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How to remove all next elements from array in javascript or lodash

I have an array of object [{a:1},{a:2},{a:3},{a:4},{a:5}].
I want to find index of {a:3} and remove all next elements from index of {a:3}.
Result should look like [{a:1},{a:2},{a:3}] and removed elements will be {a:4},{a:5}.

You can use Array#findIndex() to find that object by its number, or Array#indexOf() if you have a reference to it.
Then, you can either set the array's length (mutating) or use Array#slice() (not mutating), to reduce its length:
const array = [{a:1},{a:2},{a:3},{a:4},{a:5}]
const i = array.findIndex(({a}) => a === 3)
const arrayCopy = array.slice(0, i + 1)
//or:
array.length = i + 1
console.log(i)
console.log(arrayCopy)
console.log(array)
Or with Array#indexOf():
//If you have a reference from somewhere:
const ref = {a:3}
const array = [{a:1},{a:2},ref,{a:4},{a:5}]
const i = array.indexOf(ref)
const arrayCopy = array.slice(0, i + 1)
//or:
array.length = i + 1
console.log(i)
console.log(arrayCopy)
console.log(array)
Alternatively, if you don't have a reference, but it's guaranteed that the array is sorted, you can use a binary search, that makes this much faster for large arrays:
function binarySearch(array, target, cb) {
let start = 0, end = array.length
while(start < end){
const middle = ((end - start - 1) >>> 1) + start
const middleVal = cb(array[middle], middle)
if(middleVal === target) return middle
if(middleVal < target)
start = middle + 1
else
end = middle
}
return -1
};
const array = [{a:1},{a:2},{a:3},{a:4},{a:5}]
const i = binarySearch(array, 3, ({a}) => a)
const arrayCopy = array.slice(0, i + 1)
//or:
array.length = i + 1
console.log(i)
console.log(arrayCopy)
console.log(array)

One line vanilla JS with slice and findIndex
const data = [{ a: 1 }, { a: 2 }, { a: 3 }, { a: 4 }, { a: 5 }];
const res = data.slice(
0,
data.findIndex((el) => JSON.stringify(el) === JSON.stringify({ a: 3 })) + 1
);
console.log(res);

How to split an array in many others arrays with two specific sizes?

I'm trying to find out if there is a way in splitting an array into many others arrays, but these arrays should have a length of 4 and 8. Like:
const someArray = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,23,24,25];
output newArray = [[1,2,3,4],
[5,6,7,8,9,10,11,12],
[13,14,15,16],
[17,18,19,20,21,22,23,24],
[25]];
I've seen many solutions to chunk into specific single sizes like:
export const chunk = (array, size) => {
const chunkedArr = [];
let copiedArr = [...array];
const numOfChild = Math.ceil(copiedArr.length / size);
for (let i = 0; i < numOfChild; i++) {
chunkedArr.push(copiedArr.splice(0, size));
}
return chunkedArr;
};
which I've tried to "adapt" for my requirements, but had no success.
Any help?
Cheers!

A simple implementation consuming the array with a recursive function:
const someArray = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,23,24,25];
const iter = (sliceLen, arr, acc = []) =>
arr.length <= sliceLen
? [...acc, arr.slice(0, sliceLen)]
: iter(sliceLen === 4 ? 8 : 4, arr.slice(sliceLen), [...acc, arr.slice(0, sliceLen)])
const r = iter(4, someArray)
console.log(r)

You can use Array.slice() along with appropriate indices to achieve this.
const someArray = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,23,24,25];
let i = 0;
let step = 4;
const newArray = [];
while(i < someArray.length) {
newArray.push(someArray.slice(i, i + step));
i += step;
step = step == 4 ? 8 : 4;
}
console.log(newArray);

One approach to this would be to just swap back and forth between the chunk size for the iteration.
const someArray = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,23,24,25];
console.log(
chunk(someArray, 4, 8)
);
function chunk(input, size1, size2){
const output = [];
let chunkSize = size1;
for (let i = 0; i < input.length;) {
output.push(input.slice(i).slice(0, chunkSize));
i += chunkSize;
chunkSize = (chunkSize === size1 ? size2 : size1);
}
return output;
}

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;
}

Creating various arrays from a string using LOOPS

I have a string of values
"000111111122222223333333444455556666"
How could I use a loop to produce one array for index values from 0 to 3 (create an array of [000] and then another array of index values from 3 to 10, 10 to 17, 17 to 24, producing eg. [1111111, 2222222, 333333] and then another loop to produce an array of index values from 24 to 28, 28 to 32, 32 to 36, producing eg. [4444, 5555, 6666])?
So in total 3 different arrays have been created using three different for loops.
array1 = [000]
array2 = [1111111, 2222222, 333333]
array3 = [4444, 5555, 6666]

You may wish to try something line this (only a schematic solution!):
var l_Input = "000111111122222223333333444455556666" ;
var l_Array_1 = [] ;
var l_Array_2 = [] ;
var l_Array_3 = [] ;
var l_One_Char ;
for (var i = 0 ; i < l_Input.length ; i++) {
l_One_Char = l_Input.substring(i,i) ;
if (i < 3) {
l_Array_1.push(l_One_Char) ;
continue ;
}
if (i >= 3 && i < 10) {
l_Array_2.push(l_One_Char) ;
continue ;
}
:
:
}

I think this would work.
const str = '000111111122222223333333444455556666';
function makeArr(str, item) {
let firstIndex = str.indexOf(item);
let lastIndex = str.lastIndexOf(item) + 1;
return [ str.substring(firstIndex, lastIndex) ];
}
const first = makeArr(str, 0);
const second = [].concat(makeArr(str, 1))
.concat(makeArr(str, 2))
.concat(makeArr(str, 3));
const third = [].concat(makeArr(str, 4))
.concat(makeArr(str, 3))
.concat(makeArr(str, 3));

You could map the sub strings.
var str = '000111111122222223333333444455556666',
parts = [[3], [7, 7, 7], [4, 4, 4]],
result = parts.map((i => a => a.map(l => str.slice(i, i += l)))(0));
console.log(result);

function split(string, start, end) {
var result = [],
substring = string[start],
split;
for (var i = start + 1; i < end; i++) {
var char = string[i];
if (char === substring[0])
substring += char;
else {
result.push(substring);
substring = char;
}
}
result.push(substring);
return result;
}
split("00011122",0,8)
["000", "111", "22"]

To do this dynamically, you can use .split() and .map() methods to make an array from your string then group this array items by value.
This is how should be our code:
const str = "000111111122222223333333444455556666";
var groupArrayByValues = function(arr) {
return arr.reduce(function(a, x) {
(a[x] = a[x] || []).push(x);
return a;
}, []);
};
var arr = str.split("").map(v => +v);
var result = groupArrayByValues(arr);
This will give you an array of separate arrays with similar values each.
Demo:
const str = "000111111122222223333333444455556666";
var groupArrayByValues = function(arr) {
return arr.reduce(function(a, x) {
(a[x] = a[x] || []).push(x);
return a;
}, []);
};
var arr = str.split("").map(v => +v);
var result = groupArrayByValues(arr);
console.log(result);

Refactor variables in redux functions

I have to refactor my redux functions
In these 3 cases in am renaming the variable in a different way and i do not want this
See the example
case 'INCREMENT_LIKES' :
const index = action.index;
return [
...state.slice(0,index), // before the one we are updating
{...state[index], likes: state[index].likes + 1},
...state.slice(index + 1), // after the one we are updating
]
case 'INCREMENT_COORDINATES' :
const a = action.index;
let string = state[a].d || state[a].d2;
let array = string.split(" ");
let max = array.length;
let last = max - 2;
let i = (state[a].index || 3) + 1;
if ( i === last ) i = 3;
if (array[i] !== 'M' && array[i] !== 'L') array[i] = parseInt(array[i]) + 20;
return [
...state.slice(0,a), // before the one we are updating
{...state[a], d: array.join(' '), index: i, d2: array.join(' '), index: i }, // updating
...state.slice(a + 1), // after the one we are updating
]
case 'INCREMENT_VIEWBOX' :
const count = action.index;
let string2 = state[count].viewBox;
let array2 = string2.split(" ");
let max2 = array2.length;
let i2 = (state[count].index || 2) + 1;
if ( i2 === max2 ) i2 = 2;
array2[i2] = parseInt(array2[i2]) - 20;
return [
...state.slice(0,count), // before the one we are updating
{...state[count], viewBox: array2.join(' '), index: i2},
...state.slice(count + 1), // after the one we are updating
]
default:
return state;
These are the different variable names in the 3 cases
const index = action.index;
const a = action.index;
const count = action.index;
Same for
let string = state[a].d || state[a].d2;
let string2 = state[count].viewBox;
and
let array = string.split(" ");
let array2 = string2.split(" ");
How can i reuse the same variable name for all the 3 cases?
I mean using the same name for the 3 different cases like:
const index - let string - let array

You can add a scope to each case by wrapping them in curly brackets:
case 'INCREMENT_LIKES' : {
const index = action.index;
return [
...state.slice(0,index), // before the one we are updating
{...state[index], likes: state[index].likes + 1},
...state.slice(index + 1), // after the one we are updating
]
}
case 'SOME_OTHER_ACTION_TYPE' : {
console.log(index) // -> undefined
}