How to efficiently check conditions inside a function?

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3
down vote

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Consider a function as



f[n_]:=Module[x,y, 
x=Table[Sin[i],i,0,Pi,Pi/(n-1)];
y=Table[Cos[i],i,0,Pi,Pi/(n-1)];
Transpose[x,y]
]


Now, this function will work for all the values of n>1. However, for n=1, I want the function to return 1,1. If I use If-else inside the function, I get the result, but, it gives me the error message as well.



How can I resolve this issue?






function-construction







share|improve this question

















  • 2




    To be structurally consistent, f[1] should be 1,1, i.e., a list of points. To be efficient, f[1]=1,1; f[n_Integer]:=Sin[#], Cos[#]& /@ Range[0, Pi, Pi/(n-1)]
    – Bob Hanlon
    1 hour ago














up vote
3
down vote

favorite












Consider a function as



f[n_]:=Module[x,y, 
x=Table[Sin[i],i,0,Pi,Pi/(n-1)];
y=Table[Cos[i],i,0,Pi,Pi/(n-1)];
Transpose[x,y]
]


Now, this function will work for all the values of n>1. However, for n=1, I want the function to return 1,1. If I use If-else inside the function, I get the result, but, it gives me the error message as well.



How can I resolve this issue?






function-construction







share|improve this question

















  • 2




    To be structurally consistent, f[1] should be 1,1, i.e., a list of points. To be efficient, f[1]=1,1; f[n_Integer]:=Sin[#], Cos[#]& /@ Range[0, Pi, Pi/(n-1)]
    – Bob Hanlon
    1 hour ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











Consider a function as



f[n_]:=Module[x,y, 
x=Table[Sin[i],i,0,Pi,Pi/(n-1)];
y=Table[Cos[i],i,0,Pi,Pi/(n-1)];
Transpose[x,y]
]


Now, this function will work for all the values of n>1. However, for n=1, I want the function to return 1,1. If I use If-else inside the function, I get the result, but, it gives me the error message as well.



How can I resolve this issue?






function-construction







share|improve this question













Consider a function as



f[n_]:=Module[x,y, 
x=Table[Sin[i],i,0,Pi,Pi/(n-1)];
y=Table[Cos[i],i,0,Pi,Pi/(n-1)];
Transpose[x,y]
]


Now, this function will work for all the values of n>1. However, for n=1, I want the function to return 1,1. If I use If-else inside the function, I get the result, but, it gives me the error message as well.



How can I resolve this issue?






function-construction




function-construction






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 2 hours ago









Majis

1,288313




1,288313







  • 2




    To be structurally consistent, f[1] should be 1,1, i.e., a list of points. To be efficient, f[1]=1,1; f[n_Integer]:=Sin[#], Cos[#]& /@ Range[0, Pi, Pi/(n-1)]
    – Bob Hanlon
    1 hour ago












  • 2




    To be structurally consistent, f[1] should be 1,1, i.e., a list of points. To be efficient, f[1]=1,1; f[n_Integer]:=Sin[#], Cos[#]& /@ Range[0, Pi, Pi/(n-1)]
    – Bob Hanlon
    1 hour ago







2




2




To be structurally consistent, f[1] should be 1,1, i.e., a list of points. To be efficient, f[1]=1,1; f[n_Integer]:=Sin[#], Cos[#]& /@ Range[0, Pi, Pi/(n-1)]
– Bob Hanlon
1 hour ago




To be structurally consistent, f[1] should be 1,1, i.e., a list of points. To be efficient, f[1]=1,1; f[n_Integer]:=Sin[#], Cos[#]& /@ Range[0, Pi, Pi/(n-1)]
– Bob Hanlon
1 hour ago










3 Answers
3






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up vote
3
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ClearAll[f]
f[n_] := Transpose[Table[Sin[i], i, n /. 1 -> π / 2, _ :> 0, π, π/(n - 1)], 
 Table[Cos[i], i, 0, π, π/(n - 1)]]

f[1]



1, 1




f[3]



0, 1, 1, 0, 0, -1







share|improve this answer



























    up vote
    2
    down vote













    f[n_] := Module[x, y, 
     If[n == 1, 1, 1, 
     x = Table[Sin[i], i, 0, Pi, Pi/(n - 1)]; 
     y = Table[Cos[i], i, 0, Pi, Pi/(n - 1)]; 
     Transpose[x, y]]]

    f[1]
    (* 1,1 *)





    share|improve this answer






















    • Thanks. It's really a great way. Exactly the kind of solution I was looking for.
      – Majis
      2 hours ago

















    up vote
    2
    down vote













    Mathematica is an expression rewriting language. Use that (and eliminate unnecessary code):



    f[n_] := Transpose[Table[Sin[i], i, 0, Pi, Pi/(n - 1)],
     Table[Cos[i], i, 0, Pi, Pi/(n - 1)]]
    f[1] = 1, 1;


    When more than one rewrite is possible, Mathematica prefers the one with the more specific trigger, f[1] in this case






    share|improve this answer




















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      3 Answers
      3






      active

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      3 Answers
      3






      active

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      up vote
      3
      down vote













      ClearAll[f]
      f[n_] := Transpose[Table[Sin[i], i, n /. 1 -> π / 2, _ :> 0, π, π/(n - 1)], 
       Table[Cos[i], i, 0, π, π/(n - 1)]]

      f[1]



      1, 1




      f[3]



      0, 1, 1, 0, 0, -1







      share|improve this answer
























        up vote
        3
        down vote













        ClearAll[f]
        f[n_] := Transpose[Table[Sin[i], i, n /. 1 -> π / 2, _ :> 0, π, π/(n - 1)], 
         Table[Cos[i], i, 0, π, π/(n - 1)]]

        f[1]



        1, 1




        f[3]



        0, 1, 1, 0, 0, -1







        share|improve this answer






















          up vote
          3
          down vote










          up vote
          3
          down vote









          ClearAll[f]
          f[n_] := Transpose[Table[Sin[i], i, n /. 1 -> π / 2, _ :> 0, π, π/(n - 1)], 
           Table[Cos[i], i, 0, π, π/(n - 1)]]

          f[1]



          1, 1




          f[3]



          0, 1, 1, 0, 0, -1







          share|improve this answer












          ClearAll[f]
          f[n_] := Transpose[Table[Sin[i], i, n /. 1 -> π / 2, _ :> 0, π, π/(n - 1)], 
           Table[Cos[i], i, 0, π, π/(n - 1)]]

          f[1]



          1, 1




          f[3]



          0, 1, 1, 0, 0, -1








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 33 mins ago









          kglr

          161k8184384




          161k8184384




















              up vote
              2
              down vote













              f[n_] := Module[x, y, 
               If[n == 1, 1, 1, 
               x = Table[Sin[i], i, 0, Pi, Pi/(n - 1)]; 
               y = Table[Cos[i], i, 0, Pi, Pi/(n - 1)]; 
               Transpose[x, y]]]

              f[1]
              (* 1,1 *)





              share|improve this answer






















              • Thanks. It's really a great way. Exactly the kind of solution I was looking for.
                – Majis
                2 hours ago














              up vote
              2
              down vote













              f[n_] := Module[x, y, 
               If[n == 1, 1, 1, 
               x = Table[Sin[i], i, 0, Pi, Pi/(n - 1)]; 
               y = Table[Cos[i], i, 0, Pi, Pi/(n - 1)]; 
               Transpose[x, y]]]

              f[1]
              (* 1,1 *)





              share|improve this answer






















              • Thanks. It's really a great way. Exactly the kind of solution I was looking for.
                – Majis
                2 hours ago












              up vote
              2
              down vote










              up vote
              2
              down vote









              f[n_] := Module[x, y, 
               If[n == 1, 1, 1, 
               x = Table[Sin[i], i, 0, Pi, Pi/(n - 1)]; 
               y = Table[Cos[i], i, 0, Pi, Pi/(n - 1)]; 
               Transpose[x, y]]]

              f[1]
              (* 1,1 *)





              share|improve this answer














              f[n_] := Module[x, y, 
               If[n == 1, 1, 1, 
               x = Table[Sin[i], i, 0, Pi, Pi/(n - 1)]; 
               y = Table[Cos[i], i, 0, Pi, Pi/(n - 1)]; 
               Transpose[x, y]]]

              f[1]
              (* 1,1 *)






              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 2 hours ago

























              answered 2 hours ago









              Mariusz Iwaniuk

              6,17311026




              6,17311026











              • Thanks. It's really a great way. Exactly the kind of solution I was looking for.
                – Majis
                2 hours ago
















              • Thanks. It's really a great way. Exactly the kind of solution I was looking for.
                – Majis
                2 hours ago















              Thanks. It's really a great way. Exactly the kind of solution I was looking for.
              – Majis
              2 hours ago




              Thanks. It's really a great way. Exactly the kind of solution I was looking for.
              – Majis
              2 hours ago










              up vote
              2
              down vote













              Mathematica is an expression rewriting language. Use that (and eliminate unnecessary code):



              f[n_] := Transpose[Table[Sin[i], i, 0, Pi, Pi/(n - 1)],
               Table[Cos[i], i, 0, Pi, Pi/(n - 1)]]
              f[1] = 1, 1;


              When more than one rewrite is possible, Mathematica prefers the one with the more specific trigger, f[1] in this case






              share|improve this answer
























                up vote
                2
                down vote













                Mathematica is an expression rewriting language. Use that (and eliminate unnecessary code):



                f[n_] := Transpose[Table[Sin[i], i, 0, Pi, Pi/(n - 1)],
                 Table[Cos[i], i, 0, Pi, Pi/(n - 1)]]
                f[1] = 1, 1;


                When more than one rewrite is possible, Mathematica prefers the one with the more specific trigger, f[1] in this case






                share|improve this answer






















                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Mathematica is an expression rewriting language. Use that (and eliminate unnecessary code):



                  f[n_] := Transpose[Table[Sin[i], i, 0, Pi, Pi/(n - 1)],
                   Table[Cos[i], i, 0, Pi, Pi/(n - 1)]]
                  f[1] = 1, 1;


                  When more than one rewrite is possible, Mathematica prefers the one with the more specific trigger, f[1] in this case






                  share|improve this answer












                  Mathematica is an expression rewriting language. Use that (and eliminate unnecessary code):



                  f[n_] := Transpose[Table[Sin[i], i, 0, Pi, Pi/(n - 1)],
                   Table[Cos[i], i, 0, Pi, Pi/(n - 1)]]
                  f[1] = 1, 1;


                  When more than one rewrite is possible, Mathematica prefers the one with the more specific trigger, f[1] in this case







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 1 hour ago









                  John Doty

                  6,1491924




                  6,1491924



























                       

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