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How to correctly use Array.map() for replacing string with alphabet position


How can I format numbers as currency string in JavaScript?How to check empty/undefined/null string in JavaScript?How can I convert a string to boolean in JavaScript?How to check if a string “StartsWith” another string?How do I correctly clone a JavaScript object?How can I get query string values in JavaScript?How do I make the first letter of a string uppercase in JavaScript?How to replace all occurrences of a string?How to check whether a string contains a substring in JavaScript?Replace a letter with its alphabet position






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2















Other SO 'Replace string with alphabet positions' questions didn't utilize map, which is what I'm trying to learn how to use to solve this.



Problem:
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.



What I've tried is:
-looping over a new array instance and setting the index value to String.fromCharCode()
- taking input string making it lowercase
-splitting to array
-return array.map().join(' ')



function alphabetPosition(text) 

let alphabet = new Array(26);
for (let i = 0; i<26; ++i)
let char = String.fromCharCode(97 + i);
alphabet[i] = char;


text = text.toLowerCase();

let arr = text.split('');

return arr.map(element => return element = alphabet.indexOf(element+1) ).join(' ');



expected it to return a string of alphabet positions, but got nothing at all. What is wrong with my implementation of Array.map()?










share|improve this question

















  • 1





    "nothing at all" really?

    – Jonas Wilms
    Mar 26 at 17:49











  • I think you're at least going to need to remove the 'element = ' portion of the second-to-last line

    – Andrew Castellano
    Mar 26 at 17:51






  • 1





    I guess you are looking for alphabet.indexOf(element)+1 instead of alphabet.indexOf(element+1). Other than that, it should work.

    – Bergi
    Mar 26 at 17:52

















2















Other SO 'Replace string with alphabet positions' questions didn't utilize map, which is what I'm trying to learn how to use to solve this.



Problem:
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.



What I've tried is:
-looping over a new array instance and setting the index value to String.fromCharCode()
- taking input string making it lowercase
-splitting to array
-return array.map().join(' ')



function alphabetPosition(text) 

let alphabet = new Array(26);
for (let i = 0; i<26; ++i)
let char = String.fromCharCode(97 + i);
alphabet[i] = char;


text = text.toLowerCase();

let arr = text.split('');

return arr.map(element => return element = alphabet.indexOf(element+1) ).join(' ');



expected it to return a string of alphabet positions, but got nothing at all. What is wrong with my implementation of Array.map()?










share|improve this question

















  • 1





    "nothing at all" really?

    – Jonas Wilms
    Mar 26 at 17:49











  • I think you're at least going to need to remove the 'element = ' portion of the second-to-last line

    – Andrew Castellano
    Mar 26 at 17:51






  • 1





    I guess you are looking for alphabet.indexOf(element)+1 instead of alphabet.indexOf(element+1). Other than that, it should work.

    – Bergi
    Mar 26 at 17:52













2












2








2








Other SO 'Replace string with alphabet positions' questions didn't utilize map, which is what I'm trying to learn how to use to solve this.



Problem:
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.



What I've tried is:
-looping over a new array instance and setting the index value to String.fromCharCode()
- taking input string making it lowercase
-splitting to array
-return array.map().join(' ')



function alphabetPosition(text) 

let alphabet = new Array(26);
for (let i = 0; i<26; ++i)
let char = String.fromCharCode(97 + i);
alphabet[i] = char;


text = text.toLowerCase();

let arr = text.split('');

return arr.map(element => return element = alphabet.indexOf(element+1) ).join(' ');



expected it to return a string of alphabet positions, but got nothing at all. What is wrong with my implementation of Array.map()?










share|improve this question














Other SO 'Replace string with alphabet positions' questions didn't utilize map, which is what I'm trying to learn how to use to solve this.



Problem:
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.



What I've tried is:
-looping over a new array instance and setting the index value to String.fromCharCode()
- taking input string making it lowercase
-splitting to array
-return array.map().join(' ')



function alphabetPosition(text) 

let alphabet = new Array(26);
for (let i = 0; i<26; ++i)
let char = String.fromCharCode(97 + i);
alphabet[i] = char;


text = text.toLowerCase();

let arr = text.split('');

return arr.map(element => return element = alphabet.indexOf(element+1) ).join(' ');



expected it to return a string of alphabet positions, but got nothing at all. What is wrong with my implementation of Array.map()?







javascript






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 26 at 17:45









JimJim

776 bronze badges




776 bronze badges







  • 1





    "nothing at all" really?

    – Jonas Wilms
    Mar 26 at 17:49











  • I think you're at least going to need to remove the 'element = ' portion of the second-to-last line

    – Andrew Castellano
    Mar 26 at 17:51






  • 1





    I guess you are looking for alphabet.indexOf(element)+1 instead of alphabet.indexOf(element+1). Other than that, it should work.

    – Bergi
    Mar 26 at 17:52












  • 1





    "nothing at all" really?

    – Jonas Wilms
    Mar 26 at 17:49











  • I think you're at least going to need to remove the 'element = ' portion of the second-to-last line

    – Andrew Castellano
    Mar 26 at 17:51






  • 1





    I guess you are looking for alphabet.indexOf(element)+1 instead of alphabet.indexOf(element+1). Other than that, it should work.

    – Bergi
    Mar 26 at 17:52







1




1





"nothing at all" really?

– Jonas Wilms
Mar 26 at 17:49





"nothing at all" really?

– Jonas Wilms
Mar 26 at 17:49













I think you're at least going to need to remove the 'element = ' portion of the second-to-last line

– Andrew Castellano
Mar 26 at 17:51





I think you're at least going to need to remove the 'element = ' portion of the second-to-last line

– Andrew Castellano
Mar 26 at 17:51




1




1





I guess you are looking for alphabet.indexOf(element)+1 instead of alphabet.indexOf(element+1). Other than that, it should work.

– Bergi
Mar 26 at 17:52





I guess you are looking for alphabet.indexOf(element)+1 instead of alphabet.indexOf(element+1). Other than that, it should work.

– Bergi
Mar 26 at 17:52












4 Answers
4






active

oldest

votes


















1














Your map() last line of the function was returning the value of
an assignment.



return arr.map(element => return element = alphabet.indexOf(element+1) ).join(' ');


Just alphabet.indexOf(element) would have sufficed.



This will give you the result you want:






alphabetPosition = text => 
let alphabet = new Array(26);
for (let i = 0; i < 26; ++i)
let char = String.fromCharCode(97 + i);
alphabet[i] = char;


return text.toLowerCase().split('').map(element =>
alphabet.indexOf(element)
).join(' ');


console.log(alphabetPosition("This is a string"));





Hope this helps,






share|improve this answer






























    4














    In your map element would be a letter, "a" for example. Then you add (concat) 1 to it, which results in "a1" which is not in your alphabet. Also element = is unneccessary, returning the position is enough.






    share|improve this answer






























      2














      You've complicated the solution, the simplest approach would be to just find the charcode and return that.



      function alphabetPosition(text) 
      let str = '';
      for (var i = 0; i < text.length; i++)
      str += (text[i] + (text.charCodeAt(i) - 96));

      return str;






      share|improve this answer























      • how is str the positions of each letter of the string in the alphabet if you're using text.charCodeAt(i)?

        – Jim
        Mar 26 at 18:04











      • text.charCodeAt(i) will return the char code of the character in the position i. If your string is jim, then text.charCodeAt(0) will return the char code of j

        – varun agarwal
        Mar 26 at 18:07











      • the position of 'j' in the alphabet is 10, however...

        – Jim
        Mar 26 at 18:10






      • 1





        Yeah, hence the - 96 to normalize the values.

        – varun agarwal
        Mar 26 at 18:12


















      2














      I totally understand that is a coding challenge, interview question or likewise so if you really need to use map() you should only return the result of the callback passed to map as follows :



      return arr.map(x => alphabet.indexOf(x) + 1).join(' ')


      However reduce() seems more appropriate in your case :



      return arr.reduce((ac, cv) => ac + (alphabet.indexOf(cv) + 1) + ' ', '')





      share|improve this answer

























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        4 Answers
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        4 Answers
        4






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        active

        oldest

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        active

        oldest

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        1














        Your map() last line of the function was returning the value of
        an assignment.



        return arr.map(element => return element = alphabet.indexOf(element+1) ).join(' ');


        Just alphabet.indexOf(element) would have sufficed.



        This will give you the result you want:






        alphabetPosition = text => 
        let alphabet = new Array(26);
        for (let i = 0; i < 26; ++i)
        let char = String.fromCharCode(97 + i);
        alphabet[i] = char;


        return text.toLowerCase().split('').map(element =>
        alphabet.indexOf(element)
        ).join(' ');


        console.log(alphabetPosition("This is a string"));





        Hope this helps,






        share|improve this answer



























          1














          Your map() last line of the function was returning the value of
          an assignment.



          return arr.map(element => return element = alphabet.indexOf(element+1) ).join(' ');


          Just alphabet.indexOf(element) would have sufficed.



          This will give you the result you want:






          alphabetPosition = text => 
          let alphabet = new Array(26);
          for (let i = 0; i < 26; ++i)
          let char = String.fromCharCode(97 + i);
          alphabet[i] = char;


          return text.toLowerCase().split('').map(element =>
          alphabet.indexOf(element)
          ).join(' ');


          console.log(alphabetPosition("This is a string"));





          Hope this helps,






          share|improve this answer

























            1












            1








            1







            Your map() last line of the function was returning the value of
            an assignment.



            return arr.map(element => return element = alphabet.indexOf(element+1) ).join(' ');


            Just alphabet.indexOf(element) would have sufficed.



            This will give you the result you want:






            alphabetPosition = text => 
            let alphabet = new Array(26);
            for (let i = 0; i < 26; ++i)
            let char = String.fromCharCode(97 + i);
            alphabet[i] = char;


            return text.toLowerCase().split('').map(element =>
            alphabet.indexOf(element)
            ).join(' ');


            console.log(alphabetPosition("This is a string"));





            Hope this helps,






            share|improve this answer













            Your map() last line of the function was returning the value of
            an assignment.



            return arr.map(element => return element = alphabet.indexOf(element+1) ).join(' ');


            Just alphabet.indexOf(element) would have sufficed.



            This will give you the result you want:






            alphabetPosition = text => 
            let alphabet = new Array(26);
            for (let i = 0; i < 26; ++i)
            let char = String.fromCharCode(97 + i);
            alphabet[i] = char;


            return text.toLowerCase().split('').map(element =>
            alphabet.indexOf(element)
            ).join(' ');


            console.log(alphabetPosition("This is a string"));





            Hope this helps,






            alphabetPosition = text => 
            let alphabet = new Array(26);
            for (let i = 0; i < 26; ++i)
            let char = String.fromCharCode(97 + i);
            alphabet[i] = char;


            return text.toLowerCase().split('').map(element =>
            alphabet.indexOf(element)
            ).join(' ');


            console.log(alphabetPosition("This is a string"));





            alphabetPosition = text => 
            let alphabet = new Array(26);
            for (let i = 0; i < 26; ++i)
            let char = String.fromCharCode(97 + i);
            alphabet[i] = char;


            return text.toLowerCase().split('').map(element =>
            alphabet.indexOf(element)
            ).join(' ');


            console.log(alphabetPosition("This is a string"));






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Mar 26 at 18:00









            Miroslav GlamuzinaMiroslav Glamuzina

            2,8922 gold badges13 silver badges23 bronze badges




            2,8922 gold badges13 silver badges23 bronze badges























                4














                In your map element would be a letter, "a" for example. Then you add (concat) 1 to it, which results in "a1" which is not in your alphabet. Also element = is unneccessary, returning the position is enough.






                share|improve this answer



























                  4














                  In your map element would be a letter, "a" for example. Then you add (concat) 1 to it, which results in "a1" which is not in your alphabet. Also element = is unneccessary, returning the position is enough.






                  share|improve this answer

























                    4












                    4








                    4







                    In your map element would be a letter, "a" for example. Then you add (concat) 1 to it, which results in "a1" which is not in your alphabet. Also element = is unneccessary, returning the position is enough.






                    share|improve this answer













                    In your map element would be a letter, "a" for example. Then you add (concat) 1 to it, which results in "a1" which is not in your alphabet. Also element = is unneccessary, returning the position is enough.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Mar 26 at 17:48









                    Jonas WilmsJonas Wilms

                    75.9k7 gold badges42 silver badges67 bronze badges




                    75.9k7 gold badges42 silver badges67 bronze badges





















                        2














                        You've complicated the solution, the simplest approach would be to just find the charcode and return that.



                        function alphabetPosition(text) 
                        let str = '';
                        for (var i = 0; i < text.length; i++)
                        str += (text[i] + (text.charCodeAt(i) - 96));

                        return str;






                        share|improve this answer























                        • how is str the positions of each letter of the string in the alphabet if you're using text.charCodeAt(i)?

                          – Jim
                          Mar 26 at 18:04











                        • text.charCodeAt(i) will return the char code of the character in the position i. If your string is jim, then text.charCodeAt(0) will return the char code of j

                          – varun agarwal
                          Mar 26 at 18:07











                        • the position of 'j' in the alphabet is 10, however...

                          – Jim
                          Mar 26 at 18:10






                        • 1





                          Yeah, hence the - 96 to normalize the values.

                          – varun agarwal
                          Mar 26 at 18:12















                        2














                        You've complicated the solution, the simplest approach would be to just find the charcode and return that.



                        function alphabetPosition(text) 
                        let str = '';
                        for (var i = 0; i < text.length; i++)
                        str += (text[i] + (text.charCodeAt(i) - 96));

                        return str;






                        share|improve this answer























                        • how is str the positions of each letter of the string in the alphabet if you're using text.charCodeAt(i)?

                          – Jim
                          Mar 26 at 18:04











                        • text.charCodeAt(i) will return the char code of the character in the position i. If your string is jim, then text.charCodeAt(0) will return the char code of j

                          – varun agarwal
                          Mar 26 at 18:07











                        • the position of 'j' in the alphabet is 10, however...

                          – Jim
                          Mar 26 at 18:10






                        • 1





                          Yeah, hence the - 96 to normalize the values.

                          – varun agarwal
                          Mar 26 at 18:12













                        2












                        2








                        2







                        You've complicated the solution, the simplest approach would be to just find the charcode and return that.



                        function alphabetPosition(text) 
                        let str = '';
                        for (var i = 0; i < text.length; i++)
                        str += (text[i] + (text.charCodeAt(i) - 96));

                        return str;






                        share|improve this answer













                        You've complicated the solution, the simplest approach would be to just find the charcode and return that.



                        function alphabetPosition(text) 
                        let str = '';
                        for (var i = 0; i < text.length; i++)
                        str += (text[i] + (text.charCodeAt(i) - 96));

                        return str;







                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered Mar 26 at 17:54









                        varun agarwalvarun agarwal

                        1,0882 silver badges9 bronze badges




                        1,0882 silver badges9 bronze badges












                        • how is str the positions of each letter of the string in the alphabet if you're using text.charCodeAt(i)?

                          – Jim
                          Mar 26 at 18:04











                        • text.charCodeAt(i) will return the char code of the character in the position i. If your string is jim, then text.charCodeAt(0) will return the char code of j

                          – varun agarwal
                          Mar 26 at 18:07











                        • the position of 'j' in the alphabet is 10, however...

                          – Jim
                          Mar 26 at 18:10






                        • 1





                          Yeah, hence the - 96 to normalize the values.

                          – varun agarwal
                          Mar 26 at 18:12

















                        • how is str the positions of each letter of the string in the alphabet if you're using text.charCodeAt(i)?

                          – Jim
                          Mar 26 at 18:04











                        • text.charCodeAt(i) will return the char code of the character in the position i. If your string is jim, then text.charCodeAt(0) will return the char code of j

                          – varun agarwal
                          Mar 26 at 18:07











                        • the position of 'j' in the alphabet is 10, however...

                          – Jim
                          Mar 26 at 18:10






                        • 1





                          Yeah, hence the - 96 to normalize the values.

                          – varun agarwal
                          Mar 26 at 18:12
















                        how is str the positions of each letter of the string in the alphabet if you're using text.charCodeAt(i)?

                        – Jim
                        Mar 26 at 18:04





                        how is str the positions of each letter of the string in the alphabet if you're using text.charCodeAt(i)?

                        – Jim
                        Mar 26 at 18:04













                        text.charCodeAt(i) will return the char code of the character in the position i. If your string is jim, then text.charCodeAt(0) will return the char code of j

                        – varun agarwal
                        Mar 26 at 18:07





                        text.charCodeAt(i) will return the char code of the character in the position i. If your string is jim, then text.charCodeAt(0) will return the char code of j

                        – varun agarwal
                        Mar 26 at 18:07













                        the position of 'j' in the alphabet is 10, however...

                        – Jim
                        Mar 26 at 18:10





                        the position of 'j' in the alphabet is 10, however...

                        – Jim
                        Mar 26 at 18:10




                        1




                        1





                        Yeah, hence the - 96 to normalize the values.

                        – varun agarwal
                        Mar 26 at 18:12





                        Yeah, hence the - 96 to normalize the values.

                        – varun agarwal
                        Mar 26 at 18:12











                        2














                        I totally understand that is a coding challenge, interview question or likewise so if you really need to use map() you should only return the result of the callback passed to map as follows :



                        return arr.map(x => alphabet.indexOf(x) + 1).join(' ')


                        However reduce() seems more appropriate in your case :



                        return arr.reduce((ac, cv) => ac + (alphabet.indexOf(cv) + 1) + ' ', '')





                        share|improve this answer



























                          2














                          I totally understand that is a coding challenge, interview question or likewise so if you really need to use map() you should only return the result of the callback passed to map as follows :



                          return arr.map(x => alphabet.indexOf(x) + 1).join(' ')


                          However reduce() seems more appropriate in your case :



                          return arr.reduce((ac, cv) => ac + (alphabet.indexOf(cv) + 1) + ' ', '')





                          share|improve this answer

























                            2












                            2








                            2







                            I totally understand that is a coding challenge, interview question or likewise so if you really need to use map() you should only return the result of the callback passed to map as follows :



                            return arr.map(x => alphabet.indexOf(x) + 1).join(' ')


                            However reduce() seems more appropriate in your case :



                            return arr.reduce((ac, cv) => ac + (alphabet.indexOf(cv) + 1) + ' ', '')





                            share|improve this answer













                            I totally understand that is a coding challenge, interview question or likewise so if you really need to use map() you should only return the result of the callback passed to map as follows :



                            return arr.map(x => alphabet.indexOf(x) + 1).join(' ')


                            However reduce() seems more appropriate in your case :



                            return arr.reduce((ac, cv) => ac + (alphabet.indexOf(cv) + 1) + ' ', '')






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered Mar 26 at 17:56









                            BJRINTBJRINT

                            3761 silver badge8 bronze badges




                            3761 silver badge8 bronze badges



























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