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Deleting everything that is not a number from the array. Why isn't this working? [duplicate]

This question already has answers here:
Looping through array and removing items, without breaking for loop
(17 answers)
Closed 5 months ago.
let arr = [1, "5", 3, 27, undefined, { name: 'Steven' }, 11];
for (let i = 0; i < arr.length; i++) {
if (typeof arr[i] !== 'number') {
arr.splice(i, 1);
}
}
console.log(arr); // output is: [1, 3, 27, {…}, 11]
If I swap the places of the object and the last number output is different.
let arr = [1, "5", 3, 27, undefined, 11, { name: 'Steven' }];
for (let i = 0; i < arr.length; i++) {
if (typeof arr[i] !== 'number') {
arr.splice(i, 1);
}
}
console.log(arr); // output is: [1, 3, 27, 11]
Can anyone explain why?

You shouldn't mutate an arrays length while you are iterating over it. Would recommend you use filter

Delete numbers below a certain number in an array

Today i'm facing a really weird problem.
I'm actually trying to delete numbers below a certain number in an array.
I have this array [1, 7, 2, 3, 90, 4, 70, 20] and I only want the numbers greater than 20 !
So I should have that in output : [90, 70]
But Instead I have this array : [7, 3, 90, 70, 20] ???
Here is my code :
function superior() {
var arr = [1, 7, 2, 3, 90, 4, 70, 20]
for (var i = 0; i < arr.length; i++) {
if (arr[i] < 20) {
arr.splice(arr.indexOf(arr[i]), 1);
} else {
break;
}
}
return arr;
}
console.log(superior());

Mutating an array while you're looping through it is always going to be tricky. If you shrink the size of the array, but don't change i, then you're can end up interacting with the wrong element of the array the next time through the loop. Also, using break will stop the loop from running entirely, and so skip the rest of the array. You may have meant to use continue, but since you're on the last line of the loop anyway, continue isn't needed.
Rather than trying to change the array and loop through it at the same time, i'd recommend creating a new array. Then you can loop through the old one unhindered:
const arr = [1, 7, 2, 3, 90, 4, 70, 20]
const newArr = []
for (const i = 0; i < arr.length; i++) {
if (arr[i] >= 20) {
newArr.push(arr[i]);
}
}
console.log(newArr)
Filtering an array like this is a very common thing to do, so there are built in tools to do it for you. Every array has a .filter method. You pass into it a function describing what elements you want to keep, and it will produce the new array for you:
const arr = [1, 7, 2, 3, 90, 4, 70, 20]
const newArr = arr.filter(element => element >= 20);
console.log(newArr)

You can filter them out according to a condition.
And replace your existing array with the filtered one. Or if you don't want to replace it, use another variable and assign the filtered value to that variable.
var newArray = array.filter(item => item > 20)
Check .filter()
var array = [1, 7, 2, 3, 90, 4, 70, 20];
array = array.filter(item => item > 20)
console.log(array)

You can use Array.filter()
var arr = [1, 7, 2, 3, 90, 4, 70, 20]
var filteredArr = arr.filter(item => item > 20);
console.log(filteredArr);

You could iterate from the end and omit indexOf, because you have already the index i.
This appoach loops from the end and after splicing, the remaining lower indices remains.
function superior() {
var array = [1, 7, 2, 3, 90, 4, 70, 20],
i = array.length;
while (i--) if (array[i] < 20) array.splice(i, 1);
return array;
}
console.log(superior());

Use temporary array to push new values.
function superior() {
var arr = [1, 7, 2, 3, 90, 4, 70, 20];
temp_arr = [];
for (var i = 0; i < arr.length; i++) {
if (arr[i] > 20) {
temp_arr.push(arr[i]);
}
}
return temp_arr;
}
console.log(superior());

Returning new array with ascii values and numbers of previous array

I'm writing a function without the use of any JS built-in functions. This function accepts an array ['z','C',2,115,30] and then creates a new array with ascii values of chars in that first array plus the normal numbers that were there previously, so: [127, 67, 2, 115,30]. The problem I'm having is that I cannot properly get the numbers from the 1st array into the second array.
So far I can get the new array to [127,67,2,1,1,5,3,0]. So the chars ascii value is being inputted into the new array properly, but while iterating through the numbers it puts each digit of a number and then a comma into the array when i just want the whole number to be inputted instead of each digit.
function func(x){
var str = '';
var arr = [];
for(var i=0; i<x.length;i++){
str += x[i]
}
for(var j=0; j<str.length;j++){
if(isNaN(str[j])){
arr += str.charCodeAt(j) + ', ';
}
else if(!isNaN(str[j])){
arr += str[j] + ', ';
}
} print(arr)
}
func(['z','C',2,115,30])
I expect output of [122, 67, 2, 115, 30], current output is [122, 67, 2, 1, 1, 5, 3, 0,]

Don't combine into one string - instead, use a simple for loop:
function func(x) {
var arr = [];
for (let i = 0; i < x.length; i++) {
if (typeof x[i] == "string") arr[arr.length] = (x[i].charCodeAt(0);
else arr[arr.length] = x[i];
}
console.log(arr);
}
func(['z','C',2,115,30]);
Using array methods is much simpler though:
function func(x) {
var arr = x.map(e => typeof e == "string" ? e.charCodeAt(0) : e);
console.log(arr);
}
func(['z','C',2,115,30]);

Find the first repeated number in a Javascript array

I need to find first two numbers and show index like:
var arrWithNumbers = [2,5,5,2,3,5,1,2,4];
so the first repeated number is 2 so the variable firstIndex should have value 0. I must use for loop.
var numbers = [7, 5, 7, 6, 6, 4, 9, 10, 2, 11];
var firstIndex
for (i = numbers[0]; i <= numbers.length; i++) {
firstIndex = numbers[0]
if (numbers[i] == firstIndex) {
console.log(firstIndex);
break;
}
}

You can use Array#indexOf method with the fromIndex argument.
var numbers = [7, 5, 7, 6, 6, 4, 9, 10, 2, 11];
// iterate upto the element just before the last
for (var i = 0; i < numbers.length - 1; i++) {
// check the index of next element
if (numbers.indexOf(numbers[i], i + 1) > -1) {
// if element present log data and break the loop
console.log("index:", i, "value: ", numbers[i]);
break;
}
}
UPDATE : Use an object to refer the index of element would make it far better.
var numbers = [7, 5, 7, 6, 6, 4, 9, 10, 2, 11],
ref = {};
// iterate over the array
for (var i = 0; i < numbers.length; i++) {
// check value already defined or not
if (numbers[i] in ref) {
// if defined then log data and brek loop
console.log("index:", ref[numbers[i]], "value: ", numbers[i]);
break;
}
// define the reference of the index
ref[numbers[i]] = i;
}

Many good answers.. One might also do this job quite functionally and efficiently as follows;
var arr = [2,5,5,2,3,5,1,2,4],
frei = arr.findIndex((e,i,a) => a.slice(i+1).some(n => e === n)); // first repeating element index
console.log(frei)
If might turn out to be efficient since both .findIndex() and .some() functions will terminate as soon as the conditions are met.

You could use two for loops an check every value against each value. If a duplicate value is found, the iteration stops.
This proposal uses a labeled statement for breaking the outer loop.
var numbers = [1, 3, 6, 7, 5, 7, 6, 6, 4, 9, 10, 2, 11],
i, j;
outer: for (i = 0; i < numbers.length - 1; i++) {
for (j = i + 1; j < numbers.length; j++) {
if (numbers[i] === numbers[j]) {
console.log('found', numbers[i], 'at index', i, 'and', j);
break outer;
}
}
}

Move through each item and find if same item is found on different index, if so, it's duplicate and just save it to duplicate variable and break cycle
var numbers = [7, 5, 7, 6, 6, 4, 9, 10, 2, 11];
var duplicate = null;
for (var i = 0; i < numbers.length; i++) {
if (numbers.indexOf(numbers[i]) !== i) {
duplicate = numbers[i];
break; // stop cycle
}
}
console.log(duplicate);

var numbers = [7, 5, 7, 6, 6, 4, 9, 10, 2, 11];
var map = {};
for (var i = 0; i < numbers.length; i++) {
if (map[numbers[i]] !== undefined) {
console.log(map[numbers[i]]);
break;
} else {
map[numbers[i]] = i;
}
}
Okay so let's break this down. What we're doing here is creating a map of numbers to the index at which they first occur. So as we loop through the array of numbers, we check to see if it's in our map of numbers. If it is we've found it and return the value at that key in our map. Otherwise we add the number as a key in our map which points to the index at which it first occurred. The reason we use a map is that it is really fast O(1) so our overall runtime is O(n), which is the fastest you can do this on an unsorted array.

As an alternative, you can use indexOf and lastIndexOf and if values are different, there are multiple repetition and you can break the loop;
function getFirstDuplicate(arr) {
for (var i = 0; i < arr.length; i++) {
if (arr.indexOf(arr[i]) !== arr.lastIndexOf(arr[i]))
return arr[i];
}
}
var arrWithNumbers = [2, 5, 5, 2, 3, 5, 1, 2, 4];
console.log(getFirstDuplicate(arrWithNumbers))
var numbers = [1, 3, 6, 7, 5, 7, 6, 6, 4, 9, 10, 2, 11]
console.log(getFirstDuplicate(numbers))

I have the same task and came up with this, pretty basic solution:
var arr = [7,4,2,4,5,1,6,8,9,4];
var firstIndex = 0;
for(var i = 0; i < arr.length; i++){
for( var j = i+1; j < arr.length; j++){
if(arr[i] == arr[j]){
firstIndex = arr[i];
break;
}
}
}
console.log(firstIndex);
First for loop takes the first element from array (number 7), then the other for loop checks all other elements against it, and so on.
Important here is to define j in second loop as i+1, if not, any element would find it's equal number at the same index and firstIndex would get the value of the last one after all loops are done.

To reduce the time complexity in the aforementioned answers you can go with this solution:
function getFirstRecurringNumber(arrayOfNumbers) {
const hashMap = new Map();
for (let number of arrayOfNumbers) { // Time complexity: O(n)
const numberDuplicatesCount = hashMap.get(number);
if (numberDuplicatesCount) {
hashMap.set(number, numberDuplicatesCount + 1);
continue;
}
hashMap.set(number, 1); // Space complexity: O(n)
}
for (let entry of hashMap.entries()) { // Time complexity: O(i)
if (entry[1] > 1) {
return entry[0];
}
}
}
// Time complexity: O(n + i) instead of O(n^2)
// Space complexity: O(n)

Using the code below, I am able to get just the first '5' that appears in the array. the .some() method stops looping through once it finds a match.
let james = [5, 1, 5, 8, 2, 7, 5, 8, 3, 5];
let onlyOneFives = [];
james.some(item => {
//checking for a condition.
if(james.indexOf(item) === 0) {
//if the condition is met, then it pushes the item to a new array and then
//returns true which stop the loop
onlyOneFives.push(item);
return james.indexOf(item) === 0;
}
})
console.log(onlyOneFives)

Create a function that takes an array with numbers, inside it do the following:
First, instantiate an empty object.
Secondly, make a for loop that iterates trough every element of the array and for each one, add them to the empty object and check if the length of the object has changed, if not, well that means that you added a element that already existed so you can return it:
//Return first recurring number of given array, if there isn't return undefined.
const firstRecurringNumberOf = array =>{
objectOfArray = {};
for (let dynamicIndex = 0; dynamicIndex < array.length; dynamicIndex ++) {
const elementsBeforeAdding = (Object.keys(objectOfArray)).length;0
objectOfArray[array[dynamicIndex]] = array[dynamicIndex]
const elementsAfterAdding = (Object.keys(objectOfArray)).length;
if(elementsBeforeAdding == elementsAfterAdding){ //it means that the element already existed in the object, so it didnt was added & length doesnt change.
return array[dynamicIndex];
}
}
return undefined;
}
console.log(firstRecurringNumberOf([1,2,3,4])); //returns undefined
console.log(firstRecurringNumberOf([1,4,3,4,2,3])); //returns 4

const arr = [1,9,5,2,3,0,0];
const copiedArray = [...arr];
const index = arr.findIndex((element,i) => {
copiedArray.splice(0,1);
return copiedArray.includes(element)
})
console.log(index);

var addIndex = [7, 5, 2, 3, 4, 5, 7,6, 2];
var firstmatch = [];
for (var i = 0; i < addIndex.length; i++) {
if ($.inArray(addIndex[i], firstmatch) > -1) {
return false;
}
firstmatch.push(addIndex[i]);
}

Deleting both values from array if duplicate - JavaScript/jQuery

I have an array here:
var myArr = [1, 1, 2, 5, 5, 7, 8, 9, 9];
Now I want to remove both appearances of a duplicate. So the desired result is not:
var myArr = [1, 2, 5, 7, 8 ,9];
but
var myArr = [2, 7, 8];
Basically I know how to remove duplicates, but not in that that special way. Thats why any help would be really appreciated!
Please note: My array is filled with strings. The numbers here were only used as an example.

jsfiddle for this code:
var myArr = [1, 1, 2, 5, 5, 7, 8, 9, 9];
var newArr = myArr;
var h,i,j;
for(h = 0; h < myArr.length; h++) {
var curItem = myArr[h];
var foundCount = 0;
// search array for item
for(i = 0; i < myArr.length; i++) {
if (myArr[i] == myArr[h])
foundCount++;
}
if(foundCount > 1) {
// remove repeated item from new array
for(j = 0; j < newArr.length; j++) {
if(newArr[j] == curItem) {
newArr.splice(j, 1);
j--;
}
}
}
}

Here's my version
var a = [1, 1, 2, 5, 5, 7, 8, 9, 9];
function removeIfduplicate( arr ) {
var discarded = [];
var good = [];
var test;
while( test = arr.pop() ) {
if( arr.indexOf( test ) > -1 ) {
discarded.push( test );
continue;
} else if( discarded.indexOf( test ) == -1 ) {
good.push( test );
}
}
return good.reverse();
}
x = removeIfduplicate( a );
console.log( x ); //[2, 7, 8]

EDITED with better answer:
var myArr = [1, 1, 2, 5, 5, 7, 8, 9, 9];
function removeDuplicates(arr) {
var i, tmp;
for(i=0; i<arr.length; i++) {
tmp = arr.lastIndexOf(arr[i]);
if(tmp === i) {
//Only one of this number
} else {
//More than one
arr.splice(tmp, 1);
arr.splice(i, 1);
}
}
}

Using Hashmap
create hashmap and count occurencies
filter where hashmap.get(value) === 1 (only unique values)
const myArray = [1, 1, 2, 5, 5, 7, 8, 9, 9];
const map = new Map();
myArray.forEach(v => map.set(v, map.has(v) ? map.get(v)+1 : 1));
myArray.filter(v => map.get(v) === 1);
Old version (slower but valid too)
Heres a short version using Array.filter(). The trick is to first find all values that are NOT uniqe, and then use this array to reject all unique items in the original array.
let myArr = [1, 1, 2, 5, 5, 7, 8, 9, 9];
let duplicateValues = myArr.filter((item, indx, s) => s.indexOf(item) !== indx);
myArr.filter(item => !duplicateValues.includes(item));
// => [2, 7, 8]

Wherever removing duplicates is involved, it's not a bad idea to use a set data structure.
JavaScript doesn't have a native set implementation, but the keys of an object work just as well - and in this case help because then the values can be used to keep track of how often an item appeared in the array:
function removeDuplicates(arr) {
var counts = arr.reduce(function(counts, item) {
counts[item] = (counts[item] || 0) + 1;
return counts;
}, {});
return Object.keys(counts).reduce(function(arr, item) {
if (counts[item] === 1) {
arr.push(item);
}
return arr;
}, []);
}
var myArr = [1, 1, 2, 5, 5, 7, 8, 9, 9];
console.log(removeDuplicates(myArr), myArr);
Check out the example on jsfiddle.
Alternately, you could not use calls to reduce(), and instead use for and for(item in counts) loops:
function removeDuplicates(arr) {
var counts = {};
for(var i=0; i<arr.length; i++) {
var item = arr[i];
counts[item] = (counts[item]||0)+1;
}
var arr = [];
for(item in counts) {
if(counts[item] === 1) {
arr.push(item);
}
}
return arr;
}
Check out the example on jsfiddle.

If it's just alphanumeric, duplicates are case-sensitive, and there can be no more than two of any element, then something like this can work:
var a = [2, 1, "a", 3, 2, "A", "b", 5, 6, 6, "B", "a"],
clean_array = $.map(a.sort(), function (v,i) {
a[i] === a[i+1] && (a[i] = a[i+1] = null);
return a[i];
});
// clean_array = [1,3,5,"A","B","b"]

In this example,we are taking two arrays as function arguments, from this we are going to print only unique values of both arrays hence deleting the values that are present in both arrays.
first i am concatenating both the arrays into one. Then I taking each array value at a time and looping over the array itself searching for its no of occurrence. if no of occurrence(i.e.,count) equal to 1 then we are pushing that element into the result array. Then we can return the result array.
function diffArray(arr1, arr2) {
var newArr = [];
var myArr=arr1.concat(arr2);
var count=0;
for(i=0;i<myArr.length;i++){
for(j=0;j<myArr.length;j++){
if(myArr[j]==myArr[i]){
count++;
}
}
if(count==1){
newArr.push(myArr[i]);
}
count=0;
}
return newArr;
}

EDIT: Here is the jspref http://jsperf.com/deleting-both-values-from-array
http://jsfiddle.net/3u7FK/1/
This is the fastest way to do it in two passes without using any fancy tricks and keeping it flexible. You first spin through and find the count of every occurance and put it into and keyvalue pair. Then spin through it again and filter out the ones where the count was greater than 1. This also has the advanatage of being able to apply other filters than just "greater than 1"; as well as the having the count of occurances if you needed that as well for something else.
This should work with strings as well instead of numbers.
http://jsfiddle.net/mvBY4/1/
var myArr = [1, 1, 2, 5, 5, 7, 8, 9, 9];
var map = new Object();
for(var i = 0; i < myArr.length; i++)
{
if(map[myArr[i]] === undefined)
{
map[myArr[i]] = 1;
}
else
{
map[myArr[i]]++;
}
}
var result = new Array();
for(var i = 0; i < myArr.length; i++)
{
if(map[myArr[i]] > 1)
{
//do nothing
}
else
{
result.push(myArr[i]);
}
}
alert(result);

You can use Set (available in IE 11+) as below
const sourceArray = [1, 2, 3, 4, 5, 5, 6, 6, 7, 7, 8];
const duplicatesRemoved = new Set();
sourceArray.forEach(element => {
if (duplicatesRemoved.has(element)) {
duplicatesRemoved.delete(element)
} else {
duplicatesRemoved.add(element)
}
})
console.log(Array.from(duplicatesRemoved))
N.B. Arrow functions are not supported in older browsers. Use normal function syntax for that instead. However, Array.from can easily be polyfilled for older browsers.
Try it here.