Related
I have two arrays that I need to check the difference upon and return the index of that difference.
For example, I currently have two arrays that get updated when the input's value is changed. The newTags array gets updated whenever there is a new tag within the input, such as #testing. I need to compare the newTags array with the oldTags array and return the index of the difference.
I am currently stringifying both arrays and comparing them that way, although it is unable to return the index of the difference.
var newTags = [];
var oldTags = [];
$input.on('keyup', function () {
var newValue = $input.val();
var pattern = /#[a-zA-Z]+/ig;
var valueSearch = newValue.search(pattern);
if (valueSearch >= 0) {
newTags = newValue.match(pattern);
if ((newTags + "") != (oldTags + "")) {
//Need index of difference here
console.log(newTags, oldTags);
}
oldTags = newTags;
}
});
Working example
You can use a filter to find both the different values and indexes at the same time.
JSFiddle: https://jsfiddle.net/k0uxtnkd/
Array.prototype.diff = function(a) {
var source = this;
return this.filter(function(i) {
if (a.indexOf(i) < 0) {
diffIndexes.push(source.indexOf(i));
return true;
} else {
return false;
}
});
};
var diffIndexes = [];
var newTags = ['a','b','c'];
var oldTags = ['c'];
var diffValues = newTags.diff(oldTags);
console.log(diffIndexes); // [0, 1]
console.log(diffValues); // ['a', 'b']
To convert this to a function instead of add it to the array prototype:
JSFiddle: https://jsfiddle.net/k0uxtnkd/1/
function arrayDiff(a, b) {
return a.filter(function(i) {
if (b.indexOf(i) < 0) {
diffIndexes.push(a.indexOf(i));
return true;
} else {
return false;
}
});
};
var diffIndexes = [];
var newTags = ['a','b','c'];
var oldTags = ['c'];
var diffValues = arrayDiff(newTags, oldTags);
console.log(diffIndexes); // [0, 1]
console.log(diffValues); // ['a', 'b']
You don't need to loop through both arrays, you can simply loop through both simultaneously:
var findDivergence = function (a1, a2) {
var result = [], longerLength = a1.length >= a2.length ? a1.length : a2.length;
for (i = 0; i < longerLength; i++){
if (a1[i] !== a2[i]) {
result.push(i);
}
}
return result;
};
console.log(findDivergence(["a","b","c","d","e","f","g","h","i"], ["a","b","d","r","e","q","g"]));
//outputs [2, 3, 5, 7, 8]
This is significantly more efficient than double-looping or using indexOf (both of which will search the second array many more times than necessary). This also handles cases where the same item shows up more than once in a given array, though if one array is longer than the other and the longer one contains an element that is undefined, that index will count as a match.
for(var i=0; i < newTags.length; i++) {
for(var j=0; j < oldTags.length; j++) {
if(newTags[i] === oldTags[j]) {
console.log("match found");
console.log("Match found for value: " + newTags[i] + " at index in oldTags: " + j + );
}
else{
console.log("match not found");
}
}
}
Using 2 loops you can do a quick check, in the if statements add what you want to happen.
Below is a performance comparison of three common methods to perform the task asked in this question.
const arr1 = ['A', 'B', 'C'];
const arr2 = ['A', 'D', 'C', 'E'];
// Filter indexOf
function diffArray1(a1, a2) {
let aDiffs = [];
a1.filter((i) => {
if (a2.indexOf(i) < 0) {
aDiffs.push(a1.indexOf(i));
}
});
return aDiffs;
};
// Loop indexOf
function diffArray2(a1, a2) {
let aDiffs = [];
for (let i=0; i<a1.length; ++i) {
if (a2.indexOf(a1[i]) < 0) {
aDiffs.push(a1.indexOf(a1[i]));
}
}
return aDiffs;
};
// Loop equality
function diffArray3(a1, a2) {
let aDiffs = [];
for (let i=0; i<a1.length; ++i) {
if (a1[i] !== a2[i]) {
aDiffs.push(i);
}
}
return aDiffs;
};
diffArray1(arr2, arr1); // Returns [1, 3]
diffArray2(arr2, arr1); // Returns [1, 3]
diffArray3(arr2, arr1); // Returns [1, 3]
diffArray3() is the fastest in Chrome v102.0.5005.63 (64-bit) on my system (Intel Core i7-7700HQ 32GB RAM). diffArray1() is about 38% slower and diffArray2() is about 22.5% slower. Here's the test suite:
https://jsbench.me/59l42hhpfs/1
Feel free to fork this and add more methods; please leave the URL of the fork in the comment if you do this.
I have two arrays that I need to check the difference upon and return the index of that difference.
For example, I currently have two arrays that get updated when the input's value is changed. The newTags array gets updated whenever there is a new tag within the input, such as #testing. I need to compare the newTags array with the oldTags array and return the index of the difference.
I am currently stringifying both arrays and comparing them that way, although it is unable to return the index of the difference.
var newTags = [];
var oldTags = [];
$input.on('keyup', function () {
var newValue = $input.val();
var pattern = /#[a-zA-Z]+/ig;
var valueSearch = newValue.search(pattern);
if (valueSearch >= 0) {
newTags = newValue.match(pattern);
if ((newTags + "") != (oldTags + "")) {
//Need index of difference here
console.log(newTags, oldTags);
}
oldTags = newTags;
}
});
Working example
You can use a filter to find both the different values and indexes at the same time.
JSFiddle: https://jsfiddle.net/k0uxtnkd/
Array.prototype.diff = function(a) {
var source = this;
return this.filter(function(i) {
if (a.indexOf(i) < 0) {
diffIndexes.push(source.indexOf(i));
return true;
} else {
return false;
}
});
};
var diffIndexes = [];
var newTags = ['a','b','c'];
var oldTags = ['c'];
var diffValues = newTags.diff(oldTags);
console.log(diffIndexes); // [0, 1]
console.log(diffValues); // ['a', 'b']
To convert this to a function instead of add it to the array prototype:
JSFiddle: https://jsfiddle.net/k0uxtnkd/1/
function arrayDiff(a, b) {
return a.filter(function(i) {
if (b.indexOf(i) < 0) {
diffIndexes.push(a.indexOf(i));
return true;
} else {
return false;
}
});
};
var diffIndexes = [];
var newTags = ['a','b','c'];
var oldTags = ['c'];
var diffValues = arrayDiff(newTags, oldTags);
console.log(diffIndexes); // [0, 1]
console.log(diffValues); // ['a', 'b']
You don't need to loop through both arrays, you can simply loop through both simultaneously:
var findDivergence = function (a1, a2) {
var result = [], longerLength = a1.length >= a2.length ? a1.length : a2.length;
for (i = 0; i < longerLength; i++){
if (a1[i] !== a2[i]) {
result.push(i);
}
}
return result;
};
console.log(findDivergence(["a","b","c","d","e","f","g","h","i"], ["a","b","d","r","e","q","g"]));
//outputs [2, 3, 5, 7, 8]
This is significantly more efficient than double-looping or using indexOf (both of which will search the second array many more times than necessary). This also handles cases where the same item shows up more than once in a given array, though if one array is longer than the other and the longer one contains an element that is undefined, that index will count as a match.
for(var i=0; i < newTags.length; i++) {
for(var j=0; j < oldTags.length; j++) {
if(newTags[i] === oldTags[j]) {
console.log("match found");
console.log("Match found for value: " + newTags[i] + " at index in oldTags: " + j + );
}
else{
console.log("match not found");
}
}
}
Using 2 loops you can do a quick check, in the if statements add what you want to happen.
Below is a performance comparison of three common methods to perform the task asked in this question.
const arr1 = ['A', 'B', 'C'];
const arr2 = ['A', 'D', 'C', 'E'];
// Filter indexOf
function diffArray1(a1, a2) {
let aDiffs = [];
a1.filter((i) => {
if (a2.indexOf(i) < 0) {
aDiffs.push(a1.indexOf(i));
}
});
return aDiffs;
};
// Loop indexOf
function diffArray2(a1, a2) {
let aDiffs = [];
for (let i=0; i<a1.length; ++i) {
if (a2.indexOf(a1[i]) < 0) {
aDiffs.push(a1.indexOf(a1[i]));
}
}
return aDiffs;
};
// Loop equality
function diffArray3(a1, a2) {
let aDiffs = [];
for (let i=0; i<a1.length; ++i) {
if (a1[i] !== a2[i]) {
aDiffs.push(i);
}
}
return aDiffs;
};
diffArray1(arr2, arr1); // Returns [1, 3]
diffArray2(arr2, arr1); // Returns [1, 3]
diffArray3(arr2, arr1); // Returns [1, 3]
diffArray3() is the fastest in Chrome v102.0.5005.63 (64-bit) on my system (Intel Core i7-7700HQ 32GB RAM). diffArray1() is about 38% slower and diffArray2() is about 22.5% slower. Here's the test suite:
https://jsbench.me/59l42hhpfs/1
Feel free to fork this and add more methods; please leave the URL of the fork in the comment if you do this.
What's the fastest/best way to compare two arrays and return the difference? Much like array_diff in PHP. Is there an easy function or am I going to have to create one via each()? or a foreach loop?
I know this is an old question, but I thought I would share this little trick.
var diff = $(old_array).not(new_array).get();
diff now contains what was in old_array that is not in new_array
Working demo http://jsfiddle.net/u9xES/
Good link (Jquery Documentation): http://docs.jquery.com/Main_Page {you can search or read APIs here}
Hope this will help you if you are looking to do it in JQuery.
The alert in the end prompts the array of uncommon element Array i.e. difference between 2 array.
Please lemme know if I missed anything, cheers!
Code
var array1 = [1, 2, 3, 4, 5, 6];
var array2 = [1, 2, 3, 4, 5, 6, 7, 8, 9];
var difference = [];
jQuery.grep(array2, function(el) {
if (jQuery.inArray(el, array1) == -1) difference.push(el);
});
alert(" the difference is " + difference); // Changed variable name
use underscore as :
_.difference(array1,array2)
var arrayDiff = function (firstArr, secondArr) {
var i, o = [], fLen = firstArr.length, sLen = secondArr.length, len;
if (fLen > sLen) {
len = sLen;
} else if (fLen < sLen) {
len = fLen;
} else {
len = sLen;
}
for (i=0; i < len; i++) {
if (firstArr[i] !== secondArr[i]) {
o.push({idx: i, elem1: firstArr[i], elem2: secondArr[i]}); //idx: array index
}
}
if (fLen > sLen) { // first > second
for (i=sLen; i< fLen; i++) {
o.push({idx: i, 0: firstArr[i], 1: undefined});
}
} else if (fLen < sLen) {
for (i=fLen; i< sLen; i++) {
o.push({idx: i, 0: undefined, 1: secondArr[i]});
}
}
return o;
};
/** SUBTRACT ARRAYS **/
function subtractarrays(array1, array2){
var difference = [];
for( var i = 0; i < array1.length; i++ ) {
if( $.inArray( array1[i], array2 ) == -1 ) {
difference.push(array1[i]);
}
}
return difference;
}
You can then call the function anywhere in your code.
var I_like = ["love", "sex", "food"];
var she_likes = ["love", "food"];
alert( "what I like and she does't like is: " + subtractarrays( I_like, she_likes ) ); //returns "Naughty"!
This works in all cases and avoids the problems in the methods above. Hope that helps!
In this way you don't need to worry about if the first array is smaller than the second one.
var arr1 = [1, 2, 3, 4, 5, 6,10],
arr2 = [1, 2, 3, 4, 5, 6, 7, 8, 9];
function array_diff(array1, array2){
var difference = $.grep(array1, function(el) { return $.inArray(el,array2) < 0});
return difference.concat($.grep(array2, function(el) { return $.inArray(el,array1) < 0}));;
}
console.log(array_diff(arr1, arr2));
if you also want to compare the order of the answer you can extend the answer to something like this:
Array.prototype.compareTo = function (array2){
var array1 = this;
var difference = [];
$.grep(array2, function(el) {
if ($.inArray(el, array1) == -1) difference.push(el);
});
if( difference.length === 0 ){
var $i = 0;
while($i < array1.length){
if(array1[$i] !== array2[$i]){
return false;
}
$i++;
}
return true;
}
return false;
}
The short version can be like this:
const diff = (a, b) => b.filter((i) => a.indexOf(i) === -1);
result:
diff(['a', 'b'], ['a', 'b', 'c', 'd']);
["c", "d"]
Array operations like this is not jQuery's strongest point. You should consider a library such as Underscorejs, specifically the difference function.
This should work with unsorted arrays, double values and different orders and length, while giving you the filtered values form array1, array2, or both.
function arrayDiff(arr1, arr2) {
var diff = {};
diff.arr1 = arr1.filter(function(value) {
if (arr2.indexOf(value) === -1) {
return value;
}
});
diff.arr2 = arr2.filter(function(value) {
if (arr1.indexOf(value) === -1) {
return value;
}
});
diff.concat = diff.arr1.concat(diff.arr2);
return diff;
};
var firstArray = [1,2,3,4];
var secondArray = [4,6,1,4];
console.log( arrayDiff(firstArray, secondArray) );
console.log( arrayDiff(firstArray, secondArray).arr1 );
// => [ 2, 3 ]
console.log( arrayDiff(firstArray, secondArray).concat );
// => [ 2, 3, 6 ]
I have two arrays and I want to check if every element in arr2 is in arr1. If the value of an element is repeated in arr2, it needs to be in arr1 an equal number of times. What's the best way of doing this?
arr1 = [1, 2, 3, 4]
arr2 = [1, 2]
checkSuperbag(arr1, arr2)
> true //both 1 and 2 are in arr1
arr1 = [1, 2, 3, 4]
arr2 = [1, 2, 5]
checkSuperbag(arr1, arr2)
> false //5 is not in arr1
arr1 = [1, 2, 3]
arr2 = [1, 2, 3, 3]
checkSuperbag(arr1, arr2)
> false //3 is not in arr1 twice
Do you have to support crummy browsers? If not, the every function should make this easy.
If arr1 is a superset of arr2, then each member in arr2 must be present in arr1
var isSuperset = arr2.every(function(val) { return arr1.indexOf(val) >= 0; });
Here's a fiddle
EDIT
So you're defining superset such that for each element in arr2, it occurs in arr1 the same number of times? I think filter will help you do that (grab the shim from the preceding MDN link to support older browsers):
var isSuperset = arr2.every(function (val) {
var numIn1 = arr1.filter(function(el) { return el === val; }).length;
var numIn2 = arr2.filter(function(el) { return el === val; }).length;
return numIn1 === numIn2;
});
Updated Fiddle
END EDIT
If you do want to support older browsers, the MDN link above has a shim you can add, which I reproduce here for your convenience:
if (!Array.prototype.every)
{
Array.prototype.every = function(fun /*, thisp */)
{
"use strict";
if (this == null)
throw new TypeError();
var t = Object(this);
var len = t.length >>> 0;
if (typeof fun != "function")
throw new TypeError();
var thisp = arguments[1];
for (var i = 0; i < len; i++)
{
if (i in t && !fun.call(thisp, t[i], i, t))
return false;
}
return true;
};
}
EDIT
Note that this will be an O(N2) algorithm, so avoid running it on large arrays.
One option is to sort the two arrays, then traverse both, comparing elements. If an element in the sub-bag candidate is not found in the super-bag, the former is not a sub-bag. Sorting is generally O(n*log(n)) and the comparison is O(max(s,t)), where s and t are the array sizes, for a total time complexity of O(m*log(m)), where m=max(s,t).
function superbag(sup, sub) {
sup.sort();
sub.sort();
var i, j;
for (i=0,j=0; i<sup.length && j<sub.length;) {
if (sup[i] < sub[j]) {
++i;
} else if (sup[i] == sub[j]) {
++i; ++j;
} else {
// sub[j] not in sup, so sub not subbag
return false;
}
}
// make sure there are no elements left in sub
return j == sub.length;
}
If the elements in the actual code are integers, you can use a special-purpose integer sorting algorithm (such as radix sort) for an overall O(max(s,t)) time complexity, though if the bags are small, the built-in Array.sort will likely run faster than a custom integer sort.
A solution with potentially lesser time-complexity is to create a bag type. Integer bags are particularly easy. Flip the existing arrays for the bags: create an object or an array with the integers as keys and a repeat count for values. Using an array won't waste space by creating as arrays are sparse in Javascript. You can use bag operations for sub-bag or super-bag checks. For example, subtract the super from the sub candidate and test if the result non-empty. Alternatively, the contains operation should be O(1) (or possibly O(log(n))), so looping over the sub-bag candidate and testing if the super-bag containment exceeds the sub-bag's containment for each sub-bag element should be O(n) or O(n*log(n)).
The following is untested. Implementation of isInt left as an exercise.
function IntBag(from) {
if (from instanceof IntBag) {
return from.clone();
} else if (from instanceof Array) {
for (var i=0; i < from.length) {
this.add(from[i]);
}
} else if (from) {
for (p in from) {
/* don't test from.hasOwnProperty(p); all that matters
is that p and from[p] are ints
*/
if (isInt(p) && isInt(from[p])) {
this.add(p, from[p]);
}
}
}
}
IntBag.prototype=[];
IntBag.prototype.size=0;
IntBag.prototype.clone = function() {
var clone = new IntBag();
this.each(function(i, count) {
clone.add(i, count);
});
return clone;
};
IntBag.prototype.contains = function(i) {
if (i in this) {
return this[i];
}
return 0;
};
IntBag.prototype.add = function(i, count) {
if (!count) {
count = 1;
}
if (i in this) {
this[i] += count;
} else {
this[i] = count;
}
this.size += count;
};
IntBag.prototype.remove = function(i, count) {
if (! i in this) {
return;
}
if (!count) {
count = 1;
}
this[i] -= count;
if (this[i] > 0) {
// element is still in bag
this.size -= count;
} else {
// remove element entirely
this.size -= count + this[i];
delete this[i];
}
};
IntBag.prototype.each = function(f) {
var i;
foreach (i in this) {
f(i, this[i]);
}
};
IntBag.prototype.find = function(p) {
var result = [];
var i;
foreach (i in this.elements) {
if (p(i, this[i])) {
return i;
}
}
return null;
};
IntBag.prototype.sub = function(other) {
other.each(function(i, count) {
this.remove(i, count);
});
return this;
};
IntBag.prototype.union = function(other) {
var union = this.clone();
other.each(function(i, count) {
if (union.contains(i) < count) {
union.add(i, count - union.contains(i));
}
});
return union;
};
IntBag.prototype.intersect = function(other) {
var intersection = new IntBag();
this.each(function (i, count) {
if (other.contains(i)) {
intersection.add(i, Math.min(count, other.contains(i)));
}
});
return intersection;
};
IntBag.prototype.diff = function(other) {
var mine = this.clone();
mine.sub(other);
var others = other.clone();
others.sub(this);
mine.union(others);
return mine;
};
IntBag.prototype.subbag = function(super) {
return this.size <= super.size
&& null !== this.find(
function (i, count) {
return super.contains(i) < this.contains(i);
}));
};
See also "comparing javascript arrays" for an example implementation of a set of objects, should you ever wish to disallow repetition of elements.
No one has posted a recursive function yet and those are always fun. Call it like arr1.containsArray( arr2 ).
Demo: http://jsfiddle.net/ThinkingStiff/X9jed/
Array.prototype.containsArray = function ( array /*, index, last*/ ) {
if( arguments[1] ) {
var index = arguments[1], last = arguments[2];
} else {
var index = 0, last = 0; this.sort(); array.sort();
};
return index == array.length
|| ( last = this.indexOf( array[index], last ) ) > -1
&& this.containsArray( array, ++index, ++last );
};
Using objects (read: hash tables) in stead of sorting should reduce the amortized complexity to O(m+n):
function bagContains(arr1, arr2) {
var o = {}
var result = true;
// Count all the objects in container
for(var i=0; i < arr1.length; i++) {
if(!o[arr1[i]]) {
o[arr1[i]] = 0;
}
o[arr1[i]]++;
}
// Subtract all the objects in containee
// And exit early if possible
for(var i=0; i < arr2.length; i++) {
if(!o[arr2[i]]) {
o[arr2[i]] = 0;
}
if(--o[arr2[i]] < 0) {
result = false;
break;
}
}
return result;
}
console.log(bagContains([1, 2, 3, 4], [1, 3]));
console.log(bagContains([1, 2, 3, 4], [1, 3, 3]));
console.log(bagContains([1, 2, 3, 4], [1, 3, 7]));
Which yields true, false, false.
Found this on github lodash library. This function use built in functions to solve the problem. .includes() , .indexOf() and .every()
var array1 = ['A', 'B', 'C', 'D', 'E'];
var array2 = ['B', 'C', 'E'];
var array3 = ['B', 'C', 'Z'];
var array4 = [];
function arrayContainsArray (superset, subset) {
if (0 === subset.length) {
return false;
}
return subset.every(function (value) {
return (superset.includes(value));
});
}
function arrayContainsArray1 (superset, subset) {
if (0 === subset.length) {
return false;
}
return subset.every(function (value) {
return (superset.indexOf(value) >= 0);
});
}
console.log(arrayContainsArray(array1,array2)); //true
console.log(arrayContainsArray(array1,array3)); //false
console.log(arrayContainsArray(array1,array4)); //false
console.log(arrayContainsArray1(array1,array2)); //true
console.log(arrayContainsArray1(array1,array3)); //false
console.log(arrayContainsArray1(array1,array4)); //false
If arr2 is subset of arr1, then Length of set(arr1 + arr2) == Length of set(arr1)
var arr1 = [1, 'a', 2, 'b', 3];
var arr2 = [1, 2, 3];
Array.from(new Set(arr1)).length == Array.from(new Set(arr1.concat(arr2))).length
Here is my solution:
Array.prototype.containsIds = function (arr_ids) {
var status = true;
var current_arr = this;
arr_ids.forEach(function(id) {
if(!current_arr.includes(parseInt(id))){
status = false;
return false; // exit forEach
}
});
return status;
};
// Examples
[1,2,3].containsIds([1]); // true
[1,2,3].containsIds([2,3]); // true
[1,2,3].containsIds([3,4]); // false
As for another approach you may do as follows;
function checkIn(a,b){
return b.every(function(e){
return e === this.splice(this.indexOf(e),1)[0];
}, a.slice()); // a.slice() is the "this" in the every method
}
var arr1 = [1, 2, 3, 4],
arr2 = [1, 2],
arr3 = [1,2,3,3];
console.log(checkIn(arr1,arr2));
console.log(checkIn(arr1,arr3));
Quick solution here take two arrays if b is longer than it can't be a super set so return false. Then loop through b to see if a contains the element. If so delete it from a and move on if not return false. Worse case scenario is if b is a subset then time will b.length.
function isSuper(a,b){
var l=b.length,i=0,c;
if(l>a.length){return false}
else{
for(i;i<l;i++){
c=a.indexOf(b[i]);
if(c>-1){
a.splice(c,1);
}
else{return false}
}
return true;
}
}
This assumes that inputs will not always be in order and if a is 1,2,3 and b is 3,2,1 it will still return true.
Yet another simple solution is the following:
let a = [1,2,'a',3,'b',4,5]
let b = [1,2,4]
console.log(b.every((i) => a.includes(i)))
Hope it helps
This question already has answers here:
How to merge two arrays in JavaScript and de-duplicate items
(89 answers)
Closed 4 years ago.
Say I have an array of [34, 35, 45, 48, 49] and another array of [48, 55]. How can I get a resulting array of [34, 35, 45, 48, 49, 55]?
With the arrival of ES6 with sets and splat operator (at the time of being works only in Firefox, check compatibility table), you can write the following cryptic one liner:
var a = [34, 35, 45, 48, 49];
var b = [48, 55];
var union = [...new Set([...a, ...b])];
console.log(union);
Little explanation about this line: [...a, ...b] concatenates two arrays, you can use a.concat(b) as well. new Set() create a set out of it and thus your union. And the last [...x] converts it back to an array.
If you don't need to keep the order, and consider 45 and "45" to be the same:
function union_arrays (x, y) {
var obj = {};
for (var i = x.length-1; i >= 0; -- i)
obj[x[i]] = x[i];
for (var i = y.length-1; i >= 0; -- i)
obj[y[i]] = y[i];
var res = []
for (var k in obj) {
if (obj.hasOwnProperty(k)) // <-- optional
res.push(obj[k]);
}
return res;
}
console.log(union_arrays([34,35,45,48,49], [44,55]));
If you use the library underscore you can write like this
var unionArr = _.union([34,35,45,48,49], [48,55]);
console.log(unionArr);
<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.9.1/underscore-min.js"></script>
Ref: http://underscorejs.org/#union
I'm probably wasting time on a dead thread here. I just had to implement this and went looking to see if I was wasting my time.
I really like KennyTM's answer. That's just how I would attack the problem. Merge the keys into a hash to naturally eliminate duplicates and then extract the keys. If you actually have jQuery you can leverage its goodies to make this a 2 line problem and then roll it into an extension. The each() in jQuery will take care of not iterating over items where hasOwnProperty() is false.
jQuery.fn.extend({
union: function(array1, array2) {
var hash = {}, union = [];
$.each($.merge($.merge([], array1), array2), function (index, value) { hash[value] = value; });
$.each(hash, function (key, value) { union.push(key); } );
return union;
}
});
Note that both of the original arrays are left intact. Then you call it like this:
var union = $.union(array1, array2);
If you wants to concatenate two arrays without any duplicate value,Just try this
var a=[34, 35, 45, 48, 49];
var b=[48, 55];
var c=a.concat(b).sort();
var res=c.filter((value,pos) => {return c.indexOf(value) == pos;} );
function unique(arrayName)
{
var newArray=new Array();
label: for(var i=0; i<arrayName.length;i++ )
{
for(var j=0; j<newArray.length;j++ )
{
if(newArray[j]==arrayName[i])
continue label;
}
newArray[newArray.length] = arrayName[i];
}
return newArray;
}
var arr1 = new Array(0,2,4,4,4,4,4,5,5,6,6,6,7,7,8,9,5,1,2,3,0);
var arr2= new Array(3,5,8,1,2,32,1,2,1,2,4,7,8,9,1,2,1,2,3,4,5);
var union = unique(arr1.concat(arr2));
console.log(union);
Adapted from: https://stackoverflow.com/a/4026828/1830259
Array.prototype.union = function(a)
{
var r = this.slice(0);
a.forEach(function(i) { if (r.indexOf(i) < 0) r.push(i); });
return r;
};
Array.prototype.diff = function(a)
{
return this.filter(function(i) {return a.indexOf(i) < 0;});
};
var s1 = [1, 2, 3, 4];
var s2 = [3, 4, 5, 6];
console.log("s1: " + s1);
console.log("s2: " + s2);
console.log("s1.union(s2): " + s1.union(s2));
console.log("s2.union(s1): " + s2.union(s1));
console.log("s1.diff(s2): " + s1.diff(s2));
console.log("s2.diff(s1): " + s2.diff(s1));
// Output:
// s1: 1,2,3,4
// s2: 3,4,5,6
// s1.union(s2): 1,2,3,4,5,6
// s2.union(s1): 3,4,5,6,1,2
// s1.diff(s2): 1,2
// s2.diff(s1): 5,6
I like Peter Ajtai's concat-then-unique solution, but the code's not very clear. Here's a nicer alternative:
function unique(x) {
return x.filter(function(elem, index) { return x.indexOf(elem) === index; });
};
function union(x, y) {
return unique(x.concat(y));
};
Since indexOf returns the index of the first occurence, we check this against the current element's index (the second parameter to the filter predicate).
Shorter version of kennytm's answer:
function unionArrays(a, b) {
const cache = {};
a.forEach(item => cache[item] = item);
b.forEach(item => cache[item] = item);
return Object.keys(cache).map(key => cache[key]);
};
You can use a jQuery plugin: jQuery Array Utilities
For example the code below
$.union([1, 2, 2, 3], [2, 3, 4, 5, 5])
will return [1,2,3,4,5]
function unite(arr1, arr2, arr3) {
newArr=arr1.concat(arr2).concat(arr3);
a=newArr.filter(function(value){
return !arr1.some(function(value2){
return value == value2;
});
});
console.log(arr1.concat(a));
}//This is for Sorted union following the order :)
function unionArrays() {
var args = arguments,
l = args.length,
obj = {},
res = [],
i, j, k;
while (l--) {
k = args[l];
i = k.length;
while (i--) {
j = k[i];
if (!obj[j]) {
obj[j] = 1;
res.push(j);
}
}
}
return res;
}
var unionArr = unionArrays([34, 35, 45, 48, 49], [44, 55]);
console.log(unionArr);
Somewhat similar in approach to alejandro's method, but a little shorter and should work with any number of arrays.
function unionArray(arrayA, arrayB) {
var obj = {},
i = arrayA.length,
j = arrayB.length,
newArray = [];
while (i--) {
if (!(arrayA[i] in obj)) {
obj[arrayA[i]] = true;
newArray.push(arrayA[i]);
}
}
while (j--) {
if (!(arrayB[j] in obj)) {
obj[arrayB[j]] = true;
newArray.push(arrayB[j]);
}
}
return newArray;
}
var unionArr = unionArray([34, 35, 45, 48, 49], [44, 55]);
console.log(unionArr);
Faster
http://jsperf.com/union-array-faster
I would first concatenate the arrays, then I would return only the unique value.
You have to create your own function to return unique values. Since it is a useful function, you might as well add it in as a functionality of the Array.
In your case with arrays array1 and array2 it would look like this:
array1.concat(array2) - concatenate the two arrays
array1.concat(array2).unique() - return only the unique values. Here unique() is a method you added to the prototype for Array.
The whole thing would look like this:
Array.prototype.unique = function () {
var r = new Array();
o: for(var i = 0, n = this.length; i < n; i++)
{
for(var x = 0, y = r.length; x < y; x++)
{
if(r[x]==this[i])
{
continue o;
}
}
r[r.length] = this[i];
}
return r;
}
var array1 = [34,35,45,48,49];
var array2 = [34,35,45,48,49,55];
// concatenate the arrays then return only the unique values
console.log(array1.concat(array2).unique());
Just wrote before for the same reason (works with any amount of arrays):
/**
* Returns with the union of the given arrays.
*
* #param Any amount of arrays to be united.
* #returns {array} The union array.
*/
function uniteArrays()
{
var union = [];
for (var argumentIndex = 0; argumentIndex < arguments.length; argumentIndex++)
{
eachArgument = arguments[argumentIndex];
if (typeof eachArgument !== 'array')
{
eachArray = eachArgument;
for (var index = 0; index < eachArray.length; index++)
{
eachValue = eachArray[index];
if (arrayHasValue(union, eachValue) == false)
union.push(eachValue);
}
}
}
return union;
}
function arrayHasValue(array, value)
{ return array.indexOf(value) != -1; }
Simple way to deal with merging single array values.
var values[0] = {"id":1235,"name":"value 1"}
values[1] = {"id":4323,"name":"value 2"}
var object=null;
var first=values[0];
for (var i in values)
if(i>0)
object= $.merge(values[i],first)
You can try these:
function union(a, b) {
return a.concat(b).reduce(function(prev, cur) {
if (prev.indexOf(cur) === -1) prev.push(cur);
return prev;
}, []);
}
or
function union(a, b) {
return a.concat(b.filter(function(el) {
return a.indexOf(el) === -1;
}));
}
ES2015 version
Array.prototype.diff = function(a) {return this.filter(i => a.indexOf(i) < 0)};
Array.prototype.union = function(a) {return [...this.diff(a), ...a]}
If you want a custom equals function to match your elements, you can use this function in ES2015:
function unionEquals(left, right, equals){
return left.concat(right).reduce( (acc,element) => {
return acc.some(elt => equals(elt, element))? acc : acc.concat(element)
}, []);
}
It traverses the left+right array. Then for each element, will fill the accumulator if it does not find that element in the accumulator. At the end, there are no duplicate as specified by the equals function.
Pretty, but probably not very efficient with thousands of objects.
I think it would be simplest to create a new array, adding the unique values only as determined by indexOf.
This seems to me to be the most straightforward solution, though I don't know if it is the most efficient. Collation is not preserved.
var a = [34, 35, 45, 48, 49],
b = [48, 55];
var c = union(a, b);
function union(a, b) { // will work for n >= 2 inputs
var newArray = [];
//cycle through input arrays
for (var i = 0, l = arguments.length; i < l; i++) {
//cycle through each input arrays elements
var array = arguments[i];
for (var ii = 0, ll = array.length; ii < ll; ii++) {
var val = array[ii];
//only add elements to the new array if they are unique
if (newArray.indexOf(val) < 0) newArray.push(val);
}
}
return newArray;
}
[i for( i of new Set(array1.concat(array2)))]
Let me break this into parts for you
// This is a list by comprehension
// Store each result in an element of the array
[i
// will be placed in the variable "i", for each element of...
for( i of
// ... the Set which is made of...
new Set(
// ...the concatenation of both arrays
array1.concat(array2)
)
)
]
In other words, it first concatenates both and then it removes the duplicates (a Set, by definition cannot have duplicates)
Do note, though, that the order of the elements is not guaranteed, in this case.