Concatenate to a list of string another string





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The below code gives back a list of String but I want it work on multiple cases. The problem is that I can't create the same exact result with recursion.
The program gives back the following result:



replaceTabs 6 ["thello world"]

=> [" hello world"]


Now this should work with a longer list like:



replaceTabs 6 ["asd dsa","thello world"]

=> ["asd dsa"," hello world"]


Simple concat doesn't work, because it will give back undefined pattern.



replaceTab' :: Int -> [[Char]] -> [Char]
replaceTab' n [[x]] =
if x == 't' then replicate n ' '
else [x]

replaceTabs :: Int -> [String]-> [String]
replaceTabs n [""] = [""]
replaceTabs n (x:xs) = (return . concat $ [replaceTab' n [a] | a <- (map (:) (x))])









share|improve this question





























    0















    The below code gives back a list of String but I want it work on multiple cases. The problem is that I can't create the same exact result with recursion.
    The program gives back the following result:



    replaceTabs 6 ["thello world"]

    => [" hello world"]


    Now this should work with a longer list like:



    replaceTabs 6 ["asd dsa","thello world"]

    => ["asd dsa"," hello world"]


    Simple concat doesn't work, because it will give back undefined pattern.



    replaceTab' :: Int -> [[Char]] -> [Char]
    replaceTab' n [[x]] =
    if x == 't' then replicate n ' '
    else [x]

    replaceTabs :: Int -> [String]-> [String]
    replaceTabs n [""] = [""]
    replaceTabs n (x:xs) = (return . concat $ [replaceTab' n [a] | a <- (map (:) (x))])









    share|improve this question

























      0












      0








      0








      The below code gives back a list of String but I want it work on multiple cases. The problem is that I can't create the same exact result with recursion.
      The program gives back the following result:



      replaceTabs 6 ["thello world"]

      => [" hello world"]


      Now this should work with a longer list like:



      replaceTabs 6 ["asd dsa","thello world"]

      => ["asd dsa"," hello world"]


      Simple concat doesn't work, because it will give back undefined pattern.



      replaceTab' :: Int -> [[Char]] -> [Char]
      replaceTab' n [[x]] =
      if x == 't' then replicate n ' '
      else [x]

      replaceTabs :: Int -> [String]-> [String]
      replaceTabs n [""] = [""]
      replaceTabs n (x:xs) = (return . concat $ [replaceTab' n [a] | a <- (map (:) (x))])









      share|improve this question














      The below code gives back a list of String but I want it work on multiple cases. The problem is that I can't create the same exact result with recursion.
      The program gives back the following result:



      replaceTabs 6 ["thello world"]

      => [" hello world"]


      Now this should work with a longer list like:



      replaceTabs 6 ["asd dsa","thello world"]

      => ["asd dsa"," hello world"]


      Simple concat doesn't work, because it will give back undefined pattern.



      replaceTab' :: Int -> [[Char]] -> [Char]
      replaceTab' n [[x]] =
      if x == 't' then replicate n ' '
      else [x]

      replaceTabs :: Int -> [String]-> [String]
      replaceTabs n [""] = [""]
      replaceTabs n (x:xs) = (return . concat $ [replaceTab' n [a] | a <- (map (:) (x))])






      string list haskell concat






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      share|improve this question










      asked Nov 24 '18 at 17:30









      Akos FajsziAkos Fajszi

      445




      445
























          1 Answer
          1






          active

          oldest

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          This



          replaceTab' :: Int -> [[Char]] -> [Char]


          is the same as,



          replaceTab' :: Int -> [String] -> String


          What you should focus on is implementing a function,



          replaceTab :: Int -> String -> String


          which "fixes" a single String. Then replaceTabs is simply,



          replaceTabs :: Int -> [String] -> [String]
          replaceTabs n = map (replaceTab n)





          share|improve this answer
























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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            This



            replaceTab' :: Int -> [[Char]] -> [Char]


            is the same as,



            replaceTab' :: Int -> [String] -> String


            What you should focus on is implementing a function,



            replaceTab :: Int -> String -> String


            which "fixes" a single String. Then replaceTabs is simply,



            replaceTabs :: Int -> [String] -> [String]
            replaceTabs n = map (replaceTab n)





            share|improve this answer




























              3














              This



              replaceTab' :: Int -> [[Char]] -> [Char]


              is the same as,



              replaceTab' :: Int -> [String] -> String


              What you should focus on is implementing a function,



              replaceTab :: Int -> String -> String


              which "fixes" a single String. Then replaceTabs is simply,



              replaceTabs :: Int -> [String] -> [String]
              replaceTabs n = map (replaceTab n)





              share|improve this answer


























                3












                3








                3







                This



                replaceTab' :: Int -> [[Char]] -> [Char]


                is the same as,



                replaceTab' :: Int -> [String] -> String


                What you should focus on is implementing a function,



                replaceTab :: Int -> String -> String


                which "fixes" a single String. Then replaceTabs is simply,



                replaceTabs :: Int -> [String] -> [String]
                replaceTabs n = map (replaceTab n)





                share|improve this answer













                This



                replaceTab' :: Int -> [[Char]] -> [Char]


                is the same as,



                replaceTab' :: Int -> [String] -> String


                What you should focus on is implementing a function,



                replaceTab :: Int -> String -> String


                which "fixes" a single String. Then replaceTabs is simply,



                replaceTabs :: Int -> [String] -> [String]
                replaceTabs n = map (replaceTab n)






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 24 '18 at 17:43









                Jorge AdrianoJorge Adriano

                2,219919




                2,219919
































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