Convert integer into its character equivalent, where 0 => a, 1 => b, etc

Question

I want to convert an integer into its character equivalent based on the alphabet. For example:

0 => a
1 => b
2 => c
3 => d

etc. I could build an array and just look it up when I need it but I’m wondering if there’s a built in function to do this for me. All the examples I’ve found via Google are working with ASCII values and not a character’s position in the alphabet.

Answer 1

Assuming you want lower case letters:

var chr = String.fromCharCode(97 + n); // where n is 0, 1, 2 ...

97 is the ASCII code for lower case 'a'. If you want uppercase letters, replace 97 with 65 (uppercase 'A'). Note that if n > 25, you will get out of the range of letters.

Answer 2

Will be more portable in case of extending to other alphabets:

char='abcdefghijklmnopqrstuvwxyz'[code]

or, to be more compatible (with our beloved IE):

char='abcdefghijklmnopqrstuvwxyz'.charAt(code);

Answer 3

If you don't mind getting multi-character strings back, you can support arbitrary positive indices:

function idOf(i) {
  return (
    (i >= 26 ? idOf(((i / 26) >> 0) - 1) : "") +
    "abcdefghijklmnopqrstuvwxyz"[i % 26 >> 0]
  );
}

[0, 1, 25, 26, 27, 701, 702, 703].map(idOf);
// ['a', 'b', 'z', 'aa', 'ab', 'zz', 'aaa', 'aab']

(Not thoroughly tested for precision errors :)

Answer 4

A simple answer would be (26 characters):

String.fromCharCode(97+n);

If space is precious you could do the following (20 characters):

(10+n).toString(36);

Think about what you could do with all those extra bytes!

How this works is you convert the number to base 36, so you have the following characters:

0123456789abcdefghijklmnopqrstuvwxyz
^         ^
n        n+10

By offsetting by 10 the characters start at a instead of 0.

Not entirely sure about how fast running the two different examples client-side would compare though.

Answer 5

I don't like all the solutions that use magic numbers like 97 or 36.

const A = 'A'.charCodeAt(0);

let numberToCharacter = number => String.fromCharCode(A + number);

let characterToNumber = character => character.charCodeAt(0) - A;

this assumes uppercase letters and starts 'A' at 0.

Answer 6

Javascript's String.fromCharCode(code1, code2, ..., codeN) takes an infinite number of arguments and returns a string of letters whose corresponding ASCII values are code1, code2, ... codeN. Since 97 is 'a' in ASCII, we can adjust for your indexing by adding 97 to your index.

function indexToChar(i) {
  return String.fromCharCode(i+97); //97 in ASCII is 'a', so i=0 returns 'a', 
                                    // i=1 returns 'b', etc
}

Answer 7

There you go: (a-zA-Z)

function codeToChar( number ) {
  if ( number >= 0 && number <= 25 ) // a-z
    number = number + 97;
  else if ( number >= 26 && number <= 51 ) // A-Z
    number = number + (65-26);
  else
    return false; // range error
  return String.fromCharCode( number );
}

input: 0-51, or it will return false (range error);

OR:

var codeToChar = function() {
  var abc = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ".split("");
  return function( code ) {
    return abc[code];
  };
})();

returns undefined in case of range error. NOTE: the array will be created only once and because of closure it will be available for the the new codeToChar function. I guess it's even faster then the first method (it's just a lookup basically).

Answer 8

Use String.fromCharCode. This returns a string from a Unicode value, which matches the first 128 characters of ASCII.

var a = String.fromCharCode(97);

Answer 9

The only problem with @mikemaccana's great solution is that it uses the binary >> operator which is costly, performance-wise. I suggest this modification to his great work as a slight improvement that your colleagues can perhaps read more easily.

const getColumnName = (i) => {
     const previousLetters = (i >= 26 ? getColumnName(Math.floor(i / 26) -1 ) : '');
     const lastLetter = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'[i % 26]; 
     return previousLetters + lastLetter;
}

Or as a one-liner

const getColumnName = i => (i >= 26 ? getColumnName(Math.floor(i / 26) -1 ) : '') + 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'[i % 26];

Example:

getColumnName(0); // "A"
getColumnName(1); // "B"
getColumnName(25); // "Z"
getColumnName(26); // "AA"
getColumnName(27); // "AB"
getColumnName(80085) // "DNLF"

Answer 10

Try

(n+10).toString(36)

chr = n=>(n+10).toString(36);

for(i=0; i<26; i++) console.log(`${i} => ${ chr(i) }`);

Answer 11

If you are looking for TypeScript working functions then follow

public numericValue = (alphaChar: any) => alphaChar.toUpperCase().charCodeAt(0) - 64;

public alphaValue = (numericDigit: any) => 
   String.fromCharCode(64 + numericDigit) : '';

You can make several checks like (numericDigit >= 1 && numericDigit <= 26) ? inside function body as per the requirements.

Answer 12

Assuming you want uppercase case letters:

function numberToLetter(num){
        var alf={
            '0': 'A', '1': 'B', '2': 'C', '3': 'D', '4': 'E', '5': 'F', '6': 'G'
        };
        if(num.length== 1) return alf[num] || ' ';
        return num.split('').map(numberToLetter);
    }

Example:

numberToLetter('023') is ["A", "C", "D"]

numberToLetter('5') is "F"