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Asked: 2026-05-26T07:09:26+00:00

Please Consider : intense = Reverse[Round[Rationalize /@ N[10^Range[0, 3, 1/3]]]]; values = Range[0, 9/10,

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Please Consider :

intense = Reverse[Round[Rationalize /@ N[10^Range[0, 3, 1/3]]]];
values = Range[0, 9/10, 1/10];

intensityLegend = Column[Prepend[MapThread[
                         Function[{intensity, values},
                         Row[{Graphics[{(Lighter[Blue, values]),
                         Rectangle[{0, 0}, {4, 1}], Black, 
                         Text[Style[ToString[intensity], 16, Bold], {2, .5}]}]}]],
                        {intense, values}], Text[Style["Photons Number", Bold, 15]]]];

IntersectionDp1={{1., 588.377}, {2.15443, 580.306}, {4.64159, 573.466}, {10.,560.664}, 
                 {21.5443, 552.031}, {46.4159, 547.57}, {100.,545.051}, 
                 {215.443, 543.578}, {464.159, 542.281}, {1000., 541.346}}


FindD1=ListLogLinearPlot[Map[List, IntersectionDp1], 
                  Frame -> True, 
                  AxesOrigin -> {-1, 0}, 
                  PlotMarkers -> 
                  With[{markerSize = 0.04}, {Graphics[{Lighter[Blue, #], Disk[]}], 
                       markerSize} & /@Range[9/10, 0, -1/10]], Filling -> Axis, 
                  FillingStyle -> Opacity[0.8], 
                  PlotRange -> {{.5, 1100}, {540, 600}},
                  ImageSize->400];

Grid[{{intensityLegend, FindD1}, {intensityLegend, FindD1}}, 
      ItemSize -> {50, 20}, Frame -> True]

enter image description here

How could I get the legend Column Size to Fit the Height of the Plot Area ?

While Row adjust the size I need to use Grid. This is why I duplicated in grid.

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1 Answer

  1. Editorial Team

    Working with Image Sizes. The (* <- *) marks the important modifications to your code, the rest are mainly font size thingies:

    intense = Reverse[Round[Rationalize /@ N[10^Range[0, 3, 1/3]]]];
values = Range[0, 9/10, 1/10];
imgSize = 400;                                                          (* <- *) 
Off[Ticks::ticks]

IntersectionDp1 =  {{1., 588.377},     {2.15443, 580.306}, {4.64159, 573.466}, 
   {10., 560.664}, {21.5443, 552.031}, {46.4159, 547.57},  {100., 545.051},
   {215.443, 543.578}, {464.159, 542.281}, {1000., 541.346}}

FindD1 = ListLogLinearPlot[Map[List, IntersectionDp1], Frame -> True, 
   AxesOrigin -> {-1, 0}, 
   PlotMarkers -> 
    With[{markerSize = 0.04}, 
     {Graphics[{Lighter[Blue, #], Disk[]}], markerSize} & 
       /@ Range[9/10, 0, -1/10]], Filling -> Axis, FillingStyle -> Opacity[0.8], 
       PlotRange -> {{.5, 1100}, {540, 600}}, ImageSize -> imgSize];    (* <- *) 

intensityLegend =
  Rasterize[Column[
    Prepend[
     Reverse@MapThread[                                                 (* <- *) 
      Function[{intensity, values}, 
       Row[{Graphics[{(Lighter[Blue, values]), 
           Rectangle[{0, 0}, {4, 1}], Black, 
           Text[Style[ToString[intensity], 30, Bold], {2, .5}]}]}]],
      {intense, values}],
     Text[Style["Photons Number", Bold, 25]]]], 
   ImageSize -> {Automatic,                                             (* <- *) 
     IntegerPart@                            
      First[imgSize Cases[AbsoluteOptions[FindD1], 
         HoldPattern[AspectRatio -> x_] -> x]]}];

Grid[{{intensityLegend, FindD1}, {intensityLegend, FindD1}}, Frame -> True]

    enter image description here

    Where I reversed the intensities column for aesthetic purposes.

    Edit

    If you don’t explicitly specify the ImageSize option for the Plot, you’ll disappointingly find that AbsoluteOptions[Plot, "ImageSize"] returns "Automatic" !

    Edit Answering the @500’s comment bellow

    The expression: 

       ImageSize -> {Automatic,                                             (* <- *) 
     IntegerPart@                            
      First[imgSize Cases[AbsoluteOptions[FindD1], 
         HoldPattern[AspectRatio -> x_] -> x]]}];

    is really a working replacement for something that should work but doesn’t to get the image size of a Plot:

       ImageSize -> {Automatic, Last@AbsoluteOptions[FindD1,"ImageSize"]}

    So, what the IntegerPart[...] thing is doing is getting the vertical size of the plot image, multiplying imgSize by the AspectRatio of the Plot.
    To understand how it works, run the code and then type:

    AbsoluteOptions[FindD1]

    and you will see the Plot options there. Then the Cases[] function is just extracting the AspectRatio option.

    In fact there is a cleaner way to do what the Cases[] does. It is:

    AbsoluteOptions[FindD1,"AspectRatio"]

    but there is another bug in the AbsoluteOptions function that prevents us to use it this way.

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