ChathuraGH

This is gist. 
There are many like it, but this one is mine. 
It is my life. 
I must master it as I must master my life. 
Without me gist is useless. 
Without gist, I am useless.

ChathuraGH

var dict = {
  "x": 1,
  "y": 6,
  "z": 9,
  "a": 5,
  "b": 7,
  "c": 11,
  "d": 17,
  "t": 3
};

ChathuraGH

function sort_object(obj) {
    items = Object.keys(obj).map(function(key) {
        return [key, obj[key]];
    });
    items.sort(function(first, second) {
        return second[1] - first[1];
    });
    sorted_obj={}
    $.each(items, function(k, v) {
        use_key = v[0]

ChathuraGH

class DictUtils {
    static entries(dictionary) {

        try {

            //ECMAScript 2017 and higher, better performance if support
            return Object.entries(dictionary);

        } catch (error) {

ChathuraGH

let s = 'hello';
var result = [...s].reduce((a, e) => { a[e] = a[e] ? a[e] + 1 : 1; return a }, {}); 
console.log(result); // {h: 1, e: 1, l: 2, o: 1}

ChathuraGH

function withAMap(str) {
    // A map for the character=>count mappings
    const counts = new Map();

    // Loop through the string...
    for (const ch of str) {
        // Get the count for it, if we have one; we'll get `undefined` if we don't
        // know this character yet. Using nullish coalescing (`??`), we can turn
        // that `undefined` into a `0`. (In obsolete environments that don't

ChathuraGH

//Split the string with the spread ... operator instead of .split(''):

'🌯🌯🍣🍻'.split('')
//=> ["\ud83c", "\udf2f", "\ud83c", "\udf2f", "\ud83c", "\udf63", "\ud83c", "\udf7b"]
//vs

[...'🌯🌯🍣🍻']
//=> ["🌯", "🌯", "🍣", "🍻"]
//vs

ChathuraGH

function Char_Count(str1) {
    var chars = {};
    str1.replace(/\S/g, function(l){chars[l] = (isNaN(chars[l]) ? 1 : chars[l] + 1);});
    return chars;
}

var myString = "This is my String";
console.log(Char_Count(myString));

ChathuraGH

function countChrOccurence ('hello') {
 let charMap = new Map();
 const count = 0;
  for (const key of str) {
   charMap.set(key,count); // initialize every character with 0. this would make charMap to be 'h'=> 0, 'e' => 0, 'l' => 0, 
  }

  for (const key of str) {
    let count = charMap.get(key);
    charMap.set(key, count + 1);

ChathuraGH

function countChar(str) {
    let myObj= {};
    for (let s of str) {
        if ( myObj[s] ? myObj[s].count ++ : myObj[s] = { count : 1 } );
    }
    return myObj;
}

var charCount = countChar('abcceddd');