Vertical ranges of Column Plots in 12 Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?CustomTicks and small rangesMultiple ranges in one plotFilling between quantile rangesCombine 3D plots with different ranges into oneMake a column of two plotsTransferFunctionModel and BodePlot shows different result between Mathematica and MatlabHow do I control the ranges in plots made with LogLogPlotMultiple plots sharing a single vertical axisVertical and Horizontal Shifts of PlotsConfiguration of `PlotLayout` as “Row”/“Column”

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Vertical ranges of Column Plots in 12



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?CustomTicks and small rangesMultiple ranges in one plotFilling between quantile rangesCombine 3D plots with different ranges into oneMake a column of two plotsTransferFunctionModel and BodePlot shows different result between Mathematica and MatlabHow do I control the ranges in plots made with LogLogPlotMultiple plots sharing a single vertical axisVertical and Horizontal Shifts of PlotsConfiguration of `PlotLayout` as “Row”/“Column”










5












$begingroup$


Version 12 includes a new Layout->Column option for ListPlot which aligns multiple plots so they share a single horizontal axis. It seemed really useful at first. But data being compared in this way can have very different vertical scales.



It appears that the new capability always sets the PlotRange of all the data sets to the same value:



d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];

d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];

ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic]


enter image description here



I have tried what seems reasonable to set the ranges separately:



ListLinePlot[d1, d2, PlotLayout -> "Column", 
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic, Automatic, Automatic, Automatic]

ListLinePlot[d1, d2, PlotLayout -> "Column",
InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> Automatic, -.1, .1, Automatic, -1, 1]


But neither of these work. The first gives the same result, and the second throws an error. Does anyone have any insight into this?



EDIT:



Thank you @Szabolcs. Your solution not only solves the range problem, but permits differing scaling functions. (A cross section of a biased semiconductor junction may want to plot potentials from 0 to 10 volts, but also carrier concentrations from 0 to 1E18 on a log scale.)



d3 = Table[x, Exp[10 Cos[x]], x, 0, 4 Pi, Pi/100];
GraphicsColumn[

ListLinePlot[d1, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, Automatic, Frame -> True,
FrameTicks -> Automatic, None, None, None,
FrameLabel -> None, "Linear"],
ListLinePlot[d3, InterpolationOrder -> 2, GridLines -> Automatic,
PlotRange -> 0, 4 Pi, All, Frame -> True, FrameLabel -> "Log",
ScalingFunctions -> "Log"]

]


enter image description here










share|improve this question











$endgroup$
















    5












    $begingroup$


    Version 12 includes a new Layout->Column option for ListPlot which aligns multiple plots so they share a single horizontal axis. It seemed really useful at first. But data being compared in this way can have very different vertical scales.



    It appears that the new capability always sets the PlotRange of all the data sets to the same value:



    d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];

    d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];

    ListLinePlot[d1, d2, PlotLayout -> "Column",
    InterpolationOrder -> 2, GridLines -> Automatic,
    PlotRange -> Automatic]


    enter image description here



    I have tried what seems reasonable to set the ranges separately:



    ListLinePlot[d1, d2, PlotLayout -> "Column", 
    InterpolationOrder -> 2, GridLines -> Automatic,
    PlotRange -> Automatic, Automatic, Automatic, Automatic]

    ListLinePlot[d1, d2, PlotLayout -> "Column",
    InterpolationOrder -> 2, GridLines -> Automatic,
    PlotRange -> Automatic, -.1, .1, Automatic, -1, 1]


    But neither of these work. The first gives the same result, and the second throws an error. Does anyone have any insight into this?



    EDIT:



    Thank you @Szabolcs. Your solution not only solves the range problem, but permits differing scaling functions. (A cross section of a biased semiconductor junction may want to plot potentials from 0 to 10 volts, but also carrier concentrations from 0 to 1E18 on a log scale.)



    d3 = Table[x, Exp[10 Cos[x]], x, 0, 4 Pi, Pi/100];
    GraphicsColumn[

    ListLinePlot[d1, InterpolationOrder -> 2, GridLines -> Automatic,
    PlotRange -> 0, 4 Pi, Automatic, Frame -> True,
    FrameTicks -> Automatic, None, None, None,
    FrameLabel -> None, "Linear"],
    ListLinePlot[d3, InterpolationOrder -> 2, GridLines -> Automatic,
    PlotRange -> 0, 4 Pi, All, Frame -> True, FrameLabel -> "Log",
    ScalingFunctions -> "Log"]

    ]


    enter image description here










    share|improve this question











    $endgroup$














      5












      5








      5





      $begingroup$


      Version 12 includes a new Layout->Column option for ListPlot which aligns multiple plots so they share a single horizontal axis. It seemed really useful at first. But data being compared in this way can have very different vertical scales.



      It appears that the new capability always sets the PlotRange of all the data sets to the same value:



      d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];

      d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];

      ListLinePlot[d1, d2, PlotLayout -> "Column",
      InterpolationOrder -> 2, GridLines -> Automatic,
      PlotRange -> Automatic]


      enter image description here



      I have tried what seems reasonable to set the ranges separately:



      ListLinePlot[d1, d2, PlotLayout -> "Column", 
      InterpolationOrder -> 2, GridLines -> Automatic,
      PlotRange -> Automatic, Automatic, Automatic, Automatic]

      ListLinePlot[d1, d2, PlotLayout -> "Column",
      InterpolationOrder -> 2, GridLines -> Automatic,
      PlotRange -> Automatic, -.1, .1, Automatic, -1, 1]


      But neither of these work. The first gives the same result, and the second throws an error. Does anyone have any insight into this?



      EDIT:



      Thank you @Szabolcs. Your solution not only solves the range problem, but permits differing scaling functions. (A cross section of a biased semiconductor junction may want to plot potentials from 0 to 10 volts, but also carrier concentrations from 0 to 1E18 on a log scale.)



      d3 = Table[x, Exp[10 Cos[x]], x, 0, 4 Pi, Pi/100];
      GraphicsColumn[

      ListLinePlot[d1, InterpolationOrder -> 2, GridLines -> Automatic,
      PlotRange -> 0, 4 Pi, Automatic, Frame -> True,
      FrameTicks -> Automatic, None, None, None,
      FrameLabel -> None, "Linear"],
      ListLinePlot[d3, InterpolationOrder -> 2, GridLines -> Automatic,
      PlotRange -> 0, 4 Pi, All, Frame -> True, FrameLabel -> "Log",
      ScalingFunctions -> "Log"]

      ]


      enter image description here










      share|improve this question











      $endgroup$




      Version 12 includes a new Layout->Column option for ListPlot which aligns multiple plots so they share a single horizontal axis. It seemed really useful at first. But data being compared in this way can have very different vertical scales.



      It appears that the new capability always sets the PlotRange of all the data sets to the same value:



      d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];

      d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];

      ListLinePlot[d1, d2, PlotLayout -> "Column",
      InterpolationOrder -> 2, GridLines -> Automatic,
      PlotRange -> Automatic]


      enter image description here



      I have tried what seems reasonable to set the ranges separately:



      ListLinePlot[d1, d2, PlotLayout -> "Column", 
      InterpolationOrder -> 2, GridLines -> Automatic,
      PlotRange -> Automatic, Automatic, Automatic, Automatic]

      ListLinePlot[d1, d2, PlotLayout -> "Column",
      InterpolationOrder -> 2, GridLines -> Automatic,
      PlotRange -> Automatic, -.1, .1, Automatic, -1, 1]


      But neither of these work. The first gives the same result, and the second throws an error. Does anyone have any insight into this?



      EDIT:



      Thank you @Szabolcs. Your solution not only solves the range problem, but permits differing scaling functions. (A cross section of a biased semiconductor junction may want to plot potentials from 0 to 10 volts, but also carrier concentrations from 0 to 1E18 on a log scale.)



      d3 = Table[x, Exp[10 Cos[x]], x, 0, 4 Pi, Pi/100];
      GraphicsColumn[

      ListLinePlot[d1, InterpolationOrder -> 2, GridLines -> Automatic,
      PlotRange -> 0, 4 Pi, Automatic, Frame -> True,
      FrameTicks -> Automatic, None, None, None,
      FrameLabel -> None, "Linear"],
      ListLinePlot[d3, InterpolationOrder -> 2, GridLines -> Automatic,
      PlotRange -> 0, 4 Pi, All, Frame -> True, FrameLabel -> "Log",
      ScalingFunctions -> "Log"]

      ]


      enter image description here







      plotting version-12






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 46 mins ago







      David Keith

















      asked 2 hours ago









      David KeithDavid Keith

      1,6851314




      1,6851314




















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          There is another new feature in M12.0: GraphicsColumn will now align the frames.



          You can use this:



          d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
          d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];

          GraphicsColumn[
          ListLinePlot[#, InterpolationOrder -> 2, GridLines -> Automatic,
          PlotRange -> 0, 4 Pi, Automatic, Frame -> True] & /@ d1, d2
          ]


          enter image description here



          While I did need to set the horizontal range explicitly (so that they match), I did not need to fuss with the ImagePadding to ensure that the frames line up. This saves a lot of work.



          Here's what the same looks like in M11.3:



          enter image description here



          The grids and frames are not quite aligned.






          share|improve this answer









          $endgroup$












          • $begingroup$
            Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
            $endgroup$
            – David Keith
            44 mins ago











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          There is another new feature in M12.0: GraphicsColumn will now align the frames.



          You can use this:



          d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
          d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];

          GraphicsColumn[
          ListLinePlot[#, InterpolationOrder -> 2, GridLines -> Automatic,
          PlotRange -> 0, 4 Pi, Automatic, Frame -> True] & /@ d1, d2
          ]


          enter image description here



          While I did need to set the horizontal range explicitly (so that they match), I did not need to fuss with the ImagePadding to ensure that the frames line up. This saves a lot of work.



          Here's what the same looks like in M11.3:



          enter image description here



          The grids and frames are not quite aligned.






          share|improve this answer









          $endgroup$












          • $begingroup$
            Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
            $endgroup$
            – David Keith
            44 mins ago















          3












          $begingroup$

          There is another new feature in M12.0: GraphicsColumn will now align the frames.



          You can use this:



          d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
          d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];

          GraphicsColumn[
          ListLinePlot[#, InterpolationOrder -> 2, GridLines -> Automatic,
          PlotRange -> 0, 4 Pi, Automatic, Frame -> True] & /@ d1, d2
          ]


          enter image description here



          While I did need to set the horizontal range explicitly (so that they match), I did not need to fuss with the ImagePadding to ensure that the frames line up. This saves a lot of work.



          Here's what the same looks like in M11.3:



          enter image description here



          The grids and frames are not quite aligned.






          share|improve this answer









          $endgroup$












          • $begingroup$
            Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
            $endgroup$
            – David Keith
            44 mins ago













          3












          3








          3





          $begingroup$

          There is another new feature in M12.0: GraphicsColumn will now align the frames.



          You can use this:



          d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
          d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];

          GraphicsColumn[
          ListLinePlot[#, InterpolationOrder -> 2, GridLines -> Automatic,
          PlotRange -> 0, 4 Pi, Automatic, Frame -> True] & /@ d1, d2
          ]


          enter image description here



          While I did need to set the horizontal range explicitly (so that they match), I did not need to fuss with the ImagePadding to ensure that the frames line up. This saves a lot of work.



          Here's what the same looks like in M11.3:



          enter image description here



          The grids and frames are not quite aligned.






          share|improve this answer









          $endgroup$



          There is another new feature in M12.0: GraphicsColumn will now align the frames.



          You can use this:



          d1 = Table[x, .1 Sin[x], x, 0, 3 Pi, Pi/9];
          d2 = Table[x, Cos[x], x, 0, 4 Pi, Pi/7];

          GraphicsColumn[
          ListLinePlot[#, InterpolationOrder -> 2, GridLines -> Automatic,
          PlotRange -> 0, 4 Pi, Automatic, Frame -> True] & /@ d1, d2
          ]


          enter image description here



          While I did need to set the horizontal range explicitly (so that they match), I did not need to fuss with the ImagePadding to ensure that the frames line up. This saves a lot of work.



          Here's what the same looks like in M11.3:



          enter image description here



          The grids and frames are not quite aligned.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 1 hour ago









          SzabolcsSzabolcs

          165k14450954




          165k14450954











          • $begingroup$
            Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
            $endgroup$
            – David Keith
            44 mins ago
















          • $begingroup$
            Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
            $endgroup$
            – David Keith
            44 mins ago















          $begingroup$
          Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
          $endgroup$
          – David Keith
          44 mins ago




          $begingroup$
          Thank you, Szabolcs. That works perfectly. I edited my question to comment further.
          $endgroup$
          – David Keith
          44 mins ago

















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