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.
Related
Just need help in identifying what I am doing wrong on this codewar challenge.
I realize this may be easy for some but please note I am just a beginner with Javascript.
The challenge:
You are given an array (which will have a length of at least 3, but
could be very large) containing integers. The array is either entirely
comprised of odd integers or entirely comprised of even integers
except for a single integer N. Write a method that takes the array as
an argument and returns N.
For example:
[2, 4, 0, 100, 4, 11, 2602, 36] should return 11.
[160, 3, 1719, 19, 11, 13, -21] should return 160.
My code:
function findOutlier(integers){
var even = [];
var odd = [];
for (var i = 0; i < integers; i++) {
if (integers[i] % 2 === 0) {
even.push(integers[i]);
} else {
odd.push(integers[i]);
}
if (even.length === 1) {
return even;
} else {
return odd;
}
}
}
Another possible way:
function myFunction(integers) {
var odds = integers.filter(function(num) {return num % 2});
var evens = integers.filter(function(num) {return !(num % 2)});
return evens.length == 1 ? evens[0] : odds[0];
}
You can check out this CodePen Demo to test the function in Mocha.
I've found 2 issues inside your code block. You have to run loop over array length instead of entire array otherwise you can use foreach loop. You need to return odd/even value after finishing your loop. Please check updated code as follows, hope it'll help.
function findOutlier(integers){
var even = [];
var odd = [];
for (var i = 0; i < integers.length; i++) {
if (integers[i] % 2 === 0) {
even.push(integers[i]);
} else {
odd.push(integers[i]);
}
}
if (even.length === 1) {
console.log("OK..1");
return even;
} else {
console.log("OK..2");
return odd;
}
}
Try:
for (var i = 0; i < integers.length; i++)
instead of:
for (var i = 0; i < integers; i++)
If you want to get the first occurrence without iterating to the last item but without using an imperative code you may do as by using .find() checking inside if the rightmost bit of n is 1;
var arr = [2, 4, 0, 100, 4, 11, 2602, 36],
resodd = arr.find(n => n & 1),
reseven = arr.find(n => !(n & 1));
console.log(resodd, reseven);
push the odd and the even results into an array, and then compare what you have left
function findOutlier(integers) {
var arrO=[];
var arrE=[];
for(i=0; i<integers.length; i++){
if(integers[i] % 2==0 ){
arrE.push(integers[i])
}
else {
arrO.push(integers[i])
}
}
if(arrO.length > arrE.length){
return arrE[0]
}return arrO[0]
}
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);
}
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);
}
Hi i am working on LIME programming which is a subset of javascript.
i need to use javascript.splice to remove certain elements from my array, sad to say, LIME does not support splice function.
Any idea how do i create my own function to remove elements from an array?
Thanks for your time.
EDIT: Manage to create a simple function.
function removeElements(array, index)
{
var tempArray = new Array();
var counter = 0;
for(var i = 0; i < array.length; i++)
{
if(i != index)
{
tempArray[counter] = array[i];
counter++;
}
}
return tempArray;
}
Array.prototype.splice is fully defined in ECMA-262 §15.4.4.12, so use that as your spec and write one. e.g.
15.4.4.12 Array.prototype.splice (start, deleteCount [ , item1 [ ,item2 [ , … ] ] ] )
When the splice
method is called with two or more
arguments start, deleteCount and
(optionally) item1, item2, etc., the
deleteCount elements of the array
starting at array index start are
replaced by the arguments item1,
item2, etc. An Array object containing
the deleted elements (if any) is
returned. The following steps are
taken:...
You will probably have to create a new array, copy the members up to start from the old array, insert the new members, then copy from start + deleteCount to the end to the new array.
Edit
Here is an amended splice, the first I posted was incorrect. This one splices the array passed in and returns the removed members. It looks a bit long but I tried to keep it close to the spec and not assume support for any complex Array methods or even Math.max/min. It can be simplified quite a bit if they are.
If push isn't supported, it can be replaced fairly simply too.
function arraySplice(array, start, deleteCount) {
var result = [];
var removed = [];
var argsLen = arguments.length;
var arrLen = array.length;
var i, k;
// Follow spec more or less
start = parseInt(start, 10);
deleteCount = parseInt(deleteCount, 10);
// Deal with negative start per spec
// Don't assume support for Math.min/max
if (start < 0) {
start = arrLen + start;
start = (start > 0)? start : 0;
} else {
start = (start < arrLen)? start : arrLen;
}
// Deal with deleteCount per spec
if (deleteCount < 0) deleteCount = 0;
if (deleteCount > (arrLen - start)) {
deleteCount = arrLen - start;
}
// Copy members up to start
for (i = 0; i < start; i++) {
result[i] = array[i];
}
// Add new elements supplied as args
for (i = 3; i < argsLen; i++) {
result.push(arguments[i]);
}
// Copy removed items to removed array
for (i = start; i < start + deleteCount; i++) {
removed.push(array[i]);
}
// Add those after start + deleteCount
for (i = start + (deleteCount || 0); i < arrLen; i++) {
result.push(array[i]);
}
// Update original array
array.length = 0;
i = result.length;
while (i--) {
array[i] = result[i];
}
// Return array of removed elements
return removed;
}
If you don't care about order of the array and you're just looking for a function to perform splice, here's an example.
/**
* Time Complexity: O(count) aka: O(1)
*/
function mySplice(array, start, count) {
if (typeof count == 'undefined') count = 1
while (count--) {
var index2remove = start + count
array[index2remove] = array.pop()
}
return array
}
If you want to return the removed elements like the normal splice method does this will work:
/**
* Time Complexity: O(count) aka: O(1)
*/
function mySplice(array, index, count) {
if (typeof count == 'undefined') count = 1
var removed = []
while (count--) {
var index2remove = index + count
removed.push(array[index2remove])
array[index2remove] = array.pop()
}
// for (var i = index; i < index + count; i++) {
// removed.push(array[i])
// array[i] = array.pop()
// }
return removed
}
This modifies the original Array, and returns the items that were removed, just like the original.
Array.prototype.newSplice = function( start, toRemove, insert ) {
var remove = this.slice( start, start + toRemove );
var temp = this.slice(0,start).concat( insert, this.slice( start + toRemove ) );
this.length = 0;
this.push.apply( this, temp );
return remove;
};
Comparison test: http://jsfiddle.net/wxGDd/
var arr = [0,1,2,3,4,5,6,7,8];
var arr2 = [0,1,2,3,4,5,6,7,8];
console.log( arr.splice( 3, 2, 6 ) ); // [3, 4]
console.log( arr ); // [0, 1, 2, 6, 5, 6, 7, 8]
console.log( arr2.newSplice( 3, 2, 6 ) ); // [3, 4]
console.log( arr2 ); // [0, 1, 2, 6, 5, 6, 7, 8]
It could use a little extra detail work, but for the most part it takes care of it.
Here is a simple implement in case the Array.prototype.splice dispears
if (typeof Array.prototype.splice === 'undefined') {
Array.prototype.splice = function (index, howmany, elemes) {
howmany = typeof howmany === 'undefined' || this.length;
var elems = Array.prototype.slice.call(arguments, 2), newArr = this.slice(0, index), last = this.slice(index + howmany);
newArr = newArr.concat.apply(newArr, elems);
newArr = newArr.concat.apply(newArr, last);
return newArr;
}
}
Are there any other methods that are missing in LIME's Array implementation?
Assuming at least the most basic push() and indexOf() is available, there's several ways you could do it. How this is done would depend on whether this is destructive method or whether it should return a new array. Assuming the same input as the standard splice(index, howMany, element1, elementN) method:
Create a new array named new
push() indexes 0 to index onto the new array
Now stop at index and push() any new elements passed in. If LIME supports the standard arguments variable then you can loop through arguments with index > 2. Otherwise you'll need to pass in an array instead of a variable number of parameters.
After inserting the new objects, continue looping through the input array's elements, starting at index + howMany and going until input.length
I believe that should get you the results you're looking for.
I have used this below function as an alternative for splice()
array = mySplice(array,index,count);
above is the function call,
And this is my function mySplice()
function mySplice(array, index, count)
{
var newArray = [];
if( count > 0 )
{ count--;}
else
{ count++;}
for(i = 0; i <array.length; i++)
{
if(!((i <= index + count && i >= index) || (i <= index && i >= index + count)))
{
newArray.push(array[i])
}
}
return newArray;
}
I have done it very similar way using only one for loop
function removeElements(a,index,n){
// a=> Array , index=> index value from array to delete
// n=> number of elements you want to delete
let temp = []; // for storing deleted elements
let main_array = []; // for remaining elements which are not deleted
let k = 0;
for(let i=0;i<a.length;i++){
if((i===index) || ((index<i && i<n+index))){
temp[i]=a[i+1];
delete a[i];
}
if(a[i]!==undefined){
main_array[k] = a[i];
a[i] = main_array[k];
k++;
}
}
a=main_array;
return a;
}
a=[1,2,3,4,5];
console.log(removeElements(a,0,1));
follow link Jsfiddle
var a = [3, 2, 5, 6, 7];
var fromindex = 1
var toindex = 2;
for (var i = 0; i < a.length; i++) {
if (i >= fromindex + toindex || i < fromindex) {
console.log(a[i])
}
}
var a = [3, 2, 5, 6, 7];
var temp=[];
function splice(fromindex,toindex)
{
for (var i = 0; i < a.length; i++) {
if(i>=fromindex+toindex || i<fromindex)
{
temp.push(a[i])
}
}
return temp
}
console.log(splice(3,2))
I am not sure of how to do this, but what I want to do it create an array and be able to add new items to this array. Since the items are supposed to be a random number, when a new instance is created I would like it to be checked against the rest of the array and be sure that the number it has generated is not already in the array. How would I accomplish something like this?
I looked at Šime Vidas's answer and it seems to work, but I tried to shorten it to
var arr = [];
function add(a) {
var n = ~~(Math.random() * 100);
for (var i = 0; i < a.length; i++) {
if ( a[i] === n) { a.push(n) }
}
}
for (var i=0; i<5; i++){
add(arr)
}
document.getElementById('output').innerHTML += arr;
and I don't understand why this wouldn't work. It does pretty much the same thing, correct?
var arr = [];
function add(a) {
var n = ~~(Math.random() * 1000);
!is(a, n) && a.push(n);
}
function is(a, n) {
for (var i = 0; i < a.length; i++) {
if ( a[i] === n ) { return true; }
}
return false;
}
The add function creates a random integer number between 0 and 1000, and adds it to the array.
The is function checks whether the n number is somewhere inside the a array.
Demo: http://jsfiddle.net/kHhMp/2/
Demo 2: http://jsfiddle.net/kHhMp/3/
(Demo 2 shows that a number will only be added to the array if it's not already in it.)
btw
!is(a, n) && a.push(n);
is a short form of this:
if ( is(a, n) == false ) { a.push(n); }
The number is added to the array only if is(a, n) returns false.
UPDATE
var arr = [];
function add(a) {
var n = ~~(Math.random() * 1000),
ok = true;
for (var i = 0; i < a.length; i++) {
if ( a[i] === n ) { ok = false; }
}
ok && a.push(n);
}
If you enjoy fast code and you have many items in your array, you should use an Object rather than an Array.
Instead of doing var my_array=[]; my_array.push(my_number), use var my_object = {}; my_object[my_number] = true to add items in your structure.
With that approach, you can easily check if a new number is already in there with an if (my_object[my_number]) { /* already there */ } else { /* not there yet */ }
Once you're done, you can extract the list of numbers as an array by either using var keys = Object.keys(my_object), or if that's not available, var keys=[],i=0;for (keys[i++] in my_object);, more or less.
You may extend the built in Array object for your needs.
Array.prototype.pushUnique = function(value) {
var len = this.length;
for(var i = 0; i < len; i++) {
if(this[i]===value) return;
}
this.push(value);
}
var uniques = new Array();
uniques.pushUnique(1);
uniques.pushUnique(2);
uniques.pushUnique(1);
// array will contain only 1 and 2
The fastest, most cross-browser way is to iterate over the array using a loop:
var arr = [];
function addNum(num) {
for (i = 0, len = arr.length; i < len; i++) {
if ( arr[i] === num ) { return false; }
}
arr.push(num);
}
Be sure to get the length of the array before you run the loop so the length property isn't constantly checked.
var array = []
array[0] = 'Item'
array[0] === undefined
# returns false
array[1] === undefined
# returns true