Making a list of patterns of arbitrary length










10












$begingroup$


I would like to create a list of arbitrary length:



x1_, x2_, ...


Where each of the elements of the list has the full form:



Pattern[xi, Blank]


This answer shows how to create a list of symbols:



x1, x2, ...


but I don't know how to adapt that to obtain the above.



I intend to use this list in the definition of a function as in here.










share|improve this question











$endgroup$
















    10












    $begingroup$


    I would like to create a list of arbitrary length:



    x1_, x2_, ...


    Where each of the elements of the list has the full form:



    Pattern[xi, Blank]


    This answer shows how to create a list of symbols:



    x1, x2, ...


    but I don't know how to adapt that to obtain the above.



    I intend to use this list in the definition of a function as in here.










    share|improve this question











    $endgroup$














      10












      10








      10


      1



      $begingroup$


      I would like to create a list of arbitrary length:



      x1_, x2_, ...


      Where each of the elements of the list has the full form:



      Pattern[xi, Blank]


      This answer shows how to create a list of symbols:



      x1, x2, ...


      but I don't know how to adapt that to obtain the above.



      I intend to use this list in the definition of a function as in here.










      share|improve this question











      $endgroup$




      I would like to create a list of arbitrary length:



      x1_, x2_, ...


      Where each of the elements of the list has the full form:



      Pattern[xi, Blank]


      This answer shows how to create a list of symbols:



      x1, x2, ...


      but I don't know how to adapt that to obtain the above.



      I intend to use this list in the definition of a function as in here.







      list-manipulation functions pattern-matching






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Aug 28 '18 at 2:18









      m_goldberg

      87.6k872198




      87.6k872198










      asked Aug 28 '18 at 0:45









      WinkelriedWinkelried

      1634




      1634




















          4 Answers
          4






          active

          oldest

          votes


















          8












          $begingroup$

          Array[ToExpression["x" <> ToString @ # <> "_"] &, 5]



          x1_, x2_, x3_, x4_, x5_




          FullForm @ %



          List[Pattern[x1,Blank], Pattern[x2, Blank], Pattern[x3, Blank], Pattern[x4, Blank], Pattern[x5, Blank]]




          Also



          Thread[Pattern[Evaluate@Array[Symbol["x" <> ToString@#] &, 5], Blank]]



          x1_, x2_, x3_, x4_, x5_




          and



          ToExpression[Table["x" <> i <> "_", i, ToString /@ Range[5]]]



          x1_, x2_, x3_, x4_, x5_







          share|improve this answer











          $endgroup$








          • 1




            $begingroup$
            The middle one will fail if x1=1.
            $endgroup$
            – Kuba
            Aug 28 '18 at 7:37











          • $begingroup$
            @Kuba, good point.
            $endgroup$
            – kglr
            Aug 28 '18 at 8:09


















          8












          $begingroup$

          If you construct a list of strings instead, you can take advantage of the fact that ToExpression is listable, and supports a 3rd argument that post-processes the output. For example:



          ToExpression[
          "x1", "x2", "x3",
          StandardForm,
          Pattern[#,Blank]&
          ]



          x1_, x2_, x3_




          Or, creating the list and converting:



          ToExpression[
          Table["x" <> ToString@i, i, 5],
          StandardForm,
          Pattern[#, Blank]&
          ]



          x1_, x2_, x3_, x4_, x5_







          share|improve this answer









          $endgroup$








          • 2




            $begingroup$
            It won't work if x1 = 1.
            $endgroup$
            – Kuba
            Aug 28 '18 at 7:36










          • $begingroup$
            Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
            $endgroup$
            – Anton.Sakovich
            Aug 28 '18 at 7:44










          • $begingroup$
            @Kuba what exactly do you mean by x1 = 1 ?
            $endgroup$
            – Winkelried
            Aug 28 '18 at 8:57










          • $begingroup$
            @Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
            $endgroup$
            – Kuba
            Aug 28 '18 at 8:58


















          6












          $begingroup$

          Nothing new but shorter:



          StringTemplate["x``_"] /@ Range[10] // ToExpression



          x1_, x2_, x3_, x4_, x5_, x6_, x7_, x8_, x9_, x10_






          share|improve this answer









          $endgroup$




















            3












            $begingroup$

            Here is an example of how to make the solution in the link work for this case:



            patt = Table[
            With[
            s = Symbol["x" <> ToString[i]],
            Pattern[s, Blank]
            ], i, 10];

            Range[10] /. patt :> x5, x8



            5, 8




            Using With here is a trick to insert the symbol into Pattern. Since Pattern has the attribute HoldFirst, it would not work to write e.g.



            Pattern[Symbol["x" <> ToString[i]], Blank]


            because Symbol["x" <> ToString[i]] would not be evaluated before it was passed to Pattern, i.e. Pattern would not receive a string as is required.






            share|improve this answer











            $endgroup$












            • $begingroup$
              It won't work if x1 = 1
              $endgroup$
              – Kuba
              Aug 28 '18 at 7:36










            • $begingroup$
              @Kuba You mean if the symbols have values?
              $endgroup$
              – C. E.
              Aug 28 '18 at 9:48










            • $begingroup$
              Yes, sorry for not being clear.
              $endgroup$
              – Kuba
              Aug 28 '18 at 9:54










            • $begingroup$
              @Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
              $endgroup$
              – C. E.
              Aug 28 '18 at 10:40










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            4 Answers
            4






            active

            oldest

            votes








            4 Answers
            4






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            8












            $begingroup$

            Array[ToExpression["x" <> ToString @ # <> "_"] &, 5]



            x1_, x2_, x3_, x4_, x5_




            FullForm @ %



            List[Pattern[x1,Blank], Pattern[x2, Blank], Pattern[x3, Blank], Pattern[x4, Blank], Pattern[x5, Blank]]




            Also



            Thread[Pattern[Evaluate@Array[Symbol["x" <> ToString@#] &, 5], Blank]]



            x1_, x2_, x3_, x4_, x5_




            and



            ToExpression[Table["x" <> i <> "_", i, ToString /@ Range[5]]]



            x1_, x2_, x3_, x4_, x5_







            share|improve this answer











            $endgroup$








            • 1




              $begingroup$
              The middle one will fail if x1=1.
              $endgroup$
              – Kuba
              Aug 28 '18 at 7:37











            • $begingroup$
              @Kuba, good point.
              $endgroup$
              – kglr
              Aug 28 '18 at 8:09















            8












            $begingroup$

            Array[ToExpression["x" <> ToString @ # <> "_"] &, 5]



            x1_, x2_, x3_, x4_, x5_




            FullForm @ %



            List[Pattern[x1,Blank], Pattern[x2, Blank], Pattern[x3, Blank], Pattern[x4, Blank], Pattern[x5, Blank]]




            Also



            Thread[Pattern[Evaluate@Array[Symbol["x" <> ToString@#] &, 5], Blank]]



            x1_, x2_, x3_, x4_, x5_




            and



            ToExpression[Table["x" <> i <> "_", i, ToString /@ Range[5]]]



            x1_, x2_, x3_, x4_, x5_







            share|improve this answer











            $endgroup$








            • 1




              $begingroup$
              The middle one will fail if x1=1.
              $endgroup$
              – Kuba
              Aug 28 '18 at 7:37











            • $begingroup$
              @Kuba, good point.
              $endgroup$
              – kglr
              Aug 28 '18 at 8:09













            8












            8








            8





            $begingroup$

            Array[ToExpression["x" <> ToString @ # <> "_"] &, 5]



            x1_, x2_, x3_, x4_, x5_




            FullForm @ %



            List[Pattern[x1,Blank], Pattern[x2, Blank], Pattern[x3, Blank], Pattern[x4, Blank], Pattern[x5, Blank]]




            Also



            Thread[Pattern[Evaluate@Array[Symbol["x" <> ToString@#] &, 5], Blank]]



            x1_, x2_, x3_, x4_, x5_




            and



            ToExpression[Table["x" <> i <> "_", i, ToString /@ Range[5]]]



            x1_, x2_, x3_, x4_, x5_







            share|improve this answer











            $endgroup$



            Array[ToExpression["x" <> ToString @ # <> "_"] &, 5]



            x1_, x2_, x3_, x4_, x5_




            FullForm @ %



            List[Pattern[x1,Blank], Pattern[x2, Blank], Pattern[x3, Blank], Pattern[x4, Blank], Pattern[x5, Blank]]




            Also



            Thread[Pattern[Evaluate@Array[Symbol["x" <> ToString@#] &, 5], Blank]]



            x1_, x2_, x3_, x4_, x5_




            and



            ToExpression[Table["x" <> i <> "_", i, ToString /@ Range[5]]]



            x1_, x2_, x3_, x4_, x5_








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Aug 28 '18 at 4:10

























            answered Aug 28 '18 at 1:54









            kglrkglr

            189k10205423




            189k10205423







            • 1




              $begingroup$
              The middle one will fail if x1=1.
              $endgroup$
              – Kuba
              Aug 28 '18 at 7:37











            • $begingroup$
              @Kuba, good point.
              $endgroup$
              – kglr
              Aug 28 '18 at 8:09












            • 1




              $begingroup$
              The middle one will fail if x1=1.
              $endgroup$
              – Kuba
              Aug 28 '18 at 7:37











            • $begingroup$
              @Kuba, good point.
              $endgroup$
              – kglr
              Aug 28 '18 at 8:09







            1




            1




            $begingroup$
            The middle one will fail if x1=1.
            $endgroup$
            – Kuba
            Aug 28 '18 at 7:37





            $begingroup$
            The middle one will fail if x1=1.
            $endgroup$
            – Kuba
            Aug 28 '18 at 7:37













            $begingroup$
            @Kuba, good point.
            $endgroup$
            – kglr
            Aug 28 '18 at 8:09




            $begingroup$
            @Kuba, good point.
            $endgroup$
            – kglr
            Aug 28 '18 at 8:09











            8












            $begingroup$

            If you construct a list of strings instead, you can take advantage of the fact that ToExpression is listable, and supports a 3rd argument that post-processes the output. For example:



            ToExpression[
            "x1", "x2", "x3",
            StandardForm,
            Pattern[#,Blank]&
            ]



            x1_, x2_, x3_




            Or, creating the list and converting:



            ToExpression[
            Table["x" <> ToString@i, i, 5],
            StandardForm,
            Pattern[#, Blank]&
            ]



            x1_, x2_, x3_, x4_, x5_







            share|improve this answer









            $endgroup$








            • 2




              $begingroup$
              It won't work if x1 = 1.
              $endgroup$
              – Kuba
              Aug 28 '18 at 7:36










            • $begingroup$
              Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
              $endgroup$
              – Anton.Sakovich
              Aug 28 '18 at 7:44










            • $begingroup$
              @Kuba what exactly do you mean by x1 = 1 ?
              $endgroup$
              – Winkelried
              Aug 28 '18 at 8:57










            • $begingroup$
              @Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
              $endgroup$
              – Kuba
              Aug 28 '18 at 8:58















            8












            $begingroup$

            If you construct a list of strings instead, you can take advantage of the fact that ToExpression is listable, and supports a 3rd argument that post-processes the output. For example:



            ToExpression[
            "x1", "x2", "x3",
            StandardForm,
            Pattern[#,Blank]&
            ]



            x1_, x2_, x3_




            Or, creating the list and converting:



            ToExpression[
            Table["x" <> ToString@i, i, 5],
            StandardForm,
            Pattern[#, Blank]&
            ]



            x1_, x2_, x3_, x4_, x5_







            share|improve this answer









            $endgroup$








            • 2




              $begingroup$
              It won't work if x1 = 1.
              $endgroup$
              – Kuba
              Aug 28 '18 at 7:36










            • $begingroup$
              Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
              $endgroup$
              – Anton.Sakovich
              Aug 28 '18 at 7:44










            • $begingroup$
              @Kuba what exactly do you mean by x1 = 1 ?
              $endgroup$
              – Winkelried
              Aug 28 '18 at 8:57










            • $begingroup$
              @Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
              $endgroup$
              – Kuba
              Aug 28 '18 at 8:58













            8












            8








            8





            $begingroup$

            If you construct a list of strings instead, you can take advantage of the fact that ToExpression is listable, and supports a 3rd argument that post-processes the output. For example:



            ToExpression[
            "x1", "x2", "x3",
            StandardForm,
            Pattern[#,Blank]&
            ]



            x1_, x2_, x3_




            Or, creating the list and converting:



            ToExpression[
            Table["x" <> ToString@i, i, 5],
            StandardForm,
            Pattern[#, Blank]&
            ]



            x1_, x2_, x3_, x4_, x5_







            share|improve this answer









            $endgroup$



            If you construct a list of strings instead, you can take advantage of the fact that ToExpression is listable, and supports a 3rd argument that post-processes the output. For example:



            ToExpression[
            "x1", "x2", "x3",
            StandardForm,
            Pattern[#,Blank]&
            ]



            x1_, x2_, x3_




            Or, creating the list and converting:



            ToExpression[
            Table["x" <> ToString@i, i, 5],
            StandardForm,
            Pattern[#, Blank]&
            ]



            x1_, x2_, x3_, x4_, x5_








            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Aug 28 '18 at 1:31









            Carl WollCarl Woll

            70.5k394184




            70.5k394184







            • 2




              $begingroup$
              It won't work if x1 = 1.
              $endgroup$
              – Kuba
              Aug 28 '18 at 7:36










            • $begingroup$
              Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
              $endgroup$
              – Anton.Sakovich
              Aug 28 '18 at 7:44










            • $begingroup$
              @Kuba what exactly do you mean by x1 = 1 ?
              $endgroup$
              – Winkelried
              Aug 28 '18 at 8:57










            • $begingroup$
              @Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
              $endgroup$
              – Kuba
              Aug 28 '18 at 8:58












            • 2




              $begingroup$
              It won't work if x1 = 1.
              $endgroup$
              – Kuba
              Aug 28 '18 at 7:36










            • $begingroup$
              Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
              $endgroup$
              – Anton.Sakovich
              Aug 28 '18 at 7:44










            • $begingroup$
              @Kuba what exactly do you mean by x1 = 1 ?
              $endgroup$
              – Winkelried
              Aug 28 '18 at 8:57










            • $begingroup$
              @Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
              $endgroup$
              – Kuba
              Aug 28 '18 at 8:58







            2




            2




            $begingroup$
            It won't work if x1 = 1.
            $endgroup$
            – Kuba
            Aug 28 '18 at 7:36




            $begingroup$
            It won't work if x1 = 1.
            $endgroup$
            – Kuba
            Aug 28 '18 at 7:36












            $begingroup$
            Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
            $endgroup$
            – Anton.Sakovich
            Aug 28 '18 at 7:44




            $begingroup$
            Function[sym,Pattern[sym,Blank],HoldFirst] instead of Pattern[#,Blank]& as usual.
            $endgroup$
            – Anton.Sakovich
            Aug 28 '18 at 7:44












            $begingroup$
            @Kuba what exactly do you mean by x1 = 1 ?
            $endgroup$
            – Winkelried
            Aug 28 '18 at 8:57




            $begingroup$
            @Kuba what exactly do you mean by x1 = 1 ?
            $endgroup$
            – Winkelried
            Aug 28 '18 at 8:57












            $begingroup$
            @Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
            $endgroup$
            – Kuba
            Aug 28 '18 at 8:58




            $begingroup$
            @Winkelried if any of xi have values prior to that evaluation then you will get e.g. Pattern[1, Blank]
            $endgroup$
            – Kuba
            Aug 28 '18 at 8:58











            6












            $begingroup$

            Nothing new but shorter:



            StringTemplate["x``_"] /@ Range[10] // ToExpression



            x1_, x2_, x3_, x4_, x5_, x6_, x7_, x8_, x9_, x10_






            share|improve this answer









            $endgroup$

















              6












              $begingroup$

              Nothing new but shorter:



              StringTemplate["x``_"] /@ Range[10] // ToExpression



              x1_, x2_, x3_, x4_, x5_, x6_, x7_, x8_, x9_, x10_






              share|improve this answer









              $endgroup$















                6












                6








                6





                $begingroup$

                Nothing new but shorter:



                StringTemplate["x``_"] /@ Range[10] // ToExpression



                x1_, x2_, x3_, x4_, x5_, x6_, x7_, x8_, x9_, x10_






                share|improve this answer









                $endgroup$



                Nothing new but shorter:



                StringTemplate["x``_"] /@ Range[10] // ToExpression



                x1_, x2_, x3_, x4_, x5_, x6_, x7_, x8_, x9_, x10_







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Aug 28 '18 at 7:38









                KubaKuba

                106k12207529




                106k12207529





















                    3












                    $begingroup$

                    Here is an example of how to make the solution in the link work for this case:



                    patt = Table[
                    With[
                    s = Symbol["x" <> ToString[i]],
                    Pattern[s, Blank]
                    ], i, 10];

                    Range[10] /. patt :> x5, x8



                    5, 8




                    Using With here is a trick to insert the symbol into Pattern. Since Pattern has the attribute HoldFirst, it would not work to write e.g.



                    Pattern[Symbol["x" <> ToString[i]], Blank]


                    because Symbol["x" <> ToString[i]] would not be evaluated before it was passed to Pattern, i.e. Pattern would not receive a string as is required.






                    share|improve this answer











                    $endgroup$












                    • $begingroup$
                      It won't work if x1 = 1
                      $endgroup$
                      – Kuba
                      Aug 28 '18 at 7:36










                    • $begingroup$
                      @Kuba You mean if the symbols have values?
                      $endgroup$
                      – C. E.
                      Aug 28 '18 at 9:48










                    • $begingroup$
                      Yes, sorry for not being clear.
                      $endgroup$
                      – Kuba
                      Aug 28 '18 at 9:54










                    • $begingroup$
                      @Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
                      $endgroup$
                      – C. E.
                      Aug 28 '18 at 10:40















                    3












                    $begingroup$

                    Here is an example of how to make the solution in the link work for this case:



                    patt = Table[
                    With[
                    s = Symbol["x" <> ToString[i]],
                    Pattern[s, Blank]
                    ], i, 10];

                    Range[10] /. patt :> x5, x8



                    5, 8




                    Using With here is a trick to insert the symbol into Pattern. Since Pattern has the attribute HoldFirst, it would not work to write e.g.



                    Pattern[Symbol["x" <> ToString[i]], Blank]


                    because Symbol["x" <> ToString[i]] would not be evaluated before it was passed to Pattern, i.e. Pattern would not receive a string as is required.






                    share|improve this answer











                    $endgroup$












                    • $begingroup$
                      It won't work if x1 = 1
                      $endgroup$
                      – Kuba
                      Aug 28 '18 at 7:36










                    • $begingroup$
                      @Kuba You mean if the symbols have values?
                      $endgroup$
                      – C. E.
                      Aug 28 '18 at 9:48










                    • $begingroup$
                      Yes, sorry for not being clear.
                      $endgroup$
                      – Kuba
                      Aug 28 '18 at 9:54










                    • $begingroup$
                      @Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
                      $endgroup$
                      – C. E.
                      Aug 28 '18 at 10:40













                    3












                    3








                    3





                    $begingroup$

                    Here is an example of how to make the solution in the link work for this case:



                    patt = Table[
                    With[
                    s = Symbol["x" <> ToString[i]],
                    Pattern[s, Blank]
                    ], i, 10];

                    Range[10] /. patt :> x5, x8



                    5, 8




                    Using With here is a trick to insert the symbol into Pattern. Since Pattern has the attribute HoldFirst, it would not work to write e.g.



                    Pattern[Symbol["x" <> ToString[i]], Blank]


                    because Symbol["x" <> ToString[i]] would not be evaluated before it was passed to Pattern, i.e. Pattern would not receive a string as is required.






                    share|improve this answer











                    $endgroup$



                    Here is an example of how to make the solution in the link work for this case:



                    patt = Table[
                    With[
                    s = Symbol["x" <> ToString[i]],
                    Pattern[s, Blank]
                    ], i, 10];

                    Range[10] /. patt :> x5, x8



                    5, 8




                    Using With here is a trick to insert the symbol into Pattern. Since Pattern has the attribute HoldFirst, it would not work to write e.g.



                    Pattern[Symbol["x" <> ToString[i]], Blank]


                    because Symbol["x" <> ToString[i]] would not be evaluated before it was passed to Pattern, i.e. Pattern would not receive a string as is required.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Aug 28 '18 at 9:49

























                    answered Aug 28 '18 at 1:19









                    C. E.C. E.

                    50.8k399205




                    50.8k399205











                    • $begingroup$
                      It won't work if x1 = 1
                      $endgroup$
                      – Kuba
                      Aug 28 '18 at 7:36










                    • $begingroup$
                      @Kuba You mean if the symbols have values?
                      $endgroup$
                      – C. E.
                      Aug 28 '18 at 9:48










                    • $begingroup$
                      Yes, sorry for not being clear.
                      $endgroup$
                      – Kuba
                      Aug 28 '18 at 9:54










                    • $begingroup$
                      @Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
                      $endgroup$
                      – C. E.
                      Aug 28 '18 at 10:40
















                    • $begingroup$
                      It won't work if x1 = 1
                      $endgroup$
                      – Kuba
                      Aug 28 '18 at 7:36










                    • $begingroup$
                      @Kuba You mean if the symbols have values?
                      $endgroup$
                      – C. E.
                      Aug 28 '18 at 9:48










                    • $begingroup$
                      Yes, sorry for not being clear.
                      $endgroup$
                      – Kuba
                      Aug 28 '18 at 9:54










                    • $begingroup$
                      @Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
                      $endgroup$
                      – C. E.
                      Aug 28 '18 at 10:40















                    $begingroup$
                    It won't work if x1 = 1
                    $endgroup$
                    – Kuba
                    Aug 28 '18 at 7:36




                    $begingroup$
                    It won't work if x1 = 1
                    $endgroup$
                    – Kuba
                    Aug 28 '18 at 7:36












                    $begingroup$
                    @Kuba You mean if the symbols have values?
                    $endgroup$
                    – C. E.
                    Aug 28 '18 at 9:48




                    $begingroup$
                    @Kuba You mean if the symbols have values?
                    $endgroup$
                    – C. E.
                    Aug 28 '18 at 9:48












                    $begingroup$
                    Yes, sorry for not being clear.
                    $endgroup$
                    – Kuba
                    Aug 28 '18 at 9:54




                    $begingroup$
                    Yes, sorry for not being clear.
                    $endgroup$
                    – Kuba
                    Aug 28 '18 at 9:54












                    $begingroup$
                    @Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
                    $endgroup$
                    – C. E.
                    Aug 28 '18 at 10:40




                    $begingroup$
                    @Kuba Well, you are right. I wanted to show how to adapt the answer OP linked to, but it has this flaw.
                    $endgroup$
                    – C. E.
                    Aug 28 '18 at 10:40

















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