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remove specified value from array [duplicate]

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How can I remove a specific item from an array in JavaScript?
(142 answers)
Closed 1 year ago.
How can I check whether an array contains a value, and if so, remove it?
PS: For this exercise, I'm not allowed to use anything more than than .pop, .push and .length array functions.
My logic is the following: if the specified value is within the array, reorder the array so that the last element of it will contain this value, then remove it with .pop. But how can I find this value and reorder it without using anything more than those array functions I specified above?
This is what I managed to come up with so far:
let array_1 = [1,2,3];
if (array_1 == 2){
//reorder somehow
array_1.pop();
}
console.log(array_1);

Using this approach, you are not creating a new array but modifying it. It uses .pop().
let array_1 = [1, 2, 3];
// Iterate all array
for (let i = 0; i < array_1.length; i++) {
// While there is a 2 element in the actual index, move all elements (from i index) to the previous index
while(array_1[i] === 2) {
for (let j = i; j < array_1.length - 1; j++) {
array_1[j] = array_1[j + 1];
}
// Now remove the last element (since we move all elements to the previous index)
array_1.pop();
}
}
console.log(array_1);
Here a snippet so you can try it
let array_1 = [1, 2, 3, 4, 2, 5, 2, 6, 2];
for (let i = 0; i < array_1.length; i++) {
while(array_1[i] === 2) {
for (let j = i; j < array_1.length - 1; j++) {
array_1[j] = array_1[j + 1];
}
array_1.pop();
}
}
console.log(array_1);
This would mantain the order of the array, but without the "2" elements.

Here's another option using pop
const filter = (array, target) => {
const newArray = [];
let tmp;
while(tmp = array.pop()) {
if (tmp !== target) {
newArray.push(tmp)
}
}
console.log(newArray)
return newArray;
}
filter([1,2,3,4], 2) // [4, 3, 1] Note that it reversed the order of the array!

If you are limited to pop, push and length, you can loop over all elements, check if a given element matches the value you are looking for, and add them to a new array using push.
let array_1 = [1,2,3];
let newArray = [];
for (let i = 0; i < array_1.length; i++) {
if (array_1[i] !== 2) {
newArray.push(array_1[i]);
}
}
console.log(newArray);

// using splice
// splice(indexStart, how many, replace with)
// example :
let arr = [0,1,2,3,4,5];
// remove at index 1
arr.splice(1,1);
console.log( arr );
// replace at index 1
arr.splice(1,1,"new 1");
console.log( arr );
// merge index 2 and 3
arr.splice(2,2,"merge 2 and 3");
console.log( arr );
// create 2 new items start at index 2
arr.splice(2,2,"new 2", "new 3");
console.log( arr );

Get elements that has no double value in an array

I have a question . How do you retrieve elements that has no double value in an array?? For example: [1,1,2,2,3,4,4,5] then you retrieve [3,5] only.
Thanks in advance
for (var j = 0; j < newArr.length; j++) {
if ((arr1.indexOf(newArr[j]) === 0) && (arr2.indexOf(newArr[j]) === 0)) {
index = newArr.indexOf(j); newArr.splice(index, 1);
}
}

If the item in the array is unique then the index found from the beginning should equal the index found from the end, in other words:
var xs = [1, 1, 2, 2, 3, 4, 4, 5];
var result = xs.filter(function(x) {
return xs.indexOf(x) === xs.lastIndexOf(x);
});
console.log(result); //=> [3, 5]

sorry for the presentation its my first post !
You have to compare each element of your array to the others in order to get the number of occurence of each element
var tab = [1,1,2,2,3,4,4,5] //The array to analyze
tab = tab.sort(); // we sort the array
show(tab); // we display the array to the console (F12 to open it)
var uniqueElementTab = []; // this array will contain all single occurence
var sameElementCounter = 0;
for(x=0;x<tab.length;x++){ // for all element in the array
sameElementCounter = 0;
for(y=0;y<tab.length;y++){ // we compare it to the others
if((tab[x]==tab[y])){
sameElementCounter+=1; // +1 each time we meet the element elsewhere
}
}
if(sameElementCounter<=1){
uniqueElementTab.push(tab[x]); //if the element is unique we add it to a new array
}
}
show(uniqueElementTab); // display result
function show(tab) { // Simple function to display the content of an array
var st="";
for(i=0;i<tab.length;i++){
st += tab[i]+" ";
}
console.log(st+"\n");
}
Hope it helps.

Here is a simple "tricky" solution using Array.sort, Array.join, Array.map, String.replace and String.split functions:
var arr = [1, 1, 2, 2, 3, 4, 4, 5];
arr.sort();
var unique = arr.join("").replace(/(\d)\1+/g, "").split("").map(Number);
console.log(unique); // [3, 5]

create new array tmp,and check already value exist by indexOf .If existed delete by splice function..
var arr = [1,1,2,2,3,4,4,5];
var tmp = [];
var dup = [];
for(var i = 0; i < arr.length; i++){
var ind = tmp.indexOf(arr[i]);
if(ind == -1){
if(dup.indexOf(arr[i]) == -1){
tmp.push(arr[i]);
}
}
else{
tmp.splice(ind,1);
dup.push(arr[i]);
}
}
console.log(tmp);

This would be my way of doing this job.
var arr = [1,1,2,2,3,4,4,5],
uniques = Object.keys(arr.reduce((p,c) => (c in p ? Object.defineProperty(p, c, {enumerable : false,
writable : true,
configurable : true})
: p[c] = c,
p), {}));
console.log(uniques);

A solution for unsorted arrays with a hash table for the items. Complexity O(2n)
var array = [1, 1, 2, 2, 3, 4, 4, 5, 1],
hash = Object.create(null),
single;
array.forEach(function (a, i) {
hash[a] = a in hash ? -1 : i;
});
single = array.filter(function (a, i) {
return hash[a] === i;
});
console.log(single);

If the array is sorted, you can solve this in O(n) (see "pushUniqueSinglePass" below):
function pushUniqueSinglePass(array, unique) {
var prev; // last element seen
var run = 0; // number of times it has been seen
for (var i = 0; i < array.length; i++) {
if (array[i] != prev) {
if (run == 1) {
unique.push(prev); // "prev" appears only once
}
prev = array[i];
run = 1;
} else {
run++;
}
}
}
function pushUniqueWithSet(array, unique) {
var set = new Set();
for (var i = 0; i < array.length; i++) {
set.add(array[i]);
}
for (let e of set) {
unique.push(set);
}
}
// Utility and test functions
function randomSortedArray(n, max) {
var array = [];
for (var i = 0; i < n; i++) {
array.push(Math.floor(max * Math.random()));
}
return array.sort();
}
function runtest(i) {
var array = randomSortedArray(i, i / 2);
var r1 = [],
r2 = [];
console.log("Size: " + i);
console.log("Single-pass: " + time(
pushUniqueSinglePass, array, r1));
console.log("With set: " + time(
pushUniqueWithSet, array, r2));
// missing - assert r1 == r2
}
[10, 100, 1000, 10000,
100000, 1000000
].forEach(runtest);
function time(fun, array, unique) {
var start = new Date().getTime();
fun(array, unique);
return new Date().getTime() - start;
}
This is much more efficient than using maps or sorting (time it!). In my machine, a 1M sorted array can have its unique elements found in 18 ms; while the version that uses a set requires 10x more.

Reversing certain number of elements in an array javascript

I am working on a code where I need to reverse certain no of elements in an array and rest should remain same. For example is an array has values of 1,2,3,4,5,6 and I have to reverse 4 elements of it then output should be 4,3,2,1,5,6. I am using below code to achieve this but getting error, please suggest.
function reverseArray(n, a) {
var interimArray1 = [];
//var interimArray2=[];
//var finalArray=[];
for (var i < n; i >= 0; i--) {
interimArray1.push[a[i]];
}
for (var i = n; i < a.length; i++) {
interimArray1.push[a[i]];
}
for (var i = 0; i < interimArray1.length; i++) {
console.log(interimArray1[i]);
}
}
var arr = [1, 2, 3, 4, 5, 6];
var num = 4;
reverseArray(num, arr);

The error in your code is that you intend to call the push method on a[i] like so:
interimArray1.push(a[i]);
but instead you write:
interimArray1.push[a[i]];
You make that mistake twice. To give arguments to the push method, you must use round parenthesis ().
With that fixed, you will see that your code works perfectly.

You can use Array#slice, Array#splice as follow.
function partialReverse(arr, num, from = 0) {
var slicedArr = arr.slice(from, num + from);
arr.splice(from, num); // Remove `num` items from array
arr.splice(from, 0, ...slicedArr.reverse()); // Add `num` reversed items
return arr;
}
var arr = [1, 2, 3, 4, 5, 6];
console.log(partialReverse(arr, 4, 0)); // Reverse four items from `arr` starting from 0th index
console.log(partialReverse(arr, 4, 1)); // Reverse four items from `arr` starting from 1st index

Lots of hints but you seem to be missing them. ;-)
You need to assign an initial value to i, so:
for (var i = n; ... )
===========^
Also, you need to use () to call functions, not [], so:
interimArray1.push(a[i]);
==================^====^
Same in the following for block. Otherwise, the code works though it's more verbose than it needs to be.

This is working :
I'm sure there are faster ways of doing it. Also, it will only work for elements at the beginning of the array but you can adjust the function for what you want to achieve.
var reverseArray = function(arr,elementsToReverse) {
var tempArrayRev = [];
var tempArray = [];
for (var i=0;i<arr.length;i++) {
if (i < elementsToReverse) {
tempArrayRev[i] = arr[i];
} else {
tempArray.push(arr[i]);
}
}
return tempArrayRev.reverse().concat(tempArray);
}
var array = [1,2,3,4,5,6];
document.getElementById('arrayOutput').innerHTML += reverseArray(array,4);
<div id="arrayOutput">Array :<br></div>

This is the answer you can test it.
function reverseArray(n, a) {
var interimArray1 = [];
for (var i = 0; i < a.length; i++) {
interimArray1.push(a[i]);
}
for (var i = num; i >=0; i--) {
interimArray1[i-1] = a[n - i];
}
for (var i = 0; i < interimArray1.length; i++) {
console.log(interimArray1[i]);
}
}
var arr = [1, 2, 3, 4, 5, 6];
var num = 4;
reverseArray(num, arr);

You could use something like this.
function reverseArray(n, arrIn) {
// Splice splits the array in 2 starting at 0 index going n long
var arrOut = arrIn.splice(0,n);
// reverse is pretty straight forward
arrOut = arrOut.reverse();
// Concat joins the two together
return arrOut.concat(arrIn);
}

Returning a missing element from a string of integers

I'm trying to write a function that accepts an array of unique integers between 0 and 9 (inclusive), and returns the missing element. Here's what I've got. So near but so far.
var currentPlace = 0;
function getMissingElement(superImportantArray){
while (currentPlace < 9){
for (var i = 0; i < 10; i++) {
var arrayNum = superImportantArray[currentPlace]
if (i == arrayNum) {
currentPlace ++;
console.log("so it's not " + i);
}
else if (i !=arrayNum) {
console.log("try..." + i);
}
}
}
}
// run
var myArray = [0,5,1,3,2,9,7,6,4]; // this test should return 8
getMissingElement(myArray);
I'm not sure i'm approaching this correctly. Thanks for your time.

Just wanted to post my answer from the comments. A simpler way to handle this, in my opinion, is to loop over the original array, and flag a new array at the index that they represent. For example, if the number is 4, flag the 4th index in the new array. The reason for all this is because once this is done, one index should be left unflagged. All that would be left to do is find the unflagged index.
Here's an example (I commented the code here, not the fiddle):
function findMissing(array, min, max) {
var missing, unfilledArray, i, j;
// Array to hold the flags
unfilledArray = [];
for (i = min, j = max; i <= j; i++) {
// Flag the index in the new array with the current value
unfilledArray[array[i]] = true;
}
for (i = min, j = max; i <= j; i++) {
// Loop over new array and find the unflagged index
currentUnfilled = unfilledArray[i];
if (!currentUnfilled) {
// Current index not flagged
missing = i;
break;
}
}
return missing;
}
DEMO: http://jsfiddle.net/6GAyw/
The other little feature I added was that you explicitly specify the minimum and maximum value, which, in your case, is 0 and 9. This feature allows this solution to be used on any range of numbers (unlike my original comment/suggestion).
And not that I fully understand big O notation, but I believe this is O(2n), not O(n^2), since there aren't nested loops/indexOf.
If you were looking to get all missing numbers in a range, you can easily modify the function to return an array of unflagged indexes instead. Here's an example:
function findMissing(array, min, max) {
var missing, unfilledArray, i, j;
// Array to hold the missing numbers
missing = [];
// Array to hold the flags
unfilledArray = [];
for (i = min, j = max; i <= j; i++) {
// Flag the index in the new array with the current value
unfilledArray[array[i]] = true;
}
for (i = min, j = max; i <= j; i++) {
// Loop over new array and find the unflagged index
currentUnfilled = unfilledArray[i];
if (!currentUnfilled) {
// Current index not flagged
missing.push(i);
}
}
return missing;
}
DEMO: http://jsfiddle.net/zFS89/

function getMissingElement(superImportantArray){
var result = [], length = Math.max(10, superImportantArray.length);
for (var i = 0; i < length; i++) {
if(superImportantArray.indexOf(i) == -1){
result.push(i);
}
}
return result;
}
Try this. This will return an array of missing elements else return an empty array.
DEMO FIDDLE

So here is one way to do it: Since you know that the array only contains values from 0 to 9, you can build a "set" of numbers and remove each "seen" value in the array from the set:
function getMissingElement(superImportantArray){
var numbers = {};
for (var i = 0; i < 10; i++) {
numbers[i] = true
}
for (var i = 0, l = superImportantArray.length; i < l; i++) {
delete numbers[superImportantArray[i]];
}
return Object.keys(numbers);
}
This would return an array of all numbers that are missing. If there can always only be one missing number you can easily modify this to directly return the number instead.

This should do it.
function getMissingElement(arrayTest) {
// create an array with all digits
var digitsArray = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
var index;
for (var i=0; i<arrayTest.length; i++) {
// get the index of current digit on our array
index = digitsArray.indexOf(arrayTest[i]);
// if found, remove it from our array.
if (index >= 0) {
digitsArray.splice(index,1);
}
}
// the last remaining digit in the original array should be the one missing.
return (digitsArray[0]);
}

This one is better for the eyes.
function getMissingElement(superImportantArray) {
return superImportantArray.reduce(function (sum, value) {return sum - value;}, 45);
}

How do I extract even elements of an Array?

var arr = [4, 5, 7, 8, 14, 45, 76];
function even(a) {
var ar = [];
for (var i = 0; i < a.length; i++) {
ar.push(a[2 * i + 1]);
}
return ar;
}
alert(even(arr));
http://jsbin.com/unocar/2/edit
I have tried this code in order to output even (index) elements of an array. It works, but it also outputs some empty elements. How do I fix this code to output only existing elements?

Either use modulus:
for (var i = 0; i < a.length; i++) {
if(i % 2 === 0) { // index is even
ar.push(a[i]);
}
}
or skip every second element by incrementing i accordingly:
for(var i = 0; i < a.length; i += 2) { // take every second element
ar.push(a[i]);
}
Notice: Your code actually takes the elements with odd indexes from the array. If this is what you want you have to use i % 2 === 1 or start the loop with var i = 1 respectively.

For IE9+ use Array.filter
var arr = [4,5,7,8,14,45,76];
var filtered = arr.filter(function(element, index, array) {
return (index % 2 === 0);
});
With a fallback for older IEs, all the other browsers are OK without this fallback
if (!Array.prototype.filter)
{
Array.prototype.filter = function(fun /*, thisp */)
{
"use strict";
if (this === void 0 || this === null)
throw new TypeError();
var t = Object(this);
var len = t.length >>> 0;
if (typeof fun !== "function")
throw new TypeError();
var res = [];
var thisp = arguments[1];
for (var i = 0; i < len; i++)
{
if (i in t)
{
var val = t[i]; // in case fun mutates this
if (fun.call(thisp, val, i, t))
res.push(val);
}
}
return res;
};
}

This will work on 2018 :)
take the odd indexes and apply to filter
var arr = [4, 5, 7, 8, 14, 45, 76, 5];
let filtered = arr.filter((a,i) => i%2===1);
console.log(filtered);

Even if this question is quite old, I would like to add a one-liner filter:Odd numbers: arr.filter((e,i)=>i%2)Even numbers: arr.filter((e,i)=>i%2-1)A more 'legal' way for even numbers: arr.filter((e,i)=>!(i%2))
There's no need to check with ===1 like sumit said. mod 2 already returns a 0 or a 1, you can let them be interpreted as boolean values.
You can use i&1 instead of i%2, while it benefits performance on big arrays, it can work only on 31 bit integers.

why don't you try with the % operator. It gives you the remaining of a division.
replace the loop block with
if ((i % 2) === 0) {
ar.push(a[i])
}

var arr = [4,5,7,8,14,45,76];
function even(a)
{
var ar = [];
for (x in a)
{
if((a[x]%2)==0)
ar.push(a[x]);
}
return ar;
}
alert(even(arr));

I just wanted to explain why your result is not what you expected since everyone else shows excellent solutions. You are iterating over an array size N so your resulting array will attempt to push elements in an array that will result in size N. Since only N/2 will be found in the original array your resulting array will fill the rest with blanks to fill in the rest of N. So if you checked to see if a[2*i] exists OR checked to see if a[i] % 2 == 0 before inserting, your resulting array will contain only the even indexed values

You need to test the elements for evenness like this:
var arr = [4,5,7,8,14,45,76];
function even(a){
var ar = [];
for (var i=0; i<a.length;i++){
if (a[i] % 2 === 0)
{
ar.push(a[i]);
}
}
return ar;
}
alert(even(arr));
%2 is the modulo operator, it returns the remainder of integer division.