CODEWITHSUNDEEP
javascript
pp94
pp94

Reputation: 133

Counter the letter of a single string. Javascript

I have a problem. I'm trying to create a function that return the number of letters of frequency of a particular string. Here is an example.

var a = letter_frequency("Hello");
a["H"] == 1; a["E"] == 1; a["L"] == 2; a["A"] == undefined;

Now I have the code I wrote but I don't know how to go forward. Can you help me?

    function letter_frequency(s) {
    if (s === undefined) {
        return undefined;
    } else {
        var splitter = s.split()

    }
}

Here one test

 test( "Frequency", function() {
    deepEqual(letter_frequency() , undefined);
    var a = letter_frequency("Hello");
    equal(a["H"] , 1, "For the String 'Hello' the number of 'H' chars should be 1");
    equal(a["E"] , 1, "For the String 'Hello' the number of 'E' chars should be 1");
    equal(a["L"] , 2, "For the String 'Hello' the number of 'L' chars should be 2");
    equal(a["O"] , 1, "For the String 'Hello' the number of 'O' chars should be 1");
    deepEqual(a["A"] , undefined, "For the String 'Hello' the number of 'A' chars should be undefined");
    var a = letter_frequency("Software Atelier III");
    equal(a["I"] , 4, "For the String 'Software Atelier III' the number of 'I' chars should be 4");
    equal(a["S"] , 1, "For the String 'Software Atelier III' the number of 'S' chars should be 1");
    equal(a["A"] , 2, "For the String 'Software Atelier III' the number of 'A' chars should be 2");
    equal(a["O"] , 1, "For the String 'Software Atelier III' the number of 'O' chars should be 1");
    equal(a["T"] , 2, "For the String 'Software Atelier III' the number of 'T' chars should be 2");
    equal(a[" "] , 2, "For the String 'Software Atelier III' the number of ' ' chars should be 2");
    equal(a["F"] , 1, "For the String 'Software Atelier III' the number of 'F' chars should be 1");
    equal(a["W"] , 1, "For the String 'Software Atelier III' the number of 'W' chars should be 1");
    equal(a["T"] , 2, "For the String 'Software Atelier III' the number of 'T' chars should be 2");
    equal(a["L"] , 1, "For the String 'Software Atelier III' the number of 'L' chars should be 1");
});

SOLUTION: Thanks to h2ooooooo

So for my code I used 

function letter_frequency(s) {
if(s === undefined){
    return undefined
} else {
    var fr = {};
    for (var i = 0; i < s.length; i++) {
        var ch = s.charAt(i).toUpperCase();
        if (fr[ch]) {
            fr[ch]++;
        } else {
            fr[ch] = 1;
        }
    }
}
    return freq;
}

I linked some code from all the answer I received. The only difference is the control for the undefined question where my exercise asked for a control with undefined words

Upvotes: 0

Views: 147

Answers (2)

BenM
BenM

Reputation: 53198

Something like this should achieve what you're looking for (I based it upon the example output you provided, with the letters returned as uppercase properties of an object):

String.prototype.countLetters = function() 
{
    var arr = new Array(),
        letters = this.split('');

    for(var i = 0; i < letters.length; i++)
    {
        var letter = letters[i].toUpperCase();

        if(arr[letter] !== undefined)
        {
            arr[letter]+= 1;
        }
        else
        {
            arr[letter] = 1;
        }
    }

    return arr;
}

You can then call this using 'Hello'.countLetters();. Given this example, and using the test string of Hello, we get the following output:

[ H: 1, E: 1, L: 2, O: 1 ]

jsFiddle Demo

Upvotes: 0

h2ooooooo
h2ooooooo

Reputation: 39532

You can split up the string using .split() and loop through the letters:

function letter_frequency(str) {
    var letters = str.split(''),
        letterFrequency = {};

    for (var i = 0, len = letters.length; i < len; i++) {
        if (!letterFrequency.hasOwnProperty(letters[i])) {
            letterFrequency[letters[i]] = 1;
        } else {
            letterFrequency[letters[i]]++;
        }
    }

    return letterFrequency;
}

letter_frequency("Hello");
// Object {H: 1, e: 1, l: 2, o: 1}

If you'd rather have it all be capital letters (for some odd reason), you can do the following using .toUpperCase():

function letter_frequency(str) {
    var letters = str.split(''),
        letterFrequency = {},
        letter;

    for (var i = 0, len = letters.length; i < len; i++) {
        letter = letters[i].toUpperCase();
        if (!letterFrequency.hasOwnProperty(letter)) {
            letterFrequency[letter] = 1;
        } else {
            letterFrequency[letter]++;
        }
    }

    return letterFrequency;
}

letter_frequency("Hello");
// Object {H: 1, E: 1, L: 2, O: 1}

Upvotes: 1

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