Switch out a character with its corresponding position in the alphabet

Question

Switch out a character with its corresponding position in the alphabet

Starting out, this task seemed simple to me. However, every time I attempt to run the code on codewars, I keep getting an empty array. I'm reaching out in hopes that you can help me pinpoint the issue.

function alphabetPosition(text) {
  text.split(' ').join('');
  var chari = "";
  var arr = [];
  var alphabet = "abcdefghijklmnopqrstuvwxyz".split('');
  for(var i = 0; i < text.len; i++){
    chari = text.charAt(i).toLowerCase();
    if(alphabet.indexOf(chari) > -1){
      arr.push(alphabet.indexOf(chari));
    }
  }
  return arr;
}
console.log(alphabetPosition("Hello World"));

My approach involves extracting the text from the parameter and removing any spaces within it. I initialize an empty array variable and create an alphabet string for reference. Within the for loop, each character is converted to lowercase, and if it exists in the alphabet string, its position is added to the array (arr). Thank you for your assistance.

javascript arrays string

Answer 1

Answer №1

Check out this coding challenge that involves arrays manipulation. Give it a try:

function reverseArray(arr) {
  let reversedArr = [];
  for (let i = arr.length - 1; i >= 0; i--) {
    reversedArr.push(arr[i]);
  }
  
  return reversedArr;
}
console.log(reverseArray([1, 2, 3, 4, 5]));

Answer 2

Check out this coding challenge that involves arrays manipulation. Give it a try:

function reverseArray(arr) {
  let reversedArr = [];
  for (let i = arr.length - 1; i >= 0; i--) {
    reversedArr.push(arr[i]);
  }
  
  return reversedArr;
}
console.log(reverseArray([1, 2, 3, 4, 5]));

Answer 3

Answer №2

If you want to find the length of a string in JavaScript, use the String#length property

text.length

Don't use text.len, use the correct property instead.

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"));

Here's an ES6 solution for the same problem

function alphabetPosition(text) {
    return [...text].map(a => parseInt(a, 36) - 10).filter(a => a >= 0);
}
console.log(alphabetPosition("Hello World!!1"));

Answer 4

If you want to find the length of a string in JavaScript, use the String#length property

text.length

Don't use text.len, use the correct property instead.

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"));

Here's an ES6 solution for the same problem

function alphabetPosition(text) {
    return [...text].map(a => parseInt(a, 36) - 10).filter(a => a >= 0);
}
console.log(alphabetPosition("Hello World!!1"));

Answer 5

Answer №3

First step: removing empty spaces
Second step: assigning each character its corresponding alphabetical position
Third step: testing with the input 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"));

Answer 6

First step: removing empty spaces
Second step: assigning each character its corresponding alphabetical position
Third step: testing with the input 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"));

Answer 7

Answer №4

function getAlphabeticalPosition(text) {
  const words = text.toLowerCase().replace(/[^a-z]/g,"");
  return [...words].map(letter => letter.charCodeAt() - 96);
}

Initially, we convert the text to lowercase to eliminate any capital letters using text.toLowerCase() and then use .replace(/[^a-z]/g,"") to remove any non a-z characters.

Subsequently, we convert the string into an array using [...words] and then map it to find the ASCII character code of each a-z character.

By subtracting 96 from the ASCII value, we are able to determine the position of each letter in the alphabet (e.g., a = 1, b = 2, etc).

Answer 8

function getAlphabeticalPosition(text) {
  const words = text.toLowerCase().replace(/[^a-z]/g,"");
  return [...words].map(letter => letter.charCodeAt() - 96);
}

Initially, we convert the text to lowercase to eliminate any capital letters using text.toLowerCase() and then use .replace(/[^a-z]/g,"") to remove any non a-z characters.

Subsequently, we convert the string into an array using [...words] and then map it to find the ASCII character code of each a-z character.

By subtracting 96 from the ASCII value, we are able to determine the position of each letter in the alphabet (e.g., a = 1, b = 2, etc).

Answer 9

Answer №5

If you want to simplify things, consider utilizing the ASCII code. For example, the ASCII code for a is 97, b is 98, and so on. There is a JavaScript function called String.charCodeAt(n) that can be used to retrieve the ASCII code for a specific character. You may only need to adjust the offset for uppercase letters if you wish to include them in your support.

function alphabetPosition(text) {
var positions = [];
for (var i = 0; i < text.length; i++) {
var charCode = text.charCodeAt(i);
if (charCode >= 97 && charCode <= 122) {
positions.push(charCode - 96);
} else if (charCode >= 65 && charCode <= 90) { // remove this condition if uppercase letters are not a concern
positions.push(charCode - 64);
}
}
return positions;
}

var positions = alphabetPosition('Hello World');
console.log(positions);

Take a look at this live example

Answer 10

If you want to simplify things, consider utilizing the ASCII code. For example, the ASCII code for a is 97, b is 98, and so on. There is a JavaScript function called String.charCodeAt(n) that can be used to retrieve the ASCII code for a specific character. You may only need to adjust the offset for uppercase letters if you wish to include them in your support.

function alphabetPosition(text) {
var positions = [];
for (var i = 0; i < text.length; i++) {
var charCode = text.charCodeAt(i);
if (charCode >= 97 && charCode <= 122) {
positions.push(charCode - 96);
} else if (charCode >= 65 && charCode <= 90) { // remove this condition if uppercase letters are not a concern
positions.push(charCode - 64);
}
}
return positions;
}

var positions = alphabetPosition('Hello World');
console.log(positions);

Take a look at this live example

Answer 11

Answer №6

In this example, the function retrieves values based on a 0 indexed array using lambda expressions and filter. The original byte array is reused by splitting the input text.

function alphabetPosition(text) {
  var bytes = text.split('');
  var alphabet = "abcdefghijklmnopqrstuvwxyz".split('');
  for (var i = 0, len = text.length; i < len; i++) {
bytes[i] = alphabet.indexOf(bytes[i].toLowerCase());
  }
  return bytes.filter(n => { if(n > -1) return n; } ).join(' ');
}
console.log(alphabetPosition("Hello World"));

For a result based on a 1 indexed array, check out Kata Codewars Friendly

function alphabetPosition(text) {
  var bytes = text.split('');
  var alphabet = "abcdefghijklmnopqrstuvwxyz".split('');
  for (var i = 0, len = text.length; i < len; i++) {
bytes[i] = alphabet.indexOf(bytes[i].toLowerCase()) + 1;
  }
  return bytes.filter(n => { if(n > 0) return n; } ).join(' ');
}
console.log(alphabetPosition("Hello World"));

Answer 12

In this example, the function retrieves values based on a 0 indexed array using lambda expressions and filter. The original byte array is reused by splitting the input text.

function alphabetPosition(text) {
  var bytes = text.split('');
  var alphabet = "abcdefghijklmnopqrstuvwxyz".split('');
  for (var i = 0, len = text.length; i < len; i++) {
bytes[i] = alphabet.indexOf(bytes[i].toLowerCase());
  }
  return bytes.filter(n => { if(n > -1) return n; } ).join(' ');
}
console.log(alphabetPosition("Hello World"));

For a result based on a 1 indexed array, check out Kata Codewars Friendly

function alphabetPosition(text) {
  var bytes = text.split('');
  var alphabet = "abcdefghijklmnopqrstuvwxyz".split('');
  for (var i = 0, len = text.length; i < len; i++) {
bytes[i] = alphabet.indexOf(bytes[i].toLowerCase()) + 1;
  }
  return bytes.filter(n => { if(n > 0) return n; } ).join(' ');
}
console.log(alphabetPosition("Hello World"));

Answer 13

Answer №7

Alternatively, you can utilize the .charCodeAt method:

function convertToAlphabetPosition(text) {
    startingPoint = "a".charCodeAt(0);
    endingPoint = "z".charCodeAt(0);
    result = [];
    for (var x = 0; x < text.length; x++) {
        charIndex = text.charAt(x).toLowerCase().charCodeAt(0);
        if (charIndex >= startingPoint && charIndex <= endingPoint) {
            result.push(charIndex - startingPoint +1); // +1 to treat a as 1 instead of 0
        } else {
            result.push(0); // Or handle non-alphabetic characters in another way
        }
    }
    return result;
}
console.log(convertToAlphabetPosition("Hello World"));

Answer 14

Alternatively, you can utilize the .charCodeAt method:

function convertToAlphabetPosition(text) {
    startingPoint = "a".charCodeAt(0);
    endingPoint = "z".charCodeAt(0);
    result = [];
    for (var x = 0; x < text.length; x++) {
        charIndex = text.charAt(x).toLowerCase().charCodeAt(0);
        if (charIndex >= startingPoint && charIndex <= endingPoint) {
            result.push(charIndex - startingPoint +1); // +1 to treat a as 1 instead of 0
        } else {
            result.push(0); // Or handle non-alphabetic characters in another way
        }
    }
    return result;
}
console.log(convertToAlphabetPosition("Hello World"));

Answer 15

Answer №8

Here is an example of how you can accomplish this:

Let's create a function that will convert a given string into a coded version based on an alphabet and special characters:
var alphabet = [].reduce.call("abcdefghijklmnopqrstuvwxyz0123456789 .,!",(previous, current, index) => (previous[current] = index, previous),{}),
    inputString = "Happy 2017 whatever..!",
    codedString = [].map.call(inputString, character => alphabet[character.toLowerCase()]);
console.log(codedString);

Answer 16

Here is an example of how you can accomplish this:

Let's create a function that will convert a given string into a coded version based on an alphabet and special characters:
var alphabet = [].reduce.call("abcdefghijklmnopqrstuvwxyz0123456789 .,!",(previous, current, index) => (previous[current] = index, previous),{}),
    inputString = "Happy 2017 whatever..!",
    codedString = [].map.call(inputString, character => alphabet[character.toLowerCase()]);
console.log(codedString);

Answer 17

Answer №9

Replace len with length:

(let i = 0; i < text.length; i++)
// make the change to
(let i = 0; i < text.length; i++)

Answer 18

Replace len with length:

(let i = 0; i < text.length; i++)
// make the change to
(let i = 0; i < text.length; i++)

Answer 19

Answer №10

Check out this concise alternative that achieves the same outcome:

const getAlphabetPositions = (text) => {
  return text.split('').map((character) => character.charCodeAt(0) - 'a'.charCodeAt(0) + 1);
}

console.log(getAlphabetPositions("Hello World"));

Answer 20

Check out this concise alternative that achieves the same outcome:

const getAlphabetPositions = (text) => {
  return text.split('').map((character) => character.charCodeAt(0) - 'a'.charCodeAt(0) + 1);
}

console.log(getAlphabetPositions("Hello World"));

Answer 21

Answer №11

I just finished this CodeWars challenge and my solution worked like a charm!

const findAlphabetPosition = (sentence) => {
  let characters = Array.from(sentence.toLowerCase().replace(/[^a-z]/g,''));
  let positions = [];
  for (let j = 0; j < characters.length; j++) {
     positions.push(characters[j].charCodeAt() - 96);
  }
    return positions.join(' ');
}

Answer 22

I just finished this CodeWars challenge and my solution worked like a charm!

const findAlphabetPosition = (sentence) => {
  let characters = Array.from(sentence.toLowerCase().replace(/[^a-z]/g,''));
  let positions = [];
  for (let j = 0; j < characters.length; j++) {
     positions.push(characters[j].charCodeAt() - 96);
  }
    return positions.join(' ');
}

Answer 23

Answer №12

my solution

def convertTextToAlphabetPosition(text):
  if (text.isalpha()):                                          
    lettersOnly = ''.join(filter(str.isalpha, text)).lower()      
    alphabet = "abcdefghijklmnopqrstuvwxyz"
    result = []                                                  
    for letter in lettersOnly:
      result.append(alphabet.index(letter) + 1)
    return ' '.join(map(str, result))
  else:
    return ""                                              
}

print(convertTextToAlphabetPosition("The quick brown fox jumps over the lazy dog."))

Answer 24

my solution

def convertTextToAlphabetPosition(text):
  if (text.isalpha()):                                          
    lettersOnly = ''.join(filter(str.isalpha, text)).lower()      
    alphabet = "abcdefghijklmnopqrstuvwxyz"
    result = []                                                  
    for letter in lettersOnly:
      result.append(alphabet.index(letter) + 1)
    return ' '.join(map(str, result))
  else:
    return ""                                              
}

print(convertTextToAlphabetPosition("The quick brown fox jumps over the lazy dog."))

Answer 25

Answer №13

this example looks good

const charPositionHandler = (arr) => {
    const letters = 'abcdefghijklmnopqrstuwxyz';
    return arr.map((el) => `${letters.indexOf(el) + 1}:${el}`);
}

Answer 26

this example looks good

const charPositionHandler = (arr) => {
    const letters = 'abcdefghijklmnopqrstuwxyz';
    return arr.map((el) => `${letters.indexOf(el) + 1}:${el}`);
}

Answer 27

Answer №14

Here's a potential solution:

 function convertTextToAlphabetPosition(text) {
            const alphabet = 'abcdefghijklmnopqrstuvwxyz'
            const textWithoutSpaces = text.replace(/\s/g, '');
            const numbers = Array.from(textWithoutSpaces).map((letter) => {
                const lowerCaseLetter = letter.toLowerCase();
                const charIndex = alphabet.indexOf(lowerCaseLetter)+1
                if(charIndex) return charIndex
                return null
            })
            const withoutNullValues = numbers.filter(Boolean)
            return withoutNullValues.join(' ')
        }

Answer 28

Here's a potential solution:

 function convertTextToAlphabetPosition(text) {
            const alphabet = 'abcdefghijklmnopqrstuvwxyz'
            const textWithoutSpaces = text.replace(/\s/g, '');
            const numbers = Array.from(textWithoutSpaces).map((letter) => {
                const lowerCaseLetter = letter.toLowerCase();
                const charIndex = alphabet.indexOf(lowerCaseLetter)+1
                if(charIndex) return charIndex
                return null
            })
            const withoutNullValues = numbers.filter(Boolean)
            return withoutNullValues.join(' ')
        }

Answer 29

Answer №15

const convertToAlphabetPosition = (text) => {
  const lettersArray = text.split('');
  const alphabetIndex = lettersArray.map((letter) => {
    const charCode = letter.toUpperCase().charCodeAt() - 64;
    if (letter.length && charCode > 0 && charCode <= 26) {
      return charCode;
    }
  });
  return alphabetIndex.join(' ').replace(/\s+/g, ' ').trim();
}

I have a preference for the "Single Line Responsibility" approach, where each line of code performs only one action.

In the return statement, I remove multiple spaces for a cleaner output.

Answer 30

const convertToAlphabetPosition = (text) => {
  const lettersArray = text.split('');
  const alphabetIndex = lettersArray.map((letter) => {
    const charCode = letter.toUpperCase().charCodeAt() - 64;
    if (letter.length && charCode > 0 && charCode <= 26) {
      return charCode;
    }
  });
  return alphabetIndex.join(' ').replace(/\s+/g, ' ').trim();
}

I have a preference for the "Single Line Responsibility" approach, where each line of code performs only one action.

In the return statement, I remove multiple spaces for a cleaner output.

Answer 31

Answer №16

A more efficient iteration

 function alphabetPosition(str) {
        let split = str.split("");

        split = split.filter((el) => /^[a-zA-Z]+$/.test(el));
        let result = "";
        split.forEach(
          (el) => (result += el.toLowerCase().charCodeAt(0) - "96" + " ")
        );

        return result.trim();
      }
      console.log(alphabetPosition("The quick brown fox jumps over the lazy dog."));

Answer 32

A more efficient iteration

 function alphabetPosition(str) {
        let split = str.split("");

        split = split.filter((el) => /^[a-zA-Z]+$/.test(el));
        let result = "";
        split.forEach(
          (el) => (result += el.toLowerCase().charCodeAt(0) - "96" + " ")
        );

        return result.trim();
      }
      console.log(alphabetPosition("The quick brown fox jumps over the lazy dog."));

Answer 33

Answer №17

function getPositionOfAlphabets(input){
    //assigns numerical positions to each alphabet using an object
    const alphabetPositions = {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}

    //converts the input text to lowercase, removes non-alphabets,
    // converts the text to an array, maps each alphabet to its position in the alphabet
    // and joins them back to form a string

    const positions = input.toLowerCase()
    .replace(/[^a-z]/g, "")
    .split("")
    .map((letter) => alphabetPositions[letter])
    .join(" ")

    return positions;
}

Answer 34

function getPositionOfAlphabets(input){
    //assigns numerical positions to each alphabet using an object
    const alphabetPositions = {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}

    //converts the input text to lowercase, removes non-alphabets,
    // converts the text to an array, maps each alphabet to its position in the alphabet
    // and joins them back to form a string

    const positions = input.toLowerCase()
    .replace(/[^a-z]/g, "")
    .split("")
    .map((letter) => alphabetPositions[letter])
    .join(" ")

    return positions;
}

Switch out a character with its corresponding position in the alphabet

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

Answer №8

Answer №9

Answer №10

Answer №11

Answer №12

Answer №13

Answer №14

Answer №15

Answer №16

Answer №17

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