustard", is_selected: 1},
{id: 2, name: "Eggs", is_selected: 1},
{id: 6, name: "Fish", is_selected: 1},
{id: 9, name: "Fish", is_selected: 1},
];
let isModified = currentArr.length !== initialArr.length;
if(!isModified){
for (let index = 0; index < currentArr.length; index++) {
const id = currentArr[index].id;
isModified = initalArrMap[id] ? initalArrMap[id].is_selected != currentArr[index].is_selected : true;
if (isModified) break;
}
}
if(isModified){
alert('are you sure?');
} else {
console.log('exit the page');
}
}
cancel();
Note: I am only checking if is_selected is changed to determine if the currentArr is changed.
Edit: Added a check in case currentArr and initialArr length are not equal

The problem is using initialArr.some((o2) => o2.id !== id).
The some() method tests whether at least one element in the array passes the test implemented by the provided function. It returns true if, in the array, it finds an element for which the provided function returns true; otherwise it returns false.
What do you want to check? If initialArr does not have currentArr ids, it means my currentArr is modified, so you need to try this one:
!initialArr.some((o2) => o2.id === id)
const initialArr = [
{ id: 8, name: "Celery", is_selected: 1 },
{ id: 9, name: "Crustaceans", is_selected: 1 },
{ id: 2, name: "Eggs", is_selected: 1 },
{ id: 6, name: "Fish", is_selected: 1 }
];
const currentArr = [
{ id: 8, name: "Celery", is_selected: 1 },
{ id: 4, name: "Mustard", is_selected: 1 },
{ id: 2, name: "Eggs", is_selected: 1 },
{ id: 6, name: "Fish", is_selected: 1 }
];
let isModified = false;
for (let index = 0; index < currentArr.length; index++) {
const id = currentArr[index].id;
isModified = !initialArr.some((o2) => o2.id === id);
if (isModified)
break;
}
if (isModified) {
alert('are you sure?');
} else {
console.log('exit the page');
}

It will compare each element of first array with all elements of 2nd array
sorting key will make sure it gives correct result if sequence of keys are changed .
it will break the loop as soon as it finds the first different object.
function cancel() {
const a = [
{id: 8, name: "Celery", is_selected: 1},
{id: 9, name: "Crustaceans", is_selected: 1},
{id: 2, name: "Eggs", is_selected: 1},
{id: 6, name: "Fish", is_selected: 1},
];
const b = [
{id: 8, name: "Celery", is_selected: 1},
{id: 4, name: "Mustard", is_selected: 1},
{id: 2, name: "Eggs", is_selected: 1},
{id: 6, name: "Fish", is_selected: 1},
];
let isModified;
let ctr =0;
for (let index = 0; index < a.length; index++) {
for(let j=0; j<b.length; j++){
var aKeys = Object.keys(a[index]).sort();
var bKeys = Object.keys(b[j]).sort();
if(aKeys.toString()!== bKeys.toString()){
ctr++;
break;
}
}
}
if(ctr){
alert('Modified');
} else {
console.log('not modified');
}
}
cancel();

The reason it was returning true for every item in the array is not because of logic error but syntax error, check your array of Objects they don't have commas that separate array elements

Remove equal objects from two arrays

I have the following problem in my ReactJs application. Let's say I have two arrays like this:
var cart = [
{id: 1, name: "item1"},
{id: 2, name: "item2"},
];
var productsArr = [
{proId: 1, category: 'cat1'},
{proId: 5, category: 'cat7'},
];
Is it possible to compare these 2 arrays and find any objects in productsArr which cart's id quals productsArr's proId and remove that object from only productsArr?
(If so, as I explained in this example, productsArr[0] should be removed.)
Thanks in advance.

You can use Array#filter in conjunction with Array#some.
var cart = [
{id: 1, name: "item1"},
{id: 2, name: "item2"},
];
var productsArr = [
{proId: 1, category: 'cat1'},
{proId: 5, category: 'cat7'},
];
productsArr = productsArr.filter(({proId})=>!cart.some(({id})=>proId === id));
console.log(productsArr);

var cart = [
{id: 1, name: "item1"},
{id: 2, name: "item2"},
];
var productsArr = [
{proId: 1, category: 'cat1'},
{proId: 5, category: 'cat7'},
];
for (var i = 0; i<productsArr.length; i++) {
if (cart.find(item => item.id === productsArr[i].proId)) {
productsArr.splice(i,1);
i--;
}
}
console.log(productsArr);

Update array of objects with array of objects

My apologies if this has been addressed before, but I couldn't get it to work with anything I found.
Assume I have 2 arrays - arr1, arr2. I want to update the objects in arr1 if the the property id matches in arr1 and arr2. Objects that exist in arr2 but not in arr1 - meaning the property id does not exist in arr1 - should be pushed to arr1.
Example:
let arr1 = [
{id: 0, name: "John"},
{id: 1, name: "Sara"},
{id: 2, name: "Domnic"},
{id: 3, name: "Bravo"}
]
let arr2 = [
{id: 0, name: "Mark"},
{id: 4, name: "Sara"}
]
# Expected Outcome
let outcome = [
{id: 0, name: "Mark"},
{id: 1, name: "Sara"},
{id: 2, name: "Domnic"},
{id: 3, name: "Bravo"},
{id: 4, name: "Sara"}
]

You can use reduce and find for this:
const arr1 = [
{id: 0, name: "John"},
{id: 1, name: "Sara"},
{id: 2, name: "Domnic"},
{id: 3, name: "Bravo"}
];
const arr2 = [
{id: 0, name: "Mark"},
{id: 4, name: "Sara"}
];
arr2.reduce((res, item) => {
const existingItem = res.find(x => x.id === item.id);
if (existingItem) { existingItem.name = item.name; }
else { res.push(item); }
return res;
}, arr1);
console.log(arr1);

You could do this:
let arr1 = [
{id: 0, name: "John"},
{id: 1, name: "Sara"},
{id: 2, name: "Domnic"},
{id: 3, name: "Bravo"}
]
let arr2 = [
{id: 0, name: "Mark"},
{id: 4, name: "Sara"}
]
var res = arr1.reduce((acc, elem)=>{
var x = arr2.find(i=>i.id === elem.id);
if(x){
acc.push(x)
}else{
acc.push(elem)
}
return acc
}, []);
console.log(res)

Assuming you want to mutate the objects in arr1 rather than creating new ones, one way to do it would be using for...of to iterate the objects in arr2 and then check if there's already an object with the same id in arr1 using Array.prototype.find():
If there is one, you mutate it with Object.assign.
Otherwise, push the new object to arr1:
const arr1 = [
{ id: 0, name: 'John' },
{ id: 1, name: 'Sara' },
{ id: 2, name: 'Domnic' },
{ id: 3, name: 'Bravo' },
];
const arr2 = [
{ id: 0, name: 'Mark', sometingElse: 123 },
{ id: 2, foo: 'bar' },
{ id: 4, name: 'Sara' },
];
for (const currentElement of arr2) {
let previousElement = arr1.find(el => el.id === currentElement.id);
if (previousElement) {
Object.assign(previousElement, currentElement);
} else {
arr1.push(currentElement);
}
}
console.log(arr1);
.as-console-wrapper {
max-height: 100% !important;
}

if you want to try something different you can use foreach and filter to achieve this
let arr1 = [
{id: 0, name: "John"},
{id: 1, name: "Sara"},
{id: 2, name: "Domnic"},
{id: 3, name: "Bravo"}
]
let arr2 = [
{id: 0, name: "Mark"},
{id: 4, name: "Sara"}]
arr1.forEach(x=>{
arr2.forEach(y=>{
if(x.id==y.id){
x.name=y.name
}
})
})
arr2.filter((a)=>{if(!arr1.some(b=>a.id==b.id)) arr1.push(a)})
console.log(arr1)

You should be able to use Array.prototype.find to sort this out!
let arr1 = [
{id: 0, name: "John"},
{id: 1, name: "Sara"},
{id: 2, name: "Domnic"},
{id: 3, name: "Bravo"}
];
let arr2 = [
{id: 0, name: "Mark"},
{id: 4, name: "Sara"}
];
let updateArrayOfObjects = (arr1, arr2) => {
for (let obj of arr2) {
let item = arr1.find(v => v.id === obj.id);
if (item) item.name = obj.name;
else arr1.push({ ...obj });
}
return arr1;
};
console.log(updateArrayOfObjects(arr1, arr2));

Remove same Values from array of Object

I want to remove same object from array by comparing 2 arrays.
Sample Data:
arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
];
arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
let newArray = []; // new array with with no same values it should be unique.
arr1.map((val, i)=>{
arr2.map((val2)=>{
if(val.id == val2.id){
console.log('Matched At: '+ i) // do nothing
}else{
newArray.push(val);
}
})
})
console.log(newArray); // e.g: [{id: 2, name: "b"}, {id: 3, name: "c"},];

Array.filter combined with not Array.some.
The trick here is also to not some,..
const arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
], arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
const newArray=arr1.filter(a=>!arr2.some(s=>s.id===a.id));
console.log(newArray);
.as-console-wrapper { max-height: 100% !important; top: 0; }
As mentioned in comments the question could be interpreted slightly differently. If you also want the unqiue items from arr2, you basically just do it twice and join. IOW: check what not in arr2 is in arr1, and then check what not in arr1 that's in arr2.
eg..
const notIn=(a,b)=>a.filter(f=>!b.some(s=>f.id===s.id));
const newArray=[...notIn(arr1, arr2), ...notIn(arr2, arr1)];
Update 2:
Time complexity, as mentioned by qiAlex there is loops within loops. Although some will short circuit on finding a match, if the dataset gets large things could slow down. This is were Set and Map comes in.
So to fix this using a Set.
const notIn=(a,b)=>a.filter(a=>!b.has(a.id));
const newArray=[
...notIn(arr1, new Set(arr2.map(m=>m.id))),
...notIn(arr2, new Set(arr1.map(m=>m.id)))
];

const isInArray = (arr, id, name) => arr.reduce((result, curr) => ((curr.name === name && curr.id === id) || result), false)
const newArray = arr1.reduce((result, curr) => (isInArray(arr2, curr.id, curr.name) ? result : result.concat(curr)), [])

You can update you code using filter() method, instead of using .map() method like:
const arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
], arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
let newArray = []; // new array with with no same values it should be unique.
newArray = arr1.filter(function(a) {
for(var i=0; i < arr2.length; i++){
if(a.id == arr2[i].id) return false;
}
return true;
});
console.log(newArray);
.as-console-wrapper { max-height: 100% !important; top: 0; }

You check each element in first array whether its id lies in the second array by using Array.prototype.some. If the element is not present then only yield it.
const arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
];
const arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
const result = arr1.filter(x => !arr2.some(y => y.id === x.id));
console.log(result);

I think a simple comparer can works for getting differences and then concat them.
with this method you dont need to check which array is bigger.
arr1 = [ {id: 1, name: "a"}, {id: 2, name: "b"}, {id: 3, name: "c"}, {id: 4, name: "d"}];
arr2 = [ {id: 1, name: "a"}, {id: 4, name: "d"},];
function localComparer(b){
return function(a){
return b.filter(
function(item){
return item.id == a.id && item.name == a.name
}).length == 0;
}
}
var onlyInArr1 = arr1.filter(localComparer(arr2));
var onlyInArr2 = arr2.filter(localComparer(arr1));
console.log(onlyInArr1.concat(onlyInArr2));

We can filter values by checking whether some element is not contained in current array:
const result = arr1.reduce((a, c) => {
if (!arr2.some(a2 => a2.id === c.id))
a.push(c);
return a;
}, [])
An example:
let arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
];
let arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
const result = arr1.reduce((a, c) => {
if (!arr2.some(a2 => a2.id === c.id))
a.push(c);
return a;
}, [])
console.log(result);

Try this one -
const arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
];
const arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
const arr3 = [...arr1, ...arr2];
const mySubArray = _.uniq(arr3, 'id');
console.log(mySubArray);
<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.9.1/underscore-min.js"></script>

So many loops in every answer.
Complexity of the code my answer is 2N,
Idea is:
to merge arrays.
first loop - mark duplicates somehow
second loop - filter duplicates out
arr1 = [
{id: 1, name: "a"},
{id: 2, name: "b"},
{id: 3, name: "c"},
{id: 4, name: "d"},
];
arr2 = [
{id: 1, name: "a"},
{id: 4, name: "d"},
];
let newArray = [...arr1, ...arr2].reduce((acc, item, index) => {
acc.items.push(item);
if (typeof acc.map[item.id] !== 'undefined') {
acc.items[acc.map[item.id]] = null;
acc.items[index] = null;
}
acc.map[item.id] = index;
return acc
}, {map: {}, items: []}).items.filter(item => !!item)
console.log(newArray);

Set order to array of object in javascript

I need to find a simplest way for setting order to array of objects.
For example, there is an array:
var array = [
{id: 1, name: "Matt"},
{id: 2, name: "Jack"},
{id: 3, name: "Morgan"},
{id: 4, name: "Bruce"}
];
and I have provided
var order = [1,4,2,3];
which refers to object id property of array items.
Now I need to reorder array so it should be like:
var array = [
{id: 1, name: "Matt"},
{id: 4, name: "Bruce"},
{id: 2, name: "Jack"},
{id: 3, name: "Morgan"}
]

Use Array#sort method for sorting and inside custom sort function use Array#indexOf method to get index.
var array = [{
id: 1,
name: "Matt"
}, {
id: 2,
name: "Jack"
}, {
id: 3,
name: "Morgan"
}, {
id: 4,
name: "Bruce"
}];
var order = [1, 4, 2, 3];
array.sort(function(a, b) {
// sort based on the index in order array
return order.indexOf(a.id) - order.indexOf(b.id);
})
console.log(array);

You can also use reduce() on [1,4,2,3] array to return object where keys will be elements and values will be index of each element and then sort by that object.
var array = [
{id: 1, name: "Matt"},
{id: 2, name: "Jack"},
{id: 3, name: "Morgan"},
{id: 4, name: "Bruce"}
];
var s = [1,4,2,3].reduce((r, e, i) => {return r[e] = i, r}, {});
var result = array.sort(function(a, b) {
return s[a.id] - s[b.id];
});
console.log(result)

I guess anything that involves sort can not be more efficient than an O(2n) solution. So i would like to do this job with two reduces as follows;
var arr = [{id: 1, name: "Matt"}, {id: 2, name: "Jack"}, {id: 3, name: "Morgan"}, {id: 4, name: "Bruce"}],
order = [1,4,2,3],
lut = order.reduce((t,e,i) => (t[e] = i,t),{}),
result = arr.reduce((res,obj) => (res[lut[obj.id]] = obj, res) ,[]);
console.log(result);