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I am trying to make either a nested manipulate or just a manipulate with two windows: I need one window which functions as:
Manipulate[Graphics[Polygon[pt],
PlotRange -> 2], {{pt, {{0, 0}, {1, 0}, {1, 1}, {0, 1}, {1, -1}}},
Locator, LocatorAutoCreate -> True}]
but outputs its coordinates to another window which uses these coordinates to plot a specified graph. I am not sure if Manipulate is even the best option for this, but essentially I am trying to make a visual interface where a user can specify a polygon and then the program uses the information of those coordinates to plot a specified 3D plot.
I think I could figure out how to do this if I knew how to output the coordinates from the manipulate or how to make something that does.
For Example:
GraphicsRow[{
Manipulate[
Graphics[Polygon[rs = pt], PlotRange -> 2],
{{pt, {{0, 0}, {1, 0}, {1, 1}, {0, 1}, {1, -1}}},
Locator, LocatorAutoCreate -> True}],
Dynamic#
ParametricPlot3D[Through[(Interpolation /#
First#(Transpose /# {Append[#, 0] & /# rs}))[t]], {t, 1, Length#rs},
PlotRange -> 2]}]
I'm trying to combine 3 functions graphed on a Plot[] and 1 function graphed on a ParametricPlot[]. My equations are as follows:
plota = Plot[{-2 x, -2 Sqrt[x], -2 x^(3/5)}, {x, 0, 1}, PlotLegend -> {"-2 x", "-2 \!\(\*SqrtBox[\(x\)]\)", "-2 \!\(\*SuperscriptBox[\(x\), \(3/5\)]\)"}]
plotb = ParametricPlot[{2.4056 (u - Sin[u]), 2.4056 (Cos[u] - 1)}, {u,0, 1.40138}, PlotLegend -> {"Problem 3"}]
Show[plota, plotb]
This is the image it gives:
As yoda said, PlotLegends is terrible. However, if you don't mind setting the plot styles manually and repeating them lateron, ShowLegend can help.
plota = Plot[{-2 x, -2 Sqrt[x], -2 x^(3/5)}, {x, 0, 1},
PlotStyle -> {{Red}, {Blue}, {Orange}}];
plotb = ParametricPlot[{2.4056 (u - Sin[u]), 2.4056 (Cos[u] - 1)}, {u, 0, 1.40138},
PlotStyle -> {{Black}}];
And now
ShowLegend[Show[plota, plotb],
{{{Graphics[{Red, Line[{{0, 0}, {1, 0}}]}], Label1},
{Graphics[{Blue, Line[{{0, 0}, {1, 0}}]}], Label2},
{Graphics[{Orange, Line[{{0, 0}, {1, 0}}]}], Label3},
{Graphics[{Black, Line[{{0, 0}, {1, 0}}]}], Label4}},
LegendSize -> {0.5, 0.5}, LegendPosition -> {0.5, -0.2}}]
which will give you this:
You can also write some simple functions to make this a little less cumbersome, if you deal with this problem often.
Well, the root cause of the error is the PlotLegends package, which is a terrible, buggy package. Removing that, Show combines them correctly:
plota = Plot[{-2 x, -2 Sqrt[x], -2 x^(3/5)}, {x, 0, 1}]
plotb = ParametricPlot[{2.4056 (u - Sin[u]), 2.4056 (Cos[u] - 1)}, {u,
0, 1.40138}]
Show[plota, plotb]
You can see Simon's solution here for ideas to label your different curves without using PlotLegends. This answer by James also demonstrates why PlotLegends has the reputation it has...
You can still salvage something with the PlotLegends package. Here's an example using ShowLegends that you can modify to your tastes
colors = {Red, Green, Blue, Pink};
legends = {-2 x, -2 Sqrt[x], -2 x^(3/5), "Problem 3"};
plota = Plot[{-2 x, -2 Sqrt[x], -2 x^(3/5)}, {x, 0, 1},
PlotStyle -> colors[[1 ;; 3]]];
plotb = ParametricPlot[{2.4056 (u - Sin[u]), 2.4056 (Cos[u] - 1)}, {u,
0, 1.40138}, PlotStyle -> colors[[4]]];
ShowLegend[
Show[plota,
plotb], {Table[{Graphics[{colors[[i]], Thick,
Line[{{0, 0}, {1, 0}}]}], legends[[i]]}, {i, 4}],
LegendPosition -> {0.4, -0.15}, LegendSpacing -> 0,
LegendShadow -> None, LegendSize -> 0.6}]
As the other answers pointed out, the culprit is PlotLegend. So, sometimes is useful to be able to roll your own plot legends:
plotStyle = {Red, Green, Blue};
labls = {"a", "b", "Let's go"};
f[i_, s_] := {Graphics[{plotStyle[[i]], Line[{{0, 0}, {1, 0}}]},
ImageSize -> {15, 10}], Style[labls[[i]], s]};
Plot[{Sin[x], Sin[2 x], Sin[3 x]}, {x, 0, 2 Pi},
PlotStyle -> plotStyle,
Epilog ->
Inset[Framed[Style#Column[{Grid[Table[f[i, 15], {i, 1, 3}]]}]],
Offset[{-2, -2}, Scaled[{1, 1}]], {Right, Top}],
PlotRangePadding -> 1
]
How to get arrows for the axes when using the command Plot in Mathematica?
Thanks for any helpful answers.
For 2D plots such as generated by Plot the following works great:
Plot[Sin[x], {x, 0, 10}, AxesStyle -> Arrowheads[0.07]]
or with custom arrow heads:
h = Graphics[Line[{{-1, 1/2}, {0, 0}, {-1, -1/2}}]];
Plot[Sin[x], {x, 0, 10},
AxesStyle -> Arrowheads[{{Automatic, Automatic, h}}]]
Building on Sjoerd's answer,
a plot such as
may be obtained as follows (for example):
Plot[Sin[x], {x, -2\[Pi], 2 \[Pi]},
AxesStyle-> {
Directive[{Red,
Arrowheads[{{-0.06,0(*Xleft*),{Graphics[{
Polygon[
{{-1,0.5`},{0,0},{-1,-0.5`}}]}],0.98`}},
{0.03,.9(*Xright*),{Graphics[{
Polygon[
{{-1,0.5`},{0,0},{-1,-0.5`}}]}],0.98`}}}]}],
Directive[{Blue,
Arrowheads[{{-0.05,0(*Ydown*),{Graphics[{
Polygon[
{{-1,0.5`},{0,0},{-1,-0.5`}}]}],0.98`}},{0.03,.8(*Yup*),{Graphics[{
Polygon[
{{-1,0.5`},{0,0},{-1,-0.5`}}]}],0.98`}}}]}
]}]
There are nice examples of arrowheads given in Drawings Tools and Graphics Inspector. There are probably much better ways of getting the info but I annotate a plot with an arrow that I like and then abstract (using a suggestion from Simon):
Cases["Paste-Graphic_Here", Arrowheads[___], Infinity]
To give another example:
The code is as follows
Plot[Sin[x], {x, -2\[Pi],2 \[Pi]},
AxesStyle-> { Directive[{Red,
Arrowheads[{{-0.06,0.1(*Xleft*),
{Graphics[{arrowhead}]/.arrowhead-> arrowhead2,0.98`}},
{0.05,0.95(*Xright*),
{Graphics[{arrowhead}],0.98`}}}]/.arrowhead-> arrowhead4}],
Directive[{Blue,
Arrowheads[{{-0.05,0(*Ydown*),
{Graphics[{arrowhead}]/.arrowhead-> arrowhead3,0.98`}},{0.03,.8(*Yup*),
{Graphics[{arrowhead}]/.arrowhead-> arrowhead1,0.98`}}}]}
]}]
where
arrowhead1=Polygon[{{-1,0.5`},{0,0},{-1,-0.5`}}];
arrowhead2=Polygon[{{-1.5833333333333333`,0.4166666666666667`},{-1.5410500000000003`,0.369283333333333`},{-1.448333333333333`,0.255583333333333`},{-1.3991000000000005`,0.18721666666666673`},{-1.3564666666666663`,0.11826666666666673`},{-1.3268499999999999`,0.05408333333333341`},{-1.3166666666666667`,0.`},{-1.3268499999999999`,-0.048950000000000195`},{-1.3564666666666663`,-0.11228333333333372`},{-1.3991000000000005`,-0.18353333333333333`},{-1.448333333333333`,-0.2562833333333335`},{-1.5410500000000003`,-0.38048333333333345`},{-1.5833333333333333`,-0.43333333333333335`},{0.`,0.`},{-1.5833333333333333`,0.4166666666666667`},{-1.5833333333333333`,0.4166666666666667`}}];
arrowhead3=Polygon[{{-1,0.5`},{0,0},{-1,-0.5`},{-0.6`,0},{-1,0.5`}}];
arrowhead4={{FaceForm[GrayLevel[1]],Polygon[{{-0.6`,0},{-1.`,0.5`},{0.`,0},{-1.`,-0.5`},{-0.6`,0}}],Line[{{-0.6`,0},{-1.`,0.5`},{0.`,0},{-1.`,-0.5`},{-0.6`,0}}]}};
arrowhead5=Polygon[{{-0.6582278481012658`,-0.43037974683544306`},{0.`,0.`},{0.`,0.`},{0.`,0.`},{0.`,0.`},{0.`,0.`},{-0.6455696202531646`,0.43037974683544306`},{-0.4810126582278481`,0.`},{-0.6582278481012658`,-0.43037974683544306`},{-0.6582278481012658`,-0.43037974683544306`}}];
A list of arrowheads 1 to 5:
Here you have a solution posted in https://math.stackexchange.com/
As the solution in the reference is for Plot3D, here I modified (but not improved) it for Plot[ ]:
axes[x_, y_, f_, a_] :=
Graphics[Join[{Arrowheads[a]},
Arrow[{{0, 0}, #}] & /# {{x, 0}, {0, y}},
{Text[Style["x", FontSize -> Scaled[f]], {0.9*x, 0.1*y}],
Text[Style["y", FontSize -> Scaled[f]], {0.1 x, 0.95*y}]
}]]
Show[Plot[Exp[-x^2], {x, -2, 2},
Axes -> None,
PlotRange -> {{-2.1, 2.1}, {-.1, 1.1}}],
axes[2, 1, 0.05, 0.02]
]
When I run the following code
pMin = {-3, -3};
pMax = {3, 3};
range = {pMin, pMax};
Manipulate[
GraphicsGrid[
{
{Graphics[Locator[p], PlotRange -> range]},
{Graphics[Line[{{0, 0}, p}]]}
}, Frame -> All
],
{{p, {1, 1}}, Locator}
]
I expect the Locator control to be within the bounds of the first Graph, but instead it can be moved around the whole GraphicsGrid region. Is there an error in my code?
I also tried
{{p, {1, 1}}, pMin, pMax, Locator}
instead of
{{p, {1, 1}}, Locator}
But it behaves completely wrong.
UPDATE
Thanks to everyone, this is my final solution:
Manipulate[
distr1 = BinormalDistribution[p1, {1, 1}, \[Rho]1];
distr2 = BinormalDistribution[p2, {1, 1}, \[Rho]2];
Grid[
{
{Graphics[{Locator[p1], Locator[p2]},
PlotRange -> {{-5, 5}, {-5, 5}}]},
{Plot3D[{PDF[distr1, {x, y}], PDF[distr2, {x, y}]}, {x, -5, 5}, {y, -5, 5}, PlotRange -> All]}
}],
{{\[Rho]1, 0}, -0.9, 0.9}, {{\[Rho]2, 0}, -0.9, 0.9},
{{p1, {1, 1}}, Locator},
{{p2, {1, 1}}, Locator}
]
UPDATE
Now the problem is that I cannot resize and rotate the lower 3d graph. Does anyone know how to fix that?
I'm back to the solution with two Slider2D objects.
If you examine the InputForm you'll find that GraphicsGrid returns a Graphics object. Thus, the Locator indeed moves throughout the whole image.
GraphicsGrid[{{Graphics[Circle[]]}, {Graphics[Disk[]]}}] // InputForm
If you just change the GraphicsGrid to a Grid, the locator will be restricted to the first part but the result still looks a bit odd. Your PlotRange specification is a bit strange; it doesn't seem to correspond to any format specified in the Documentation center. Perhaps you want something like the following.
Manipulate[
Grid[{
{Graphics[Locator[p], Axes -> True,
PlotRange -> {{-3, 3}, {-3, 3}}]},
{Graphics[Line[{{0, 0}, p}], Axes -> True,
PlotRange -> {{-3, 3}, {-3, 3}}]}},
Frame -> All],
{{p, {1, 1}}, Locator}]
LocatorPane[] does a nice job of confining the locator to a region.
This is a variation on the method used by Mr. Wizard.
Column[{ LocatorPane[Dynamic[pt3],
Framed#Graphics[{}, ImageSize -> 150, PlotRange -> 3]],
Framed#Graphics[{Line[{{-1, 0}, Dynamic#pt3}]}, ImageSize -> {150, 150},
PlotRange -> 3]}]
I would have assumed that you'd want the locator to share the space with the line it controls. In fact, to be "attached" to the line. This turns out to be even easier to implement.
Column[{LocatorPane[Dynamic[pt3],Framed#Graphics[{Line[{{-1, 0}, Dynamic#pt3}]},
ImageSize -> 150, PlotRange -> 3]]}]
I am not sure what you are trying to achieve. There are a number of problems I see, but I don't know what to address. Perhaps you just want a simple Slider2D construction?
DynamicModule[{p = {1, 1}},
Column#{Slider2D[Dynamic[p], {{-3, -3}, {3, 3}},
ImageSize -> {200, 200}],
Graphics[Line[{{0, 0}, Dynamic[p]}],
PlotRange -> {{-3, 3}, {-3, 3}}, ImageSize -> {200, 200}]}]
This is a reply to the updated question about 3D graphic rotation.
I believe that LocatorPane as suggested by David is a good way to approach this. I just put in a generic function since your example would not run on Mathematica 7.
DynamicModule[{pt = {{-1, 3}, {1, 1}}},
Column[{
LocatorPane[Dynamic[pt],
Framed#Graphics[{}, PlotRange -> {{-5, 5}, {-5, 5}}]],
Dynamic#
Plot3D[{x^2 pt[[1, 1]] + y^2 pt[[1, 2]],
-x^2 pt[[2, 1]] - y^2 pt[[2, 1]]},
{x, -5, 5}, {y, -5, 5}]
}]
]
How can I make it such that plotting the following function
ListPointPlot3D[points, PlotStyle -> PointSize[0.05]];
the points I see are green or yellow, for instance, instead of the typical dark blue ones?
Thanks
Use Directive to combine styles, ie
ListPointPlot3D[points, PlotStyle -> Directive[{PointSize[0.05], Green}]]
Edit I give you below two possible solutions in a context related to your previous question. Nevertheless, please note that #Yaroslav's code is much better.
f[x_, y_] := x^2 + y^2;
t = Graphics3D[{PointSize[Large], Red, Point#
Flatten[Table[{x, y, f[x, y]}, {x, 0, 10, 1}, {y, 1, 2, 1}], 1]}];
b = Plot3D[f[x, y], {x, -10, 10}, {y, -10, 10},
ColorFunction -> "MintColors"];
Show[{b, t}]
Or
f[x_, y_] := x^2 + y^2;
points = Flatten[Table[{x, y, f[x, y]}, {x, 0, 10, 1}, {y, 1, 2, 1}],
1];
a = ListPointPlot3D[points,
PlotStyle -> Table[{Red, PointSize[0.05]}, {Length#t}]];
b = Plot3D[f[x, y], {x, -10, 10}, {y, -10, 10},
ColorFunction -> "MintColors"];
Show[{b, a}]
Sometimes I find the following approach useful, as it allows me to
manipulate the plot symbol (PlotMarkers does not seem to work with ListPointPlot3D,
at least in Mathematica 7) [originally suggested by Jens-Peer Kuska]:
ListPointPlot3D[{{1,1,1},{2,2,2},{3,3,3}}]/.Point[xy_]:>(Style[Text["\[FilledUpTriangle]",#],Red,FontSize-> 20]&/#xy)