Giving Plot options defined outside of the Plot expression












3












$begingroup$


How can I give Plot formating expressions on a separate line just like ListPlot?



When I use the following code with ListPlot, it produces a plot without any errors:



graphs = {ImageSize -> Full, Frame -> True};
ListPlot[Table[x, {x, 1, 2, .01}], graphs]


However, the same thing doesn't work for Plot:



graphs = {ImageSize -> Full, Frame -> True};
Plot[x, {x, 1, 2}, graphs]


Why? What is the simple notation change that I need to make it work?










share|improve this question











$endgroup$

















    3












    $begingroup$


    How can I give Plot formating expressions on a separate line just like ListPlot?



    When I use the following code with ListPlot, it produces a plot without any errors:



    graphs = {ImageSize -> Full, Frame -> True};
    ListPlot[Table[x, {x, 1, 2, .01}], graphs]


    However, the same thing doesn't work for Plot:



    graphs = {ImageSize -> Full, Frame -> True};
    Plot[x, {x, 1, 2}, graphs]


    Why? What is the simple notation change that I need to make it work?










    share|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      How can I give Plot formating expressions on a separate line just like ListPlot?



      When I use the following code with ListPlot, it produces a plot without any errors:



      graphs = {ImageSize -> Full, Frame -> True};
      ListPlot[Table[x, {x, 1, 2, .01}], graphs]


      However, the same thing doesn't work for Plot:



      graphs = {ImageSize -> Full, Frame -> True};
      Plot[x, {x, 1, 2}, graphs]


      Why? What is the simple notation change that I need to make it work?










      share|improve this question











      $endgroup$




      How can I give Plot formating expressions on a separate line just like ListPlot?



      When I use the following code with ListPlot, it produces a plot without any errors:



      graphs = {ImageSize -> Full, Frame -> True};
      ListPlot[Table[x, {x, 1, 2, .01}], graphs]


      However, the same thing doesn't work for Plot:



      graphs = {ImageSize -> Full, Frame -> True};
      Plot[x, {x, 1, 2}, graphs]


      Why? What is the simple notation change that I need to make it work?







      plotting options






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 6 hours ago







      axsvl77

















      asked 10 hours ago









      axsvl77axsvl77

      383212




      383212






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          You can use



          Plot[x, {x, 1, 2}, Evaluate@graphs]


          Why?



          The reason Plot[x, {x, 1, 2}, graphs] doesn't work and ListPlot[Table[x, {x, 1, 2, .01}], graphs]does is that Plot has attribute HoldAll ("all arguments (..) maintained in an unevaluated form")



          Attributes[Plot]



          {HoldAll, Protected, ReadProtected}




          whereas ListPlot doesn't:



          Attributes[ListPlot]



          {Protected, ReadProtected}







          share|improve this answer











          $endgroup$













          • $begingroup$
            Alternatively, you can inject graph using With as in m_goldberg's answer or using Plot[x, {x, 1, 2}, #] &@graphs.
            $endgroup$
            – kglr
            6 hours ago



















          1












          $begingroup$

          You can also use With because it makes the needed substitution before Plot sees any of its arguments.



          options = {ImageSize -> Full, Frame -> True};
          With[{opts = options}, Plot[x, {x, 1, 2}, opts]


          plot






          share|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            You can use



            Plot[x, {x, 1, 2}, Evaluate@graphs]


            Why?



            The reason Plot[x, {x, 1, 2}, graphs] doesn't work and ListPlot[Table[x, {x, 1, 2, .01}], graphs]does is that Plot has attribute HoldAll ("all arguments (..) maintained in an unevaluated form")



            Attributes[Plot]



            {HoldAll, Protected, ReadProtected}




            whereas ListPlot doesn't:



            Attributes[ListPlot]



            {Protected, ReadProtected}







            share|improve this answer











            $endgroup$













            • $begingroup$
              Alternatively, you can inject graph using With as in m_goldberg's answer or using Plot[x, {x, 1, 2}, #] &@graphs.
              $endgroup$
              – kglr
              6 hours ago
















            3












            $begingroup$

            You can use



            Plot[x, {x, 1, 2}, Evaluate@graphs]


            Why?



            The reason Plot[x, {x, 1, 2}, graphs] doesn't work and ListPlot[Table[x, {x, 1, 2, .01}], graphs]does is that Plot has attribute HoldAll ("all arguments (..) maintained in an unevaluated form")



            Attributes[Plot]



            {HoldAll, Protected, ReadProtected}




            whereas ListPlot doesn't:



            Attributes[ListPlot]



            {Protected, ReadProtected}







            share|improve this answer











            $endgroup$













            • $begingroup$
              Alternatively, you can inject graph using With as in m_goldberg's answer or using Plot[x, {x, 1, 2}, #] &@graphs.
              $endgroup$
              – kglr
              6 hours ago














            3












            3








            3





            $begingroup$

            You can use



            Plot[x, {x, 1, 2}, Evaluate@graphs]


            Why?



            The reason Plot[x, {x, 1, 2}, graphs] doesn't work and ListPlot[Table[x, {x, 1, 2, .01}], graphs]does is that Plot has attribute HoldAll ("all arguments (..) maintained in an unevaluated form")



            Attributes[Plot]



            {HoldAll, Protected, ReadProtected}




            whereas ListPlot doesn't:



            Attributes[ListPlot]



            {Protected, ReadProtected}







            share|improve this answer











            $endgroup$



            You can use



            Plot[x, {x, 1, 2}, Evaluate@graphs]


            Why?



            The reason Plot[x, {x, 1, 2}, graphs] doesn't work and ListPlot[Table[x, {x, 1, 2, .01}], graphs]does is that Plot has attribute HoldAll ("all arguments (..) maintained in an unevaluated form")



            Attributes[Plot]



            {HoldAll, Protected, ReadProtected}




            whereas ListPlot doesn't:



            Attributes[ListPlot]



            {Protected, ReadProtected}








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 9 hours ago

























            answered 10 hours ago









            kglrkglr

            188k10204422




            188k10204422












            • $begingroup$
              Alternatively, you can inject graph using With as in m_goldberg's answer or using Plot[x, {x, 1, 2}, #] &@graphs.
              $endgroup$
              – kglr
              6 hours ago


















            • $begingroup$
              Alternatively, you can inject graph using With as in m_goldberg's answer or using Plot[x, {x, 1, 2}, #] &@graphs.
              $endgroup$
              – kglr
              6 hours ago
















            $begingroup$
            Alternatively, you can inject graph using With as in m_goldberg's answer or using Plot[x, {x, 1, 2}, #] &@graphs.
            $endgroup$
            – kglr
            6 hours ago




            $begingroup$
            Alternatively, you can inject graph using With as in m_goldberg's answer or using Plot[x, {x, 1, 2}, #] &@graphs.
            $endgroup$
            – kglr
            6 hours ago











            1












            $begingroup$

            You can also use With because it makes the needed substitution before Plot sees any of its arguments.



            options = {ImageSize -> Full, Frame -> True};
            With[{opts = options}, Plot[x, {x, 1, 2}, opts]


            plot






            share|improve this answer









            $endgroup$


















              1












              $begingroup$

              You can also use With because it makes the needed substitution before Plot sees any of its arguments.



              options = {ImageSize -> Full, Frame -> True};
              With[{opts = options}, Plot[x, {x, 1, 2}, opts]


              plot






              share|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                You can also use With because it makes the needed substitution before Plot sees any of its arguments.



                options = {ImageSize -> Full, Frame -> True};
                With[{opts = options}, Plot[x, {x, 1, 2}, opts]


                plot






                share|improve this answer









                $endgroup$



                You can also use With because it makes the needed substitution before Plot sees any of its arguments.



                options = {ImageSize -> Full, Frame -> True};
                With[{opts = options}, Plot[x, {x, 1, 2}, opts]


                plot







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 6 hours ago









                m_goldbergm_goldberg

                87.5k872198




                87.5k872198






























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