Replace a letter with its alphabet position
First : deleting space
Second : mapping each char with its alphabet rank
Third : test with the string Happy new year
var alphabet = "abcdefghijklmnopqrstuvwxyz".split('');
var alphabetPosition = text =>
text.split('').map(x => alphabet.indexOf(x) + 1);
console.log(alphabetPosition("happy new year"));function alphabetPosition(text) {
const words = text.toLowerCase().replace(/[^a-z]/g,"");
return [...words].map(v=> v.charCodeAt() - 96);
}
First we take the text and transform it into lowercase to get rid of the capital letters using text.toLowerCase() and then we do .replace(/[^a-z]/g,"") to replace all the non a-z characters with nothing.
The next step is to spread the string out into an array using [...words] and then mapping it to get the ascii character code of each a-z character.
Since a = 97 and b = 98 etc we will subtract 96 so that we get a = 1 and b = 2 etc (the position of the letters in the alphabet)
The Kata works with this code. Try with this one:
function alphabetPosition(text) {
var result = "";
for (var i = 0; i < text.length; i++) {
var code = text.toUpperCase().charCodeAt(i)
if (code > 64 && code < 91) result += (code - 64) + " ";
}
return result.slice(0, result.length - 1);
}
console.log(alphabetPosition("The sunset sets at twelve o' clock."));You need the String#length property
text.length
instead of text.len.
function alphabetPosition(text) {
var chari,
arr = [],
alphabet = "abcdefghijklmnopqrstuvwxyz",
i;
for (var i = 0; i < text.length; i++){
chari = text[i].toLowerCase();
if (alphabet.indexOf(chari) !== -1){
arr.push(alphabet.indexOf(chari));
}
}
return arr;
}
console.log(alphabetPosition("Hello World!!1"));A solution with ES6
function alphabetPosition(text) {
return [...text].map(a => parseInt(a, 36) - 10).filter(a => a >= 0);
}
console.log(alphabetPosition("Hello World!!1"));