Develop Reference

Issue with random sampling

I'm trying to pick a random element from an array without picking the same element twice.
After a few iterations it is undefined.
It works with return(pick_random_letter()). But WHY?
Why does pick_random_letter() not go back and return later?
let letters = [
{ name: 'A' },
{ name: 'B' },
{ name: 'C' },
{ name: 'D' },
]
let used_letters = [];
function pick_random_letter(){
let random = Math.floor(Math.random() * letters.length);
let next_letter = letters[random].name;
if (used_letters.includes(next_letter)){
if (used_letters.length == letters.length) used_letters = [];
pick_random_letter(); // return(pick_random_letter()) works
}else{
used_letters.push(next_letter);
return next_letter;
}
}
for (let i = 0; i < 4; i++){
let random_letter = pick_random_letter();
console.log(random_letter);
console.log(used_letters);
}
Working example:
let letters = [
{ name: 'A' },
{ name: 'B' },
{ name: 'C' },
{ name: 'D' },
]
let used_letters = [];
function pick_random_letter(){
let random = Math.floor(Math.random() * letters.length);
let next_letter = letters[random].name;
if (used_letters.includes(next_letter)){
if (used_letters.length == letters.length) used_letters = [];
return(pick_random_letter());
}else{
used_letters.push(next_letter);
return next_letter;
}
}
for (let i = 0; i < 4; i++){
let random_letter = pick_random_letter();
console.log(random_letter);
console.log(used_letters);
}
Thank you

I think you would be better off using a shuffling approach. Create a shuffled list of your letters and pick from these. This avoids any need to check if we've returned a letter previously.
We'd start by creating a shuffledLetters array, then each time we call pick_random_letter() we return the next value.
The shuffle() function here is a basic Fisher–Yates / Knuth shuffle.
function shuffle(arr) {
for (let i = arr.length - 1; i > 0; i--) {
const j = Math.floor(Math.random() * (i + 1));
[arr[i], arr[j]] = [arr[j], arr[i]];
}
return arr;
}
let letters = [
{ name: 'A' },
{ name: 'B' },
{ name: 'C' },
{ name: 'D' },
]
// Create a shuffled copy of the letters array
let shuffledLetters = shuffle([...letters]);
function pick_random_letter() {
return shuffledLetters.pop();
}
for (let i = 0; i < letters.length; i++) {
let random_letter = pick_random_letter();
console.log(random_letter);
}
.as-console-wrapper { max-height: 100% !important; }

Javascript: object array mapping and matching with IE11

I'm looking for a javascript implementation (for IE11) for this problem; my inputs are two arrays like these:
var array1 = [{id: 1, param:"bon jour"}, {id: 2, param:"Hi"}, {id: 3, param:"Hello"}];
var array2 = [{item: "Peter", values:"1,2", singlevalue:"2"},
{item: "Mark", values:"1,2,3", singlevalue:"3"},
{item: "Lou", values:"2", singlevalue:"2"}];
and I should create a new array (array3) with array2 data plus 2 new fields ("params" and "singleparam"), using matching between array1[i].id and array2[x].values to evaluate "params" and between array1[i].id and array2[x].singlevalue to evaluate "singleparam", with this kind of result:
array3 = [{item: "Peter", values:"1,2", singlevalue:"2", params:"bon jour,Hi", singleparam:"Hi"},
{item: "Mark", values:"1,2,3", singlevalue:"3", params:"bon jour,Hi,Hello", singleparam:"Hello"},
{item: "Lou", values:"2", singlevalue:"2", params:"Hi", singleparam:"Hi"}];
I'm a javascript newbie and I've tried this kind of solution:
var array3 = array2.map(function(x, array1)
{
const newOb = {};
newOb.item = x.item;
newOb.values = x.values;
newOb.singlevalue = x.singlevalue;
newOb.params = function(x.values, array1)
{
var str = "";
var idArray = x.values.split(",");
for(i = 0; i < idArray.lenght; i++)
{
for(j = 0; i < array1.lenght; j++)
{
if(idArray[i] == array1[j].id)
{
str += array1[j].param + ",";
break;
}
}
}
return str;
};
newOb.singleparam = function(x.singlevalue, array1)
{
var val;
for(j = 0; i < array1.lenght; j++)
{
if(array1[j].id == x.singlevalue)
val = array1[j].param;
}
return val;
}
return newOb;
});
console.log(array3);
with this error: Error: Unexpected token '.'
I'd like to find an efficient solution considering that array1 has less than 10 elements, but array2 could contains more than 1000 objects.
Thanks in advance for your support

I will skip the functions stop and singlevalues and there were also some syntax errors,
for example the correct one is length and not lenght
var array1 = [{id: 1, param:"bon jour"}, {id: 2, param:"Hi"}, {id: 3, param:"Hello"}];
var array2 = [{item: "Peter", values:"1,2", singlevalue:"2"},
{item: "Mark", values:"1,2,3", singlevalue:"3"},
{item: "Lou", values:"2", singlevalue:"2"}];
function newArray3() {
return array2.map(x => {
const newOb = {};
newOb.item = x.item;
newOb.values = x.values;
newOb.singlevalue = x.singlevalue;
newOb.params = paramsFunction(x.values, array1);
newOb.singleparam = singleParamFunction(x.singlevalue, array1);
return newOb;
})
}
function singleParamFunction(x, array1) {
var val;
for(i = 0; i < array1.length; i++) {
if(array1[i].id.toString() == x) {
val = array1[i].param;
}
}
return val;
}
function paramsFunction(x, array1) {
var str = "";
var idArray = x.split(",");
for(i = 0; i < idArray.length; i++)
{
for(j = 0; j < array1.length; j++)
{
if(idArray[i] == array1[j].id.toString())
{
str += array1[j].param + ",";
break;
}
}
}
return str;
}
array3 = newArray3();
console.log(array3)

The solution provided by the #Walteann Costa can show the desired results in other browsers but it will not work for the IE browser as his code sample uses the => Arrow functions that is not supported in the IE browser.
As your question asks the solution for the IE browser, I tried to modify the code sample provided by the #Walteann Costa. Below modified code can work with the IE 11 browser.
<!doctype html>
<html>
<head>
<script>
"use strict";
var array1 = [{
id: 1,
param: "bon jour"
}, {
id: 2,
param: "Hi"
}, {
id: 3,
param: "Hello"
}];
var array2 = [{
item: "Peter",
values: "1,2",
singlevalue: "2"
}, {
item: "Mark",
values: "1,2,3",
singlevalue: "3"
}, {
item: "Lou",
values: "2",
singlevalue: "2"
}];
function newArray3() {
return array2.map(function (x) {
var newOb = {};
newOb.item = x.item;
newOb.values = x.values;
newOb.singlevalue = x.singlevalue;
newOb.params = paramsFunction(x.values, array1);
newOb.singleparam = singleParamFunction(x.singlevalue, array1);
return newOb;
});
}
function singleParamFunction(x, array1) {
var val,i,j;
for (i = 0; i < array1.length; i++) {
if (array1[i].id.toString() == x) {
val = array1[i].param;
}
}
return val;
}
function paramsFunction(x, array1) {
var str = "";
var idArray = x.split(",");
var i,j;
for (i = 0; i < idArray.length; i++) {
for (j = 0; j < array1.length; j++) {
if (idArray[i] == array1[j].id.toString()) {
str += array1[j].param + ",";
break;
}
}
}
return str;
}
var array3 = newArray3();
console.log(array3[0]);
console.log(array3[1]);
console.log(array3[2]);
</script>
</head>
<body>
</body>
</html>
Output in the IE 11:

from an array of objects how do I find which value comes up most often, in javascript? [duplicate]

I'm looking for an elegant way of determining which element has the highest occurrence (mode) in a JavaScript array.
For example, in
['pear', 'apple', 'orange', 'apple']
the 'apple' element is the most frequent one.

This is just the mode. Here's a quick, non-optimized solution. It should be O(n).
function mode(array)
{
if(array.length == 0)
return null;
var modeMap = {};
var maxEl = array[0], maxCount = 1;
for(var i = 0; i < array.length; i++)
{
var el = array[i];
if(modeMap[el] == null)
modeMap[el] = 1;
else
modeMap[el]++;
if(modeMap[el] > maxCount)
{
maxEl = el;
maxCount = modeMap[el];
}
}
return maxEl;
}

There have been some developments in javascript since 2009 - I thought I'd add another option. I'm less concerned with efficiency until it's actually a problem so my definition of "elegant" code (as stipulated by the OP) favours readability - which is of course subjective...
function mode(arr){
return arr.sort((a,b) =>
arr.filter(v => v===a).length
- arr.filter(v => v===b).length
).pop();
}
mode(['pear', 'apple', 'orange', 'apple']); // apple
In this particular example, should two or more elements of the set have equal occurrences then the one that appears latest in the array will be returned. It's also worth pointing out that it will modify your original array - which can be prevented if you wish with an Array.slice call beforehand.
Edit: updated the example with some ES6 fat arrows because 2015 happened and I think they look pretty... If you are concerned with backwards compatibility you can find this in the revision history.

As per George Jempty's request to have the algorithm account for ties, I propose a modified version of Matthew Flaschen's algorithm.
function modeString(array) {
if (array.length == 0) return null;
var modeMap = {},
maxEl = array[0],
maxCount = 1;
for (var i = 0; i < array.length; i++) {
var el = array[i];
if (modeMap[el] == null) modeMap[el] = 1;
else modeMap[el]++;
if (modeMap[el] > maxCount) {
maxEl = el;
maxCount = modeMap[el];
} else if (modeMap[el] == maxCount) {
maxEl += "&" + el;
maxCount = modeMap[el];
}
}
return maxEl;
}
This will now return a string with the mode element(s) delimited by a & symbol. When the result is received it can be split on that & element and you have your mode(s).
Another option would be to return an array of mode element(s) like so:
function modeArray(array) {
if (array.length == 0) return null;
var modeMap = {},
maxCount = 1,
modes = [];
for (var i = 0; i < array.length; i++) {
var el = array[i];
if (modeMap[el] == null) modeMap[el] = 1;
else modeMap[el]++;
if (modeMap[el] > maxCount) {
modes = [el];
maxCount = modeMap[el];
} else if (modeMap[el] == maxCount) {
modes.push(el);
maxCount = modeMap[el];
}
}
return modes;
}
In the above example you would then be able to handle the result of the function as an array of modes.

Based on Emissary's ES6+ answer, you could use Array.prototype.reduce to do your comparison (as opposed to sorting, popping and potentially mutating your array), which I think looks quite slick.
const mode = (myArray) =>
myArray.reduce(
(a,b,i,arr)=>
(arr.filter(v=>v===a).length>=arr.filter(v=>v===b).length?a:b),
null)
I'm defaulting to null, which won't always give you a truthful response if null is a possible option you're filtering for, maybe that could be an optional second argument
The downside, as with various other solutions, is that it doesn't handle 'draw states', but this could still be achieved with a slightly more involved reduce function.

a=['pear', 'apple', 'orange', 'apple'];
b={};
max='', maxi=0;
for(let k of a) {
if(b[k]) b[k]++; else b[k]=1;
if(maxi < b[k]) { max=k; maxi=b[k] }
}

As I'm using this function as a quiz for the interviewers, I post my solution:
const highest = arr => (arr || []).reduce( ( acc, el ) => {
acc.k[el] = acc.k[el] ? acc.k[el] + 1 : 1
acc.max = acc.max ? acc.max < acc.k[el] ? el : acc.max : el
return acc
}, { k:{} }).max
const test = [0,1,2,3,4,2,3,1,0,3,2,2,2,3,3,2]
console.log(highest(test))

Trying out a declarative approach here. This solution builds an object to tally up the occurrences of each word. Then filters the object down to an array by comparing the total occurrences of each word to the highest value found in the object.
const arr = ['hello', 'world', 'hello', 'again'];
const tally = (acc, x) => {
if (! acc[x]) {
acc[x] = 1;
return acc;
}
acc[x] += 1;
return acc;
};
const totals = arr.reduce(tally, {});
const keys = Object.keys(totals);
const values = keys.map(x => totals[x]);
const results = keys.filter(x => totals[x] === Math.max(...values));

This solution has O(n) complexity:
function findhighestOccurenceAndNum(a) {
let obj = {};
let maxNum, maxVal;
for (let v of a) {
obj[v] = ++obj[v] || 1;
if (maxVal === undefined || obj[v] > maxVal) {
maxNum = v;
maxVal = obj[v];
}
}
console.log(maxNum + ' has max value = ' + maxVal);
}
findhighestOccurenceAndNum(['pear', 'apple', 'orange', 'apple']);

For the sake of really easy to read, maintainable code I share this:
function getMaxOcurrences(arr = []) {
let item = arr[0];
let ocurrencesMap = {};
for (let i in arr) {
const current = arr[i];
if (ocurrencesMap[current]) ocurrencesMap[current]++;
else ocurrencesMap[current] = 1;
if (ocurrencesMap[item] < ocurrencesMap[current]) item = current;
}
return {
item: item,
ocurrences: ocurrencesMap[item]
};
}
Hope it helps someone ;)!

Here’s the modern version using built-in maps (so it works on more than things that can be converted to unique strings):
'use strict';
const histogram = iterable => {
const result = new Map();
for (const x of iterable) {
result.set(x, (result.get(x) || 0) + 1);
}
return result;
};
const mostCommon = iterable => {
let maxCount = 0;
let maxKey;
for (const [key, count] of histogram(iterable)) {
if (count > maxCount) {
maxCount = count;
maxKey = key;
}
}
return maxKey;
};
console.log(mostCommon(['pear', 'apple', 'orange', 'apple']));

Time for another solution:
function getMaxOccurrence(arr) {
var o = {}, maxCount = 0, maxValue, m;
for (var i=0, iLen=arr.length; i<iLen; i++) {
m = arr[i];
if (!o.hasOwnProperty(m)) {
o[m] = 0;
}
++o[m];
if (o[m] > maxCount) {
maxCount = o[m];
maxValue = m;
}
}
return maxValue;
}
If brevity matters (it doesn't), then:
function getMaxOccurrence(a) {
var o = {}, mC = 0, mV, m;
for (var i=0, iL=a.length; i<iL; i++) {
m = a[i];
o.hasOwnProperty(m)? ++o[m] : o[m] = 1;
if (o[m] > mC) mC = o[m], mV = m;
}
return mV;
}
If non–existent members are to be avoided (e.g. sparse array), an additional hasOwnProperty test is required:
function getMaxOccurrence(a) {
var o = {}, mC = 0, mV, m;
for (var i=0, iL=a.length; i<iL; i++) {
if (a.hasOwnProperty(i)) {
m = a[i];
o.hasOwnProperty(m)? ++o[m] : o[m] = 1;
if (o[m] > mC) mC = o[m], mV = m;
}
}
return mV;
}
getMaxOccurrence([,,,,,1,1]); // 1
Other answers here will return undefined.

Here is another ES6 way of doing it with O(n) complexity
const result = Object.entries(
['pear', 'apple', 'orange', 'apple'].reduce((previous, current) => {
if (previous[current] === undefined) previous[current] = 1;
else previous[current]++;
return previous;
}, {})).reduce((previous, current) => (current[1] >= previous[1] ? current : previous))[0];
console.log("Max value : " + result);

function mode(arr){
return arr.reduce(function(counts,key){
var curCount = (counts[key+''] || 0) + 1;
counts[key+''] = curCount;
if (curCount > counts.max) { counts.max = curCount; counts.mode = key; }
return counts;
}, {max:0, mode: null}).mode
}

Another JS solution from: https://www.w3resource.com/javascript-exercises/javascript-array-exercise-8.php
Can try this too:
let arr =['pear', 'apple', 'orange', 'apple'];
function findMostFrequent(arr) {
let mf = 1;
let m = 0;
let item;
for (let i = 0; i < arr.length; i++) {
for (let j = i; j < arr.length; j++) {
if (arr[i] == arr[j]) {
m++;
if (m > mf) {
mf = m;
item = arr[i];
}
}
}
m = 0;
}
return item;
}
findMostFrequent(arr); // apple

This solution can return multiple elements of an array in case of a tie. For example, an array
arr = [ 3, 4, 3, 6, 4, ];
has two mode values: 3 and 6.
Here is the solution.
function find_mode(arr) {
var max = 0;
var maxarr = [];
var counter = [];
var maxarr = [];
arr.forEach(function(){
counter.push(0);
});
for(var i = 0;i<arr.length;i++){
for(var j=0;j<arr.length;j++){
if(arr[i]==arr[j])counter[i]++;
}
}
max=this.arrayMax(counter);
for(var i = 0;i<arr.length;i++){
if(counter[i]==max)maxarr.push(arr[i]);
}
var unique = maxarr.filter( this.onlyUnique );
return unique;
};
function arrayMax(arr) {
var len = arr.length, max = -Infinity;
while (len--) {
if (arr[len] > max) {
max = arr[len];
}
}
return max;
};
function onlyUnique(value, index, self) {
return self.indexOf(value) === index;
}

const frequence = (array) =>
array.reduce(
(acc, item) =>
array.filter((v) => v === acc).length >=
array.filter((v) => v === item).length
? acc
: item,
null
);
frequence([1, 1, 2])

var array = [1, 3, 6, 6, 6, 6, 7, 7, 12, 12, 17],
c = {}, // counters
s = []; // sortable array
for (var i=0; i<array.length; i++) {
c[array[i]] = c[array[i]] || 0; // initialize
c[array[i]]++;
} // count occurrences
for (var key in c) {
s.push([key, c[key]])
} // build sortable array from counters
s.sort(function(a, b) {return b[1]-a[1];});
var firstMode = s[0][0];
console.log(firstMode);

Here is my solution to this problem but with numbers and using the new 'Set' feature. Its not very performant but i definitely had a lot of fun writing this and it does support multiple maximum values.
const mode = (arr) => [...new Set(arr)]
.map((value) => [value, arr.filter((v) => v === value).length])
.sort((a,b) => a[1]-b[1])
.reverse()
.filter((value, i, a) => a.indexOf(value) === i)
.filter((v, i, a) => v[1] === a[0][1])
.map((v) => v[0])
mode([1,2,3,3]) // [3]
mode([1,1,1,1,2,2,2,2,3,3,3]) // [1,2]
By the way do not use this for production this is just an illustration of how you can solve it with ES6 and Array functions only.

const mode = (str) => {
return str
.split(' ')
.reduce((data, key) => {
let counter = data.map[key] + 1 || 1
data.map[key] = counter
if (counter > data.counter) {
data.counter = counter
data.mode = key
}
return data
}, {
counter: 0,
mode: null,
map: {}
})
.mode
}
console.log(mode('the t-rex is the greatest of them all'))

Here is my solution :-
function frequent(number){
var count = 0;
var sortedNumber = number.sort();
var start = number[0], item;
for(var i = 0 ; i < sortedNumber.length; i++){
if(start === sortedNumber[i] || sortedNumber[i] === sortedNumber[i+1]){
item = sortedNumber[i]
}
}
return item
}
console.log( frequent(['pear', 'apple', 'orange', 'apple']))

Try it too, this does not take in account browser version.
function mode(arr){
var a = [],b = 0,occurrence;
for(var i = 0; i < arr.length;i++){
if(a[arr[i]] != undefined){
a[arr[i]]++;
}else{
a[arr[i]] = 1;
}
}
for(var key in a){
if(a[key] > b){
b = a[key];
occurrence = key;
}
}
return occurrence;
}
alert(mode(['segunda','terça','terca','segunda','terça','segunda']));
Please note that this function returns latest occurence in the array
when 2 or more entries appear same number of times!

With ES6, you can chain the method like this:
function findMostFrequent(arr) {
return arr
.reduce((acc, cur, ind, arr) => {
if (arr.indexOf(cur) === ind) {
return [...acc, [cur, 1]];
} else {
acc[acc.indexOf(acc.find(e => e[0] === cur))] = [
cur,
acc[acc.indexOf(acc.find(e => e[0] === cur))][1] + 1
];
return acc;
}
}, [])
.sort((a, b) => b[1] - a[1])
.filter((cur, ind, arr) => cur[1] === arr[0][1])
.map(cur => cur[0]);
}
console.log(findMostFrequent(['pear', 'apple', 'orange', 'apple']));
console.log(findMostFrequent(['pear', 'apple', 'orange', 'apple', 'pear']));
If two elements have the same occurrence, it will return both of them. And it works with any type of element.

// O(n)
var arr = [1, 2, 3, 2, 3, 3, 5, 6];
var duplicates = {};
max = '';
maxi = 0;
arr.forEach((el) => {
duplicates[el] = duplicates[el] + 1 || 1;
if (maxi < duplicates[el]) {
max = el;
maxi = duplicates[el];
}
});
console.log(max);

I came up with a shorter solution, but it's using lodash. Works with any data, not just strings. For objects can be used:
const mostFrequent = _.maxBy(Object.values(_.groupBy(inputArr, el => el.someUniqueProp)), arr => arr.length)[0];
This is for strings:
const mostFrequent = _.maxBy(Object.values(_.groupBy(inputArr, el => el)), arr => arr.length)[0];
Just grouping data under a certain criteria, then finding the largest group.

Here is my way to do it so just using .filter.
var arr = ['pear', 'apple', 'orange', 'apple'];
function dup(arrr) {
let max = { item: 0, count: 0 };
for (let i = 0; i < arrr.length; i++) {
let arrOccurences = arrr.filter(item => { return item === arrr[i] }).length;
if (arrOccurences > max.count) {
max = { item: arrr[i], count: arrr.filter(item => { return item === arrr[i] }).length };
}
}
return max.item;
}
console.log(dup(arr));

Easy solution !
function mostFrequentElement(arr) {
let res = [];
for (let x of arr) {
let count = 0;
for (let i of arr) {
if (i == x) {
count++;
}
}
res.push(count);
}
return arr[res.indexOf(Math.max(...res))];
}
array = [13 , 2 , 1 , 2 , 10 , 2 , 1 , 1 , 2 , 2];
let frequentElement = mostFrequentElement(array);
console.log(`The frequent element in ${array} is ${frequentElement}`);
Loop on all element and collect the Count of each element in the array that is the idea of the solution

Here is my solution :-
const arr = [
2, 1, 10, 7, 10, 3, 10, 8, 7, 3, 10, 5, 4, 6, 7, 9, 2, 2, 2, 6, 3, 7, 6, 9, 8,
9, 10, 8, 8, 8, 4, 1, 9, 3, 4, 5, 8, 1, 9, 3, 2, 8, 1, 9, 6, 3, 9, 2, 3, 5, 3,
2, 7, 2, 5, 4, 5, 5, 8, 4, 6, 3, 9, 2, 3, 3, 10, 3, 3, 1, 4, 5, 4, 1, 5, 9, 6,
2, 3, 10, 9, 4, 3, 4, 5, 7, 2, 7, 2, 9, 8, 1, 8, 3, 3, 3, 3, 1, 1, 3,
];
function max(arr) {
let newObj = {};
arr.forEach((d, i) => {
if (newObj[d] != undefined) {
++newObj[d];
} else {
newObj[d] = 0;
}
});
let nwres = {};
for (let maxItem in newObj) {
if (newObj[maxItem] == Math.max(...Object.values(newObj))) {
nwres[maxItem] = newObj[maxItem];
}
}
return nwres;
}
console.log(max(arr));

I guess you have two approaches. Both of which have advantages.
Sort then Count or Loop through and use a hash table to do the counting for you.
The hashtable is nice because once you are done processing you also have all the distinct elements. If you had millions of items though, the hash table could end up using a lot of memory if the duplication rate is low. The sort, then count approach would have a much more controllable memory footprint.

var mode = 0;
var c = 0;
var num = new Array();
var value = 0;
var greatest = 0;
var ct = 0;
Note: ct is the length of the array.
function getMode()
{
for (var i = 0; i < ct; i++)
{
value = num[i];
if (i != ct)
{
while (value == num[i + 1])
{
c = c + 1;
i = i + 1;
}
}
if (c > greatest)
{
greatest = c;
mode = value;
}
c = 0;
}
}

You can try this:
// using splice()
// get the element with the highest occurence in an array
function mc(a) {
var us = [], l;
// find all the unique elements in the array
a.forEach(function (v) {
if (us.indexOf(v) === -1) {
us.push(v);
}
});
l = us.length;
while (true) {
for (var i = 0; i < l; i ++) {
if (a.indexOf(us[i]) === -1) {
continue;
} else if (a.indexOf(us[i]) != -1 && a.length > 1) {
// just delete it once at a time
a.splice(a.indexOf(us[i]), 1);
} else {
// default to last one
return a[0];
}
}
}
}
// using string.match method
function su(a) {
var s = a.join(),
uelms = [],
r = {},
l,
i,
m;
a.forEach(function (v) {
if (uelms.indexOf(v) === -1) {
uelms.push(v);
}
});
l = uelms.length;
// use match to calculate occurance times
for (i = 0; i < l; i ++) {
r[uelms[i]] = s.match(new RegExp(uelms[i], 'g')).length;
}
m = uelms[0];
for (var p in r) {
if (r[p] > r[m]) {
m = p;
} else {
continue;
}
}
return m;
}

Combination of all possible values from first array?

How do a combine the array of arrays based on the first array1 or basically group by array1.
Below is the four Array, where i have to form objects based on A and then based on B.
var array1=["A","B"];
var array2=["1","2","3", "4"];
var array3=["N","O","P", "Q"];
var array4=["R"];
Below is how i need :
[ {
'take': 'A',
'take2': '1',
'take3': 'N',
'take4': 'R'
}, {
'take': 'A',
'take2': '2',
'take3': 'N',
'take4': 'R'
}, {
'take': 'A',
'take2': '3',
'take3': 'N',
'take4': 'R'
}, {
'take': 'A',
'take2': '4',
'take3': 'N',
'take4': 'R'
}, {
'take': 'A',
'take2': '1',
'take3': 'O',
'take4': 'R'
}]
This is something i have tried, but not sure how can i loop n number of n arrays
var result = array1.reduce( (a, v) =>
[...a, ...array2.map(x=>v+x)],
[]);

here is a keep it simple solution (if you know how many arrays do you have) :
const possibilities = [];
const ar1length = array1.length;
const ar2length = array2.length;
const ar3length = array3.length;
const ar4length = array4.length;
// Not cleanest solution available but it does the job
for ( let i = 0; i < ar1length; i++) {
for (let j = 0; j < ar2length; j++) {
for (let k = 0; k < ar3length; k++) {
for (let l = 0; l < ar4length; l++) {
possibilities.push({
"take": array1[i],
"take1": array2[j],
"take2": array3[k],
"take3": array4[l]
});
}
}
}
}
Oh and if you want an unknown number of arrays, you may add all of these arrays to an array in order to iterate over it I guess

I've written a function for this task a while ago, takes an arbitrary amount of Arrays and non-arrays and computes all possible combinations
var array1 = ["A", "B"];
var array2 = ["1", "2", "3", "4"];
var array3 = ["N", "O", "P", "Q"];
var array4 = ["R"];
console.log(combinations(array1, array2, array3, array4).join("\n"));
function combinations(...columns) {
const state = [], combinations = [state];
let head = null;
for (let column = 0; column < columns.length; ++column) {
let value = columns[column];
if (Array.isArray(value)) {
if (value.length > 1) {
head = {
next: head,
column,
row: 0
};
}
value = value[0];
}
state[column] = value;
}
let todo = head;
while(todo) {
if (++todo.row === columns[todo.column].length) {
todo.row = 0;
state[todo.column] = columns[todo.column][todo.row];
todo = todo.next;
} else {
state[todo.column] = columns[todo.column][todo.row];
combinations.push(state.slice());
todo = head;
}
}
return combinations;
}
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Here's a recursive approach which works for every number of arrays, you just have to call combine(array1, array2, ..., arrayn):
var array1=["A","B"];
var array2=["1","2","3", "4"];
var array3=["N","O","P", "Q"];
var array4=["R"];
function combine(arr1, ...arr2) {
if(arr2.length === 0) return Array.from(arr1, (x) => x.reduce((obj, y, i) => (obj[`take${i}`] = y, obj), {}));
return combine(arr1.flatMap(d => arr2[0].map(v => {
return [...Object.values(d), ...Object.values(v)]
})), ...arr2.slice(1));
}
console.log(combine(array1, array2, array3, array4));

Insert two consecutive elements in array at specific Index in Javascript

I have two arrays like this,
var firstArray = ['one','two','three','four','five','six','seven'];
var secondArray =['1','2','3','4','5','6','7','8'];
I have to insert second array elements into first array like this,
var combinedArray =['one','two','three','1','2','four','five','six','3','4','seven','5','6','7','8']
I know that I could splice and insert at specific index for one element. However I am confused how exactly to achieve this pattern.
Could any one help me out with this?

You could use a pattern for the chunks and slice the wanted length for a new array.
var firstArray = ['one', 'two', 'three', 'four', 'five', 'six', 'seven'],
secondArray = ['1', '2', '3', '4', '5', '6', '7', '8'],
data = [firstArray, secondArray],
pattern = [3, 2],
result = [],
i = 0,
l = data.reduce(function (r, a) { return Math.max(r, a.length); }, 0);
while (i < l) {
pattern.forEach(function (a, j) {
result = result.concat(data[j].slice(i * a, (i + 1) * a));
});
i++;
}
console.log(result);
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You can actually splice more than one item:
firstArray.splice(3, 0, "1", "2");

You can create two variables i and j and increment first one by 3 that you will use to slice first array and increment the second one by 2 and you will use that one to slice second array. If the i + 3 > a.length you will concat rest of elements in b array to result.
var a = ['one','two','three','four','five','six','seven'];
var b = ['1','2','3','4','5','6','7','8'];
var r = [], i = 0, j = 0
while(i < a.length) {
r.push(...a.slice(i, i + 3), ...b.slice(j, i + 3 < a.length ? j + 2 : b.length))
i += 3, j += 2
}
console.log(r)

a more granular approach:
var firstArray = ['one','two','three','four','five','six','seven'];
var secondArray =['1','2','3','4','5','6','7','8'];
var combinedArray = flatten(zip(
toGroupsOf(3, firstArray),
toGroupsOf(2, secondArray)
));
console.log(combinedArray);
//a the utilities for that
function isUint(value){
return value === (value >>> 0)
}
function toGroupsOf(length, arrayOrString){
if( !isUint(length) || !length )
throw new Error("invalid length " + JSON.stringify(length));
return Array.from(
{ length: Math.ceil(arrayOrString.length / length) },
(v,i) => arrayOrString.slice(i*length, (i+1)*length)
);
}
function zip(...arraysOrStrings){
var numColumns = arraysOrStrings.length,
lengths = arraysOrStrings.map(item => (item && +item.length) || 0),
x=0, y=0;
return Array.from(
{ length: lengths.reduce((a,b)=>a+b, 0) },
function(v,i){
for(var safety = numColumns+1; safety--;){
if(y < lengths[x])
return arrays[x++][y];
else if(++x >= numColumns)
x=0, ++y;
}
throw new Error("something went wrong, this line should have never been reached");
}
)
}
function flatten(array){
return [].concat.apply([], array);
}