Develop Reference

Sort object array based on another array of keys

I have two arrays containing objects. One contains keys in some order and another has data and I need to sort the data array in order against the given sorted key array. How can I do this?
var a = ['d','a','b','c'] ;
var b = [{a:1},{c:3},{d:4},{b:2}];
The result should be:
result = [{d:4},{a:1},{b:2},{c:3]

Try this
var a = ['d','a','b','c'] ;
var b = [{a:1},{c:3},{d:4},{b:2}];
b.sort(function(x,y){
var xkey = a.indexOf(Object.keys(x)[0]);
var ykey = a.indexOf(Object.keys(y)[0]);
return xkey - ykey;
})
document.body.innerHTML += JSON.stringify(b,0,4);

A different approach from above ones would be, using Lodash Javascript Library.
var a = ['d','a','b','c'] ;
var b = [{a:1},{c:3},{d:4},{b:2}];
var results = _.map(a,function(av) {
var obj = {};
obj[av] = _.find(b, av)[av];
return obj
});
document.body.innerHTML += JSON.stringify(results);
<script src="https://cdn.jsdelivr.net/lodash/4.11.1/lodash.min.js"></script>

This approach respects the keys in the objects of the array for sorting.
Only the items of a are used for lookup and their respective order.
In this case d gets all sort value of the item of b, so d looks like
[ 1, 3, 0, 2 ]
While sorting with indices, we need e, which has simply the indices of b
[ 0, 1, 2, 3 ]
after sorting it became
[ 2, 0, 3, 1 ]
the wanted sort order. Now the original array is remapped to the new order.
But why?
Usually objects contains more than one property. If you use Object.keys and take just the first element, then you could go wrong, because this element is not the wanted element for getting the sort order.
To make it save, you need a different approach, which does not use Array#indexOf in combination with a fixed item of Object.keys.
var a = ['d', 'a', 'b', 'c'],
b = [{ a: 1 }, { c: 3 }, { d: 4 }, { b: 2 }],
d = b.map(function (bb) {
var k = -1;
a.some(function (aa, i) {
if (aa in bb) {
k = i;
return true;
}
});
return k;
}),
e = b.map(function (_, i) { return i; });
e.sort(function (a, b) {
return d[a] - d[b];
});
b = e.map(function (a) {
return b[a];
});
document.write('<pre> ' + JSON.stringify(b, 0, 4) + '</pre>');

This should do the trick
result = a.map(function(key) {
for(var i=0; i<b.length; ++i) {
if(key in b[i]) return b[i];
}
});

Brute force approach is to loop through each of a array and check the b array for it's presence.
var a = ['d','a','b','c'] ;
var b = [{a:1},{c:3},{d:4},{b:2}];
var ans = [];
for(var i = 0; i < a.length; ++i)
{
for(var j = 0; j < b.length; ++j)
{
if(b[j][a[i]])
ans.push(b[j]);
}
}
document.write(JSON.stringify(ans, 0, 4));

Modify an object to a new Array in Javascript

sorry, i m a beginner in javascript.
Can someone explain me how to modify this Object
{toto:[12,13,15],titi:[45,12,34]}
to this Array
newArray = [
{
toto:12,
titi:45
},{
toto:13,
titi:12
},{
toto:15,
titi:34}
]
Also, what the solution if the toto and titi doesn't have the same lenght
Thanks for support!

Here's how I did it. In this way, you don't need to know the names of the keys or the size of the array, but it does require a few loops.
obj = {toto:[12,13,15],titi:[45,12,34]};
newArray = [];
// Find the longest array in your data set
longest = 0;
Object.keys(obj).forEach(function(key) {
if (obj[key].length > longest) {
longest = obj[key].length;
}
});
// Loop through the existing data set to create new objects
for (i = 0; i<longest; i++) {
newObject = {};
Object.keys(obj).forEach(function(key) {
newObject[key] = obj[key][i];
});
newArray.push(newObject);
}
console.log(newArray);
plnkr.co demo in the script.js file.
If you want to ignore keys that would have undefined values for uneven loops, you can add a conditional inside the forEach loop that creates a new object:
Object.keys(obj).forEach(function(key) {
if (obj[key][i] !== undefined) {
newObject[key] = obj[key][i];
}
});

Assuming lengths of toto and titi are the same:
Obj = {toto:[12,13,15],titi:[45,12,34]};
newArray = [];
for (var k in Obj["toto"]) {
newArray.push({ toto:Obj["toto"][k],titi:Obj["titi"][k] });
}

Since the lengths of your inner arrays are equal, you should be able to simply loop through them and add a value from each array (for each iteration) into a new array :
// Your input
var input = {toto:[12,13,15],titi:[45,12,34]};
// An array to store your output
var output = [];
// Since your inner arrays are of equal size, you can loop through them
// as follows
for(var i = 0; i < input.toto.length; i++){
output.push({ toto: input.toto[i], titi: input.titi[i]});
}
You can see a working example of this here and what the output array looks like below :

A more generic approach
var object = { toto: [12, 13, 15], titi: [45, 12, 34] },
newArray = function (o) {
var keys = Object.keys(o),
l = keys.reduce(function (r, a) { return Math.max(r, o[a].length); }, 0),
i = 0,
t,
result = [];
while (i < l) {
t = {};
keys.forEach(function (k) { t[k] = o[k][i]; });
result.push(t);
i++;
}
return result;
}(object);
document.write('<pre>' + JSON.stringify(newArray, 0, 4) + '</pre>');

Array.prototype.indexOf() cannot find array inside of multi-dimensional array [duplicate]

Let's say we have the following js array
var ar = [
[2,6,89,45],
[3,566,23,79],
[434,677,9,23]
];
var val = [3,566,23,79];
Is there a js builtin function or jQuery one with which you can search the array ar for val?
Thanks
***UPDATE*************
Taking fusion's response I created this prototype
Array.prototype.containsArray = function(val) {
var hash = {};
for(var i=0; i<this.length; i++) {
hash[this[i]] = i;
}
return hash.hasOwnProperty(val);
}

you could create a hash.
var ar = [
[2,6,89,45],
[3,566,23,79],
[434,677,9,23]
];
var hash = {};
for(var i = 0 ; i < ar.length; i += 1) {
hash[ar[i]] = i;
}
var val = [434,677,9,23];
if(hash.hasOwnProperty(val)) {
document.write(hash[val]);
}

You can also use a trick with JSON serializing. It is short and simple, but kind of hacky.
It works, because "[0,1]" === "[0,1]".
Here is the working demo snippet:
Array.prototype.indexOfForArrays = function(search)
{
var searchJson = JSON.stringify(search); // "[3,566,23,79]"
var arrJson = this.map(JSON.stringify); // ["[2,6,89,45]", "[3,566,23,79]", "[434,677,9,23]"]
return arrJson.indexOf(searchJson);
};
var arr = [
[2,6,89,45],
[3,566,23,79],
[434,677,9,23]
];
document.body.innerText = arr.indexOfForArrays([3,566,23,79]);

function indexOfArray(val, array) {
var hash = {};
for (var i = 0; i < array.length; i++) {
hash[array[i]] = i;
}
return (hash.hasOwnProperty(val)) ? hash[val] : -1;
};
I consider this more useful for than containsArray(). It solves the same problem (using a hash table) but returns the index (rather than only boolean true/false).

Can you try this?
var ar = [
[2,6,89,45],
[3,566,23,79],
[434,677,9,23]
];
var val = [3,566,23,79];
var sval = val.join("");
for(var i in ar)
{
var sar = ar[i].join("");
if (sar==sval)
{
alert("found!");
break;
}
}

Why don't you use javascript array functions?
function filterArrayByValues(array, values) {
return array.filter(function (arrayItem) {
return values.some(function (value) {
return value === arrayItem;
});
});
}
Or if your array is more complicated, and you want compare only one property but as result return whole object:
function filterArrayByValues(array, values, propertyName) {
return array.filter(function (arrayItem) {
return values.some(function (value) {
return value === arrayItem[propertyName];
});
});
}
More about used functions: filter() and some()

You can use Array.prototype.some(), Array.prototype.every() to check each element of each array.
var ar = [
[2, 6, 89, 45],
[3, 566, 23, 79],
[434, 677, 9, 23]
];
var val = [3, 566, 23, 79];
var bool = ar.some(function(arr) {
return arr.every(function(prop, index) {
return val[index] === prop
})
});
console.log(bool);

I guess there is no such JS functionality available. but you can create one
function arrEquals( one, two )
{
if( one.length != two.length )
{
return false;
}
for( i = 0; i < one.length; i++ )
{
if( one[i] != two[i] )
{
return false;
}
}
return true;
}

The problem with this is that of object/array equality in Javascript. Essentially, the problem is that two arrays are not equal, even if they have the same values. You need to loop through the array and compare the members to your search key (val), but you'll need a way of accurately comparing arrays.
The easiest way round this is to use a library that allows array/object comparison. underscore.js has a very attractive method to do this:
for (var i = 0; i < ar.length; i++) {
if (_.isEqual(ar[i], val)) {
// value is present
}
}
If you don't want to use another library (though I would urge you to -- or at least borrow the message from the Underscore source), you could do this with JSON.stringify...
var valJSON = JSON.stringify(val);
for (var i = 0; i < ar.length; i++) {
if (valJSON === JSON.stringify(ar[i]) {
// value is present
}
}
This will almost certainly be significantly slower, however.

You can use toString convertion to compare elements
var ar = [
[2,6,89,45],
[3,566,23,79],
[434,677,9,23]
];
var val = [3,566,23,79];
s = !ar.every(a => (a.toString() != val.toString()));
console.log(s) // true

Use this instead
if (ar.join(".").indexOf(val) > -1) {
return true;
} else {
return false;
}

Use lodash isEqual
const isValIncludedInAr = ar.some(element => isEqual(element, val))

const arrayOne = [2,6,89,45];
const arrayTwo = [3,566,23,79];
const arrayThree = [434,677,9,23];
const data = new Set([arrayOne, arrayTwo, arrayThree]);
// Check array if exist
console.log( data.has(arrayTwo) ); // It will return true.
// If you want to make a set into array it's simple
const arrayData = [...data];
console.log(arrayData); // It will return [[2,6,89,45], [3,566,23,79], [434,677,9,23]]

Group array values that are in the same range of multiples of 10

I have an array, for example:
var arr = [2,4,7,11,25,608,65,109,99,100,504,606,607];
I need to make it so each value that is within range of its multiple of ten below and and multiple of ten above it is grouped together.
For example, 2,4,7 are between 0 and 10 so they must be together.
11 would be alone in its group, like 25,65 etc.
606,607,608 would be together.
The array above should become:
[ [2,4,7],[11],[25],[65],[99],[101],[504],[100,109],[606,607,608] ]
I've been thinking about it for a couple of hours and I wasn't able to come up with anything yet. It's really bad not much worth mentioning but so far I'm playing with Math.round (http://jsfiddle.net/40napnyx/2/)
Edit:
I'd like to add another issue (although I will not pick the correct answer based on this if the actual answers don't include the solution of this new problem).
The array may contain values with letters. Those values should all be in the same group, in alphabetical order (only the first letter is taken to determine the order). This group should be the last value in the resulting array.
So for instance
var arr = [2, 4, 11,'a3', 25, 7, 'j', 'bzy4];
Would be [[2,4,7],[11],[25],['a3','bzy4', 'j']]

Here's a generic function:
function groupBy(ary, keyFunc) {
var r = {};
ary.forEach(function(x) {
var y = keyFunc(x);
r[y] = (r[y] || []).concat(x);
});
return Object.keys(r).map(function(y) {
return r[y];
});
}
// usage:
var arr = [2,4,7,11,25,608,65,99,101,504,606,607];
g = groupBy(arr, function(x) { return Math.floor(x / 10) });
document.write(JSON.stringify(g));
For your bonus question, just apply groupBy twice:
function groupBy(ary, keyFunc) {
var r = {};
ary.forEach(function(x) {
var y = keyFunc(x);
r[y] = (r[y] || []).concat(x);
});
return Object.keys(r).map(function(y) {
return r[y];
});
}
var arr = [2, 4, 11,'a3', 25, 7, 'j', 'bzy4'];
g = groupBy(arr, isNaN);
g = [].concat(
groupBy(g[0], function(x) { return Math.floor(x / 10) }),
[g[1].sort()]
);
document.write(JSON.stringify(g));

This is what I came up with:
var arr = [2,4,7,11,25,65,99,101,504,606,607,608];
arr.sort(function(a, b) {
if(a > b) return 1;
if(a < b) return -1;
else return 0;
});
var grouped_array = [];
for(var i = 0; i < arr.length; i++) {
var group = [];
var curr_group = Math.floor(arr[i]/10);
while(Math.floor(arr[i]/10) == curr_group) {
group.push(arr[i]);
i++;
}
grouped_array.push(group);
}
http://jsfiddle.net/40napnyx/3/

How to get unique values in an array [duplicate]

This question already has answers here:
Get all unique values in a JavaScript array (remove duplicates)
(91 answers)
Closed 1 year ago.
How can I get a list of unique values in an array? Do I always have to use a second array or is there something similar to java's hashmap in JavaScript?
I am going to be using JavaScript and jQuery only. No additional libraries can be used.

Here's a much cleaner solution for ES6 that I see isn't included here. It uses the Set and the spread operator: ...
var a = [1, 1, 2];
[... new Set(a)]
Which returns [1, 2]

Or for those looking for a one-liner (simple and functional) compatible with current browsers:
let a = ["1", "1", "2", "3", "3", "1"];
let unique = a.filter((item, i, ar) => ar.indexOf(item) === i);
console.log(unique);
Update 2021
I would recommend checking out Charles Clayton's answer, as of recent changes to JS there are even more concise ways to do this.
Update 18-04-2017
It appears as though 'Array.prototype.includes' now has widespread support in the latest versions of the mainline browsers (compatibility)
Update 29-07-2015:
There are plans in the works for browsers to support a standardized 'Array.prototype.includes' method, which although does not directly answer this question; is often related.
Usage:
["1", "1", "2", "3", "3", "1"].includes("2"); // true
Pollyfill (browser support, source from mozilla):
// https://tc39.github.io/ecma262/#sec-array.prototype.includes
if (!Array.prototype.includes) {
Object.defineProperty(Array.prototype, 'includes', {
value: function(searchElement, fromIndex) {
// 1. Let O be ? ToObject(this value).
if (this == null) {
throw new TypeError('"this" is null or not defined');
}
var o = Object(this);
// 2. Let len be ? ToLength(? Get(O, "length")).
var len = o.length >>> 0;
// 3. If len is 0, return false.
if (len === 0) {
return false;
}
// 4. Let n be ? ToInteger(fromIndex).
// (If fromIndex is undefined, this step produces the value 0.)
var n = fromIndex | 0;
// 5. If n ≥ 0, then
// a. Let k be n.
// 6. Else n < 0,
// a. Let k be len + n.
// b. If k < 0, let k be 0.
var k = Math.max(n >= 0 ? n : len - Math.abs(n), 0);
// 7. Repeat, while k < len
while (k < len) {
// a. Let elementK be the result of ? Get(O, ! ToString(k)).
// b. If SameValueZero(searchElement, elementK) is true, return true.
// c. Increase k by 1.
// NOTE: === provides the correct "SameValueZero" comparison needed here.
if (o[k] === searchElement) {
return true;
}
k++;
}
// 8. Return false
return false;
}
});
}

Since I went on about it in the comments for #Rocket's answer, I may as well provide an example that uses no libraries. This requires two new prototype functions, contains and unique
Array.prototype.contains = function(v) {
for (var i = 0; i < this.length; i++) {
if (this[i] === v) return true;
}
return false;
};
Array.prototype.unique = function() {
var arr = [];
for (var i = 0; i < this.length; i++) {
if (!arr.contains(this[i])) {
arr.push(this[i]);
}
}
return arr;
}
var duplicates = [1, 3, 4, 2, 1, 2, 3, 8];
var uniques = duplicates.unique(); // result = [1,3,4,2,8]
console.log(uniques);
For more reliability, you can replace contains with MDN's indexOf shim and check if each element's indexOf is equal to -1: documentation

One Liner, Pure JavaScript
With ES6 syntax
list = list.filter((x, i, a) => a.indexOf(x) === i)
x --> item in array
i --> index of item
a --> array reference, (in this case "list")
With ES5 syntax
list = list.filter(function (x, i, a) {
return a.indexOf(x) === i;
});
Browser Compatibility: IE9+

Using EcmaScript 2016 you can simply do it like this.
var arr = ["a", "a", "b"];
var uniqueArray = Array.from(new Set(arr)); // Unique Array ['a', 'b'];
Sets are always unique, and using Array.from() you can convert a Set to an array. For reference have a look at the documentations.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/from
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set

These days, you can use ES6's Set data type to convert your array to a unique Set. Then, if you need to use array methods, you can turn it back into an Array:
var arr = ["a", "a", "b"];
var uniqueSet = new Set(arr); // {"a", "b"}
var uniqueArr = Array.from(uniqueSet); // ["a", "b"]
//Then continue to use array methods:
uniqueArr.join(", "); // "a, b"

If you want to leave the original array intact,
you need a second array to contain the uniqe elements of the first-
Most browsers have Array.prototype.filter:
const unique = array1.filter((item, index, array) => array.indexOf(item) === index);
//if you need a 'shim':
Array.prototype.filter= Array.prototype.filter || function(fun, scope){
var T= this, A= [], i= 0, itm, L= T.length;
if(typeof fun== 'function'){
while(i<L){
if(i in T){
itm= T[i];
if(fun.call(scope, itm, i, T)) A[A.length]= itm;
}
++i;
}
}
return A;
}
Array.prototype.indexOf= Array.prototype.indexOf || function(what, i){
if(!i || typeof i!= 'number') i= 0;
var L= this.length;
while(i<L){
if(this[i]=== what) return i;
++i;
}
return -1;
}

Fast, compact, no nested loops, works with any object not just strings and numbers, takes a predicate, and only 5 lines of code!!
function findUnique(arr, predicate) {
var found = {};
arr.forEach(d => {
found[predicate(d)] = d;
});
return Object.keys(found).map(key => found[key]);
}
Example: To find unique items by type:
var things = [
{ name: 'charm', type: 'quark'},
{ name: 'strange', type: 'quark'},
{ name: 'proton', type: 'boson'},
];
var result = findUnique(things, d => d.type);
// [
// { name: 'charm', type: 'quark'},
// { name: 'proton', type: 'boson'}
// ]
If you want it to find the first unique item instead of the last add a found.hasOwnPropery() check in there.

Not native in Javascript, but plenty of libraries have this method.
Underscore.js's _.uniq(array) (link) works quite well (source).

If you don't need to worry so much about older browsers, this is exactly what Sets are designed for.
The Set object lets you store unique values of any type, whether
primitive values or object references.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set
const set1 = new Set([1, 2, 3, 4, 5, 1]);
// returns Set(5) {1, 2, 3, 4, 5}

Using jQuery, here's an Array unique function I made:
Array.prototype.unique = function () {
var arr = this;
return $.grep(arr, function (v, i) {
return $.inArray(v, arr) === i;
});
}
console.log([1,2,3,1,2,3].unique()); // [1,2,3]

Short and sweet solution using second array;
var axes2=[1,4,5,2,3,1,2,3,4,5,1,3,4];
var distinct_axes2=[];
for(var i=0;i<axes2.length;i++)
{
var str=axes2[i];
if(distinct_axes2.indexOf(str)==-1)
{
distinct_axes2.push(str);
}
}
console.log("distinct_axes2 : "+distinct_axes2); // distinct_axes2 : 1,4,5,2,3

Majority of the solutions above have a high run time complexity.
Here is the solution that uses reduce and can do the job in O(n) time.
Array.prototype.unique = Array.prototype.unique || function() {
var arr = [];
this.reduce(function (hash, num) {
if(typeof hash[num] === 'undefined') {
hash[num] = 1;
arr.push(num);
}
return hash;
}, {});
return arr;
}
var myArr = [3,1,2,3,3,3];
console.log(myArr.unique()); //[3,1,2];
Note:
This solution is not dependent on reduce. The idea is to create an object map and push unique ones into the array.

You only need vanilla JS to find uniques with Array.some and Array.reduce. With ES2015 syntax it's only 62 characters.
a.reduce((c, v) => b.some(w => w === v) ? c : c.concat(v)), b)
Array.some and Array.reduce are supported in IE9+ and other browsers. Just change the fat arrow functions for regular functions to support in browsers that don't support ES2015 syntax.
var a = [1,2,3];
var b = [4,5,6];
// .reduce can return a subset or superset
var uniques = a.reduce(function(c, v){
// .some stops on the first time the function returns true
return (b.some(function(w){ return w === v; }) ?
// if there's a match, return the array "c"
c :
// if there's no match, then add to the end and return the entire array
c.concat(v)}),
// the second param in .reduce is the starting variable. This is will be "c" the first time it runs.
b);
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/some
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce

Another thought of this question. Here is what I did to achieve this with fewer code.
var distinctMap = {};
var testArray = ['John', 'John', 'Jason', 'Jason'];
for (var i = 0; i < testArray.length; i++) {
var value = testArray[i];
distinctMap[value] = '';
};
var unique_values = Object.keys(distinctMap);
console.log(unique_values);

Array.prototype.unique = function () {
var dictionary = {};
var uniqueValues = [];
for (var i = 0; i < this.length; i++) {
if (dictionary[this[i]] == undefined){
dictionary[this[i]] = i;
uniqueValues.push(this[i]);
}
}
return uniqueValues;
}

I have tried this problem in pure JS.
I have followed following steps 1. Sort the given array, 2. loop through the sorted array, 3. Verify previous value and next value with current value
// JS
var inpArr = [1, 5, 5, 4, 3, 3, 2, 2, 2,2, 100, 100, -1];
//sort the given array
inpArr.sort(function(a, b){
return a-b;
});
var finalArr = [];
//loop through the inpArr
for(var i=0; i<inpArr.length; i++){
//check previous and next value
if(inpArr[i-1]!=inpArr[i] && inpArr[i] != inpArr[i+1]){
finalArr.push(inpArr[i]);
}
}
console.log(finalArr);
Demo

You can enter array with duplicates and below method will return array with unique elements.
function getUniqueArray(array){
var uniqueArray = [];
if (array.length > 0) {
uniqueArray[0] = array[0];
}
for(var i = 0; i < array.length; i++){
var isExist = false;
for(var j = 0; j < uniqueArray.length; j++){
if(array[i] == uniqueArray[j]){
isExist = true;
break;
}
else{
isExist = false;
}
}
if(isExist == false){
uniqueArray[uniqueArray.length] = array[i];
}
}
return uniqueArray;
}

Here is an approach with customizable equals function which can be used for primitives as well as for custom objects:
Array.prototype.pushUnique = function(element, equalsPredicate = (l, r) => l == r) {
let res = !this.find(item => equalsPredicate(item, element))
if(res){
this.push(element)
}
return res
}
usage:
//with custom equals for objects
myArrayWithObjects.pushUnique(myObject, (left, right) => left.id == right.id)
//with default equals for primitives
myArrayWithPrimitives.pushUnique(somePrimitive)

I was just thinking if we can use linear search to eliminate the duplicates:
JavaScript:
function getUniqueRadios() {
var x=document.getElementById("QnA");
var ansArray = new Array();
var prev;
for (var i=0;i<x.length;i++)
{
// Check for unique radio button group
if (x.elements[i].type == "radio")
{
// For the first element prev will be null, hence push it into array and set the prev var.
if (prev == null)
{
prev = x.elements[i].name;
ansArray.push(x.elements[i].name);
} else {
// We will only push the next radio element if its not identical to previous.
if (prev != x.elements[i].name)
{
prev = x.elements[i].name;
ansArray.push(x.elements[i].name);
}
}
}
}
alert(ansArray);
}
HTML:
<body>
<form name="QnA" action="" method='post' ">
<input type="radio" name="g1" value="ANSTYPE1"> good </input>
<input type="radio" name="g1" value="ANSTYPE2"> avg </input>
<input type="radio" name="g2" value="ANSTYPE3"> Type1 </input>
<input type="radio" name="g2" value="ANSTYPE2"> Type2 </input>
<input type="submit" value='SUBMIT' onClick="javascript:getUniqueRadios()"></input>
</form>
</body>