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)
Related
I can plot the curve corresponding to an implicit equation:
ContourPlot[x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1}]
But I cannot find a way to color the contour line depending on the location of the point. More precisely, I want to color the curve in 2 colors, depending on whether x² + y² < k or not.
I looked into ColorFunction but this is only for coloring the region between the contour lines.
And I was not able to get ContourStyle to accept a location-dependent expression.
you could use RegionFunction to split the plot in two:
Show[{
ContourPlot[x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1},
RegionFunction -> Function[{x, y, z}, x^2 + y^2 < .5],
ContourStyle -> Red],
ContourPlot[x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1},
RegionFunction -> Function[{x, y, z}, x^2 + y^2 >= .5],
ContourStyle -> Green]
}]
Maybe something like this
pl = ContourPlot[x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1}]
points = pl[[1, 1]];
colorf[{x_, y_}] := ColorData["Rainbow"][Rescale[x, {-1, 1}]]
pl /. {Line[a_] :> {Line[a, VertexColors -> colorf /# points[[a]]]}}
which produces
This does not provide a direct solution to your question but I believe it is of interest.
It is possible to color a line progressively from within ContourPlot using what I think is an undocumented format, namely a Function that surrounds the Line object. Internally this is similar to what Heike did, but her solution uses the vertex numbers to then find the matching coordinates allowing styling by spacial position, rather than position along the line.
ContourPlot[
x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1},
BaseStyle -> {12, Thickness[0.01]},
ContourStyle ->
(Line[#, VertexColors -> ColorData["DeepSeaColors"] /# Rescale##] & ## # &)
]
For some of the less adept, less information is more. Time was wasted browsing for a way to set the color of contour lines until I chanced onto Roelig's edited answer. I just needed ContourStyle[].
Show[{ContourPlot[
x^2 + 2 x y Tan[2 # ] - y^2 == 1, {x, -3, 3}, {y, -3.2, 3.2},
ContourStyle -> Green] & /# Range[-Pi/4, Pi/4, .1]},
Background -> Black]
Can I plot and deal with implicit functions in Mathematica?
for example :-
x^3 + y^3 = 6xy
Can I plot a function like this?
ContourPlot[x^3 + y^3 == 6*x*y, {x, -2.7, 5.7}, {y, -7.5, 5}]
Two comments:
Note the double equals sign and the multiplication symbols.
You can find this exact input via the WolframAlpha interface. This interface is more forgiving and accepts your input almost exactly - although, I did need to specify that I wanted some type of plot.
Yes, using ContourPlot.
And it's even possible to plot the text x^3 + y^3 = 6xy along its own curve, by replacing the Line primitive with several Text primitives:
ContourPlot[x^3 + y^3 == 6 x y, {x, -4, 4}, {y, -4, 4},
Background -> Black, PlotPoints -> 7, MaxRecursion -> 1, ImageSize -> 500] /.
{
Line[s_] :>
Map[
Text[Style["x^3+y^3 = 6xy", 16, Hue[RandomReal[]]], #, {0, 0}, {1, 1}] &,
s]
}
Or you can animate the equation along the curve, like so:
res = Table[ Normal[
ContourPlot[x^3 + y^3 == 6 x y, {x, -4, 4}, {y, -4, 4},
Background -> Black,
ImageSize -> 600]] /.
{Line[s_] :> {Line[s],
Text[Style["x^3+y^3 = 6xy", 16, Red], s[[k]], {0, 0},
s[[k + 1]] - s[[k]]]}},
{k, 1, 448, 3}];
ListAnimate[res]
I'm guessing this is what you need:
http://reference.wolfram.com/mathematica/Compatibility/tutorial/Graphics/ImplicitPlot.html
ContourPlot[x^3 + y^3 == 6 x*y, {x, -10, 10}, {y, -10, 10}]
I was initially attempting visualize a 4 parameter function with Plot3D and Manipulate sliders (with two params controlled by sliders and the other vary in the "x-y" plane). However, I'm not getting any output when my non-plotted parameters are Manipulate controlled?
The following 1d plot example replicates what I'm seeing in the more complex plot attempt:
Clear[g, mu]
g[ x_] = (x Sin[mu])^2
Manipulate[ Plot[ g[x], {x, -10, 10}], {{mu, 1}, 0, 2 \[Pi]}]
Plot[ g[x] /. mu -> 1, {x, -10, 10}]
The Plot with a fixed value of mu has the expected parabolic output in the {0,70} automatically selected plotrange, whereas the Manipulate plot is blank in the {0, 1} range.
I was suspecting that the PlotRange wasn't selected with good defaults when the mu slider control was used, but adding in a PlotRange manually also shows no output:
Manipulate[ Plot[ g[x], {x, -10, 10}, PlotRange -> {0, 70}], {{mu, 1}, 0, 2 \[Pi]}]
This is because the Manipulate parameters are local.
The mu in Manipulate[ Plot[ g[x], {x, -10, 10}], {{mu, 1}, 0, 2 \[Pi]}] is different from the global mu you clear on the previous line.
I suggest using
g[x_, mu_] := (x Sin[mu])^2
Manipulate[Plot[g[x, mu], {x, -10, 10}], {{mu, 1}, 0, 2 \[Pi]}]
The following works too, but it keeps changing the value of a global variable, which may cause surprises later unless you pay attention, so I don't recommend it:
g[x_] := (x Sin[mu])^2
Manipulate[
mu = mu2;
Plot[g[x], {x, -10, 10}],
{{mu2, 1}, 0, 2 \[Pi]}
]
It may happen that you Clear[mu], but find that it gets a value the moment the Manipulate object is scrolled into view.
Another way to overcome Manipulate's localization is to bring the function inside the Manipulate[]:
Manipulate[Module[{x,g},
g[x_]=(x Sin[mu])^2;
Plot[g[x], {x, -10, 10}]], {{mu, 1}, 0, 2 \[Pi]}]
or even
Manipulate[Module[{x,g},
g=(x Sin[mu])^2;
Plot[g, {x, -10, 10}]], {{mu, 1}, 0, 2 \[Pi]}]
Both of which give
Module[{x,g},...] prevents unwanted side-effects from the global context. This enables a simple definition of g: I've had Manipulate[]ed plots with dozens of adjustable parameters, which can be cumbersome when passing all those parameters as arguments to the function.
I want to look at both the real and imaginary parts of some functions that depend on a parameter n. Individually (with set values of n), I get perfectly nice graphs, but when putting them in a Manipulate they become very small.
Here is the exact code I'm using; remove the manipulate and the graphs display at a good size, but with it they are too small to be legible.
Manipulate[
Plot3D[Im[Sqrt[-1 + (x + I y)^2 n]], {x, -2, 2}, {y, -1, 1},
AxesLabel -> Automatic]
Plot3D[Re[Sqrt[-1 + (x + I y)^2 n]], {x, -2, 2}, {y, -2, 2},
AxesLabel -> Automatic]
, {n, 1, 10, 1}]
Why is it doing this, and how can I fix it?
Manipulate[
Row[{
Plot3D[Im[Sqrt[-1 + (x + I y)^2 n]], {x, -2, 2}, {y, -1, 1},
AxesLabel -> Automatic, ImageSize -> 300] ,
Plot3D[Re[Sqrt[-1 + (x + I y)^2 n]], {x, -2, 2}, {y, -2, 2},
AxesLabel -> Automatic, ImageSize -> 300]}],
{n, 1, 10, 1}]
Edit
Remember that you may also do something like:
a = Sequence ##{{x, -2, 2}, {y, -1, 1}, AxesLabel-> Automatic, ImageSize-> 200};
Manipulate[
Row[{
Plot3D[Im[Sqrt[-1 + (x + I y)^2 n]], Evaluate#a],
Plot3D[Re[Sqrt[-1 + (x + I y)^2 n]], Evaluate#a]}],
{n, 1, 10, 1},
PreserveImageOptions -> False]
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]
]