How to reverse X-axis in Graphics?
4
$begingroup$
The following
Graphics[Circle, Axes -> {True, True}, ScalingFunctions -> {Identity, "Reverse"}]
doesn't work
I need "minus" direction go to the right...
plotting graphics
asked Feb 10 at 11:14
DimsDims
30117
$endgroup$
add a comment |
4
$begingroup$
The following
Graphics[Circle, Axes -> {True, True}, ScalingFunctions -> {Identity, "Reverse"}]
doesn't work
I need "minus" direction go to the right...
plotting graphics
asked Feb 10 at 11:14
DimsDims
30117
$endgroup$
add a comment |
4
4
4
$begingroup$
The following
Graphics[Circle, Axes -> {True, True}, ScalingFunctions -> {Identity, "Reverse"}]
doesn't work
I need "minus" direction go to the right...
plotting graphics
asked Feb 10 at 11:14
DimsDims
30117
$endgroup$
The following
Graphics[Circle, Axes -> {True, True}, ScalingFunctions -> {Identity, "Reverse"}]
doesn't work
I need "minus" direction go to the right...
plotting graphics
plotting graphics
asked Feb 10 at 11:14
DimsDims
30117
asked Feb 10 at 11:14
DimsDims
30117
asked Feb 10 at 11:14
DimsDims
30117
asked Feb 10 at 11:14
DimsDims
30117
asked Feb 10 at 11:14
DimsDims
30117
30117
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
6
$begingroup$
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}]
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}, ]
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}]}]
answered Feb 10 at 11:43
kglrkglr
190k10206424
$endgroup$
$begingroup$
So it is not possible without cheating? Send shame to Stephen!
$endgroup$
– Dims
Feb 10 at 12:30
$begingroup$
@Dims, I think it is not possible to use the optionScalingFunctionsinGraphics.
$endgroup$
– kglr
Feb 10 at 12:35
add a comment |
1
$begingroup$
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}]
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}]
answered Feb 10 at 13:37
Bob HanlonBob Hanlon
61.4k33598
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
$begingroup$
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}]
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}, ]
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}]}]
answered Feb 10 at 11:43
kglrkglr
190k10206424
$endgroup$
$begingroup$
So it is not possible without cheating? Send shame to Stephen!
$endgroup$
– Dims
Feb 10 at 12:30
$begingroup$
@Dims, I think it is not possible to use the optionScalingFunctionsinGraphics.
$endgroup$
– kglr
Feb 10 at 12:35
add a comment |
6
$begingroup$
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}]
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}, ]
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}]}]
answered Feb 10 at 11:43
kglrkglr
190k10206424
$endgroup$
$begingroup$
So it is not possible without cheating? Send shame to Stephen!
$endgroup$
– Dims
Feb 10 at 12:30
$begingroup$
@Dims, I think it is not possible to use the optionScalingFunctionsinGraphics.
$endgroup$
– kglr
Feb 10 at 12:35
add a comment |
6
6
6
$begingroup$
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}]
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}, ]
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}]}]
answered Feb 10 at 11:43
kglrkglr
190k10206424
$endgroup$
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}]
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}, ]
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}]}]
answered Feb 10 at 11:43
kglrkglr
190k10206424
edited Feb 10 at 12:37
answered Feb 10 at 11:43
kglrkglr
190k10206424
answered Feb 10 at 11:43
kglrkglr
190k10206424
answered Feb 10 at 11:43
kglrkglr
190k10206424
190k10206424
$begingroup$
So it is not possible without cheating? Send shame to Stephen!
$endgroup$
– Dims
Feb 10 at 12:30
$begingroup$
@Dims, I think it is not possible to use the optionScalingFunctionsinGraphics.
$endgroup$
– kglr
Feb 10 at 12:35
add a comment |
$begingroup$
So it is not possible without cheating? Send shame to Stephen!
$endgroup$
– Dims
Feb 10 at 12:30
$begingroup$
@Dims, I think it is not possible to use the optionScalingFunctionsinGraphics.
$endgroup$
– kglr
Feb 10 at 12:35
$begingroup$
So it is not possible without cheating? Send shame to Stephen!
$endgroup$
– Dims
Feb 10 at 12:30
$begingroup$
So it is not possible without cheating? Send shame to Stephen!
$endgroup$
– Dims
Feb 10 at 12:30
$begingroup$
@Dims, I think it is not possible to use the option
ScalingFunctions in Graphics.$endgroup$
– kglr
Feb 10 at 12:35
$begingroup$
@Dims, I think it is not possible to use the option
ScalingFunctions in Graphics.$endgroup$
– kglr
Feb 10 at 12:35
add a comment |
1
$begingroup$
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}]
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}]
answered Feb 10 at 13:37
Bob HanlonBob Hanlon
61.4k33598
$endgroup$
add a comment |
1
$begingroup$
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}]
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}]
answered Feb 10 at 13:37
Bob HanlonBob Hanlon
61.4k33598
$endgroup$
add a comment |
1
1
1
$begingroup$
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}]
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}]
answered Feb 10 at 13:37
Bob HanlonBob Hanlon
61.4k33598
$endgroup$
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}]
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}]
answered Feb 10 at 13:37
Bob HanlonBob Hanlon
61.4k33598
answered Feb 10 at 13:37
Bob HanlonBob Hanlon
61.4k33598
answered Feb 10 at 13:37
Bob HanlonBob Hanlon
61.4k33598
answered Feb 10 at 13:37
Bob HanlonBob Hanlon
61.4k33598
61.4k33598
add a comment |
add a comment |
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