6
From Removing unwanted appearance of underlying mesh, which is a duplicate:
ContourPlot[x^2 + y^2, {x, -10, 10}, {y, -10, 10},
Contours -> {1.0, 4.0, 10.5}, ContourStyle -> None,
ContourShading -> {RGBColor[1, .2, 0, .1], RGBColor[1, .2, 0, .3],
RGBColor[1, .2, 0, .5], RGBColor[1, .2, 0, .7]},
Method -> {"TransparentPolygonMesh&...
5
Maybe you could use ContourPlot to obtain some good automatic spacing for the contour labels.
Clear[f, a, b, cp, plt, crd, txts]
f[x_, y_] = (x^3 + y^3)/(x^2 + y^2);
cp = ContourPlot[f[x, y], {x, -1, 1}, {y, -1, 1}, Contours -> 20,
ContourLabels -> All]
plt = Plot3D[f[x, y], {x, -1, 1}, {y, -1, 1}, MeshFunctions -> {#3 &},
Mesh -> 19]
...
4
I suspect this is a bug.
Note:
DiscretePlot3D[
PDF[DiscreteUniformDistribution[{{1, 6}, {1, 6}}], {x, y}], {x, 0,
6}, {y, 0, 6}, PlotMarkers -> {"Sphere", Large},
PlotRange -> {{0.5, 6.5}, {0.5, 6.5}, {0, .04}}]
Yet:
DiscretePlot3D[
PDF[DiscreteUniformDistribution[{{1, 6}, {1, 6}}], {x, y}], {x, 1,
6}, {y, 1, 6}, PlotMarkers -> {...
4
Use ImagePadding and PlotRangeClipping:
Plot[
{Piecewise[{{(5 - .5*q^(1/2))^2, q <= 100/9}, {(10 - 2*q^(1/2))^2,
100/9 < q < 25}}],
Piecewise[{{(10 - 2*q^(1/2))^2, q <= 100/9}, {(5 - .5*q^(1/2))^2,
100/9 < q < 30}}], 25 - q}, {q, 0, 27},
PlotRange -> {{0, 27}, {0, 27}},
AspectRatio -> 1,
Exclusions -> None,
...
3
I refer to this similar post to solve your problem(just use the plotrange as bounding box via the second argument of DiscretizeRegion).
a = Disk[{0, 1}, 0.7];
b = Disk[{-0.5, 0}, 1.3];
c = Disk[{0.5, 0}];
subsets = Subsets[{a, b, c}, {1, 3}];
subsetscolors =
Map[Function[{c},
Blend[Flatten[
Map[Table[Map[Append[#, 1.5/Length[c]] &, c], 2] ...
3
This does what you want, but I am not sure that it is better than a ContourPlot approach in any meaningful way:
Plot[Evaluate[y /. First@Solve[#, y] & /@ Thread@eqs], {x, -10, 10}]
3
The colors are not determined by the intensity but rather by the position of the function in the list of functions given as an argument to ParametricPlot3D. To make the colors correspond to the intensity(last column) use the option PlotStyle
Clear["Global`*"]
rotz[Ω_] := {{Cos[Ω], -Sin[Ω],
0}, {Sin[Ω], Cos[Ω], 0}, {0, 0, 1}}
rotx[Ω_] := {{1, 0,...
3
You can solve for $y$ to get the two level sets. Note that the plot in the question is inaccurate, even for a high number of plot points, and it seems to think they are left/right, but actually they are top/bottom:
{f1, f2} = y /. Solve[f[x, y] == 0, y] /. C[1] -> 0;
Plot[{f1, f2}, {x, 3.45, 3.5}]
These expressions are both ArcTan of a large inner ...
2
Your question is rather murky because your two examples aren't compatible in dimension. However, I think your main problem is that your dataplots needs to be a 2D array of array of plots, not a 2D array of tooltips. Consider the following code and see if it doesn't help you deal with your problem.
SeedRandom[5];
data1 = Table[RandomInteger[{1, 5}], {i, 1, 4}...
2
It's just a simple test.
{p, q, r} = {x^2 + (y - 1)^2 - 0.7^2 > 0, (x + 0.5)^2 + y^2 - 1.3^2 >
0, (x - 0.5)^2 + y^2 - 1^2 > 0};
boolean =
Reverse[List @@
Distribute[And[Or[p, ! p], Or[q, ! q], Or[r, ! r]], Or, And]];
(* boolean={(!a&&!b&&!c),(a&&!b&&!c),(!a&&b&&!c),(!a&&!b&&...
2
Our purpose is the follolwing graphics:
The ordinary differential equation at hand can be solved exactly in terms of elliptic functions. The solution will be periodic (with singularities) and so the numerical approach is not satisfactory since the solution cannot be continued past singularity unless one takes into account basic properties of elliptic ...
2
solve ode
Y = NDSolveValue[{(Pi y[x]^2) == Sqrt[ 1 + y'[x]^2] , y[0] == 1 },y, {x, -1, 1}, Method -> "StiffnessSwitching"]
Solution is only real for -.23<x<.42
revolute around x-axis
ParametricPlot3D[ {x, Y[x] Cos[t], Y[x] Sin[t]}, {x, -.3, 1}, {t, 0,2 Pi}, AxesLabel -> {x, y, z}, BoxRatios -> {1, 1, 1} ]
alternativly ...
2
When you override the Automatic setting you need to specify all of the ticks that you want.
Clear["Global`*"]
majorTick := {#1, #1, {0.0125`, 0.}, {GrayLevel[0.],
AbsoluteThickness[0.5]}} & ;
minorTick = {#1, "", {0.0075`, 0.}, {GrayLevel[0.],
AbsoluteThickness[0.25]}} &;
minorXticks = minorTick /@ ...
2
There two methods which can speed up the generation of the two circles.
Method I
Clear["`*"];
h = .3;
c = .9;
surf = y^2 + (Sqrt[x^2 + z^2] - 2)^2 - c^2;
T = ParametricPlot3D[{(2 + c Cos[2 Pi v]) Sin[
2 Pi u], (2 + c Cos[2 Pi v]) Cos[2 Pi u], c Sin[2 Pi v]}, {u, 0,
1}, {v, 0, 1},
RegionFunction ->
Function[{x, y, z, u, v}, ...
2
The mesh comes from reducing the opacity. To avoid the mesh, use fully opaque colors.
$Version
(* "12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)" *)
Clear["Global`*"]
ContourPlot[x^2 + y^2, {x, -10, 10}, {y, -10, 10},
Contours -> {1.0, 4.0, 10.5},
ContourStyle -> None,
ContourShading ->
{LightRed, Pink, Red, Darker[Red]...
2
Using the option MaxRecursion reduces the number of PlotPoints needed.
Clear["Global`*"]
f[x_, y_] :=
2 (-4 + x^2) Sinh[(π x)/3] +
1/16 (((4 + x^2)^2 + 64 (-4 + x^2) Cos[y] Cosh[(2 π x)/3] +
256 x Sin[y] Sinh[(2 π x)/3]) Sinh[π x] -
2 (4 + x^2)^2 Sinh[(5 π x)/3] + (-12 + x^2)^2 Sinh[(7 π x)/3]);
ContourPlot[f[x, y], {x, 3....
1
Not an answer,just a another thinking.
Clear["`*"];
m = {{1, 1}, {-1, 1}};
vars = {x, y};
b = {1, 2};
eqs = m.vars - b // Evaluate;
ParametricPlot[{u, v}, {u, -10, 10}, {v, -10, 10},
MeshFunctions -> (Function[{x, y}, #] & /@ eqs), Mesh -> {{0}},
MeshShading -> {{LightYellow, LightGreen}, {LightCyan, LightBrown}},
PlotStyle -> ...
1
There are ComplexPlot and ComplexPlot3D commands since version 12 of Mathematica. See http://reference.wolfram.com/language/ref/ComplexPlot.html and http://reference.wolfram.com/language/ref/ComplexPlot3D.html for more info.
1
I recommend using the AxesOrigin -> {0, 0} option. Like so:
Plot[
{Piecewise[{{(5 - .5*q^(1/2))^2, q <= 100/9}, {(10 - 2*q^(1/2))^2, 100/9 < q < 25}}],
Piecewise[{{(10 - 2*q^(1/2))^2, q <= 100/9}, {(5 - .5*q^(1/2))^2, 100/9 < q < 30}}], 25 - q}, {q, 0, 30},
PlotRange -> {{0, 27}, {0, 27}},
PlotRangePadding -> Scaled[.02],...
1
If you do not want part of the axes to show, draw them manually:
Plot[{
Piecewise[{{(5 - .5*q^(1/2))^2, q <= 100/9}, {(10 - 2*q^(1/2))^2,
100/9 < q < 25}}],
Piecewise[{{(10 - 2*q^(1/2))^2, q <= 100/9}, {(5 - .5*q^(1/2))^2,
100/9 < q < 30}}], 25 - q}, {q, 0, 30},
PlotRange -> {{0, 27}, {0, 27}},
PlotRangePadding -> 0....
1
You could use ListPlot to plot the points and then combine the two plots using Show.
Show[
ListPlot[{{25, 0}, {0, 25}},
PlotStyle -> {Black, PointSize[0.025]}, AspectRatio -> 1, Ticks -> None],
Plot[{Piecewise[{{(5 - .5*q^(1/2))^2, q <= 100/9}, {(10 - 2*q^(1/2))^2, 100/9 < q < 25}}],
Piecewise[{{(10 - 2*q^(1/2))^2, q <= 100/9},...
1
Plot[Sin[Exp[ Sin[x]]] - .6, {x, 0, 4 Pi}, Filling -> Axis,
FillingStyle -> {LightRed, LightBlue}]
Plot[Sin[Exp[ Sin[x]]] - .6, {x, 0, 4 Pi},
ColorFunction -> Function[{x, y}, Hue[y]], Filling -> Axis,
FillingStyle -> Automatic]
Show[Plot[Sin[Exp[ Sin[x]]] - .6, {x, 0, 4 Pi},
ColorFunction -> Function[{x, y}, Hue[y]], Filling -&...
1
According to the comment above posting this as an answer.
To get the desired behaviour you can wrap the elements of your chart in the symbolic wrapper (Tooltip in this case). Here's a crude approximation:
BarChart3D @ Table[
Tooltip[i^2,
ListLinePlot[Table[{x, Sin[x]}, {x, 0, i, .1}]
]
]
, {i, 1, 5}
]
Tooltip can contain anything in it's ...
1
Better practice is
ContourPlot[x y, {x, -2, 2}, {y, -2, 2}, ContourShading -> False,
Contours -> 10,
ContourLabels -> (Rotate[
Text[" " <> ToString[#3] <> " ", {#1, #2},
Background -> White], -ArcTan[#2/#1]] &)]
This reaches the level of for example Matlab contour plots that are very popular.
...
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