Number of Occurances

Question

var exampleString = 'hi, my name is Lisa Kudrov';
var stringBase = [
 'a','b','c','d','e',
 'f','g','h','i','j',
 'k','l','m','n','o',
 'p','q','r','s','t',
 'u','v','w','x','y',
 'z','_'
];
var countObject = {} ; 
function characterCount(word, character) {
 var count = 0;
 for (var i = 0; i < word.length; i++) {
   if (word[i] === character) {
       count++;
   }
 }
 return count;
 }
 for (var i = 0, l = stringBase.length; i < l; i++) {
    var currentChar = stringBase[i];
    countObject[currentChar] = characterCount(exampleString, currentChar);
 }
 console.log(countObject);

I need my output to show the letters that exist only. For example: a:1,c:1 and etc... also how can I shortcut this program in a way that when i call the function, i'll be calling characterCount('hi, my name is Lisa Kudrov')

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score 1 · Answer 1 · answered Jan 29 '20 at 14:19

1

You can simply group characters using Array.reduce

let characterCount = str => {
 return Object.fromEntries([...[...str.toLowerCase()].filter(c => /[a-z_]/.test(c)).reduce((acc, curr) => (acc.set(curr, acc.get(curr) +1 || 1), acc), new Map())])
}

console.log(characterCount('hi, my name is Lisa Kudrov'));

answered Jan 29 '20 at 14:19

AZ_

3,094
1
9
19

I think only reduce can have extra braces for more readability, if required I can add. :) – AZ_ Jan 29 '20 at 14:25

Mister Jojo · Answer 2 · 2020-01-29T17:31:59.733

I done this..

exampleString = 'hi, my name is Lisa_Kudrov';

function countLetters( inTxt)
{
  let counters = [...inTxt]
                      .filter(l=>/[a-z_]/.test(l))
                      .reduce((r,letter)=>
                        {
                        if(!Boolean(r[letter] )) r[letter] = 0
                        r[letter]++
                        return r
                        },{})

  console.log ('resp', counters)
}
countLetters( exampleString)

.as-console-wrapper { max-height: 100% !important; top: 0; }

Number of Occurances

2 Answers2