How to access this object? it keeps returning a string of undefined

Question

In this kata you are required to, given a string, replace every letter with its position in the alphabet.

If anything in the text isn't a letter, ignore it and don't return it.

"a" = 1, "b" = 2, etc.

Example

alphabet_position("The sunset sets at twelve o' clock.")

Should return "20 8 5 19 21 14 19 5 20 19 5 20 19 1 20 20 23 5 12 22 5 15 3 12 15 3 11" (as a string)

How to access this object? it keeps returning a string of undefined.

function alphabetPosition(text) {
  var alphabet = {
    a: 1,
    b: 2,
    c: 3,
    d: 4,
    e: 5,
    f: 6,
    g: 7,
    h: 8,
    i: 9,
    j: 10,
    k: 11,
    l: 12,
    m: 13,
    n: 14,
    o: 15,
    p: 16,
    q: 17,
    r: 18,
    s: 19,
    t: 20,
    u: 21,
    v: 22,
    w: 23,
    x: 24,
    y: 25,
    z: 26
  }
  var number = 0;
  var string = "";
  var letter = "";

  for (i = 0; i < text.length; i++) {
    letter = text.charAt(i);
    number = alphabet.letter;
    string += number + " ";

  }
  return string;
}

var res = alphabetPosition("The sunset sets at twelve o' clock.");

console.log(res, res === "20 8 5 19 21 14 19 5 20 19 5 20 19 1 20 20 23 5 12 22 5 15 3 12 15 3 11");

`number = alphabet[letter]` What you have is looking up the value at the literal key `"letter"`, which doesn't exist - hence `undefined` — Robin Zigmond, Nov 13 '18 at 13:09
https://repl.it/@johntduong/Code-Wars-Replace-With-Alphabet-Position — mplungjan, Nov 13 '18 at 13:10
you may also want to consider what happens when you have an uppercase letter in your string (hint: convert to lowercase) — Federico klez Culloca, Nov 13 '18 at 13:10

score 1 · Answer 1 · answered Nov 13 '18 at 13:13

Try following

alphabet.letter should be alphabet[letter.toLowerCase()] - for ignore case
For spaces and other characters, place a check for if(number)

function alphabetPosition(text) {
    var alphabet = {a: 1,b: 2,c: 3,d: 4,e: 5,f: 6,g: 7,h: 8,i: 9,j: 10,k: 11,l: 12,m: 13,n: 14,o: 15,p: 16,q: 17,r: 18,s: 19,t: 20,u: 21,v: 22,w: 23,x: 24,y: 25,z: 26};
    var number = 0;
    var string = "";
    var letter ="";

    for (i=0; i<text.length; i++) {
      letter = text.charAt(i);
      number = alphabet[letter.toLowerCase()];
      if(number) string += number + " "; 

    }
    return string;
}
console.log(alphabetPosition("The sunset sets at twelve o' clock."));

Mihai Alexandru-Ionut · Answer 2 · 2018-11-13T13:19:53.460

There are some mistakes in your code. Please follow below steps in order to solve the problem.

Use bracket notation for accessing dynamically properties.
Use toLowerCase() method in order to convert upper case letters also.
Treat the special space case, because there is not converting rule for it.

function alphabetPosition(text) {
var alphabet ={
a: 1,
b: 2,
c: 3,
d: 4,
e: 5,
f: 6,
g: 7,
h: 8,
i: 9,
j: 10,
k: 11,
l: 12,
m: 13,
n: 14,
o: 15,
p: 16,
q: 17,
r: 18,
s: 19,
t: 20,
u: 21,
v: 22,
w: 23,
x: 24,
y: 25,
z: 26
}
var number = 0;
var string = "";
var letter ="";

for (i=0; i< text.length; i++) {
  letter = text.charAt(i).toLowerCase();
  number = alphabet[letter] || '';
  string += number + " "; 

}
  return string;
}

console.log(alphabetPosition("The sunset sets at twelve o' clock."))

Also, you can simplify your method by using map method.

return text.split('').map(c => alphabet[c.toLowerCase()] || '').join(' ');

Tornike Shavishvili · Answer 3 · 2018-11-13T13:37:11.697

You had issue with your code: when storing object property name in a variable, you should use [] notation to access property value. in your example you had alphabet.letter which should have been alphabet[letter]

function alphabetPosition(text) {
  var alphabet = {
    a: 1,
    b: 2,
    c: 3,
    d: 4,
    e: 5,
    f: 6,
    g: 7,
    h: 8,
    i: 9,
    j: 10,
    k: 11,
    l: 12,
    m: 13,
    n: 14,
    o: 15,
    p: 16,
    q: 17,
    r: 18,
    s: 19,
    t: 20,
    u: 21,
    v: 22,
    w: 23,
    x: 24,
    y: 25,
    z: 26
  }
  var number = 0;
  var string = "";
  var letter = "";

  for (i = 0; i < text.length; i++) {
    letter = text.charAt(i).toLowerCase();
    number = alphabet[letter];
    if (typeof number !== 'undefined') {
       string += number + " ";
    }
    
  }
  return string.trim();
}

var res = alphabetPosition("The sunset sets at twelve o' clock.");

console.log(res, res === "20 8 5 19 21 14 19 5 20 19 5 20 19 1 20 20 23 5 12 22 5 15 3 12 15 3 11");

score 0 · Answer 4 · answered Nov 13 '18 at 13:23

You can skip that object altogether if you consider that the index of each letter is its ascii/utf-16 code minus 97 (the ascii code of the letter 'a') plus 1 (because your letters start at 1 instead of 0). So you can solve it like this:

function alphabetPosition(text) {
  var str = "";
  for (i = 0; i < text.length; i++) {
    var c = text.charAt(i).toLowerCase();
    var code = c.charCodeAt(0);
    if (code >= 97 && code <= 122) str += ((code - 97 + 1) + " ")
  }
  return str;
}

console.log(alphabetPosition("The sunset sets at twelve o' clock."));

Prasad Telkikar · Answer 5 · 2018-11-13T13:30:57.183

0

Use ASCII codes, your code will not required alphabet array

var str = "The sunset sets at twelve o' clock.";

function CharToAsciiConversion(str){
  var result = "";
  str = str.toLowerCase();
  for (var i = 0; i < str.length; i++) {
  var ascii = str.charCodeAt(i) - 96;
  if(ascii > 0 && ascii < 26)
     result += ascii + " ";
  }
  return result;
}

console.log(CharToAsciiConversion(str));

edited Nov 13 '18 at 13:30

answered Nov 13 '18 at 13:24

Prasad Telkikar

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How to access this object? it keeps returning a string of undefined

5 Answers5