5
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The following

Graphics[Circle[], Axes -> {True, True}, ScalingFunctions -> {Identity, "Reverse"}]

doesn't work

enter image description here

I need "minus" direction go to the right...

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2 Answers 2

6
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For Circle[] (or any graphics primitive symmetric around the origin), you can use custom ticks:

Graphics[Circle[], 
 Axes -> True, TicksStyle -> 16,
 Ticks -> {Charting`ScaledTicks["Reverse"], Automatic}]

enter image description here

You can also cheat by using the graphics primitives as Epilog in a plotting function that accepts ScalingFunctions(say, Plot):

Plot[x, {x, -1, 1}, AspectRatio -> 1, TicksStyle -> 16, 
 PlotStyle -> None, 
 PlotRange -> {{-1, 1}, {-1, 1}}, Axes -> True, 
 ScalingFunctions -> {"Reverse", Identity},
 Epilog -> {Circle[]}, ]

enter image description here

In general, you can use ScalingTransform[{-1, 1}] or ReflectionTransform[{-1, 0}] on graphics primitives and use custom ticks:

SeedRandom[1]
pnts = RandomReal[{-5, 5}, {10, 2}];
Row[{Graphics[{Opacity[.5], Blue, 
    Polygon[pnts[[FindShortestTour[pnts][[2]]]]]}, 
    Axes -> {True, True}, TicksStyle -> 16, ImageSize -> 300],
  Graphics[{Opacity[.5], Green, 
     GeometricTransformation[
      Polygon[pnts[[FindShortestTour[pnts][[2]]]]], ScalingTransform[{-1, 1}]]}, 
    Axes -> {True, True}, TicksStyle -> 16, ImageSize -> 300]
   Graphics[{Opacity[.5], Red, 
     GeometricTransformation[
      Polygon[pnts[[FindShortestTour[pnts][[2]]]]], ScalingTransform[{-1, 1}]]}, 
    Axes -> {True, True}, TicksStyle -> 16, ImageSize -> 300,
    Ticks -> {Charting`ScaledTicks["Reverse"], Automatic}]}]

enter image description here

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2
  • $\begingroup$ So it is not possible without cheating? Send shame to Stephen! $\endgroup$
    – Dims
    Commented Feb 10, 2019 at 12:30
  • $\begingroup$ @Dims, I think it is not possible to use the option ScalingFunctions in Graphics. $\endgroup$
    – kglr
    Commented Feb 10, 2019 at 12:35
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For objects like circles that are readily converted to mathematical functions:

r = 1;

ContourPlot[
 x^2 + y^2 == r^2,
 {x, -1.05, 1.05}, {y, -1.05, 1.05},
 Frame -> False,
 Axes -> True,
 ScalingFunctions -> {"Reverse", Identity}]

enter image description here

Or

Plot[
 {Sqrt[r^2 - x^2], -Sqrt[r^2 - x^2]},
 {x, -1.05, 1.05},
 PlotStyle -> ColorData[97][1],
 AspectRatio -> 1,
 ScalingFunctions -> {"Reverse", Identity}]
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