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JavaScript - find the same first element using for loop

I've got to create var with few elements:
var arrNum = [4,7,5,3,4,5,6,7,8,10]
I need to find first number that is the same in array using for loop. So it will be "4" and "4"
I need to create var sameIndex and adjust the same number to sameIndex and print after for loop
So I did loop
for(var i = 0; i < arrNum.length; i++){
console.log("")
console.log("Loop number is " + i)
if(arrNum[i] === arrNum[i]{
break
sameIndex = going[i]
}
}
console.log(sameIndex)
It's not working.

One way would be to use Array#indexOf. It returns the first index of the given element in the array:
var arrNum = [4, 7, 5, 3, 4, 5, 6, 7, 8, 10];
var i;
var sameIndex = -1;
for (i = 0; i < arrNum.length; i++) {
if (arrNum.indexOf(arrNum[i]) !== i) {
console.log('This is the second occurrence of', arrNum[i]);
sameIndex = arrNum.indexOf(arrNum[i]);
break;
}
}
console.log('The indices are', sameIndex, 'and', i);

If you're looking to get the first duplicate, then use :
var arrNum = [4,7,5,3,4,5,6,7,8,10];
function firstDuplicate(array){
var history = [];
for(var element of array){
if(history.indexOf(element)<0)
//not in the history yet
history.push(element);
else
return element;
}
return null; //or any distinctive value
}
var fDup = firstDuplicate(arrNum);

You could take a hash table and display the value.
var array = [4, 7, 5, 3, 4, 5, 6, 7, 8, 10],
hash = Object.create(null),
i;
for (i = 0; i < array.length; i++) {
if (hash[array[i]]) {
console.log('first dupe: ' + array[i]);
break;
}
hash[array[i]] = true;
}

A working script:
var arrNum = [4,7,5,3,4,5,6,7,8,10];
var sameIndex = -1, going = new Array();
for(var j = 0; j < arrNum.length; j++) {
for(var i = 1; i < arrNum.length; i++){
console.log("Loop number is " + j + ", " + i)
if(arrNum[i] === arrNum[j]){
sameIndex = i;
break;
}
}
if(sameIndex > 0) {
console.log(j, sameIndex)
break;
}
}

var arrNum = [4,7,5,3,4,5,6,7,8,10];
for(var i = 0; i < arrNum.length; i++){
for(var j = i+1; j < arrNum.length; j++) {
if(arrNum[i] === arrNum[j]) {
console.log('value: '+arrNum[i]+' index1: '+i+' index2: '+j);
}
}
}

Iterate over the original array and if the value is not in a comparison array - push it into a common numbers array. The first item in that common numbers array is the target.
var origArray = [4,7,5,3,4,5,6,7,8,10];
var newArray = [];
var commonNums = [];
origArray.forEach(function(item,i) {
newArray.indexOf(item) == -1
? newArray.push(item)
: commonNums.push({item: item, index:i});
})
if(commonNums.length > 0) {
var firstCommonNum = commonNums[0].item;
var firstCommonNumIndex = commonNums[0].index;
console.log("common number " + firstCommonNum);
console.log("Loop number is " + firstCommonNumIndex);
}

How to retrieve elements in an array after filtered by other array using Javascript>

just a quick question. How can I retrieve elements in an array given that the elements cannot be divided by other elements in an array? for example= arr =[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16] and collection =[2,3,5,7], then the result would be result =[11,13]
I have tried with this code , but it didn't work
for(var i=0; i<arr.length;i++){
for (var j=0; j<collection.length; j++){
if (arr[i]/collection[j] === 0){
arr.splice(i,1);
}
}
}

Something like this
1) using filter and forEach:
var arr =[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16];
var collection =[2,3,5,7];
function notDivided(arr1, arr2) {
return arr1.filter( function(item) {
var can_be_divided = false;
arr2.forEach( function(item_to_divide) {
if (item % item_to_divide === 0) can_be_divided = true;
});
return !can_be_divided;
});
}
var new_array = notDivided(arr,collection);
console.log(new_array);
2) using filter and some (#torazaburo's suggestion)
function notDivided(arr1, arr2) {
return arr1.filter( function(item) {
return !arr2.some( function(item_to_divide) {
return (item % item_to_divide === 0);
});
});
}
Or, yet, using arrow functions:
notDivided = (arr1, arr2) => arr1.filter( (item) => !arr2.some( (item_to_divide) => item % item_to_divide === 0 ) );

declare new array to collect results
use % symbol to check
example 4 % 2 = 0
this mean 4 can be divided by 2 because result is 0 so you can make
if (!(arr[i]%collection[j] === 0)){
newarr.push(arr[i])
}
edited
if you wanna example without think here return what you post too
var arr =[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16];
var collection =[2,3,5,7];
var newarr=[];
var addtonew = 0;
for(var i=0; i < arr.length; i++){
addtonew = -1;
for (var j=0; j<collection.length; j++){
if (arr[i]%collection[j] === 0)
{
addtonew = -1;
break;
}
addtonew = arr[i];
}
if (addtonew > -1 )
newarr.push(addtonew);
}
#[11, 13]

show most frequently occuring input value [duplicate]

var store = ['1','2','2','3','4'];
I want to find out that 2 appear the most in the array. How do I go about doing that?

I would do something like:
var store = ['1','2','2','3','4'];
var frequency = {}; // array of frequency.
var max = 0; // holds the max frequency.
var result; // holds the max frequency element.
for(var v in store) {
frequency[store[v]]=(frequency[store[v]] || 0)+1; // increment frequency.
if(frequency[store[v]] > max) { // is this frequency > max so far ?
max = frequency[store[v]]; // update max.
result = store[v]; // update result.
}
}

Solution with emphasis to Array.prototype.forEach and the problem of getting more than one key if the max count is shared among more items.
Edit: Proposal with one loop, only.
var store = ['1', '2', '2', '3', '4', '5', '5'],
distribution = {},
max = 0,
result = [];
store.forEach(function (a) {
distribution[a] = (distribution[a] || 0) + 1;
if (distribution[a] > max) {
max = distribution[a];
result = [a];
return;
}
if (distribution[a] === max) {
result.push(a);
}
});
console.log('max: ' + max);
console.log('key/s with max count: ' + JSON.stringify(result));
console.log(distribution);

arr.sort();
var max=0,result,freq = 0;
for(var i=0; i < arr.length; i++){
if(arr[i]===arr[i+1]){
freq++;
}
else {
freq=0;
}
if(freq>max){
result = arr[i];
max = freq;
}
}
return result;

Make a histogram, find the key for the maximum number in the histogram.
var hist = [];
for (var i = 0; i < store.length; i++) {
var n = store[i];
if (hist[n] === undefined) hist[n] = 0;
else hist[n]++;
}
var best_count = hist[store[0]];
var best = store[0];
for (var i = 0; i < store.length; i++) {
if (hist[store[i]] > best_count) {
best_count = hist[store[i]];
best = store[i];
}
}
alert(best + ' occurs the most at ' + best_count + ' occurrences');
This assumes either there are no ties, or you don't care which is selected.

Another ES6 option. Works with strings or numbers.
function mode(arr) {
const store = {}
arr.forEach((num) => store[num] ? store[num] += 1 : store[num] = 1)
return Object.keys(store).sort((a, b) => store[b] - store[a])[0]
}

If the array is sorted this should work:
function popular(array) {
if (array.length == 0) return [null, 0];
var n = max = 1, maxNum = array[0], pv, cv;
for(var i = 0; i < array.length; i++, pv = array[i-1], cv = array[i]) {
if (pv == cv) {
if (++n >= max) {
max = n; maxNum = cv;
}
} else n = 1;
}
return [maxNum, max];
};
popular([1,2,2,3,4,9,9,9,9,1,1])
[9, 4]
popular([1,2,2,3,4,9,9,9,9,1,1,10,10,10,10,10])
[10, 5]

This version will quit looking when the count exceeds the number of items not yet counted.
It works without sorting the array.
Array.prototype.most= function(){
var L= this.length, freq= [], unique= [],
tem, max= 1, index, count;
while(L>= max){
tem= this[--L];
if(unique.indexOf(tem)== -1){
unique.push(tem);
index= -1, count= 0;
while((index= this.indexOf(tem, index+1))!= -1){
++count;
}
if(count> max){
freq= [tem];
max= count;
}
else if(count== max) freq.push(tem);
}
}
return [freq, max];
}
//test
var A= ["apples","oranges","oranges","oranges","bananas",
"bananas","oranges","bananas"];
alert(A.most()) // [oranges,4]
A.push("bananas");
alert(A.most()) // [bananas,oranges,4]

I solved it this way for finding the most common integer
function mostCommon(arr) {
// finds the first most common integer, doesn't account for 2 equally common integers (a tie)
freq = [];
// set all frequency counts to 0
for(i = 0; i < arr[arr.length-1]; i++) {
freq[i] = 0;
}
// use index in freq to represent the number, and the value at the index represent the frequency count
for(i = 0; i < arr.length; i++) {
freq[arr[i]]++;
}
// find biggest number's index, that's the most frequent integer
mostCommon = freq[0];
for(i = 0; i < freq.length; i++) {
if(freq[i] > mostCommon) {
mostCommon = i;
}
}
return mostCommon;
}

This is my solution.
var max_frequent_elements = function(arr){
var a = [], b = [], prev;
arr.sort();
for ( var i = 0; i < arr.length; i++ ) {
if ( arr[i] !== prev ) {
a.push(arr[i]);
b.push(1);
} else {
b[b.length-1]++;
}
prev = arr[i];
}
var max = b[0]
for(var p=1;p<b.length;p++){
if(b[p]>max)max=b[p]
}
var indices = []
for(var q=0;q<a.length;q++){
if(b[q]==max){indices.push(a[q])}
}
return indices;
};

All the solutions above are iterative.
Here's a ES6 functional mutation-less version:
Array.prototype.mostRepresented = function() {
const indexedElements = this.reduce((result, element) => {
return result.map(el => {
return {
value: el.value,
count: el.count + (el.value === element ? 1 : 0),
};
}).concat(result.some(el => el.value === element) ? [] : {value: element, count: 1});
}, []);
return (indexedElements.slice(1).reduce(
(result, indexedElement) => (indexedElement.count > result.count ? indexedElement : result),
indexedElements[0]) || {}).value;
};
It could be optimized in specific situations where performance is the bottleneck, but it has a great advantage of working with any kind of array elements.
The last line could be replaced with:
return (indexedElements.maxBy(el => el.count) || {}).value;
With:
Array.prototype.maxBy = function(fn) {
return this.slice(1).reduce((result, element) => (fn(element) > fn(result) ? element : result), this[0]);
};
for clarity

If the array contains strings try this solution
function GetMaxFrequency (array) {
var store = array;
var frequency = []; // array of frequency.
var result; // holds the max frequency element.
for(var v in store) {
var target = store[v];
var numOccurences = $.grep(store, function (elem) {
return elem === target;
}).length;
frequency.push(numOccurences);
}
maxValue = Math.max.apply(this, frequency);
result = store[$.inArray(maxValue,frequency)];
return result;
}
var store = ['ff','cc','cc','ff','ff','ff','ff','ff','ff','yahya','yahya','cc','yahya'];
alert(GetMaxFrequency(store));

A fairly short solution.
function mostCommon(list) {
var keyCounts = {};
var topCount = 0;
var topKey = {};
list.forEach(function(item, val) {
keyCounts[item] = keyCounts[item] + 1 || 1;
if (keyCounts[item] > topCount) {
topKey = item;
topCount = keyCounts[item];
}
});
return topKey;
}
document.write(mostCommon(['AA', 'AA', 'AB', 'AC']))

This solution returns an array of the most appearing numbers in an array, in case multiple numbers appear at the "max" times.
function mode(numbers) {
var counterObj = {};
var max = 0;
var result = [];
for(let num in numbers) {
counterObj[numbers[num]] = (counterObj[numbers[num]] || 0) + 1;
if(counterObj[numbers[num]] >= max) {
max = counterObj[numbers[num]];
}
}
for (let num in counterObj) {
if(counterObj[num] == max) {
result.push(parseInt(num));
}
}
return result;
}

Remove duplicate item from array Javascript [duplicate]

This question already has answers here:
Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array
(97 answers)
Closed 8 years ago.
I'm looking for an easy way of removing a duplicate value from an array. I figured out how to detect if there is a duplicate or not, just I don't know how to "push" it from the value. For example, if you go to the link provided, and then type, "abca" (press return/enter key after each letter).. it will alert "duplicate!"
But I also want to figure out how to remove that duplicate from the textarea?
http://jsfiddle.net/P3gpp/
This is the part that seems to not be working ::
sort = sort.push(i);
textVal = sort;
return textVal;

Why do it the hard way, it can be done more easily using javascript filter function which is specifically for this kind of operations:
var arr = ["apple", "bannana", "orange", "apple", "orange"];
arr = arr.filter( function( item, index, inputArray ) {
return inputArray.indexOf(item) == index;
});
---------------------
Output: ["apple", "bannana", "orange"]

Based on user2668376 solution, this will return a new array without duplicates.
Array.prototype.removeDuplicates = function () {
return this.filter(function (item, index, self) {
return self.indexOf(item) == index;
});
};
After that you can do:
[1, 3, 3, 7].removeDuplicates();
Result will be; [1, 3, 7].

These are the functions I created/use for removing duplicates:
var removeDuplicatesInPlace = function (arr) {
var i, j, cur, found;
for (i = arr.length - 1; i >= 0; i--) {
cur = arr[i];
found = false;
for (j = i - 1; !found && j >= 0; j--) {
if (cur === arr[j]) {
if (i !== j) {
arr.splice(i, 1);
}
found = true;
}
}
}
return arr;
};
var removeDuplicatesGetCopy = function (arr) {
var ret, len, i, j, cur, found;
ret = [];
len = arr.length;
for (i = 0; i < len; i++) {
cur = arr[i];
found = false;
for (j = 0; !found && (j < len); j++) {
if (cur === arr[j]) {
if (i === j) {
ret.push(cur);
}
found = true;
}
}
}
return ret;
};
So using the first one, this is how your code could look:
function cleanUp() {
var text = document.getElementById("fld"),
textVal = text.value,
array;
textVal = textVal.replace(/\r/g, " ");
array = textVal.split(/\n/g);
text.value = removeDuplicatesInPlace(array).join("\n");
}
DEMO: http://jsfiddle.net/VrcN6/1/

You can use Array.reduce() to remove the duplicates. You need a helper object to keep track of how many times an item has been seen.
function cleanUp()
{
var textBox = document.getElementById("fld"),
array = textBox.value.split(/\r?\n/g),
o = {},
output;
output = array.reduce(function(prev, current) {
var key = '$' + current;
// have we seen this value before?
if (o[key] === void 0) {
prev.push(current);
o[key] = true;
}
return prev;
}, []);
// write back the result
textBox.value = output.join("\n");
}
The output of the reduce() step can be used directly to populate the text area again, without affecting the original sort order.
Demo

You can do this easily with just an object:
function removeDuplicates(text) {
var seen = {};
var result = '';
for (var i = 0; i < text.length; i++) {
var char = text.charAt(i);
if (char in seen) {
continue;
} else {
seen[char] = true;
result += char;
}
}
return result;
}
function cleanUp() {
var elem = document.getElementById("fld");
elem.value = removeDuplicates(elem.value);
}

arr3 = [1, 2, 3, 2, 4, 5];
unique = [];
function findUnique(val)
{
status = '0';
unique.forEach(function(itm){
if(itm==val){
status=1;
}
})
return status;
}
arr3.forEach(function(itm){
rtn = findUnique(itm);
if(rtn==0)
unique.push(itm);
});
console.log(unique); // [1, 2, 3, 4, 5]

Get the item that appears the most times in an array

var store = ['1','2','2','3','4'];
I want to find out that 2 appear the most in the array. How do I go about doing that?

I would do something like:
var store = ['1','2','2','3','4'];
var frequency = {}; // array of frequency.
var max = 0; // holds the max frequency.
var result; // holds the max frequency element.
for(var v in store) {
frequency[store[v]]=(frequency[store[v]] || 0)+1; // increment frequency.
if(frequency[store[v]] > max) { // is this frequency > max so far ?
max = frequency[store[v]]; // update max.
result = store[v]; // update result.
}
}

Solution with emphasis to Array.prototype.forEach and the problem of getting more than one key if the max count is shared among more items.
Edit: Proposal with one loop, only.
var store = ['1', '2', '2', '3', '4', '5', '5'],
distribution = {},
max = 0,
result = [];
store.forEach(function (a) {
distribution[a] = (distribution[a] || 0) + 1;
if (distribution[a] > max) {
max = distribution[a];
result = [a];
return;
}
if (distribution[a] === max) {
result.push(a);
}
});
console.log('max: ' + max);
console.log('key/s with max count: ' + JSON.stringify(result));
console.log(distribution);

arr.sort();
var max=0,result,freq = 0;
for(var i=0; i < arr.length; i++){
if(arr[i]===arr[i+1]){
freq++;
}
else {
freq=0;
}
if(freq>max){
result = arr[i];
max = freq;
}
}
return result;

Make a histogram, find the key for the maximum number in the histogram.
var hist = [];
for (var i = 0; i < store.length; i++) {
var n = store[i];
if (hist[n] === undefined) hist[n] = 0;
else hist[n]++;
}
var best_count = hist[store[0]];
var best = store[0];
for (var i = 0; i < store.length; i++) {
if (hist[store[i]] > best_count) {
best_count = hist[store[i]];
best = store[i];
}
}
alert(best + ' occurs the most at ' + best_count + ' occurrences');
This assumes either there are no ties, or you don't care which is selected.

Another ES6 option. Works with strings or numbers.
function mode(arr) {
const store = {}
arr.forEach((num) => store[num] ? store[num] += 1 : store[num] = 1)
return Object.keys(store).sort((a, b) => store[b] - store[a])[0]
}

If the array is sorted this should work:
function popular(array) {
if (array.length == 0) return [null, 0];
var n = max = 1, maxNum = array[0], pv, cv;
for(var i = 0; i < array.length; i++, pv = array[i-1], cv = array[i]) {
if (pv == cv) {
if (++n >= max) {
max = n; maxNum = cv;
}
} else n = 1;
}
return [maxNum, max];
};
popular([1,2,2,3,4,9,9,9,9,1,1])
[9, 4]
popular([1,2,2,3,4,9,9,9,9,1,1,10,10,10,10,10])
[10, 5]

This version will quit looking when the count exceeds the number of items not yet counted.
It works without sorting the array.
Array.prototype.most= function(){
var L= this.length, freq= [], unique= [],
tem, max= 1, index, count;
while(L>= max){
tem= this[--L];
if(unique.indexOf(tem)== -1){
unique.push(tem);
index= -1, count= 0;
while((index= this.indexOf(tem, index+1))!= -1){
++count;
}
if(count> max){
freq= [tem];
max= count;
}
else if(count== max) freq.push(tem);
}
}
return [freq, max];
}
//test
var A= ["apples","oranges","oranges","oranges","bananas",
"bananas","oranges","bananas"];
alert(A.most()) // [oranges,4]
A.push("bananas");
alert(A.most()) // [bananas,oranges,4]

I solved it this way for finding the most common integer
function mostCommon(arr) {
// finds the first most common integer, doesn't account for 2 equally common integers (a tie)
freq = [];
// set all frequency counts to 0
for(i = 0; i < arr[arr.length-1]; i++) {
freq[i] = 0;
}
// use index in freq to represent the number, and the value at the index represent the frequency count
for(i = 0; i < arr.length; i++) {
freq[arr[i]]++;
}
// find biggest number's index, that's the most frequent integer
mostCommon = freq[0];
for(i = 0; i < freq.length; i++) {
if(freq[i] > mostCommon) {
mostCommon = i;
}
}
return mostCommon;
}

This is my solution.
var max_frequent_elements = function(arr){
var a = [], b = [], prev;
arr.sort();
for ( var i = 0; i < arr.length; i++ ) {
if ( arr[i] !== prev ) {
a.push(arr[i]);
b.push(1);
} else {
b[b.length-1]++;
}
prev = arr[i];
}
var max = b[0]
for(var p=1;p<b.length;p++){
if(b[p]>max)max=b[p]
}
var indices = []
for(var q=0;q<a.length;q++){
if(b[q]==max){indices.push(a[q])}
}
return indices;
};

All the solutions above are iterative.
Here's a ES6 functional mutation-less version:
Array.prototype.mostRepresented = function() {
const indexedElements = this.reduce((result, element) => {
return result.map(el => {
return {
value: el.value,
count: el.count + (el.value === element ? 1 : 0),
};
}).concat(result.some(el => el.value === element) ? [] : {value: element, count: 1});
}, []);
return (indexedElements.slice(1).reduce(
(result, indexedElement) => (indexedElement.count > result.count ? indexedElement : result),
indexedElements[0]) || {}).value;
};
It could be optimized in specific situations where performance is the bottleneck, but it has a great advantage of working with any kind of array elements.
The last line could be replaced with:
return (indexedElements.maxBy(el => el.count) || {}).value;
With:
Array.prototype.maxBy = function(fn) {
return this.slice(1).reduce((result, element) => (fn(element) > fn(result) ? element : result), this[0]);
};
for clarity

If the array contains strings try this solution
function GetMaxFrequency (array) {
var store = array;
var frequency = []; // array of frequency.
var result; // holds the max frequency element.
for(var v in store) {
var target = store[v];
var numOccurences = $.grep(store, function (elem) {
return elem === target;
}).length;
frequency.push(numOccurences);
}
maxValue = Math.max.apply(this, frequency);
result = store[$.inArray(maxValue,frequency)];
return result;
}
var store = ['ff','cc','cc','ff','ff','ff','ff','ff','ff','yahya','yahya','cc','yahya'];
alert(GetMaxFrequency(store));

A fairly short solution.
function mostCommon(list) {
var keyCounts = {};
var topCount = 0;
var topKey = {};
list.forEach(function(item, val) {
keyCounts[item] = keyCounts[item] + 1 || 1;
if (keyCounts[item] > topCount) {
topKey = item;
topCount = keyCounts[item];
}
});
return topKey;
}
document.write(mostCommon(['AA', 'AA', 'AB', 'AC']))

This solution returns an array of the most appearing numbers in an array, in case multiple numbers appear at the "max" times.
function mode(numbers) {
var counterObj = {};
var max = 0;
var result = [];
for(let num in numbers) {
counterObj[numbers[num]] = (counterObj[numbers[num]] || 0) + 1;
if(counterObj[numbers[num]] >= max) {
max = counterObj[numbers[num]];
}
}
for (let num in counterObj) {
if(counterObj[num] == max) {
result.push(parseInt(num));
}
}
return result;
}