Related
const source = [2, 9, 9, 1, 6];
const ans = source.filter((item, index, arr)=> arr.indexOf(item) === index);
console.log(ans);
here i'm able to remove the duplicate elements but how to make 9 which is repeated highest to come first in the new array??
any help would be appreciated
ans should be [9, 2, 1, 6]
This should work for all cases where array should be sorted by most number of reoccurrences.
const source = [2,1,9,9,6];
const indexes = [];
var ans = source.filter((item,index,arr)=>{
if(arr.indexOf(item) === index){
indexes.push({item:item,count:1})
return true;
}
else if(arr.indexOf(item) !== index){
indexes[indexes.findIndex(object=> object.item === item)].count++
return false;
}
return false;
})
ans =(indexes.sort((a,b)=>{return b.count - a.count}).map(obj =>obj.item))
console.log(ans)
If using more space is okay, you can use a hash map for counting elements and then convert it to an array.
Something like this.
let arr = [2, 9, 9, 1, 6];
// count elements
const map = arr.reduce((acc, e) => acc.set(e, (acc.get(e) || 0) + 1), new Map());
// sort by values and convert back to array
const res = [...map.entries()].sort((a, b) => b[0] - a[0]).map((a) => {
return a[0]
});
console.log(res)
Try This
function removeAndSort(arr) {
var order = {};
for (var i = 0; i < arr.length; i++) {
var value = arr[i];
if (value in order) {
order[value]++;
} else {
order[value] = 1;
}
}
var result = [];
for (value in order) {
result.push(value);
}
function sort(a, b) {
return order[b] - order[a];
}
return result.sort(sort);
}
console.log(removeAndSort([2, 9, 9, 1, 6]));
It's Absolutely Working, Check It
Instead of removing the duplicates with the code you have you need to find a way to create a frequency map to save the duplicates information. In this example I create a map using an object like this...
const freq = { 9: [9, 2], 1: [1, 1] ... };
which uses the current iterated element as a key in an object, and the value is the element along with its duplicate value. You can grab those arrays using Object.values, sort them by that duplicate value, and then map over that sorted nested array to produce the result.
Note, however, due to the sorting process this results in [9, 1, 2, 6].
const source = [2, 9, 9, 1, 6];
// `reduce` over the array creating a key/value pair where
// the value is an array - the first element is the element value,
// and the second value is its duplicate value
const nested = source.reduce((acc, c) => {
// If the object property as defined by the element
// value doesn't exist assign an array to it initialising
// the duplicate value to zero
acc[c] ??= [c, 0];
// Increment the duplicate value
++acc[c][1];
// Return the object for the next iteration
return acc;
}, {});
// We take the `Object.values` of the object (a series of
// nested arrays and sort them by their duplicate value,
// finally `mapping` over the sorted arrays and extracting
// the first element of each
const out = Object.values(nested).sort((a, b) => {
return b[1] - a[1];
}).map(arr => arr[0]);
console.log(out);
Additional documentation
Logical nullish assignment
function sortAndFilter(source) {
let duplicates = {};
//count the duplications
source.filter((item, index, arr) => {
if(arr.indexOf(item) != index)
return duplicates[item] = (duplicates[item] || 0) + 1;
duplicates[item] = 0;
})
//sort the numbers based on the amount of duplications
return Object.keys(duplicates).map(a => parseInt(a)).sort((a, b) => duplicates[b] - duplicates[a]);
}
Output: [ 9, 6, 2, 1 ]
This could do the job
this is best answer for your question
const source = [2, 9, 9, 1, 6];
function duplicate(array) {
let duplicates = array.filter((item, index) => array.indexOf(item) !== index);
return duplicates.concat(array.filter((item) => !duplicates.includes(item)));
}
console.log(duplicate(source));
function myFunction() {
const source = [2, 9, 9, 1, 6];
const ans = source.filter((item, index, arr)=> arr.indexOf(item) === index);
ans.sort((a, b) => b-a);
console.log(ans);
}
Output: [ 9, 6, 2, 1 ]
I need to check a JavaScript array to see if there are any duplicate values. What's the easiest way to do this? I just need to find what the duplicated values are - I don't actually need their indexes or how many times they are duplicated.
I know I can loop through the array and check all the other values for a match, but it seems like there should be an easier way.
Similar question:
Get all unique values in a JavaScript array (remove duplicates)
You could sort the array and then run through it and then see if the next (or previous) index is the same as the current. Assuming your sort algorithm is good, this should be less than O(n2):
const findDuplicates = (arr) => {
let sorted_arr = arr.slice().sort(); // You can define the comparing function here.
// JS by default uses a crappy string compare.
// (we use slice to clone the array so the
// original array won't be modified)
let results = [];
for (let i = 0; i < sorted_arr.length - 1; i++) {
if (sorted_arr[i + 1] == sorted_arr[i]) {
results.push(sorted_arr[i]);
}
}
return results;
}
let duplicatedArray = [9, 9, 111, 2, 3, 4, 4, 5, 7];
console.log(`The duplicates in ${duplicatedArray} are ${findDuplicates(duplicatedArray)}`);
In case, if you are to return as a function for duplicates. This is for similar type of case.
Reference: https://stackoverflow.com/a/57532964/8119511
If you want to elimate the duplicates, try this great solution:
function eliminateDuplicates(arr) {
var i,
len = arr.length,
out = [],
obj = {};
for (i = 0; i < len; i++) {
obj[arr[i]] = 0;
}
for (i in obj) {
out.push(i);
}
return out;
}
console.log(eliminateDuplicates([1,6,7,3,6,8,1,3,4,5,1,7,2,6]))
Source:
http://dreaminginjavascript.wordpress.com/2008/08/22/eliminating-duplicates/
This is my answer from the duplicate thread (!):
When writing this entry 2014 - all examples were for-loops or jQuery. JavaScript has the perfect tools for this: sort, map and reduce.
Find duplicate items
var names = ['Mike', 'Matt', 'Nancy', 'Adam', 'Jenny', 'Nancy', 'Carl']
const uniq = names
.map((name) => {
return {
count: 1,
name: name
};
})
.reduce((result, b) => {
result[b.name] = (result[b.name] || 0) + b.count;
return result;
}, {});
const duplicates = Object.keys(uniq).filter((a) => uniq[a] > 1);
console.log(duplicates); // [ 'Nancy' ]
More functional syntax:
#Dmytro-Laptin pointed out some code that can be removed. This is a more compact version of the same code. Using some ES6 tricks and higher-order functions:
const names = ['Mike', 'Matt', 'Nancy', 'Adam', 'Jenny', 'Nancy', 'Carl'];
const count = names =>
names.reduce((result, value) => ({ ...result,
[value]: (result[value] || 0) + 1
}), {}); // don't forget to initialize the accumulator
const duplicates = dict =>
Object.keys(dict).filter((a) => dict[a] > 1);
console.log(count(names)); // { Mike: 1, Matt: 1, Nancy: 2, Adam: 1, Jenny: 1, Carl: 1 }
console.log(duplicates(count(names))); // [ 'Nancy' ]
UPDATED: Short one-liner to get the duplicates:
[1, 2, 2, 4, 3, 4].filter((e, i, a) => a.indexOf(e) !== i) // [2, 4]
To get the array without duplicates simply invert the condition:
[1, 2, 2, 4, 3, 4].filter((e, i, a) => a.indexOf(e) === i) // [1, 2, 3, 4]
Note that this answer’s main goal is to be short. If you need something performant for a big array, one possible solution is to sort your array first (if it is sortable) then do the following to get the same kind of results as above:
myHugeSortedArray.filter((e, i, a) => a[i-1] === e)
Here is an example for a 1 000 000 integers array:
const myHugeIntArrayWithDuplicates =
[...Array(1_000_000).keys()]
// adding two 0 and four 9 duplicates
.fill(0, 2, 4).fill(9, 10, 14)
console.time("time")
console.log(
myHugeIntArrayWithDuplicates
// a possible sorting method for integers
.sort((a, b) => a > b ? 1 : -1)
.filter((e, i, a) => a[i-1] === e)
)
console.timeEnd("time")
On my AMD Ryzen 7 5700G dev machine it outputs:
[ 0, 0, 9, 9, 9, 9 ]
time: 22.738ms
As pointed out in the comments both the short solution and the performant solution will return an array with several time the same duplicate if it occurs more than once in the original array:
[1, 1, 1, 2, 2, 2, 2].filter((e, i, a) => a.indexOf(e) !== i) // [1, 1, 2, 2, 2]
If unique duplicates are wanted then a function like
function duplicates(arr) {
return [...new Set(arr.filter((e, i, a) => a.indexOf(e) !== i))]
}
can be used so that duplicates([1, 1, 1, 2, 2, 2, 2]) returns [1, 2].
When all you need is to check that there are no duplicates as asked in this question you can use the every() method:
[1, 2, 3].every((e, i, a) => a.indexOf(e) === i) // true
[1, 2, 1].every((e, i, a) => a.indexOf(e) === i) // false
Note that every() doesn't work for IE 8 and below.
Find duplicate values in an array
This should be one of the shortest ways to actually find duplicate values in an array. As specifically asked for by the OP, this does not remove duplicates but finds them.
var input = [1, 2, 3, 1, 3, 1];
var duplicates = input.reduce(function(acc, el, i, arr) {
if (arr.indexOf(el) !== i && acc.indexOf(el) < 0) acc.push(el); return acc;
}, []);
document.write(duplicates); // = 1,3 (actual array == [1, 3])
This doesn't need sorting or any third party framework. It also doesn't need manual loops. It works with every value indexOf() (or to be clearer: the strict comparision operator) supports.
Because of reduce() and indexOf() it needs at least IE 9.
You can add this function, or tweak it and add it to Javascript's Array prototype:
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])
{
alert('this is a DUPE!');
continue o;
}
}
r[r.length] = this[i];
}
return r;
}
var arr = [1,2,2,3,3,4,5,6,2,3,7,8,5,9];
var unique = arr.unique();
alert(unique);
UPDATED: The following uses an optimized combined strategy. It optimizes primitive lookups to benefit from hash O(1) lookup time (running unique on an array of primitives is O(n)). Object lookups are optimized by tagging objects with a unique id while iterating through so so identifying duplicate objects is also O(1) per item and O(n) for the whole list. The only exception is items that are frozen, but those are rare and a fallback is provided using an array and indexOf.
var unique = function(){
var hasOwn = {}.hasOwnProperty,
toString = {}.toString,
uids = {};
function uid(){
var key = Math.random().toString(36).slice(2);
return key in uids ? uid() : uids[key] = key;
}
function unique(array){
var strings = {}, numbers = {}, others = {},
tagged = [], failed = [],
count = 0, i = array.length,
item, type;
var id = uid();
while (i--) {
item = array[i];
type = typeof item;
if (item == null || type !== 'object' && type !== 'function') {
// primitive
switch (type) {
case 'string': strings[item] = true; break;
case 'number': numbers[item] = true; break;
default: others[item] = item; break;
}
} else {
// object
if (!hasOwn.call(item, id)) {
try {
item[id] = true;
tagged[count++] = item;
} catch (e){
if (failed.indexOf(item) === -1)
failed[failed.length] = item;
}
}
}
}
// remove the tags
while (count--)
delete tagged[count][id];
tagged = tagged.concat(failed);
count = tagged.length;
// append primitives to results
for (i in strings)
if (hasOwn.call(strings, i))
tagged[count++] = i;
for (i in numbers)
if (hasOwn.call(numbers, i))
tagged[count++] = +i;
for (i in others)
if (hasOwn.call(others, i))
tagged[count++] = others[i];
return tagged;
}
return unique;
}();
If you have ES6 Collections available, then there is a much simpler and significantly faster version. (shim for IE9+ and other browsers here: https://github.com/Benvie/ES6-Harmony-Collections-Shim)
function unique(array){
var seen = new Set;
return array.filter(function(item){
if (!seen.has(item)) {
seen.add(item);
return true;
}
});
}
var a = ["a","a","b","c","c"];
a.filter(function(value,index,self){ return (self.indexOf(value) !== index )})
This should get you what you want, Just the duplicates.
function find_duplicates(arr) {
var len=arr.length,
out=[],
counts={};
for (var i=0;i<len;i++) {
var item = arr[i];
counts[item] = counts[item] >= 1 ? counts[item] + 1 : 1;
if (counts[item] === 2) {
out.push(item);
}
}
return out;
}
find_duplicates(['one',2,3,4,4,4,5,6,7,7,7,'pig','one']); // -> ['one',4,7] in no particular order.
Find non-unique values from 3 arrays (or more):
ES2015
// 🚩🚩 🚩 🚩 🚩
var arr = [1,2,2,3,3,4,5,6,2,3,7,8,5,22],
arr2 = [1,2,511,12,50],
arr3 = [22,0],
merged,
nonUnique;
// Combine all the arrays to a single one
merged = arr.concat(arr2, arr3)
// create a new (dirty) Array with only the non-unique items
nonUnique = merged.filter((item,i) => merged.includes(item, i+1))
// Cleanup - remove duplicate & empty items items
nonUnique = [...new Set(nonUnique)]
console.log(nonUnique)
PRE-ES2015:
In the below example I chose to superimpose a unique method on top of the Array prototype, allowing access from everywhere and has more "declarative" syntax. I do not recommend this approach on large projects, since it might very well collide with another method with the same custom name.
Array.prototype.unique = function () {
var arr = this.sort(), i=arr.length; // input must be sorted for this to work
while(i--)
arr[i] === arr[i-1] && arr.splice(i,1) // remove duplicate item
return arr
}
Array.prototype.nonunique = function () {
var arr = this.sort(), i=arr.length, res = []; // input must be sorted for this to work
while(i--)
arr[i] === arr[i-1] && (res.indexOf(arr[i]) == -1) && res.push(arr[i])
return res
}
// 🚩🚩 🚩 🚩 🚩
var arr = [1,2,2,3,3,4,5,6,2,3,7,8,5,22],
arr2 = [1,2,511,12,50],
arr3 = [22,0],
// merge all arrays & call custom Array Prototype - "unique"
unique = arr.concat(arr2, arr3).unique(),
nonunique = arr.concat(arr2, arr3).nonunique()
console.log(unique) // [1,12,2,22,3,4,5,50,511,6,7,8]
console.log(nonunique) // [1,12,2,22,3,4,5,50,511,6,7,8]
using underscore.js
function hasDuplicate(arr){
return (arr.length != _.uniq(arr).length);
}
The simplest and quickest way is to use the Set object:
const numbers = [1, 2, 3, 2, 4, 5, 5, 6];
const set = new Set(numbers);
const duplicates = numbers.filter(item => {
if (set.has(item)) {
set.delete(item);
return false;
} else {
return true;
}
});
// OR more concisely
const duplicates = numbers.filter(item => !set.delete(item));
console.log(duplicates);
// [ 2, 5 ]
This is my proposal (ES6):
let a = [1, 2, 3, 4, 2, 2, 4, 1, 5, 6]
let b = [...new Set(a.sort().filter((o, i) => o !== undefined && a[i + 1] !== undefined && o === a[i + 1]))]
// b is now [1, 2, 4]
Here's the simplest solution I could think of:
const arr = [-1, 2, 2, 2, 0, 0, 0, 500, -1, 'a', 'a', 'a']
const filtered = arr.filter((el, index) => arr.indexOf(el) !== index)
// => filtered = [ 2, 2, 0, 0, -1, 'a', 'a' ]
const duplicates = [...new Set(filtered)]
console.log(duplicates)
// => [ 2, 0, -1, 'a' ]
That's it.
Note:
It works with any numbers including 0, strings and negative numbers e.g. -1 -
Related question: Get all unique values in a JavaScript array (remove duplicates)
The original array arr is preserved (filter returns the new array instead of modifying the original)
The filtered array contains all duplicates; it can also contain more than 1 same value (e.g. our filtered array here is [ 2, 2, 0, 0, -1, 'a', 'a' ])
If you want to get only values that are duplicated (you don't want to have multiple duplicates with the same value) you can use [...new Set(filtered)] (ES6 has an object Set which can store only unique values)
Hope this helps.
Here is mine simple and one line solution.
It searches not unique elements first, then makes found array unique with the use of Set.
So we have array of duplicates in the end.
var array = [1, 2, 2, 3, 3, 4, 5, 6, 2, 3, 7, 8, 5, 22, 1, 2, 511, 12, 50, 22];
console.log([...new Set(
array.filter((value, index, self) => self.indexOf(value) !== index))]
);
Shortest vanilla JS:
[1,1,2,2,2,3].filter((v,i,a) => a.indexOf(v) !== i) // [1, 2, 2]
one liner simple way
var arr = [9,1,2,4,3,4,9]
console.log(arr.filter((ele,indx)=>indx!==arr.indexOf(ele))) //get the duplicates
console.log(arr.filter((ele,indx)=>indx===arr.indexOf(ele))) //remove the duplicates
var a = [324,3,32,5,52,2100,1,20,2,3,3,2,2,2,1,1,1].sort();
a.filter(function(v,i,o){return i&&v!==o[i-1]?v:0;});
or when added to the prototyp.chain of Array
//copy and paste: without error handling
Array.prototype.unique =
function(){return this.sort().filter(function(v,i,o){return i&&v!==o[i-1]?v:0;});}
See here: https://gist.github.com/1305056
Fast and elegant way using es6 object destructuring and reduce
It runs in O(n) (1 iteration over the array) and doesn't repeat values that appear more than 2 times
const arr = ['hi', 'hi', 'hi', 'bye', 'bye', 'asd']
const {
dup
} = arr.reduce(
(acc, curr) => {
acc.items[curr] = acc.items[curr] ? acc.items[curr] += 1 : 1
if (acc.items[curr] === 2) acc.dup.push(curr)
return acc
}, {
items: {},
dup: []
},
)
console.log(dup)
// ['hi', 'bye']
You can use filter method and indexOf() to get all the duplicate values
function duplicate(arr) {
return duplicateArray = arr.filter((item, index) => arr.indexOf(item) !== index)
}
arr.indexOf(item) will always return the first index at which a given element can be
found
ES5 only (i.e., it needs a filter() polyfill for IE8 and below):
var arrayToFilter = [ 4, 5, 5, 5, 2, 1, 3, 1, 1, 2, 1, 3 ];
arrayToFilter.
sort().
filter( function(me,i,arr){
return (i===0) || ( me !== arr[i-1] );
});
Here is a very light and easy way:
var codes = dc_1.split(',');
var i = codes.length;
while (i--) {
if (codes.indexOf(codes[i]) != i) {
codes.splice(i,1);
}
}
ES6 offers the Set data structure which is basically an array that doesn't accept duplicates.
With the Set data structure, there's a very easy way to find duplicates in an array (using only one loop).
Here's my code
function findDuplicate(arr) {
var set = new Set();
var duplicates = new Set();
for (let i = 0; i< arr.length; i++) {
var size = set.size;
set.add(arr[i]);
if (set.size === size) {
duplicates.add(arr[i]);
}
}
return duplicates;
}
With ES6 (or using Babel or Typescipt) you can simply do:
var duplicates = myArray.filter(i => myArray.filter(ii => ii === i).length > 1);
https://es6console.com/j58euhbt/
Simple code with ES6 syntax (return sorted array of duplicates):
let duplicates = a => {d=[]; a.sort((a,b) => a-b).reduce((a,b)=>{a==b&&!d.includes(a)&&d.push(a); return b}); return d};
How to use:
duplicates([1,2,3,10,10,2,3,3,10]);
I have just figured out a simple way to achieve this using an Array filter
var list = [9, 9, 111, 2, 3, 4, 4, 5, 7];
// Filter 1: to find all duplicates elements
var duplicates = list.filter(function(value,index,self) {
return self.indexOf(value) !== self.lastIndexOf(value) && self.indexOf(value) === index;
});
console.log(duplicates);
This answer might also be helpful, it leverages js reduce operator/method to remove duplicates from array.
const result = [1, 2, 2, 3, 3, 3, 3].reduce((x, y) => x.includes(y) ? x : [...x, y], []);
console.log(result);
Higher ranked answers have a few inherent issues including the use of legacy javascript, incorrect ordering or with only support for 2 duplicated items.
Here's a modern solution which fixes those problems:
const arrayNonUniq = array => {
if (!Array.isArray(array)) {
throw new TypeError("An array must be provided!")
}
return array.filter((value, index) => array.indexOf(value) === index && array.lastIndexOf(value) !== index)
}
arrayNonUniq([1, 1, 2, 3, 3])
//=> [1, 3]
arrayNonUniq(["foo", "foo", "bar", "foo"])
//=> ['foo']
You can also use the npm package array-non-uniq.
The following function (a variation of the eliminateDuplicates function already mentioned) seems to do the trick, returning test2,1,7,5 for the input ["test", "test2", "test2", 1, 1, 1, 2, 3, 4, 5, 6, 7, 7, 10, 22, 43, 1, 5, 8]
Note that the problem is stranger in JavaScript than in most other languages, because a JavaScript array can hold just about anything. Note that solutions that use sorting might need to provide an appropriate sorting function--I haven't tried that route yet.
This particular implementation works for (at least) strings and numbers.
function findDuplicates(arr) {
var i,
len=arr.length,
out=[],
obj={};
for (i=0;i<len;i++) {
if (obj[arr[i]] != null) {
if (!obj[arr[i]]) {
out.push(arr[i]);
obj[arr[i]] = 1;
}
} else {
obj[arr[i]] = 0;
}
}
return out;
}
var arr = [2, 1, 2, 2, 4, 4, 2, 5];
function returnDuplicates(arr) {
return arr.reduce(function(dupes, val, i) {
if (arr.indexOf(val) !== i && dupes.indexOf(val) === -1) {
dupes.push(val);
}
return dupes;
}, []);
}
alert(returnDuplicates(arr));
This function avoids the sorting step and uses the reduce() method to push duplicates to a new array if it doesn't already exist in it.
I have an array a = [1,1,1,2,2,3,3,3,4,4,4,6,6,6,7,7]
I want to fetch all the duplicate pair in this array list.
Since there are pairs of 2 and 7 the output should be -
Output: [2, 7]
I tried writing my own logic but I am very weak in that area. Can somebody help?
function getDuplicateArrayElements(arr){
let sorted_arr = arr.slice().sort();
let results = [];
for (let i = 0; i < sorted_arr.length; i++) {
let matchingElementCount = 1;
for (let j = i + 1; j < sorted_arr.length - i; j++) {
if (sorted_arr[j] === sorted_arr[i]) {
++matchingElementCount;
} else {
if(matchingElementCount % 2 === 0) {
results.push(sorted_arr[i]);
}
i = j - 1;
break;
}
}
}
return results; } var a = [1,1,1,2,2,3,3,3,4,6,6,6,7,7]; var duplicateValues= getDuplicateArrayElements(a);
You can achieve your result by using reduce and forEach.
const arr = [1,1,1,1,2,2,3,3,3,4,4,4,6,6,6,7,7];
// Generate a hashmap from the given array for counting the frequency.
const hashMap = arr.reduce((a, c) => {
a[c] = (a[c] || 0) + 1;
return a;
}, {});
const pair = [];
// If the frequency is divided by 2 then push the key of the hashMap into pair array.
Object.entries(hashMap).forEach(([k, v]) => {
if (v % 2 === 0) {
[...Array(Math.floor(v / 2))].forEach(_ => pair.push(k));
}
})
console.log(pair);
You can grab the frequency of each number, and then filter out any which have an odd frequency. You can then .flatMap() the frequencies to an array containing your number for each pair you found like so:
const a = [1,1,1,2,2,3,3,3,4,4,4,6,6,6,7,7];
const freq = a.reduce((m, n) => m.set(n, (m.get(n) || 0) + 1), new Map);
const res = [...freq].filter(([n, count]) => count % 2 == 0).flatMap(([n, c]) => Array(c/2).fill(n));
console.log(res);
This way, if you have four 1s (ie: two pairs of 1s), the filter will pick up on that, allowing you to flat-map the [1, 4] array to an array of [1, 1], which is merged into the larger resulting array.
You could create a helper map and keep the counts of each number as the values and the numbers itself as the keys. After iterating through the array, you only need to find the ones with a count divisible by 2:
var a = [1, 1, 1, 2, 2, 3, 3, 3, 4, 6, 6, 6, 7, 7]
function findDuplicates(arr) {
const map = {};
for (const curr of arr) {
if (!map[curr]) {
map[curr] = 0;
}
map[curr]++;
}
const res = [];
for (const key in map) {
if (map.hasOwnProperty(key) && map[key] % 2 === 0) {
res.push(Number.parseInt(key));
}
}
return res;
}
console.log(findDuplicates(a));
You can first count the occurrence of each numbers and if it is greater than 0 and divisible by 2 then add these to final result else don't
function getDuplicateArrayElements(arr) {
let map = {}
let results = [];
for (let num of arr) {
map[num] = map[num] || 0
map[num]++
}
return Object.keys(map)
.filter(v => map[v] && map[v] % 2 === 0)
.map(v => new Array(map[v]/2).fill(+v))
.flat()
.sort()
}
var a = [1, 1, 1, 2, 2, 3, 3, 3, 4, 6, 6, 6, 7, 7,8,8,8,8];
var duplicateValues = getDuplicateArrayElements(a);
console.log(duplicateValues)
const a = {};
[1,1,1,2,2,3,3,3,4,6,6,6,7,7].forEach(v => {a[v] = a[v] ? a[v] + 1 : 1});
const l = [];
Object.keys(a).forEach(k => !(a[k] % 2) && l.push(k));
Here you go:
function getDuplicateArrayElements(arr){
var dupilcates=arr.filter(x => arr.filter(y=>y==x).length==2);
var found=[];
for(var i=0;i<dupilcates.length;i=i+2)
found.push(dupilcates[i]);
return found;
}
This will give you the desired pairs. with [1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 6, 6, 6, 6, 7, 7] input it will return [1,1,2,6,6,7]:
function getDuplicateArrayElements(arr){
let sorted_arr = arr.slice().sort();
let results = [];
let i = 0;
while (i < sorted_arr.length) {
let counter = 1;
let j = i;
while (sorted_arr[j] === sorted_arr[j+1]) {
counter++;
j++;
}
if (counter%2 == 0) {
results.push(...Array(counter/2).fill(sorted_arr[i]))
}
i += counter;
}
return results;
}
var a = [1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 6, 6, 6, 6, 7, 7];
console.log(getDuplicateArrayElements(a));
Another rather concise solution:
a = [1,1,1,2,2,3,3,3,4,4,4,6,6,6,7,7];
uniques = new Set(a); //filter out duplicates
res = [];
uniques.forEach((key)=>{
if(a.filter(elem => elem === key).length === 2){res.push(key)};
//filter out only the elements which match the key being tested
//if there are 2, push to result
})
Edit: even more concise, but perhaps less efficient:
a = [1,1,1,2,2,3,3,3,4,4,4,6,6,6,7,7];
res = Array.from(new Set(a.filter(elem => a.filter(el => el === elem).length === 2)));
Javascript has awesome JSON object, in my opinion, you can use json as a dictionary;
{ key: _value }.
Loop throw array one times, no sort, no slice
key is array's element value, _value is frequency
var frequencies = {};
for (let i = 0; i < a.length; a++) {
if (result[a[i]] == 'over') continue;
if (result[a[i]] == undefined) { // First times
result[a[i]] = 1
} else if (result[a[i]] == 1) { // Second times
result[a[i]] = 2
} else { // Ignore if > 2
result[a[i]] = 'over'
}
}
// result: {1: "over", 2: 2, 3: "over", 4: "over", 6: "over", 7: 2}
so now pick keys have value equal 2
function getDuplicateArrayElements(numbers: number[]): number[] {
const occurences = new Map<number, number>();
for (let number of numbers) {
if (occurences.has(number)) {
const current = occurences.get(number)!;
occurences.set(number, current + 1);
} else
occurences.set(number, 1)
}
return (
Array
.from(occurences.entries())
.reduce<number[]>(
(accumulator, [key, value]) => {
if (value === 2) {
return accumulator.concat(key)
}
return accumulator
},
[]
)
)
}
const a = [1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 6, 6, 6, 7, 7];
getDuplicateArrayElements(a); // [2, 7]
This question already has answers here:
Check if an array is descending, ascending or not sorted?
(10 answers)
Check if array values are ascending or descending
(1 answer)
Closed 4 years ago.
I need to create a program that checks the list in the array is sorted. I have three input data:
1,2,3,4,5
1,2,8,9,9
1,2,2,3,2
So here is my code:
let sorts = +gets(); // 3
let list = [];
for (let i = 0; i < sorts; i++) {
list[i] = gets().split(',').map(Number); // The Array will be: [ [ 1, 2, 3, 4, 5 ], [ 1, 2, 8, 9, 9 ], [ 1, 2, 2, 3, 2 ] ]
}
for (let i = 0; i < list[i][i].length; i++){
if (list[i][i] < list[i][i +1]) {
print('true');
} else {
print('false');
}
}
I need to print for all lists on new line true or false. For this example my output needs to be:
true
true
false
I have no idea how to resolve this.
You can use array#every to check if each value is greater than the previous value.
const isSorted = arr => arr.every((v,i,a) => !i || a[i-1] <= v);
console.log(isSorted([1,2,3,4,5]));
console.log(isSorted([1,2,8,9,9]));
console.log(isSorted([1,2,2,3,2]));
How about something like this:
!![1,2,3,4,5].reduce((n, item) => n !== false && item >= n && item)
// true
!![1,2,8,9,9].reduce((n, item) => n !== false && item >= n && item)
// true
!![1,2,2,3,2].reduce((n, item) => n !== false && item >= n && item)
// false
Reduce will literally reduce the array down to a single value - a boolean in our case.
Here, we are calling a function per iteration, the (n, item) is our function signature, it's body being n !== false && item >- n && item - we are making sure that n exists (n is our accumulator - read up!), testing if item is greater than n, and making sure item exists.
This happens for every element in your array. We then use !! to force the result into a tru boolean.
Simply try this way by using slice method : It will check if previous element is less than the next element.If the condition is true for every element then it will return true else false
arr.slice(1).every((item, i) => arr[i] <= item);
Checkout this below sample as Demo
var arr = [[1,2,3,4,5],[1,2,8,9,9],[1,2,2,3,2],[0,1,2,3,4,5]];
function isArrayIsSorted (arr) {
return arr.slice(1).every((item, i) => arr[i] <= item)
}
var result= [];
for (var i = 0; i < arr.length; i++){
result.push(isArrayIsSorted(arr[i]))
}
console.log(result);
Sorted Number Lists
Including Negative Numbers, Zeros, and Adjacent Duplicates
Use every() method which will return true should all of the numbers be in order otherwise it will return false. The conditions are as follows:
(num <= arr[idx + 1]) || (idx === arr.length - 1)
if the current number is less than or equal to the next number...
OR...
if the current index is equal to the last index...
return 1 (truthy)
Demo
var arr0 = [1, 2, 3, 4, 5];
var arr1 = [1, 2, 8, 9, 9];
var arr2 = [1, 2, 2, 3, 2];
var arr3 = [0, 0, 0, 1, 3];
var arr4 = [-3, 0, 1, 3, 3];
var arr5 = [-4, -2, 0, 0, -4];
function sorted(array) {
return array.every(function(num, idx, arr) {
return (num <= arr[idx + 1]) || (idx === arr.length - 1) ? 1 : 0;
});
}
console.log(arr0 +' | '+sorted(arr0));
console.log(arr1 +' | '+sorted(arr1));
console.log(arr2 +' | '+sorted(arr2));
console.log(arr3 +' | '+sorted(arr3));
console.log(arr4 +' | '+sorted(arr4));
console.log(arr5 +' | '+sorted(arr5));
var str = ["1,2,3,4,5", "1,2,8,9,9", "1,2,2,3,2"];
for (var i in str){
var list = str[i].split(',').map(Number);
console.log(list);
var isSorted = true;
for(var j = 0 ; j < list.length - 1 ; j++){
if(list[j] > list[j+1]) {
isSorted = false;
break;
}
}
console.log(isSorted);
}
Maybe you can use this helping method that checks if is sorted correctly:
var arr1 = [1, 2, 3, 4, 4];
var arr2 = [3, 2, 1];
console.log(checkList(arr1));
console.log(checkList(arr2));
function checkList(arr) {
for (var i = 0; i < arr.length; i++) {
if (arr[i + 1]) {
if (arr[i] > arr[i + 1]) {
return false;
}
}
}
return true;
}
There are plenty of ways how to do that. Here is mine
const isArraySorted = array =>
array
.slice(0) // clone array
.sort((a, b) => a - b) // sort it
.every((el, i) => el === array[i]) // compare with initial value)
You can check if stringified sorted copy of original array has same value as the original one. Might not be the most cool or performant one, but I like it's simplicity and clarity.
const arraysToCheck = [
[1, 2, 3, 4, 5],
[1, 2, 8, 9, 9],
[1, 2, 2, 3, 2]
]
const isSorted = arraysToCheck.map(
item => JSON.stringify([...item].sort((a, b) => a - b)) === JSON.stringify(item)
);
console.log(isSorted);
If i get what you mean, you want to know if an array is sorted or not. This is an example of such a solution, try it. I pasted some codes below.
var myArray=[1,4,3,6];
if(isSorted(myArray)){
console.log("List is sorted");
}else{
console.log("List is not sorted");
}
function isSorted(X){
var sorted=false;
for(var i=0;i<X.length;i++){
var next=i+1;
if (next<=X.length-1){
if(X[i]>X[next]){
sorted=false;
break;
}else{
sorted=true;
}
}
}
return sorted;
}
I need to check a JavaScript array to see if there are any duplicate values. What's the easiest way to do this? I just need to find what the duplicated values are - I don't actually need their indexes or how many times they are duplicated.
I know I can loop through the array and check all the other values for a match, but it seems like there should be an easier way.
Similar question:
Get all unique values in a JavaScript array (remove duplicates)
You could sort the array and then run through it and then see if the next (or previous) index is the same as the current. Assuming your sort algorithm is good, this should be less than O(n2):
const findDuplicates = (arr) => {
let sorted_arr = arr.slice().sort(); // You can define the comparing function here.
// JS by default uses a crappy string compare.
// (we use slice to clone the array so the
// original array won't be modified)
let results = [];
for (let i = 0; i < sorted_arr.length - 1; i++) {
if (sorted_arr[i + 1] == sorted_arr[i]) {
results.push(sorted_arr[i]);
}
}
return results;
}
let duplicatedArray = [9, 9, 111, 2, 3, 4, 4, 5, 7];
console.log(`The duplicates in ${duplicatedArray} are ${findDuplicates(duplicatedArray)}`);
In case, if you are to return as a function for duplicates. This is for similar type of case.
Reference: https://stackoverflow.com/a/57532964/8119511
If you want to elimate the duplicates, try this great solution:
function eliminateDuplicates(arr) {
var i,
len = arr.length,
out = [],
obj = {};
for (i = 0; i < len; i++) {
obj[arr[i]] = 0;
}
for (i in obj) {
out.push(i);
}
return out;
}
console.log(eliminateDuplicates([1,6,7,3,6,8,1,3,4,5,1,7,2,6]))
Source:
http://dreaminginjavascript.wordpress.com/2008/08/22/eliminating-duplicates/
This is my answer from the duplicate thread (!):
When writing this entry 2014 - all examples were for-loops or jQuery. JavaScript has the perfect tools for this: sort, map and reduce.
Find duplicate items
var names = ['Mike', 'Matt', 'Nancy', 'Adam', 'Jenny', 'Nancy', 'Carl']
const uniq = names
.map((name) => {
return {
count: 1,
name: name
};
})
.reduce((result, b) => {
result[b.name] = (result[b.name] || 0) + b.count;
return result;
}, {});
const duplicates = Object.keys(uniq).filter((a) => uniq[a] > 1);
console.log(duplicates); // [ 'Nancy' ]
More functional syntax:
#Dmytro-Laptin pointed out some code that can be removed. This is a more compact version of the same code. Using some ES6 tricks and higher-order functions:
const names = ['Mike', 'Matt', 'Nancy', 'Adam', 'Jenny', 'Nancy', 'Carl'];
const count = names =>
names.reduce((result, value) => ({ ...result,
[value]: (result[value] || 0) + 1
}), {}); // don't forget to initialize the accumulator
const duplicates = dict =>
Object.keys(dict).filter((a) => dict[a] > 1);
console.log(count(names)); // { Mike: 1, Matt: 1, Nancy: 2, Adam: 1, Jenny: 1, Carl: 1 }
console.log(duplicates(count(names))); // [ 'Nancy' ]
UPDATED: Short one-liner to get the duplicates:
[1, 2, 2, 4, 3, 4].filter((e, i, a) => a.indexOf(e) !== i) // [2, 4]
To get the array without duplicates simply invert the condition:
[1, 2, 2, 4, 3, 4].filter((e, i, a) => a.indexOf(e) === i) // [1, 2, 3, 4]
Note that this answer’s main goal is to be short. If you need something performant for a big array, one possible solution is to sort your array first (if it is sortable) then do the following to get the same kind of results as above:
myHugeSortedArray.filter((e, i, a) => a[i-1] === e)
Here is an example for a 1 000 000 integers array:
const myHugeIntArrayWithDuplicates =
[...Array(1_000_000).keys()]
// adding two 0 and four 9 duplicates
.fill(0, 2, 4).fill(9, 10, 14)
console.time("time")
console.log(
myHugeIntArrayWithDuplicates
// a possible sorting method for integers
.sort((a, b) => a > b ? 1 : -1)
.filter((e, i, a) => a[i-1] === e)
)
console.timeEnd("time")
On my AMD Ryzen 7 5700G dev machine it outputs:
[ 0, 0, 9, 9, 9, 9 ]
time: 22.738ms
As pointed out in the comments both the short solution and the performant solution will return an array with several time the same duplicate if it occurs more than once in the original array:
[1, 1, 1, 2, 2, 2, 2].filter((e, i, a) => a.indexOf(e) !== i) // [1, 1, 2, 2, 2]
If unique duplicates are wanted then a function like
function duplicates(arr) {
return [...new Set(arr.filter((e, i, a) => a.indexOf(e) !== i))]
}
can be used so that duplicates([1, 1, 1, 2, 2, 2, 2]) returns [1, 2].
When all you need is to check that there are no duplicates as asked in this question you can use the every() method:
[1, 2, 3].every((e, i, a) => a.indexOf(e) === i) // true
[1, 2, 1].every((e, i, a) => a.indexOf(e) === i) // false
Note that every() doesn't work for IE 8 and below.
Find duplicate values in an array
This should be one of the shortest ways to actually find duplicate values in an array. As specifically asked for by the OP, this does not remove duplicates but finds them.
var input = [1, 2, 3, 1, 3, 1];
var duplicates = input.reduce(function(acc, el, i, arr) {
if (arr.indexOf(el) !== i && acc.indexOf(el) < 0) acc.push(el); return acc;
}, []);
document.write(duplicates); // = 1,3 (actual array == [1, 3])
This doesn't need sorting or any third party framework. It also doesn't need manual loops. It works with every value indexOf() (or to be clearer: the strict comparision operator) supports.
Because of reduce() and indexOf() it needs at least IE 9.
You can add this function, or tweak it and add it to Javascript's Array prototype:
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])
{
alert('this is a DUPE!');
continue o;
}
}
r[r.length] = this[i];
}
return r;
}
var arr = [1,2,2,3,3,4,5,6,2,3,7,8,5,9];
var unique = arr.unique();
alert(unique);
UPDATED: The following uses an optimized combined strategy. It optimizes primitive lookups to benefit from hash O(1) lookup time (running unique on an array of primitives is O(n)). Object lookups are optimized by tagging objects with a unique id while iterating through so so identifying duplicate objects is also O(1) per item and O(n) for the whole list. The only exception is items that are frozen, but those are rare and a fallback is provided using an array and indexOf.
var unique = function(){
var hasOwn = {}.hasOwnProperty,
toString = {}.toString,
uids = {};
function uid(){
var key = Math.random().toString(36).slice(2);
return key in uids ? uid() : uids[key] = key;
}
function unique(array){
var strings = {}, numbers = {}, others = {},
tagged = [], failed = [],
count = 0, i = array.length,
item, type;
var id = uid();
while (i--) {
item = array[i];
type = typeof item;
if (item == null || type !== 'object' && type !== 'function') {
// primitive
switch (type) {
case 'string': strings[item] = true; break;
case 'number': numbers[item] = true; break;
default: others[item] = item; break;
}
} else {
// object
if (!hasOwn.call(item, id)) {
try {
item[id] = true;
tagged[count++] = item;
} catch (e){
if (failed.indexOf(item) === -1)
failed[failed.length] = item;
}
}
}
}
// remove the tags
while (count--)
delete tagged[count][id];
tagged = tagged.concat(failed);
count = tagged.length;
// append primitives to results
for (i in strings)
if (hasOwn.call(strings, i))
tagged[count++] = i;
for (i in numbers)
if (hasOwn.call(numbers, i))
tagged[count++] = +i;
for (i in others)
if (hasOwn.call(others, i))
tagged[count++] = others[i];
return tagged;
}
return unique;
}();
If you have ES6 Collections available, then there is a much simpler and significantly faster version. (shim for IE9+ and other browsers here: https://github.com/Benvie/ES6-Harmony-Collections-Shim)
function unique(array){
var seen = new Set;
return array.filter(function(item){
if (!seen.has(item)) {
seen.add(item);
return true;
}
});
}
var a = ["a","a","b","c","c"];
a.filter(function(value,index,self){ return (self.indexOf(value) !== index )})
This should get you what you want, Just the duplicates.
function find_duplicates(arr) {
var len=arr.length,
out=[],
counts={};
for (var i=0;i<len;i++) {
var item = arr[i];
counts[item] = counts[item] >= 1 ? counts[item] + 1 : 1;
if (counts[item] === 2) {
out.push(item);
}
}
return out;
}
find_duplicates(['one',2,3,4,4,4,5,6,7,7,7,'pig','one']); // -> ['one',4,7] in no particular order.
Find non-unique values from 3 arrays (or more):
ES2015
// 🚩🚩 🚩 🚩 🚩
var arr = [1,2,2,3,3,4,5,6,2,3,7,8,5,22],
arr2 = [1,2,511,12,50],
arr3 = [22,0],
merged,
nonUnique;
// Combine all the arrays to a single one
merged = arr.concat(arr2, arr3)
// create a new (dirty) Array with only the non-unique items
nonUnique = merged.filter((item,i) => merged.includes(item, i+1))
// Cleanup - remove duplicate & empty items items
nonUnique = [...new Set(nonUnique)]
console.log(nonUnique)
PRE-ES2015:
In the below example I chose to superimpose a unique method on top of the Array prototype, allowing access from everywhere and has more "declarative" syntax. I do not recommend this approach on large projects, since it might very well collide with another method with the same custom name.
Array.prototype.unique = function () {
var arr = this.sort(), i=arr.length; // input must be sorted for this to work
while(i--)
arr[i] === arr[i-1] && arr.splice(i,1) // remove duplicate item
return arr
}
Array.prototype.nonunique = function () {
var arr = this.sort(), i=arr.length, res = []; // input must be sorted for this to work
while(i--)
arr[i] === arr[i-1] && (res.indexOf(arr[i]) == -1) && res.push(arr[i])
return res
}
// 🚩🚩 🚩 🚩 🚩
var arr = [1,2,2,3,3,4,5,6,2,3,7,8,5,22],
arr2 = [1,2,511,12,50],
arr3 = [22,0],
// merge all arrays & call custom Array Prototype - "unique"
unique = arr.concat(arr2, arr3).unique(),
nonunique = arr.concat(arr2, arr3).nonunique()
console.log(unique) // [1,12,2,22,3,4,5,50,511,6,7,8]
console.log(nonunique) // [1,12,2,22,3,4,5,50,511,6,7,8]
using underscore.js
function hasDuplicate(arr){
return (arr.length != _.uniq(arr).length);
}
The simplest and quickest way is to use the Set object:
const numbers = [1, 2, 3, 2, 4, 5, 5, 6];
const set = new Set(numbers);
const duplicates = numbers.filter(item => {
if (set.has(item)) {
set.delete(item);
return false;
} else {
return true;
}
});
// OR more concisely
const duplicates = numbers.filter(item => !set.delete(item));
console.log(duplicates);
// [ 2, 5 ]
This is my proposal (ES6):
let a = [1, 2, 3, 4, 2, 2, 4, 1, 5, 6]
let b = [...new Set(a.sort().filter((o, i) => o !== undefined && a[i + 1] !== undefined && o === a[i + 1]))]
// b is now [1, 2, 4]
Here's the simplest solution I could think of:
const arr = [-1, 2, 2, 2, 0, 0, 0, 500, -1, 'a', 'a', 'a']
const filtered = arr.filter((el, index) => arr.indexOf(el) !== index)
// => filtered = [ 2, 2, 0, 0, -1, 'a', 'a' ]
const duplicates = [...new Set(filtered)]
console.log(duplicates)
// => [ 2, 0, -1, 'a' ]
That's it.
Note:
It works with any numbers including 0, strings and negative numbers e.g. -1 -
Related question: Get all unique values in a JavaScript array (remove duplicates)
The original array arr is preserved (filter returns the new array instead of modifying the original)
The filtered array contains all duplicates; it can also contain more than 1 same value (e.g. our filtered array here is [ 2, 2, 0, 0, -1, 'a', 'a' ])
If you want to get only values that are duplicated (you don't want to have multiple duplicates with the same value) you can use [...new Set(filtered)] (ES6 has an object Set which can store only unique values)
Hope this helps.
Here is mine simple and one line solution.
It searches not unique elements first, then makes found array unique with the use of Set.
So we have array of duplicates in the end.
var array = [1, 2, 2, 3, 3, 4, 5, 6, 2, 3, 7, 8, 5, 22, 1, 2, 511, 12, 50, 22];
console.log([...new Set(
array.filter((value, index, self) => self.indexOf(value) !== index))]
);
Shortest vanilla JS:
[1,1,2,2,2,3].filter((v,i,a) => a.indexOf(v) !== i) // [1, 2, 2]
one liner simple way
var arr = [9,1,2,4,3,4,9]
console.log(arr.filter((ele,indx)=>indx!==arr.indexOf(ele))) //get the duplicates
console.log(arr.filter((ele,indx)=>indx===arr.indexOf(ele))) //remove the duplicates
var a = [324,3,32,5,52,2100,1,20,2,3,3,2,2,2,1,1,1].sort();
a.filter(function(v,i,o){return i&&v!==o[i-1]?v:0;});
or when added to the prototyp.chain of Array
//copy and paste: without error handling
Array.prototype.unique =
function(){return this.sort().filter(function(v,i,o){return i&&v!==o[i-1]?v:0;});}
See here: https://gist.github.com/1305056
Fast and elegant way using es6 object destructuring and reduce
It runs in O(n) (1 iteration over the array) and doesn't repeat values that appear more than 2 times
const arr = ['hi', 'hi', 'hi', 'bye', 'bye', 'asd']
const {
dup
} = arr.reduce(
(acc, curr) => {
acc.items[curr] = acc.items[curr] ? acc.items[curr] += 1 : 1
if (acc.items[curr] === 2) acc.dup.push(curr)
return acc
}, {
items: {},
dup: []
},
)
console.log(dup)
// ['hi', 'bye']
You can use filter method and indexOf() to get all the duplicate values
function duplicate(arr) {
return duplicateArray = arr.filter((item, index) => arr.indexOf(item) !== index)
}
arr.indexOf(item) will always return the first index at which a given element can be
found
ES5 only (i.e., it needs a filter() polyfill for IE8 and below):
var arrayToFilter = [ 4, 5, 5, 5, 2, 1, 3, 1, 1, 2, 1, 3 ];
arrayToFilter.
sort().
filter( function(me,i,arr){
return (i===0) || ( me !== arr[i-1] );
});
Here is a very light and easy way:
var codes = dc_1.split(',');
var i = codes.length;
while (i--) {
if (codes.indexOf(codes[i]) != i) {
codes.splice(i,1);
}
}
ES6 offers the Set data structure which is basically an array that doesn't accept duplicates.
With the Set data structure, there's a very easy way to find duplicates in an array (using only one loop).
Here's my code
function findDuplicate(arr) {
var set = new Set();
var duplicates = new Set();
for (let i = 0; i< arr.length; i++) {
var size = set.size;
set.add(arr[i]);
if (set.size === size) {
duplicates.add(arr[i]);
}
}
return duplicates;
}
With ES6 (or using Babel or Typescipt) you can simply do:
var duplicates = myArray.filter(i => myArray.filter(ii => ii === i).length > 1);
https://es6console.com/j58euhbt/
Simple code with ES6 syntax (return sorted array of duplicates):
let duplicates = a => {d=[]; a.sort((a,b) => a-b).reduce((a,b)=>{a==b&&!d.includes(a)&&d.push(a); return b}); return d};
How to use:
duplicates([1,2,3,10,10,2,3,3,10]);
I have just figured out a simple way to achieve this using an Array filter
var list = [9, 9, 111, 2, 3, 4, 4, 5, 7];
// Filter 1: to find all duplicates elements
var duplicates = list.filter(function(value,index,self) {
return self.indexOf(value) !== self.lastIndexOf(value) && self.indexOf(value) === index;
});
console.log(duplicates);
This answer might also be helpful, it leverages js reduce operator/method to remove duplicates from array.
const result = [1, 2, 2, 3, 3, 3, 3].reduce((x, y) => x.includes(y) ? x : [...x, y], []);
console.log(result);
Higher ranked answers have a few inherent issues including the use of legacy javascript, incorrect ordering or with only support for 2 duplicated items.
Here's a modern solution which fixes those problems:
const arrayNonUniq = array => {
if (!Array.isArray(array)) {
throw new TypeError("An array must be provided!")
}
return array.filter((value, index) => array.indexOf(value) === index && array.lastIndexOf(value) !== index)
}
arrayNonUniq([1, 1, 2, 3, 3])
//=> [1, 3]
arrayNonUniq(["foo", "foo", "bar", "foo"])
//=> ['foo']
You can also use the npm package array-non-uniq.
The following function (a variation of the eliminateDuplicates function already mentioned) seems to do the trick, returning test2,1,7,5 for the input ["test", "test2", "test2", 1, 1, 1, 2, 3, 4, 5, 6, 7, 7, 10, 22, 43, 1, 5, 8]
Note that the problem is stranger in JavaScript than in most other languages, because a JavaScript array can hold just about anything. Note that solutions that use sorting might need to provide an appropriate sorting function--I haven't tried that route yet.
This particular implementation works for (at least) strings and numbers.
function findDuplicates(arr) {
var i,
len=arr.length,
out=[],
obj={};
for (i=0;i<len;i++) {
if (obj[arr[i]] != null) {
if (!obj[arr[i]]) {
out.push(arr[i]);
obj[arr[i]] = 1;
}
} else {
obj[arr[i]] = 0;
}
}
return out;
}
var arr = [2, 1, 2, 2, 4, 4, 2, 5];
function returnDuplicates(arr) {
return arr.reduce(function(dupes, val, i) {
if (arr.indexOf(val) !== i && dupes.indexOf(val) === -1) {
dupes.push(val);
}
return dupes;
}, []);
}
alert(returnDuplicates(arr));
This function avoids the sorting step and uses the reduce() method to push duplicates to a new array if it doesn't already exist in it.