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How to get the nth value of a function with arbitrary number of arguments


How to distribute a generic function of two arguments (without evaluating the arguments)How to take the derivative w.r.t. an arbitrary function?How can I get the number of slots in Function?General function taking general number of argumentsOptional arguments with lists as function parameterFunction with lazy and non-lazy arguments






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;








3














$begingroup$


I know that I can define a function of a List and get the nth argument like:



f[x_List]:=x[[1]]+x[[2]]


What if the argument of f is not a list? I.e. how to define an equivalent function for



f[x__Integer]:=?


This is probably extremely basic, but I couldn't figure it out from the Slot documentation.










share|improve this question










$endgroup$





















    3














    $begingroup$


    I know that I can define a function of a List and get the nth argument like:



    f[x_List]:=x[[1]]+x[[2]]


    What if the argument of f is not a list? I.e. how to define an equivalent function for



    f[x__Integer]:=?


    This is probably extremely basic, but I couldn't figure it out from the Slot documentation.










    share|improve this question










    $endgroup$

















      3












      3








      3





      $begingroup$


      I know that I can define a function of a List and get the nth argument like:



      f[x_List]:=x[[1]]+x[[2]]


      What if the argument of f is not a list? I.e. how to define an equivalent function for



      f[x__Integer]:=?


      This is probably extremely basic, but I couldn't figure it out from the Slot documentation.










      share|improve this question










      $endgroup$




      I know that I can define a function of a List and get the nth argument like:



      f[x_List]:=x[[1]]+x[[2]]


      What if the argument of f is not a list? I.e. how to define an equivalent function for



      f[x__Integer]:=?


      This is probably extremely basic, but I couldn't figure it out from the Slot documentation.







      functions






      share|improve this question














      share|improve this question











      share|improve this question




      share|improve this question










      asked Oct 14 at 11:40









      user366202user366202

      755 bronze badges




      755 bronze badges























          2 Answers
          2






          active

          oldest

          votes


















          4
















          $begingroup$

          With f[x__Integer] := ... you can use



          x[[i]]


          to get the ith argument.






          share|improve this answer










          $endgroup$






















            3
















            $begingroup$

            Szabolcs's answer (wrapping the sequence x with List and using Part) is the way.



            An alternative is to use Indexed



            ClearAll[f2]

            f2[x__Integer] := Indexed[x, 1] + Indexed[x, 2]
            f2[1, 2, 3, 4, 5]



            3




            You can also use Slot (#):



            ClearAll[f3, f4]

            f3[x__Integer] := Slot[1] + Slot[2] &[x]
            f3[1, 2, 3, 4, 5]



            3




            f4[x__Integer] := #1 + #2 &[x]
            f4[1, 2, 3, 4, 5]



            3




            but you can use a pure function directly to define your function:



            f5 = #1 + #2 &





            share|improve this answer










            $endgroup$














            • $begingroup$
              Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
              $endgroup$
              – user366202
              Oct 14 at 14:24










            • $begingroup$
              @user366202, I am not sure.
              $endgroup$
              – kglr
              Oct 14 at 14:26












            Your Answer








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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            4
















            $begingroup$

            With f[x__Integer] := ... you can use



            x[[i]]


            to get the ith argument.






            share|improve this answer










            $endgroup$



















              4
















              $begingroup$

              With f[x__Integer] := ... you can use



              x[[i]]


              to get the ith argument.






              share|improve this answer










              $endgroup$

















                4














                4










                4







                $begingroup$

                With f[x__Integer] := ... you can use



                x[[i]]


                to get the ith argument.






                share|improve this answer










                $endgroup$



                With f[x__Integer] := ... you can use



                x[[i]]


                to get the ith argument.







                share|improve this answer













                share|improve this answer




                share|improve this answer










                answered Oct 14 at 11:43









                SzabolcsSzabolcs

                173k18 gold badges473 silver badges1008 bronze badges




                173k18 gold badges473 silver badges1008 bronze badges


























                    3
















                    $begingroup$

                    Szabolcs's answer (wrapping the sequence x with List and using Part) is the way.



                    An alternative is to use Indexed



                    ClearAll[f2]

                    f2[x__Integer] := Indexed[x, 1] + Indexed[x, 2]
                    f2[1, 2, 3, 4, 5]



                    3




                    You can also use Slot (#):



                    ClearAll[f3, f4]

                    f3[x__Integer] := Slot[1] + Slot[2] &[x]
                    f3[1, 2, 3, 4, 5]



                    3




                    f4[x__Integer] := #1 + #2 &[x]
                    f4[1, 2, 3, 4, 5]



                    3




                    but you can use a pure function directly to define your function:



                    f5 = #1 + #2 &





                    share|improve this answer










                    $endgroup$














                    • $begingroup$
                      Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
                      $endgroup$
                      – user366202
                      Oct 14 at 14:24










                    • $begingroup$
                      @user366202, I am not sure.
                      $endgroup$
                      – kglr
                      Oct 14 at 14:26















                    3
















                    $begingroup$

                    Szabolcs's answer (wrapping the sequence x with List and using Part) is the way.



                    An alternative is to use Indexed



                    ClearAll[f2]

                    f2[x__Integer] := Indexed[x, 1] + Indexed[x, 2]
                    f2[1, 2, 3, 4, 5]



                    3




                    You can also use Slot (#):



                    ClearAll[f3, f4]

                    f3[x__Integer] := Slot[1] + Slot[2] &[x]
                    f3[1, 2, 3, 4, 5]



                    3




                    f4[x__Integer] := #1 + #2 &[x]
                    f4[1, 2, 3, 4, 5]



                    3




                    but you can use a pure function directly to define your function:



                    f5 = #1 + #2 &





                    share|improve this answer










                    $endgroup$














                    • $begingroup$
                      Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
                      $endgroup$
                      – user366202
                      Oct 14 at 14:24










                    • $begingroup$
                      @user366202, I am not sure.
                      $endgroup$
                      – kglr
                      Oct 14 at 14:26













                    3














                    3










                    3







                    $begingroup$

                    Szabolcs's answer (wrapping the sequence x with List and using Part) is the way.



                    An alternative is to use Indexed



                    ClearAll[f2]

                    f2[x__Integer] := Indexed[x, 1] + Indexed[x, 2]
                    f2[1, 2, 3, 4, 5]



                    3




                    You can also use Slot (#):



                    ClearAll[f3, f4]

                    f3[x__Integer] := Slot[1] + Slot[2] &[x]
                    f3[1, 2, 3, 4, 5]



                    3




                    f4[x__Integer] := #1 + #2 &[x]
                    f4[1, 2, 3, 4, 5]



                    3




                    but you can use a pure function directly to define your function:



                    f5 = #1 + #2 &





                    share|improve this answer










                    $endgroup$



                    Szabolcs's answer (wrapping the sequence x with List and using Part) is the way.



                    An alternative is to use Indexed



                    ClearAll[f2]

                    f2[x__Integer] := Indexed[x, 1] + Indexed[x, 2]
                    f2[1, 2, 3, 4, 5]



                    3




                    You can also use Slot (#):



                    ClearAll[f3, f4]

                    f3[x__Integer] := Slot[1] + Slot[2] &[x]
                    f3[1, 2, 3, 4, 5]



                    3




                    f4[x__Integer] := #1 + #2 &[x]
                    f4[1, 2, 3, 4, 5]



                    3




                    but you can use a pure function directly to define your function:



                    f5 = #1 + #2 &






                    share|improve this answer













                    share|improve this answer




                    share|improve this answer










                    answered Oct 14 at 12:24









                    kglrkglr

                    223k10 gold badges253 silver badges511 bronze badges




                    223k10 gold badges253 silver badges511 bronze badges














                    • $begingroup$
                      Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
                      $endgroup$
                      – user366202
                      Oct 14 at 14:24










                    • $begingroup$
                      @user366202, I am not sure.
                      $endgroup$
                      – kglr
                      Oct 14 at 14:26
















                    • $begingroup$
                      Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
                      $endgroup$
                      – user366202
                      Oct 14 at 14:24










                    • $begingroup$
                      @user366202, I am not sure.
                      $endgroup$
                      – kglr
                      Oct 14 at 14:26















                    $begingroup$
                    Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
                    $endgroup$
                    – user366202
                    Oct 14 at 14:24




                    $begingroup$
                    Thanks. Is the pure function method likely to be faster than the Indexed/List methods?
                    $endgroup$
                    – user366202
                    Oct 14 at 14:24












                    $begingroup$
                    @user366202, I am not sure.
                    $endgroup$
                    – kglr
                    Oct 14 at 14:26




                    $begingroup$
                    @user366202, I am not sure.
                    $endgroup$
                    – kglr
                    Oct 14 at 14:26


















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