Develop Reference

Find combination of nested JSON keys

I am having some trouble solving this issue without using nested for-loops. I would like to do it using recursion.
Given the following object:
{
"Color": {
"Black": [],
"White": []
},
"Effect": {
"30W": [],
"40W": [],
"60W": []
}
}
The code should compute each combination of Color and Effect and add a number in the list such that the following is produced:
{
"Color": {
"Black": [
1,
2,
3
],
"White": [
4,
5,
6
]
},
"Effect": {
"30W": [
1,
4
],
"40W": [
2,
5
],
"60W": [
3,
6
]
}
}
My attempt is as follows:
const func = (object, entries) => {
for (let prop in object) {
let counter = 0;
const objLength = Object.keys(object[prop]).length;
for (let key in object[prop]) {
console.log(key + counter)
for (let i = 0; i < entries.length / objLength; i++) {
object[prop][key].push(entries[counter]);
counter++;
}
}
}
return object;
}
However, this does not return the desired output. I think it is because of the inner-most for loop condition.

The best way to handle this is to create your JavaScript object and convert it to a string.
// creating your object with attributes. Objects in objects or whatever you
// need
var obj = new Object();
obj.name = "Dale";
obj.age = 30;
obj.married = true;
dataToAdd.forEach(function(item, index) {
item.married = false;
})
// Then convert it to a string using the following code
var jsonString= JSON.stringify(obj);
console.log(jsonString);

I solved the question by considering the space of each attribute key. Then it is just a matter of finding the cartesian, and adding values accordingly:
const cartesian =(...a) => a.reduce((a, b) => a.flatMap(d => b.map(e => [d, e].flat())));
function diy(jsonObj, counter) {
let permObj = []
let keys = Object.keys(jsonObj)
keys.forEach(key => {
permObj.push(Object.keys(jsonObj[key]))
});
permObj = cartesian(...permObj)
for(let i = 0; i < keys.length; i++) {
for(let j = 0; j < permObj.length; j++) {
jsonObj[keys[i]][permObj[j][i]].push(j + counter);
}
}
return jsonObj;
}

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;
}

javascript leave max 3 occurrences of the same obj in an array

If I populate an array of pairs (inside a get json) like so:
var arrItems = [];
for (let y=0; y<50 ;y++){
var titl = data.item[y].name;
var img = data.item[y].image[0].url;
arrItems.push({t:titl,i:img});
}
How can I then filter it to leave only 3 pairs where value is the same?
Example:
arrItems = [
{t:one,i:square.jpg},
{t:two,i:square.jpg},
{t:three,i:square.jpg},
{t:four,i:square.jpg},
{t:five,i:triangle.jpg}
];
Becomes
arrItems = [
{t:one,i:square.jpg},
{t:two,i:square.jpg},
{t:three,i:square.jpg},
{t:five,i:triangle.jpg}
];
Both JavaScript or jQuery are OK.

You could take a hash table and count the occurences of the wanted property and filter with a max value.
var items = [{ t: 'one', i: 'square.jpg' }, { t: 'two', i: 'square.jpg' }, { t: 'three', i: 'square.jpg' }, { t: 'four', i: 'square.jpg' }, { t: 'five', i: 'triangle.jpg' }],
count = {},
result = items.filter(({ i }) => {
count[i] = (count[i] || 0) + 1;
return count[i] <= 3;
});
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

There might be some more efficient ways to write it, but I would think the easiest to understand is the straightforward iteration with counting:
var counts = {};
for (var i = 0; i < arrItems.length; i++) {
var item = arrItems[i];
var count = counts[item.i] || 0;
if (count >= 3) {
arrItems.splice(i, 1);
i--;
} else {
counts[item.i] = ++count;
}
}

You can use reduce function
var arrItems = [
{t:'one',i:'square.jpg'},
{t:'two',i:'square.jpg'},
{t:'three',i:'square.jpg'},
{t:'four',i:'square.jpg'},
{t:'five',i:'triangle.jpg'}
];
var output = arrItems.reduce((acc, {t,i})=>{
acc['mem'][i] = (acc['mem'][i] || 0) + 1;
acc['mem'][i] <= 3 ? acc['output'].push({t, i}) : '';
return acc;
}, {'mem':{}, 'output':[]});
console.log(output);

Need an algorithm to manipulate array structure in javascript

In javascript, here is my start array:
[{
name: 'aaa',
value: 1
},
{
name: 'bbb',
value: 0
},
{
name: 'bbb',
value: 1
}]
I want to transform it into this array as result:
[{
name: 'aaa',
value: 1
},
{
name: 'bbb',
value: [0, 1]
}]
I need a good and simple algorithm to do this

How about:
var array = [{
name: 'aaa',
value: 1
},
{
name: 'bbb',
value: 0
},
{
name: 'bbb',
value: 1
}];
var map = {};
for(var i = 0; i < array.length; i++) {
var name = array[i].name;
if (map[name] === undefined) {
map[name] = [];
}
map[name].push(array[i].value);
}
var result = [];
for(var key in map) {
var value = map[key];
result.push({
name: key,
value: value.length === 1 ? value[0] : value
});
}
Easiest way is to create a map to keep track of which names are used. Then convert this map back to an array of objects.
If you want to use Arrays for value then change it to:
result.push({
name: key,
value: value
});

here's pseudocode for simplest implementation
hash = {}
for(pair in array) {
hash[pair.name] ||= []
hash[pair.name] << pair.value
}
result = []
for(k, v in hash) {
result << {name: k, value: v}
}

This function does the trick
function consolidate(var arrayOfObjects)
{
// create a dictionary of values first
var dict = {};
for(var i = 0; i < arrayOfObjects.length; i++)
{
var n = arrayOfObjects[i].name;
if (!dict[n])
{
dict[n] = [];
}
dict[n].push(arrayOfObjects[i].value);
}
// convert dictionary to array again
var result = [];
for(var key in dict)
{
result.push({
name: key,
value: dict[key].length == 1 ? dict[key][0] : dict[key]
});
}
return result;
}

An alternative solution:
function convert(arr) {
var res = [];
var map = {};
for (var i=0;i<arr.length;i++) {
var arrObj = arr[i];
var oldObj = map[arrObj.name];
if (oldObj == undefined) {
oldObj = {name:arrObj.name, value:arrObj.value};
map[arrObj.name] = oldObj;
res.push(oldObj);
} else {
if( typeof oldObj.value === 'number' ) {
oldObj.value = [oldObj.value];
}
oldObj.value.push(arrObj.value);
}
}
return res;
}
In theory it should work a bit faster and use less memory. Basically it creates a result array and a map which is an index for the same array (no duplicate objects). So it fills the result in one iteration instead of two and does not need to convert map to array (which saves several CPU cycles :P ).
Added:
Here is a variation of that function in case value: [1] is acceptable:
function convert(arr) {
var res = [];
var map = {};
for (var i=0;i<arr.length;i++) {
var arrObj = arr[i];
var oldObj = map[arrObj.name];
if (oldObj == undefined) {
oldObj = {name:arrObj.name, value:[arrObj.value]};
map[arrObj.name] = oldObj;
res.push(oldObj);
} else {
oldObj.value.push(arrObj.value);
}
}
return res;
}