remove first occurance of element from list












1












$begingroup$


As explained here i'm learning a limited subset of scheme.



My task has been to sort a numerical list. To implement a simple insertion sort, i need to remove a single element from a list.



I've done the following (in the spirit of The Little Schemer)



(define remove-elem
(lambda (ll elem)
(cond
((null? ll) '())
((eq? (car ll) elem) (cdr ll))
(else
(cons (car ll) (remove-elem (cdr ll) elem))))
))


What are the other ways of implementing this function?










share|improve this question











$endgroup$

















    1












    $begingroup$


    As explained here i'm learning a limited subset of scheme.



    My task has been to sort a numerical list. To implement a simple insertion sort, i need to remove a single element from a list.



    I've done the following (in the spirit of The Little Schemer)



    (define remove-elem
    (lambda (ll elem)
    (cond
    ((null? ll) '())
    ((eq? (car ll) elem) (cdr ll))
    (else
    (cons (car ll) (remove-elem (cdr ll) elem))))
    ))


    What are the other ways of implementing this function?










    share|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      As explained here i'm learning a limited subset of scheme.



      My task has been to sort a numerical list. To implement a simple insertion sort, i need to remove a single element from a list.



      I've done the following (in the spirit of The Little Schemer)



      (define remove-elem
      (lambda (ll elem)
      (cond
      ((null? ll) '())
      ((eq? (car ll) elem) (cdr ll))
      (else
      (cons (car ll) (remove-elem (cdr ll) elem))))
      ))


      What are the other ways of implementing this function?










      share|improve this question











      $endgroup$




      As explained here i'm learning a limited subset of scheme.



      My task has been to sort a numerical list. To implement a simple insertion sort, i need to remove a single element from a list.



      I've done the following (in the spirit of The Little Schemer)



      (define remove-elem
      (lambda (ll elem)
      (cond
      ((null? ll) '())
      ((eq? (car ll) elem) (cdr ll))
      (else
      (cons (car ll) (remove-elem (cdr ll) elem))))
      ))


      What are the other ways of implementing this function?







      scheme






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Apr 13 '17 at 12:40









      Community

      1




      1










      asked Oct 29 '13 at 22:17









      ricardoricardo

      185210




      185210






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          That looks mostly right to me. I'll suggest two things.



          First, there is nice syntatic sugar for defining functions:



          (define (remove-elem xs elem) 
          (cond ...)
          )


          Save yourself the lambda.



          Second, you are using eq?. That is the wrong checker for this problem. That function returns true if the two objects are the exact same object in memory. It could be true for primitives like ints, but it's not necessarily true. What you want to use is eqv? which will guaranteed work for numbers.






          share|improve this answer









          $endgroup$





















            0












            $begingroup$

            How would you change this code using foldr, map, and filter?





            share








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            Farzad Saeidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              That looks mostly right to me. I'll suggest two things.



              First, there is nice syntatic sugar for defining functions:



              (define (remove-elem xs elem) 
              (cond ...)
              )


              Save yourself the lambda.



              Second, you are using eq?. That is the wrong checker for this problem. That function returns true if the two objects are the exact same object in memory. It could be true for primitives like ints, but it's not necessarily true. What you want to use is eqv? which will guaranteed work for numbers.






              share|improve this answer









              $endgroup$


















                2












                $begingroup$

                That looks mostly right to me. I'll suggest two things.



                First, there is nice syntatic sugar for defining functions:



                (define (remove-elem xs elem) 
                (cond ...)
                )


                Save yourself the lambda.



                Second, you are using eq?. That is the wrong checker for this problem. That function returns true if the two objects are the exact same object in memory. It could be true for primitives like ints, but it's not necessarily true. What you want to use is eqv? which will guaranteed work for numbers.






                share|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  That looks mostly right to me. I'll suggest two things.



                  First, there is nice syntatic sugar for defining functions:



                  (define (remove-elem xs elem) 
                  (cond ...)
                  )


                  Save yourself the lambda.



                  Second, you are using eq?. That is the wrong checker for this problem. That function returns true if the two objects are the exact same object in memory. It could be true for primitives like ints, but it's not necessarily true. What you want to use is eqv? which will guaranteed work for numbers.






                  share|improve this answer









                  $endgroup$



                  That looks mostly right to me. I'll suggest two things.



                  First, there is nice syntatic sugar for defining functions:



                  (define (remove-elem xs elem) 
                  (cond ...)
                  )


                  Save yourself the lambda.



                  Second, you are using eq?. That is the wrong checker for this problem. That function returns true if the two objects are the exact same object in memory. It could be true for primitives like ints, but it's not necessarily true. What you want to use is eqv? which will guaranteed work for numbers.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Oct 29 '13 at 23:06









                  BarryBarry

                  17.2k13087




                  17.2k13087

























                      0












                      $begingroup$

                      How would you change this code using foldr, map, and filter?





                      share








                      New contributor




                      Farzad Saeidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      $endgroup$


















                        0












                        $begingroup$

                        How would you change this code using foldr, map, and filter?





                        share








                        New contributor




                        Farzad Saeidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.






                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          How would you change this code using foldr, map, and filter?





                          share








                          New contributor




                          Farzad Saeidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          $endgroup$



                          How would you change this code using foldr, map, and filter?






                          share








                          New contributor




                          Farzad Saeidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.








                          share


                          share






                          New contributor




                          Farzad Saeidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          answered 1 min ago









                          Farzad SaeidiFarzad Saeidi

                          1




                          1




                          New contributor




                          Farzad Saeidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.





                          New contributor





                          Farzad Saeidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          Farzad Saeidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






























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