I want to compare 2 arrays of objects in JavaScript code. The objects have 8 total properties, but each object will not have a value for each, and the arrays are never going to be any larger than 8 items each, so maybe the brute force method of traversing each and then looking at the values of the 8 properties is the easiest way to do what I want to do, but before implementing, I wanted to see if anyone had a more elegant solution. Any thoughts?
As serialization doesn't work generally (only when the order of properties matches: JSON.stringify({a:1,b:2}) !== JSON.stringify({b:2,a:1})) you have to check the count of properties and compare each property as well:
const objectsEqual = (o1, o2) =>
Object.keys(o1).length === Object.keys(o2).length
&& Object.keys(o1).every(p => o1[p] === o2[p]);
const obj1 = { name: 'John', age: 33};
const obj2 = { age: 33, name: 'John' };
const obj3 = { name: 'John', age: 45 };
console.log(objectsEqual(obj1, obj2)); // true
console.log(objectsEqual(obj1, obj3)); // false
If you need a deep comparison, you can call the function recursively:
const obj1 = { name: 'John', age: 33, info: { married: true, hobbies: ['sport', 'art'] } };
const obj2 = { age: 33, name: 'John', info: { hobbies: ['sport', 'art'], married: true } };
const obj3 = { name: 'John', age: 33 };
const objectsEqual = (o1, o2) =>
typeof o1 === 'object' && Object.keys(o1).length > 0
? Object.keys(o1).length === Object.keys(o2).length
&& Object.keys(o1).every(p => objectsEqual(o1[p], o2[p]))
: o1 === o2;
console.log(objectsEqual(obj1, obj2)); // true
console.log(objectsEqual(obj1, obj3)); // false
Then it's easy to use this function to compare objects in arrays:
const arr1 = [obj1, obj1];
const arr2 = [obj1, obj2];
const arr3 = [obj1, obj3];
const arraysEqual = (a1, a2) =>
a1.length === a2.length && a1.every((o, idx) => objectsEqual(o, a2[idx]));
console.log(arraysEqual(arr1, arr2)); // true
console.log(arraysEqual(arr1, arr3)); // false
EDIT: You cannot overload operators in current, common browser-based implementations of JavaScript interpreters.
To answer the original question, one way you could do this, and mind you, this is a bit of a hack, simply serialize the two arrays to JSON and then compare the two JSON strings. That would simply tell you if the arrays are different, obviously you could do this to each of the objects within the arrays as well to see which ones were different.
Another option is to use a library which has some nice facilities for comparing objects - I use and recommend MochiKit.
EDIT: The answer kamens gave deserves consideration as well, since a single function to compare two given objects would be much smaller than any library to do what I suggest (although my suggestion would certainly work well enough).
Here is a naïve implemenation that may do just enough for you - be aware that there are potential problems with this implementation:
function objectsAreSame(x, y) {
var objectsAreSame = true;
for(var propertyName in x) {
if(x[propertyName] !== y[propertyName]) {
objectsAreSame = false;
break;
}
}
return objectsAreSame;
}
The assumption is that both objects have the same exact list of properties.
Oh, and it is probably obvious that, for better or worse, I belong to the only-one-return-point camp. :)
Honestly, with 8 objects max and 8 properties max per object, your best bet is to just traverse each object and make the comparisons directly. It'll be fast and it'll be easy.
If you're going to be using these types of comparisons often, then I agree with Jason about JSON serialization...but otherwise there's no need to slow down your app with a new library or JSON serialization code.
I know this is an old question and the answers provided work fine ... but this is a bit shorter and doesn't require any additional libraries ( i.e. JSON ):
function arraysAreEqual(ary1,ary2){
return (ary1.join('') == ary2.join(''));
}
I have worked a bit on a simple algorithm to compare contents of two objects and return an intelligible list of difference. Thought I would share. It borrows some ideas for jQuery, namely the map function implementation and the object and array type checking.
It returns a list of "diff objects", which are arrays with the diff info. It's very simple.
Here it is:
// compare contents of two objects and return a list of differences
// returns an array where each element is also an array in the form:
// [accessor, diffType, leftValue, rightValue ]
//
// diffType is one of the following:
// value: when primitive values at that index are different
// undefined: when values in that index exist in one object but don't in
// another; one of the values is always undefined
// null: when a value in that index is null or undefined; values are
// expressed as boolean values, indicated wheter they were nulls
// type: when values in that index are of different types; values are
// expressed as types
// length: when arrays in that index are of different length; values are
// the lengths of the arrays
//
function DiffObjects(o1, o2) {
// choose a map() impl.
// you may use $.map from jQuery if you wish
var map = Array.prototype.map?
function(a) { return Array.prototype.map.apply(a, Array.prototype.slice.call(arguments, 1)); } :
function(a, f) {
var ret = new Array(a.length), value;
for ( var i = 0, length = a.length; i < length; i++ )
ret[i] = f(a[i], i);
return ret.concat();
};
// shorthand for push impl.
var push = Array.prototype.push;
// check for null/undefined values
if ((o1 == null) || (o2 == null)) {
if (o1 != o2)
return [["", "null", o1!=null, o2!=null]];
return undefined; // both null
}
// compare types
if ((o1.constructor != o2.constructor) ||
(typeof o1 != typeof o2)) {
return [["", "type", Object.prototype.toString.call(o1), Object.prototype.toString.call(o2) ]]; // different type
}
// compare arrays
if (Object.prototype.toString.call(o1) == "[object Array]") {
if (o1.length != o2.length) {
return [["", "length", o1.length, o2.length]]; // different length
}
var diff =[];
for (var i=0; i<o1.length; i++) {
// per element nested diff
var innerDiff = DiffObjects(o1[i], o2[i]);
if (innerDiff) { // o1[i] != o2[i]
// merge diff array into parent's while including parent object name ([i])
push.apply(diff, map(innerDiff, function(o, j) { o[0]="[" + i + "]" + o[0]; return o; }));
}
}
// if any differences were found, return them
if (diff.length)
return diff;
// return nothing if arrays equal
return undefined;
}
// compare object trees
if (Object.prototype.toString.call(o1) == "[object Object]") {
var diff =[];
// check all props in o1
for (var prop in o1) {
// the double check in o1 is because in V8 objects remember keys set to undefined
if ((typeof o2[prop] == "undefined") && (typeof o1[prop] != "undefined")) {
// prop exists in o1 but not in o2
diff.push(["[" + prop + "]", "undefined", o1[prop], undefined]); // prop exists in o1 but not in o2
}
else {
// per element nested diff
var innerDiff = DiffObjects(o1[prop], o2[prop]);
if (innerDiff) { // o1[prop] != o2[prop]
// merge diff array into parent's while including parent object name ([prop])
push.apply(diff, map(innerDiff, function(o, j) { o[0]="[" + prop + "]" + o[0]; return o; }));
}
}
}
for (var prop in o2) {
// the double check in o2 is because in V8 objects remember keys set to undefined
if ((typeof o1[prop] == "undefined") && (typeof o2[prop] != "undefined")) {
// prop exists in o2 but not in o1
diff.push(["[" + prop + "]", "undefined", undefined, o2[prop]]); // prop exists in o2 but not in o1
}
}
// if any differences were found, return them
if (diff.length)
return diff;
// return nothing if objects equal
return undefined;
}
// if same type and not null or objects or arrays
// perform primitive value comparison
if (o1 != o2)
return [["", "value", o1, o2]];
// return nothing if values are equal
return undefined;
}
I tried JSON.stringify() and worked for me.
let array1 = [1,2,{value:'alpha'}] , array2 = [{value:'alpha'},'music',3,4];
JSON.stringify(array1) // "[1,2,{"value":"alpha"}]"
JSON.stringify(array2) // "[{"value":"alpha"},"music",3,4]"
JSON.stringify(array1) === JSON.stringify(array2); // false
There is a optimized code for case when function needs to equals to empty arrays (and returning false in that case)
const objectsEqual = (o1, o2) => {
if (o2 === null && o1 !== null) return false;
return o1 !== null && typeof o1 === 'object' && Object.keys(o1).length > 0 ?
Object.keys(o1).length === Object.keys(o2).length &&
Object.keys(o1).every(p => objectsEqual(o1[p], o2[p]))
: (o1 !== null && Array.isArray(o1) && Array.isArray(o2) && !o1.length &&
!o2.length) ? true : o1 === o2;
}
Here is my attempt, using Node's assert module + npm package object-hash.
I suppose that you would like to check if two arrays contain the same objects, even if those objects are ordered differently between the two arrays.
var assert = require('assert');
var hash = require('object-hash');
var obj1 = {a: 1, b: 2, c: 333},
obj2 = {b: 2, a: 1, c: 444},
obj3 = {b: "AAA", c: 555},
obj4 = {c: 555, b: "AAA"};
var array1 = [obj1, obj2, obj3, obj4];
var array2 = [obj3, obj2, obj4, obj1]; // [obj3, obj3, obj2, obj1] should work as well
// calling assert.deepEquals(array1, array2) at this point FAILS (throws an AssertionError)
// even if array1 and array2 contain the same objects in different order,
// because array1[0].c !== array2[0].c
// sort objects in arrays by their hashes, so that if the arrays are identical,
// their objects can be compared in the same order, one by one
var array1 = sortArrayOnHash(array1);
var array2 = sortArrayOnHash(array2);
// then, this should output "PASS"
try {
assert.deepEqual(array1, array2);
console.log("PASS");
} catch (e) {
console.log("FAIL");
console.log(e);
}
// You could define as well something like Array.prototype.sortOnHash()...
function sortArrayOnHash(array) {
return array.sort(function(a, b) {
return hash(a) > hash(b);
});
}
My practice implementation with sorting, tested and working.
const obj1 = { name: 'John', age: 33};
const obj2 = { age: 33, name: 'John' };
const obj3 = { name: 'John', age: 45 };
const equalObjs = ( obj1, obj2 ) => {
let keyExist = false;
for ( const [key, value] of Object.entries(obj1) ) {
// Search each key in reference object and attach a callback function to
// compare the two object keys
if( Object.keys(obj2).some( ( e ) => e == key ) ) {
keyExist = true;
}
}
return keyExist;
}
console.info( equalObjs( obj1, obj2 ) );
Compare your arrays
// Sort Arrays
var arr1 = arr1.sort(( a, b ) => {
var fa = Object.keys(a);
var fb = Object.keys(b);
if (fa < fb) {
return -1;
}
if (fa > fb) {
return 1;
}
return 0;
});
var arr2 = arr2.sort(( a, b ) => {
var fa = Object.keys(a);
var fb = Object.keys(b);
if (fa < fb) {
return -1;
}
if (fa > fb) {
return 1;
}
return 0;
});
const equalArrays = ( arr1, arr2 ) => {
// If the arrays are different length we an eliminate immediately
if( arr1.length !== arr2.length ) {
return false;
} else if ( arr1.every(( obj, index ) => equalObjs( obj, arr2[index] ) ) ) {
return true;
} else {
return false;
}
}
console.info( equalArrays( arr1, arr2 ) );
I am sharing my compare function implementation as it might be helpful for others:
/*
null AND null // true
undefined AND undefined // true
null AND undefined // false
[] AND [] // true
[1, 2, 'test'] AND ['test', 2, 1] // true
[1, 2, 'test'] AND ['test', 2, 3] // false
[undefined, 2, 'test'] AND ['test', 2, 1] // false
[undefined, 2, 'test'] AND ['test', 2, undefined] // true
[[1, 2], 'test'] AND ['test', [2, 1]] // true
[1, 'test'] AND ['test', [2, 1]] // false
[[2, 1], 'test'] AND ['test', [2, 1]] // true
[[2, 1], 'test'] AND ['test', [2, 3]] // false
[[[3, 4], 2], 'test'] AND ['test', [2, [3, 4]]] // true
[[[3, 4], 2], 'test'] AND ['test', [2, [5, 4]]] // false
[{x: 1, y: 2}, 'test'] AND ['test', {x: 1, y: 2}] // true
1 AND 1 // true
{test: 1} AND ['test', 2, 1] // false
{test: 1} AND {test: 1} // true
{test: 1} AND {test: 2} // false
{test: [1, 2]} AND {test: [1, 2]} // true
{test: [1, 2]} AND {test: [1]} // false
{test: [1, 2], x: 1} AND {test: [1, 2], x: 2} // false
{test: [1, { z: 5 }], x: 1} AND {x: 1, test: [1, { z: 5}]} // true
{test: [1, { z: 5 }], x: 1} AND {x: 1, test: [1, { z: 6}]} // false
*/
function is_equal(x, y) {
const
arr1 = x,
arr2 = y,
is_objects_equal = function (obj_x, obj_y) {
if (!(
typeof obj_x === 'object' &&
Object.keys(obj_x).length > 0
))
return obj_x === obj_y;
return Object.keys(obj_x).length === Object.keys(obj_y).length &&
Object.keys(obj_x).every(p => is_objects_equal(obj_x[p], obj_y[p]));
}
;
if (!( Array.isArray(arr1) && Array.isArray(arr2) ))
return (
arr1 && typeof arr1 === 'object' &&
arr2 && typeof arr2 === 'object'
)
? is_objects_equal(arr1, arr2)
: arr1 === arr2;
if (arr1.length !== arr2.length)
return false;
for (const idx_1 of arr1.keys())
for (const idx_2 of arr2.keys())
if (
(
Array.isArray(arr1[idx_1]) &&
this.is_equal(arr1[idx_1], arr2[idx_2])
) ||
is_objects_equal(arr1[idx_1], arr2[idx_2])
)
{
arr2.splice(idx_2, 1);
break;
}
return !arr2.length;
}
Please try this one:
function used_to_compare_two_arrays(a, b)
{
// This block will make the array of indexed that array b contains a elements
var c = a.filter(function(value, index, obj) {
return b.indexOf(value) > -1;
});
// This is used for making comparison that both have same length if no condition go wrong
if (c.length !== a.length) {
return 0;
} else{
return 1;
}
}
The objectsAreSame function mentioned in #JasonBunting's answer works fine for me. However, there's a little problem: If x[propertyName] and y[propertyName] are objects (typeof x[propertyName] == 'object'), you'll need to call the function recursively in order to compare them.
not sure about the performance ... will have to test on big objects .. however, this works great for me.. the advantage it has compared to the other solutions is, the objects/array do not have to be in the same order ....
it practically takes the first object in the first array, and scans the second array for every objects .. if it's a match, it will proceed to another
there is absolutely a way for optimization but it's working :)
thx to #ttulka I got inspired by his work ... just worked on it a little bit
const objectsEqual = (o1, o2) => {
let match = false
if(typeof o1 === 'object' && Object.keys(o1).length > 0) {
match = (Object.keys(o1).length === Object.keys(o2).length && Object.keys(o1).every(p => objectsEqual(o1[p], o2[p])))
}else {
match = (o1 === o2)
}
return match
}
const arraysEqual = (a1, a2) => {
let finalMatch = []
let itemFound = []
if(a1.length === a2.length) {
finalMatch = []
a1.forEach( i1 => {
itemFound = []
a2.forEach( i2 => {
itemFound.push(objectsEqual(i1, i2))
})
finalMatch.push(itemFound.some( i => i === true))
})
}
return finalMatch.every(i => i === true)
}
const ar1 = [
{ id: 1, name: "Johnny", data: { body: "Some text"}},
{ id: 2, name: "Jimmy"}
]
const ar2 = [
{name: "Jimmy", id: 2},
{name: "Johnny", data: { body: "Some text"}, id: 1}
]
console.log("Match:",arraysEqual(ar1, ar2))
jsfiddle: https://jsfiddle.net/x1pubs6q/
or just use lodash :))))
const _ = require('lodash')
const isArrayEqual = (x, y) => {
return _.isEmpty(_.xorWith(x, y, _.isEqual));
};
using _.some from lodash: https://lodash.com/docs/4.17.11#some
const array1AndArray2NotEqual =
_.some(array1, (a1, idx) => a1.key1 !== array2[idx].key1
|| a1.key2 !== array2[idx].key2
|| a1.key3 !== array2[idx].key3);
There`s my solution. It will compare arrays which also have objects and arrays. Elements can be stay in any positions.
Example:
const array1 = [{a: 1}, {b: 2}, { c: 0, d: { e: 1, f: 2, } }, [1,2,3,54]];
const array2 = [{a: 1}, {b: 2}, { c: 0, d: { e: 1, f: 2, } }, [1,2,3,54]];
const arraysCompare = (a1, a2) => {
if (a1.length !== a2.length) return false;
const objectIteration = (object) => {
const result = [];
const objectReduce = (obj) => {
for (let i in obj) {
if (typeof obj[i] !== 'object') {
result.push(`${i}${obj[i]}`);
} else {
objectReduce(obj[i]);
}
}
};
objectReduce(object);
return result;
};
const reduceArray1 = a1.map(item => {
if (typeof item !== 'object') return item;
return objectIteration(item).join('');
});
const reduceArray2 = a2.map(item => {
if (typeof item !== 'object') return item;
return objectIteration(item).join('');
});
const compare = reduceArray1.map(item => reduceArray2.includes(item));
return compare.reduce((acc, item) => acc + Number(item)) === a1.length;
};
console.log(arraysCompare(array1, array2));
This is work for me to compare two array of objects without taking into consideration the order of the items
const collection1 = [
{ id: "1", name: "item 1", subtitle: "This is a subtitle", parentId: "1" },
{ id: "2", name: "item 2", parentId: "1" },
{ id: "3", name: "item 3", parentId: "1" },
]
const collection2 = [
{ id: "3", name: "item 3", parentId: "1" },
{ id: "2", name: "item 2", parentId: "1" },
{ id: "1", name: "item 1", subtitle: "This is a subtitle", parentId: "1" },
]
const contains = (arr, obj) => {
let i = arr.length;
while (i--) {
if (JSON.stringify(arr[i]) === JSON.stringify(obj)) {
return true;
}
}
return false;
}
const isEqual = (obj1, obj2) => {
let n = 0
if (obj1.length !== obj2.length) {
return false;
}
for (let i = 0; i < obj1.length; i++) {
if (contains(obj2, obj1[i])) {
n++
}
}
return n === obj1.length
}
console.log(isEqual(collection1,collection2))
if you take into consideration the order of the items use built in function in lodash isEqual
comparing with json is pretty bad. try this package to compare nested arrays and get the difference.
https://www.npmjs.com/package/deep-object-diff
If you stringify them...
type AB = {
nome: string;
}
const a: AB[] = [{ nome: 'Célio' }];
const b: AB[] = [{ nome: 'Célio' }];
console.log(a === b); // false
console.log(JSON.stringify(a) === JSON.stringify(b)); // true
I would like to merge an array with another array. The only catch is that each array is within an object.
Intuitively I tried {...arrObj, ...newArrObj} however this leads newArrObj overwriting items in the arrObj.
const array = ['an', 'array'];
const newArray = [, , 'new', 'ehrray'];
const obj = {
key: { ...array
}
};
const newObj = {
key: { ...newArray
}
};
const merged = { ...obj,
...newObj
};
console.log(merged);
I would expect merged to be:
{
"key": {
"0": "an",
"1": "array",
"2": "new",
"3": "ehrray"
}
}
but receive
{
"key": {
"2": "new",
"3": "ehrray"
}
}
This might be useful
const a0 = ['1', '2', undefined , undefined, '5', '6', '7'];
const a1 = [undefined, undefined, '3', '4'];
function merge(a, b) {
return a.map(function(v,i){ return v?v:b[i]});
}
console.log(a0 > a1?merge(a0, a1):merge(a1, a0));
I wanted to updated that I ended up going with a recursive merge to get the nested object containing an array merged.
const array = ['an', 'array'];
const newArray = [, , 'new', 'ehrray'];
const obj = {
key: { ...array
}
};
const newObj = {
key: { ...newArray
}
};
const merge = (obj1, obj2) => {
const recursiveMerge = (obj, entries) => {
for (const [key, value] of entries) {
if (typeof value === "object") {
obj[key] = obj[key] ? { ...obj[key]
} : {};
recursiveMerge(obj[key], Object.entries(value))
} else {
obj[key] = value;
}
}
return obj;
}
return recursiveMerge(obj1, Object.entries(obj2))
}
console.log(merge(obj, newObj));
The idea is that there are unset values with only a few set. eg. const newArray = new Array(4); newArray[2] = 'new';
{ value: null }, even { value: undefined } is not the same thing as { foo: 42 } with no value at all. That's the reason that in your example "an" and "array" are overwritten with the nulls from the newArray.
This particular example you can solve by swapping the order in which you add the arrays to the result, but as soon as both arrays contain null-values there is no way to do it with spread-syntax / Object.assign alone. You have to implement the behaviour:
const array = new Array('an', 'array', null, null, "and", "more", "from", "array");
const newArray = new Array(null, null, 'new', 'ehrray');
function merge(a, b) {
const result = [];
for (let i = 0; i < a.length || i < b.length; ++i) {
result[i] = b[i] == null ? a[i] : b[i];
}
return result;
}
console.log(merge(array, newArray));
I'm trying to compare two arrays of objects and returns a list of updated objects. I don't want to use lodash just the javascript data structures and functions.
E.g:
I have a first array which named arr1 = [
{
name: 'attribute 1',
id: 12,
value: 40,
docs:[],
version: 1,
},
{
name: 'attribute 41',
id: 12,
value: 6,
version: 1,
}
]
And another array:
array2 = [
{
name: 'attribute 1',
attributeTypeId: 12,
value: 65,
docs: ['bla bla']
}
]
I'm trying to iterate through the two arrays and detect the differences and returns an array like that:
result = [
{
name: 'attribute 1',
id: 12,
value: 65,
docs:['bla bla'],
version: 1,
},
{
name: 'attribute 41',
id: 12,
value: 6,
version: 1,
}]
I wrote some uncomplete function (not optimized yet just a brute force solution):
const filterProperties = (e) => {
return e.toLowerCase() !== 'name' && e.toLowerCase() !== 'id'
}
// function sort
const sortProperties = (a, b) => a < b ? -1 : 1;
let result = []
attributesUpdate.forEach(attr => {
const attrProps = Object.getOwnPropertyNames(attr);
// iterate the attributes
for (let i = 0; i < attributes.length; i++) {
let attribute = attributes[i];
// check if the attribute to update has a different name or attributeTypeId
if (attribute.name !== attr.name) {
result = result.concat(attr);
}
// check if the attribute to update has the same name, id
// of the originalOne
if (attribute.name === attr.name && attribute.id=== attr.id) {
let obj = {
name: attribute.name,
id: attribute.id,
}
// get the properties of the attribute
const attributeProps = Object.getOwnPropertyNames(attribute);
// extract the name and id from the list
const filtredAttributeProps = attributeProps.filter(filterProperties);
const filteredattrProps = attrProps.filter(filterProperties);
// returns the length of each array of properties
const attrLength = filteredattrProps.length;
const attributeLength = filtredAttributeProps.length;
if (attrLength === attributeLength) {
for (let j = 0; j < attrLength; j++) {
const propName = filteredattrProps[j];
obj[propName] = attr[propName];
}
result = result.filter(e => e.name === attr.name
&& e.id=== attr.id)
.map(e => Object.assign(e, {obj}))
}
if (attrLength !== attributeLength) {
// sort the array of properties
const sortedAttrProps = filteredattrProps.sort(sortProperties);
const sortedAttributeProps = filtredAttributeProps.sort(sortProperties);
// check the shortest object
const min = attrLength < attributeLength ? attrLength : attributeLength;
// get the biggest object
const longestObjProps = attrLength === min ? sortedAttributeProps : sortedAttrProps;
const longestObj = attrLength === min ? attribute : attr
const shortestProps = attrLength === min ? sortedAttrProps: sortedAttributeProps;
const shortestObj = attrLength === min ? attr : attribute
// fill the object with attr properties
for(let j = 0; j < min; j++) {
const propName = shortestProps[j];
obj[propName] = shortestObj[propName];
}
// fill the remaining properties in the object
const remainingProperties = longestObjProps.filter(e => !shortestProps.includes(e));
for (let j = 0; j < remainingProperties.length; j++) {
const propName = remainingProperties[j];
obj[propName] = longestObj[propName]
}
if (!result.length || result.filter(e => e.name !== attr.name &&
e.id!== attr.id).length === 0) {
result.concat(obj);
}
}
}
}
})
console.log('result: ', result);
I got such a result :
[
{
name: 'attribute 1',
attributeTypeId: 12,
value: 65,
docs: ['bla bla']
}
]
How can I fix this code to get the desired results? I hope that my question will not be downvoted. Any suggestion will be welcome.
What this code does is loop through the objects in array2, and then when it finds that there is a matching name/id in arr1, it simply updates the properties of that object. If not found, it will add the object to arr1.
arr1 = [{
name: 'attribute 1',
id: 12,
value: 40,
docs: [],
version: 1,
},
{
name: 'attribute 41',
id: 12,
value: 6,
version: 1,
}
];
array2 = [{
name: 'attribute 1',
attributeTypeId: 12,
value: 65,
docs: ['bla bla']
}];
updateArray(arr1, array2);
console.log(arr1);
function updateArray(arrayToUpdate, dataToUpdateWith) {
dataToUpdateWith.forEach(function(obj) {
var objToUpdate = checkIfNameIdExists(arrayToUpdate, obj.name, obj.attributeTypeId);
if (objToUpdate === false) {
objToUpdate = obj;
arrayToUpdate.push(objToUpdate);
} else {
for (var prop in obj) {
if (objToUpdate.hasOwnProperty(prop)) {
var nameInFinalObject = prop;
if (prop === "attributeTypeId") {
nameInFinalObject = "id";
}
objToUpdate[nameInFinalObject] = obj[prop];
}
}
}
});
}
function checkIfNameIdExists(arrOfObj, name, id) {
if (name === null) {
return false;
}
var output = false;
arrOfObj.forEach(function(obj) {
if (obj.name === name) {
output = obj;
return true;
}
});
return output;
}
Assumptions:
The values in each of the objects are same type and values are not nested so there is a need to recursively traverse the tree to compare equality etc.
The first array is the source and the subsequent (with the same name) is the mutated form.
We are not handling removals of properties from the source object. From what is given by the OP we are only accounting for value changes.
const d1 = [{ name: 'attribute 1', id: 12, value: 40, docs: [], version: 1, }, { name: 'attribute 41', id: 12, value: 6, version: 1, } ]
const d2 = [{ name: 'attribute 1', attributeTypeId: 12, value: 65, docs: ['bla bla'] }]
const isChanged = (a, b) =>
Array.isArray(a) ? !a.every(x => b.includes(x)) : a !== b
const compare = (o1, o2) => Object.entries(o1).reduce((r, [k,v]) => {
if(k in o2 && isChanged(o2[k], v))
Object.assign(r, {[k]: o2[k]})
return r
}, o1)
const group = (a, b) => [...a, ...b].reduce((r,c) =>
(r[c.name] = [...r[c.name] || [], c], r), {})
const result = Object.values(group(d1,d2)).reduce((r,c) =>
(r.push(c.length == 2 ? compare(...c) : c[0]), r), [])
console.log(result)
The idea is to merge the objects in one array, group them by name and if there ware any changes the groups with length of 2 would be compared by the compare function. Otherwise just added to the end result.
My array object is like below example.
[
{'email':'test#gmail.com', 'name':'abc'},
{'email':'test1#gmail.com', 'name':'bbc'},
{'email':'test2#gmail.com', 'name':'aaa'},
{'email':'test3#gmail.com', 'name':'cba'},
{'email':'test3#gmail.com', 'name':'cab'},
]
So my new array will have key value pare of alphabet A as a key and values are all object that's name start from A alphabet and so on.one more thing is if alphabet A has 2 objects that start from a then I also want to sort then in ascending order as I show in final output example below
final output I want is like this.
[
"a" : [{'email':'test#gmail.com', 'name':'aaa'},{'email':'test2#gmail.com', 'name':'abc'}],
"b" : [{'email':'test1#gmail.com', 'name':'bbc'}],
"c" : [{'email':'test3#gmail.com', 'name':'cab'},{'email':'test3#gmail.com', 'name':'cba'}]
]
You could use reduce method to create an object and inside sort method to sort values by name.
const data = [{'email':'test#gmail.com', 'name':'Abc'},{'email':'test1#gmail.com', 'name':'bbc'},{'email':'test2#gmail.com', 'name':'aaa'},{'email':'test3#gmail.com', 'name':'cba'},{'email':'test3#gmail.com', 'name':'cab'},]
const sorted = data.reduce((r, o) => {
let key = o.name.slice(0, 1).toLowerCase();
r[key] = (r[key] || []).concat(o);
r[key].sort((a, b) => a.name.localeCompare(b.name));
return r;
}, {})
console.log(sorted)
You could sort the array and then group it.
var array = [{ email: 'test#gmail.com', name: 'abc' }, { email: 'test1#gmail.com', name: 'bbc' }, { email: 'test2#gmail.com', name: 'aaa' }, { email: 'test3#gmail.com', name: 'cba' }, { email: 'test3#gmail.com', name: 'cab' }],
grouped = array
.sort(({ name: a }, { name: b }) => a.localeCompare(b))
.reduce((r, o) => {
var group = o.name[0].toLowerCase();
(r[group] = r[group] || []).push(o);
return r;
}, Object.create(null));
console.log(grouped);
.as-console-wrapper { max-height: 100% !important; top: 0; }
var array = [
{'email':'test#gmail.com', 'name':'abc'},
{'email':'test1#gmail.com', 'name':'bbc'},
{'email':'test2#gmail.com', 'name':'aaa'},
{'email':'test3#gmail.com', 'name':'cba'},
{'email':'test3#gmail.com', 'name':'cab'},
]
function getArray(array=[]){
let newObject = {};
array.forEach(i=>{
let key = i['name'].slice(0,1)
if( key && newObject[key] ){
newObject[key].push(i)
}else{
newObject[key] = Array(i)
}
})
return newObject
}
console.log(getArray(array))
You can make it using array reduce method:
const list = [
{'email':'test#gmail.com', 'name':'abc'},
{'email':'test1#gmail.com', 'name':'bbc'},
{'email':'test2#gmail.com', 'name':'aaa'},
{'email':'test3#gmail.com', 'name':'cba'},
{'email':'test3#gmail.com', 'name':'cab'},
]
const newList = list.reduce((acc, currVal) => {
const firstLetter = currVal.name.charAt(0);
if(!acc[firstLetter]){
acc[firstLetter] = [];
}
acc[firstLetter].push(currVal);
return acc
}, {})
console.log(newList)
Loop trough the array, create an object based on it and sort it as is being filled.
const myArr = [
{'email':'test#gmail.com', 'name':'abc'},
{'email':'test1#gmail.com', 'name':'bbc'},
{'email':'test2#gmail.com', 'name':'aaa'},
{'email':'test3#gmail.com', 'name':'cba'},
{'email':'test3#gmail.com', 'name':'cab'}
];
const finalObj = {};
function compare(a,b) {
if (a.name < b.name)
return -1;
if (a.name > b.name)
return 1;
return 0;
}
myArr.forEach(item => {
const alph = item.name.substr(0, 1);
if (finalObj[alph]) {
finalObj[alph] = [...finalObj[alph], item].sort(compare);
} else {
finalObj[alph] = [item];
}
});
console.log(finalObj);
You can use lodash groupBy method to achieve desired result.
var collection =[
{'email':'test#gmail.com', 'name':'abc'},
{'email':'test1#gmail.com', 'name':'bbc'},
{'email':'test2#gmail.com', 'name':'aaa'},
{'email':'test3#gmail.com', 'name':'cba'},
{'email':'test3#gmail.com', 'name':'cab'},
]
console.log(_.groupBy(collection, (item) => {
return item.name[0]
}))