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i was coding and i found this problem, the goal is turn the items array into a object with property key/value, counting the items that appear more than once like that:
{
cookie:{
MILK: 1,
CHOCOLATE: 2,
DELUXE: 1
},
bread:{
BIG: 2
},
beer:{
NEW: 1,
OLD: 1
}
}
I tried this code below
const items = [
"cookie-MILK",
"cookie-CHOCOLATE",
"cookie-CHOCOLATE",
"cookie-DELUXE",
"bread-BIG",
"bread-BIG",
"beer-NEW",
"beer-OLD"
]
let newArray = [];
items.forEach((e) => {
let splitArray = e.split("-");
newArray.push([splitArray[0], splitArray[1]]);
});
let result = newArray.reduce((acc, val) => {
if (!acc[val[0]] && !acc[val[1]] ) acc[val[0]] = {
[val[1]]: 1,
};
else acc[val[0]][val[1]]++;
return acc;
}, {});
But this code returns it and i don't know how to solve this question
{
cookie:{
MILK: 1,
CHOCOLATE: NaN,
DELUXE: NaN
},
bread:{
BIG: 2
},
beer:{
NEW: 1,
OLD: NaN
}
}
You could take a logical nullish assignment ??= for assigning an object or zero and increment the value.
const
items = ["cookie-MILK", "cookie-CHOCOLATE", "cookie-CHOCOLATE", "cookie-DELUXE", "bread-BIG", "bread-BIG", "beer-NEW", "beer-OLD"],
result = items.reduce((acc, val) => {
const [left, right] = val.split("-");
(acc[left] ??= {})[right] ??= 0;
acc[left][right]++;
return acc;
}, {});
console.log(result);
I think is beter solution:
const items = [
"cookie-MILK",
"cookie-CHOCOLATE",
"cookie-CHOCOLATE",
"cookie-DELUXE",
"bread-BIG",
"bread-BIG",
"beer-NEW",
"beer-OLD"
];
let res = {};
items.forEach(item => {
let itemParsed = item.split("-");
if(typeof res[itemParsed[0]] == "undefined")
res[itemParsed[0]] = {}
if(typeof res[itemParsed[0]][itemParsed[1]] == "undefined")
res[itemParsed[0]][itemParsed[1]] = 0;
res[itemParsed[0]][itemParsed[1]]++;
})
console.log(res)
I have got the following code that returns an object based on a filter
i want to convert this to functional programming using map, filter.
var records_object = {
"record": [
"analog",
"laser",
"monochrome",
"digital"
],
"vcd": [
12,
3,
6,
0
],
"dvd": [
1,
0,
0,
16
]
}
var arr_idx = [];
for (i = 0; i < records_object.record.length; i++) {
if (records_object.record[i].match(/digital/i) != null||
records_object.record[i].match(/analog/i) != null) {
arr_idx.push(i);
}
}
for (el in records_object) {
records_object[el] = records_object[el].filter(function (x, i) {
return arr_idx.indexOf(i) != -1;
});
}
console.log(records_object);
so far i was able to do this , now i am stuck
const getIndex = (data) => {
return data.record
.map((e, i) =>
e.includes("analog") || e.includes("digital") ? i : undefined
)
.filter((x) => x !== undefined);
};
You can do this,
var records_object = {
"record": [
"analog",
"laser",
"monochrome",
"digital"
],
"vcd": [
12,
3,
6,
0
],
"dvd": [
1,
0,
0,
16
]
}
let arrayIndexes = records_object.record.map((item, index) => {
if(item.match(/digital/i) != null || item.match(/analog/i) !== null) {
return index;
}
}).filter(item => item !== undefined);
let newObject = Object.keys(records_object).reduce((prev, key) => {
prev[key] = records_object[key].filter((item, index) => arrayIndexes.includes(index));
return prev;
}, {});
console.log(newObject);
The problem was with filter.
when you are running map it returns array [0, undefined, undefined, 3] and that array is being filtered and as you are using filter(x => x), this will iterate through the returned array and remove all the falsy values and return the resulting array.
In [0, undefined, undefined, 3]'s case, only 3 is the truthy value and that's why you are getting only [3] as 0 too is falsy.
You can modify your code slightly to get this resolved.
var records_object = {
record: ["analog", "laser", "monochrome", "digital"],
vcd: [12, 3, 6, 0],
dvd: [1, 0, 0, 16],
};
const getIndex = (data) => {
return data.record
.map((e, i) =>
e.includes("analog") || e.includes("digital") ? i : undefined
)
.filter((x) => x !== undefined);
};
console.log(getIndex(records_object));
Here is the solution using reduce and filter function. I've saved the result in new object.
var records_object = {
"record": [
"analog",
"laser",
"monochrome",
"digital"
],
"vcd": [
12,
3,
6,
0
],
"dvd": [
1,
0,
0,
16
]
};
const matchByString = ['analog', 'digital'];
const isMatch = (el, stringElements) => stringElements.some((strEl) => el.match(new RegExp(strEl, 'i')) != null);
const filterByIndex = records_object.record.reduce((acc, el, index) => isMatch(el, matchByString) ? [...acc, index] : acc, [])
const result = {};
Object.keys(records_object).forEach(i => result[i] = records_object[i].filter((el, i) => filterByIndex.includes(i)));
console.log(result)
I am facing trival problem to flatten simple object with nested one inside.
Tried sollution fro SO, but it throws error:
const newWeather = Object.assign({}, ...function _flatten(o) { return [].concat(...Object.keys(o).map(k => typeof o[k] === 'object' ? _flatten(o[k]) : ({[k]: o[k]})))}({id: 1}))
// also tried these ones:
console.log(Object.keys(weatherDetails).reduce((a, b, c) => {
return Object.assign(a, {
a: b
})
}, {}));
// another one
let newWeather = Object.assign({}, (function() {
var obj = {}
for (var i = 0; i < Object.keys(weatherDetails).length; i++) {
console.log(i, Object.keys(weatherDetails))
obj[Object.keys(weatherDetails)] = weatherDetails[Object.keys(weatherDetails)]
}
return obj
})())
Here is my object I need to flatten, so we need to turn this:
{
temperature: null,
humidity: null,
pressure: null,
windspeed: null,
pollution: {
PM1: 1,
PM10: 2,
PM25: 3
}
}
Into this:
{
temperature: null,
humidity: null,
pressure: null,
windspeed: null,
PM1: 1,
PM10: 2,
PM25: 3
}
Assuming you want to have a generic solution, not one that's custom-tailored to your pollution example with static keys, here's a quick way of achieving that:
You just iterate through your object's property keys. If a property is an object (let's call it child object), you'll copy your child object's properties to your main object.
const obj = {
temperature: null,
humidity: null,
pressure: null,
windspeed: null,
pollution: {
PM1: 1,
PM10: 2,
PM25: 3
}
};
function flatten(object) {
for (const key in object) {
if (!object.hasOwnProperty(key)) {
continue;
}
if (typeof object[key] === 'object' && !Array.isArray(object[key]) && object[key] != null) {
const childObject = object[key];
delete object[key];
object = {...object, ...childObject};
}
}
return object;
}
console.log(flatten(obj));
Just to share a different approach (elegant enough, maybe), here is a solution relying on function generators to recursively flatten an object.
Because it relies on function generators, you can eventually build the object dynamically and skip undesired keys due to the fact the the result is iterable.
The below example has intentionally been made slightly more complex to handle arrays and null values as well, though not required in the original question.
const original = {
temperature: null,
humidity: null,
pressure: null,
windspeed: null,
arrayKey: [1,2,3,'star!'],
fnKey: function(i) {
return i * 3;
},
pollution: {
PM1: 1,
PM10: 2,
PM25: 3
}
};
// Flattens an object.
function* flattenObject(obj, flattenArray = false) {
// Loop each key -> value pair entry in the provided object.
for (const [key, value] of Object.entries(obj)) {
// If the target value is an object and it's not null (because typeof null is 'object'), procede.
if (typeof(value) === 'object' && value !== null) {
// if the targeted value is an array and arrays should be flattened, flatten the array.
if (Array.isArray(value) && flattenArray) yield* flattenObject(value);
// Otherwise, if the value is not an array, flatten it (it must be an object-like or object type).
else if (!Array.isArray(value)) yield* flattenObject(value);
// otherwise, just yield the key->value pair.
else yield [key, value];
}
// otherwise, the value must be something which is not an object, hence, just yield it.
else yield [key, value];
}
}
// usage: assign to a new object all the flattened properties, using the spread operator (...) to assign the values progressively.
const res = Object.fromEntries(flattenObject(original));
console.log(res);
// sample usage by flattening arrays as well.
const res_flattened_arrays = Object.fromEntries(flattenObject(original, true));
console.log(res_flattened_arrays);
// custom object building by skipping a desired key
const resWithoutTemperature = {};
for (const [key, value] of flattenObject(original)) {
if (key !== 'temperature') resWithoutTemperature[key] = value;
}
console.log(resWithoutTemperature);
This would be easier with the Object.entries() method
You loop over the object keys and values, delete all entries that have an object as value and assign the entries off that value to the object.
let a = {
temperature: null,
humidity: null,
pressure: null,
windspeed: null,
pollution: {
PM1: 1,
PM10: 2,
PM25: 3
}
}
Object.entries(a).map(([key, value]) => {
if(value && typeof value === 'object') {
delete a[key]; // Delete entry
Object.assign(a, value); // Add values from entry to object
}
});
console.log(a)
One liner:
Object.entries(a).map(([key, value]) => value && typeof value === 'object' && delete a[key] && Object.assign(a, value));
Also here's a immutable functional approach:
Object.fromEntries(Object.entries(a).map(([key, value]) =>
value && typeof value === 'object' ?
Object.entries(value) : [[key, value]]
).flat());
Personally I prefer this last approach since it doesn't mutate the original or any object.
Just merge and delete every child property that's an instanceof Object.
let obj =
{
temperature: null,
humidity: null,
pressure: null,
windspeed: null,
pollution: {
PM1: 1,
PM10: 2,
PM25: 3,
pollution: 4
}
};
function flatten(obj)
{
obj = Object.assign({}, obj);
for (let i in obj)
if (obj[i] instanceof Object)
{
obj = Object.assign(obj, obj[i]);
// Prevent deletion of property i/"pollution", if it was not replaced by one of the child object's properties
if (obj[i] === obj[i][i])
delete obj[i];
}
return obj;
}
let obj_flattened = flatten(obj);
console.log(obj_flattened);
I usually use Lodash for these kinds of transformations.
With it, it's very straightforward to do so.
Check out the following code sample:
const data = {
temperature: null,
humidity: null,
pressure: null,
windspeed: null,
pollution: {
PM1: 1,
PM10: 2,
PM25: 3
}
};
let flat = _.merge(data, data.pollution);
delete flat.pollution;
console.log(flat); // returns {"temperature":null,"humidity":null,"pressure":null,"windspeed":null,"PM1":1,"PM10":2,"PM25":3}
Try this (it will flatten any object contained in any object) iterating through the object attributes and identifying if an attribute is another object to flat and add to the "root" one:
var o = {
temperature: null,
humidity: null,
pressure: null,
windspeed: null,
pollution: {
PM1: 1,
PM10: 2,
PM25: 3,
newobject:{
a:1,
b:2,
c: {
x:3,
y:4,
z:5
}
}
}
}
function flatten(obj){
let retObj = {};
let objConst = {}.constructor;
for (el in obj){
if(obj[el] !== null && obj[el].constructor === objConst){
retObj = Object.assign({}, retObj, flatten(obj[el]));
} else {
retObj[el] = obj[el];
}
}
return retObj;
}
console.log(flatten(o));
I have an extremely large JSON object structured like this:
{A : 1, B : 2, C : 3, D : 4}
I need a function that can swap the values with keys in my object and I don't know how to do it. I would need an output like this:
{1 : A, 2 : B, 3 : C, 4 : D}
Is there any way that I can do this would manually created a new object where everything is swapped?
Thanks
function swap(json){
var ret = {};
for(var key in json){
ret[json[key]] = key;
}
return ret;
}
Example here FIDDLE don't forget to turn on your console to see the results.
ES6 versions:
static objectFlip(obj) {
const ret = {};
Object.keys(obj).forEach(key => {
ret[obj[key]] = key;
});
return ret;
}
Or using Array.reduce() & Object.keys()
static objectFlip(obj) {
return Object.keys(obj).reduce((ret, key) => {
ret[obj[key]] = key;
return ret;
}, {});
}
Or using Array.reduce() & Object.entries()
static objectFlip(obj) {
return Object.entries(obj).reduce((ret, entry) => {
const [ key, value ] = entry;
ret[ value ] = key;
return ret;
}, {});
}
Now that we have Object.fromEntries:
const f = obj => Object.fromEntries(Object.entries(obj).map(a => a.reverse()))
console.log(
f({A : 'a', B : 'b', C : 'c'})
) // => {a : 'A', b : 'B', c : 'C'}
or:
const f = obj => Object.fromEntries(Object.entries(obj).map(([k, v]) => [v, k]))
console.log(
f({A : 'a', B : 'b', C : 'c'})
) // => {a : 'A', b : 'B', c : 'C'}
(Updated to remove superfluous parentheses - thanks #devin-g-rhode)
you can use lodash function _.invert it also can use multivlaue
var object = { 'a': 1, 'b': 2, 'c': 1 };
_.invert(object);
// => { '1': 'c', '2': 'b' }
// with `multiValue`
_.invert(object, true);
// => { '1': ['a', 'c'], '2': ['b'] }
Using ES6:
const obj = { a: "aaa", b: "bbb", c: "ccc", d: "ddd" };
Object.assign({}, ...Object.entries(obj).map(([a,b]) => ({ [b]: a })))
Get the keys of the object, and then use the Array's reduce function to go through each key and set the value as the key, and the key as the value.
const data = {
A: 1,
B: 2,
C: 3,
D: 4
}
const newData = Object.keys(data).reduce(function(obj, key) {
obj[data[key]] = key;
return obj;
}, {});
console.log(newData);
In ES6/ES2015 you can combine use of Object.keys and reduce with the new Object.assign function, an arrow function, and a computed property name for a pretty straightforward single statement solution.
const foo = { a: 1, b: 2, c: 3 };
const bar = Object.keys(foo)
.reduce((obj, key) => Object.assign({}, obj, { [foo[key]]: key }), {});
If you're transpiling using the object spread operator (stage 3 as of writing this) that will simplify things a bit further.
const foo = { a: 1, b: 2, c: 3 };
const bar = Object.keys(foo)
.reduce((obj, key) => ({ ...obj, [foo[key]]: key }), {});
Finally, if you have Object.entries available (stage 4 as of writing), you can clean up the logic a touch more (IMO).
const foo = { a: 1, b: 2, c: 3 };
const bar = Object.entries(foo)
.reduce((obj, [key, value]) => ({ ...obj, [value]: key }), {});
2021's answer
The concise way by using ES6 syntax like this.
const obj = {A : 1, B : 2, C : 3, D : 4}
console.log(
Object.entries(obj).reduce((acc, [key, value]) => (acc[value] = key, acc), {})
);
Explain:
(acc[value] = key, acc)
Using Comma operator (,) syntax.
The comma operator (,) evaluates each of its operands (from left to
right) and returns the value of the last operand.
As a complement of #joslarson and #jPO answers:
Without ES6 needed, you can use Object.keys Array.reduce and the Comma Operator:
Object.keys(foo).reduce((obj, key) => (obj[foo[key]] = key, obj), {});
Some may find it ugly, but it's "kinda" quicker as the reduce doesn't spread all the properties of the obj on each loop.
Using Ramda:
const swapKeysWithValues =
R.pipe(
R.keys,
R.reduce((obj, k) => R.assoc(source[k], k, obj), {})
);
const result = swapKeysWithValues(source);
Try
let swap = (o,r={})=> Object.keys(o).map(k=> r[o[k]]=k) && r;
let obj = {A : 1, B : 2, C : 3, D : 4};
let swap = (o,r={})=> Object.keys(o).map(k=> r[o[k]]=k) && r;
console.log(swap(obj));
With pure Ramda in a pure and point-free style:
const swapKeysAndValues = R.pipe(
R.toPairs,
R.map(R.reverse),
R.fromPairs,
);
Or, with a little more convoluted ES6 version, still pure functional:
const swapKeysAndValues2 = obj => Object
.entries(obj)
.reduce((newObj, [key, value]) => ({...newObj, [value]: key}), {})
Shortest one I came up with using ES6..
const original = {
first: 1,
second: 2,
third: 3,
fourth: 4,
};
const modified = Object
.entries(original)
.reduce((all, [key, value]) => ({ ...all, [value]: key }), {});
console.log('modified result:', modified);
var data = {A : 1, B : 2, C : 3, D : 4}
var newData = {};
Object.keys(data).forEach(function(key){newData[data[key]]=key});
console.log(newData);
Here is a pure functional implementation of flipping keys and values in ES6:
TypeScript
const flipKeyValues = (originalObj: {[key: string]: string}): {[key: string]: string} => {
if(typeof originalObj === "object" && originalObj !== null ) {
return Object
.entries(originalObj)
.reduce((
acc: {[key: string]: string},
[key, value]: [string, string],
) => {
acc[value] = key
return acc;
}, {})
} else {
return {};
}
}
JavaScript
const flipKeyValues = (originalObj) => {
if(typeof originalObj === "object" && originalObj !== null ) {
return Object
.entries(originalObj)
.reduce((acc, [key, value]) => {
acc[value] = key
return acc;
}, {})
} else {
return {};
}
}
const obj = {foo: 'bar'}
console.log("ORIGINAL: ", obj)
console.log("FLIPPED: ", flipKeyValues(obj))
function swapKV(obj) {
const entrySet = Object.entries(obj);
const reversed = entrySet.map(([k, v])=>[v, k]);
const result = Object.fromEntries(reversed);
return result;
}
This can make your object, {A : 1, B : 2, C : 3, D : 4}, array-like, so you can have
const o = {A : 1, B : 2, C : 3, D : 4}
const arrayLike = swapKV(o);
arrayLike.length = 5;
const array = Array.from(arrayLike);
array.shift(); // undefined
array; // ["A", "B", "C", "D"]
Here is an option that will swap keys with values but not lose duplicates, if your object is : { a: 1, b: 2, c: 2}, it will always return an array in the output :
function swapMap(map) {
const invertedMap = {};
for (const key in map) {
const value = map[key];
invertedMap[value] = invertedMap[value] || [];
invertedMap[value].push(key);
}
return invertedMap;
}
swapMap({a: "1", b: "2", c: "2"})
// Returns => {"1": ["a"], "2":["b", "c"]}
A simple TypeScript variant:
const reverseMap = (map: { [key: string]: string }) => {
return Object.keys(map).reduce((prev, key) => {
const value = map[key];
return { ...prev, [value]: [...(prev.value || []), key] };
}, {} as { [key: string]: [string] })
}
Usage:
const map = { "a":"1", "b":"2", "c":"2" };
const reversedMap = reverseMap(map);
console.log(reversedMap);
Prints:
{ "1":["a"], "2":["b", "c"] }
Rewriting answer of #Vaidd4, but using Object.assign (instead of comma operator):
/**
* Swap object keys and values
* #param {Object<*>} obj
* #returns {Object<string>}
*/
function swapObject(obj) {
return Object.keys(obj).reduce((r, key) => (Object.assign(r, {
[obj[key]]: key,
})), {});
}
Or, shorter:
Object.keys(obj).reduce((r, key) => (Object.assign(r, {[obj[key]]: key})), {});
function myFunction(obj) {
return Object.keys(obj).reduce((acc, cur) => {
return { ...acc, [obj[cur]]: cur };
}, {});
}
This is the solution that I'm using:
function objSwap(obj, tl = false) {
return Object.entries(obj).reduce((a, [k, v]) => (a[v = tl ? v.toLowerCase() : v] = k = tl ? k.toLowerCase() : k, a), {});
}
As a bonus: if you need to swap then check some values I added the possibility to lowercase keys and values. Simply you've to set tl = true, else if you don't need it ...
function objSwap(obj) {
return Object.entries(obj).reduce((a, [k, v]) => (a[v] = k, a), {});
}
Using a for...of loop:
let obj = {A : 1, B : 2, C : 3, D : 4}
for (let [key, value] of Object.entries(obj)){
obj[value] = key
delete obj[key]
}
console.log(obj) // {1: 'A', 2: 'B', 3: 'C', 4: 'D'}
ONE OF THE ES6 WAYS IS HERE
const invertObject = (object) =>Object.entries(object).reduce((result, value) => ({...result, [value[1]]: value[0] }), {});
let obj = invertObject({A : 1, B : 2, C : 3, D : 4});
Here's a type-safe way using TypeScript that has not been suggested before. This solution takes two generics that means the return type will be typed as expected. It's faster than doing methods with .reduce or Object.entries.
// Easier way to type `symbol | number | string` (the only valid keys of an object)
export type AnyKey = keyof any;
export function mirror<K extends AnyKey, V extends AnyKey>(
object: Record<K, V>,
) {
const ret: Partial<Record<V, K>> = {};
for (const key in object) {
ret[object[key]] = key;
}
return ret as Record<V, K>;
}
Usage:
const obj = mirror({
a: 'b',
c: 'd',
});
// {b: 'a', d: 'c'}
obj;
Modern JS solution:
const swapKeyValue = (object) =>
Object.entries(object).reduce((swapped, [key, value]) => (
{ ...swapped, [value]: key }
), {});
Typescript:
type ValidKey = number | string;
const swapKeyValue = <K extends ValidKey, V extends ValidKey>(
object: Record<K, V>
): Record<V, K> =>
Object.entries(object)
.reduce((swapped, [key, value]) => (
{ ...swapped, [value as ValidKey]: key }
), {} as Record<V, K>);
I believe it's better to do this task by using an npm module, like invert-kv.
invert-kv: Invert the key/value of an object. Example: {foo: 'bar'} → {bar: 'foo'}
https://www.npmjs.com/package/invert-kv
const invertKv = require('invert-kv');
invertKv({foo: 'bar', unicorn: 'rainbow'});
//=> {bar: 'foo', rainbow: 'unicorn'}
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