Hot answers tagged

375

I have to confess that I see this as a proper challenge, as I am usually quite creative in finding/combining functions to provide a desired behavior. So I will give it another try. which is generated using box[x_, x1_, x2_, a_, b_] := Tanh[a (x - x1)] + Tanh[-b (x - x2)]; ex[z_, z0_, s_] := Exp[-(z - z0)^2/s] and r[z_, x_] := (*body*).4 (1.0 - .4 ...


362

The code below attempts to apply the XKCD style to a variety of plots and charts. The idea is to first apply cartoon-like styles to the graphics objects (thick lines, silly font etc), and then to apply a distortion using image processing. The final function is xkcdConvert which is simply applied to a standard plot or chart. The font style and size are set ...


306

Mostly thanks to Belisarius's elegant wrapping, you can do h[fun_, divisor_, color_, at_] := Module[{k}, k = BSplineFunction[Table[fun@x + RandomReal[{-0.1, 0.1}/divisor], {x, 0.01, 10, .1}]]; ParametricPlot[k[x], {x,0.1,0.9}, PlotStyle->{color, AbsoluteThickness@at}, Axes-> None]]; Show[{ h[{#, 1.5 + 10 (Sin[#]^2/Sqrt[#]) Exp[-(# - 5)^2/2]} ...


198

This might get me suspended from the site butt I cannot resist. The shape you are looking for can probably be approximated (depending how anal you want to be about the outcome) by two assymetric probability distributions. The obvious choices would be a Poasson or a log normal distribution. I will use the latter as it is continuous. Now the bummer is that ...


107

Completely redesigned at 2015 12 19! Simplified interface, more functionality, robust performance. Plots can now be easily uploaded to StackExchange from within PlotExplorer thanks to halirutan! Code is at end of post. Functionality: Works with any graphics object, plots, charts, etc. (e.g. Plot, ListPlot, ArrayPlot, RegionPlot, GeoGraphics, BarChart, ...


103

Parametric Buttocks Manipulator Manipulate[ ParametricPlot3D[{ (e u^p + (1 + (c - a u) (u - 1)) Cos[t]^2) Sin[t], (e u^p + (1 + (d - b u) (u - 1)) Cos[t]^2) Cos[t], 2 u}, {t, -0.2, Pi + 0.2}, {u, 0, 1.1}, Lighting -> "Neutral", Mesh -> None, PlotStyle -> Directive[Specularity[0], RGBColor[0.92, 0.85, 0.73]], Axes -> False], {{a, ...


92

This can be done with Overlay if the ImagePadding and the horizontal range for each plot is the same. For example, plot1 = ListLinePlot[ Accumulate[RandomReal[{0, 1}, {100}]], PlotStyle -> Blue, ImagePadding -> 25, Frame -> {True, True, True, False}, FrameStyle -> {Automatic, Blue, Automatic, Automatic} ] plot2 = ...


86

In case you want more flexibility, it's also possible to design your own legends, for example along the lines of this MathGroup post. For your example, the process would start with the function legendMaker. Instead of repeating the same definition as in the above post, I've overhauled legendMaker in response to image_doctor's answer, to separate out the ...


84

Time to join in the fun. version 2 Result Method I produce the basic plot with ticks and labels: Plot[{x/2, (x + Sin[x])/2.2}, {x, 0, 2 Pi}, MaxRecursion -> 0, PlotPoints -> 30, Axes -> False, Frame -> {True, True, False, False}, FrameTicks -> {{{0.2, "Start", 0.07}, {3, "lunch", 0.05}, {6, "Finish", 0.06}}, None}, PlotLabel -> ...


81

Building on Heike's ColorFunction, I came up with this: The white bits are the trickiest - you need to make sure the brightness is high where the saturation is low, otherwise the black lines appear on top of the white ones. The code is below. The functions defined are: complexGrid[max,n] simply generates an $n\times n$ grid of complex numbers ranging ...


75

I'm very late to the party, but here's a convenient xkcd guy generator: This was generated as: With[{ h = xkcdGuy[-10, "hat", 0.2, {20, -90}, {-57, -10}, {-20, 0}, {20, 0}], n = xkcdGuy[0, "none", -0.2, {-10, 0}, {50, 10}, {-20, 0}, {20, 0}]}, Graphics[{First@n, Rotate[Translate[First@h, {3.3, 0}], 10 Degree]}] ] // xkcdConvert using ...


75

Yes we can. The following DashedGraphics3D[ ] function is designed to convert ordinary Graphics3D object to the "line-drawing" style raster image. Clear[DashedGraphics3D] DashedGraphics3D::optx = "Invalid options for Graphics3D are omitted: `1`."; Off[OptionValue::nodef]; Options[DashedGraphics3D] = {ViewAngle -> 0.4, ViewPoint ...


68

To implement datenwolf's suggestion to perturb curves with Perlin noise to give that "hand-drawn" look and feel, here's one way to use one-dimensional Perlin noise for the perturbation: fBm = With[{permutations = Apply[Join, ConstantArray[RandomSample[Range[0, 255]], 2]]}, Compile[{{x, _Real}}, Module[{xf = Floor[x], xi, xa, u, i, j}, xi = ...


61

Basic method There appears to be a mechanism for doing just that, though I have yet to map its capabilities. As a basic example for the time being: Themes`AddThemeRules["wizard", DefaultPlotStyle -> Thread@Directive[{Purple, Orange, Hue[0.6]}, Thick], LabelStyle -> 18, AxesStyle -> White, TicksStyle -> LightGray, Background -> ...


58

Strategy is simple texture map 2D plot on a rectangle under your 3D surface. I took a liberty with some styling that I like - you can always come back to yours. contourPotentialPlot1 = ContourPlot[-3600. h^2 + 0.02974 h^4 - 5391.90 s^2 + 0.275 h^2 s^2 + 0.125 s^4, {h, -400, 400}, {s, -300, 300}, PlotRange -> {-1.4*10^8, 2*10^7}, Contours -> 15, ...


56

The list structure is not manifest to Plot as it has the attribute HoldAll (to get a function's attributes, either use Attributes[func] or ??func). Hence Plot evaluates the Table functions as one unit and it appears as if there is only one function, not four. Evaluate will make the list structure manifest and each function will be plotted with a separate ...


53

Edit: Added the reversal and some refinements ω = 1; posP[t_, φ_] := Sin[ω t + φ] {Cos[φ], Sin[φ]} posL[φ_] := {-#, #} &@{Cos[φ], Sin[φ]} Animate[ Graphics[{PointSize[0.02], Table[{Black, Line[posL[π i]], Hue[i], Point[posP[t, π i]]}, {i, 0, 1, 1/(3π-Abs[9.43-t])}] }, PlotRange -> {{-1.5, 1.5}, {-1.5, 1.5}} ], {t, 0, 6π, 0.2} ]


51

In this answer, I will concentrate on the colors only to create something like this Copying the colors from python is a very fast way to get similar results. Nevertheless, the best way to understand what's happening is still to read the underlying publication that was used in seaborn: A colour scheme for the display of astronomical intensity images ...


50

I take zero credit for this. It is a method I learned from Maxim Rytin. ContourPlot3D[{(x^2 + y^2 + z^2 + 8)^2 - 36 (x^2 + y^2), y^2 + (z - 2)^2 - 4}, {x, -4, 4}, {y, -4, 4}, {z, -2, 2}, Contours -> {0}, ContourStyle -> Opacity[0], Mesh -> None, BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> {{Green, Tube[.03]}}}, Boxed ...


49

If you want to make several sequences of the Collatz function for turning it into a graph, you probably want to memorize, which parts you already calculated. What we try to do is to create a graph like this (image from xkcd) When we would calculate the whole chain for each number until it (hopefully) reaches the end sequence 8,4,1 we do a lot of work over ...


48

The cure is undocumented, unfortunately now documented, as of version 10. Try adding Method -> {"AxesInFront" -> False}, like so: Plot[2 x - 2, {x, -10, 10}, PlotRange -> {{-10, 10}, {-10, 10}}, PlotStyle -> Directive[Black, AbsoluteThickness[2]], ImageSize -> 300, AxesStyle -> Directive[RGBColor[.8, .8, .8], ...


48

Consider this: ParametricPlot3D[ RotationTransform[a, {0, 1, 0}][{0, 0, Sin[3 a] + 5/4}], {a, 0, 2 Pi}, Evaluated -> True] Now rotate this around a circle, while rotating it at the same time around its' origin: ParametricPlot3D[ RotationTransform[b, {0, 0, 1}][{6, 0, 0} + RotationTransform[a + 3 b, {0, 1, 0}][{0, 0, Sin[3 a] + 5/4}]], ...


46

The colors alone are indexed color scheme #97: ColorData[97, "ColorList"] Update: further digging in reveals these PlotTheme indexed color relationships: {"Default" -> 97, "Earth" -> 98, "Garnet" -> 99, "Opal" -> 100, "Sapphire" -> 101, "Steel" -> 102, "Sunrise" -> 103, "Textbook" -> 104, ...


44

I'm posting this as a second answer, as it's really a completely different approach. It's also been substantially expanded as of April 25, 2012. While this still doesn't specifically address the question of adding a region, it does plot the countries separately. Of course, each country could be viewed as a region in itself. Our objective is to make a ...


43

Of course I can't know all the details of what you really want, but here is a simple starting point, plotGrid, that takes care of the relative sizing and the tick label display automatically. Instead of using panels or grids, I'm placing everything into a single Graphics by using a separate Inset for each plot in the list l. The dimensions of this list l ...


43

Needs["PolyhedronOperations`"] poly = Geodesate[PolyhedronData["Dodecahedron", "Faces"], 4]; amplitude = 0.15; twist = 4; verts = poly[[1]]; faces = poly[[2]]; phases = RandomReal[2 Pi, Length[verts]]; newverts[t_] := MapIndexed[{Sequence @@ (RotationMatrix[twist Last[#1]].Most[#1]), Last[#1]} (1 + amplitude Sin[t + phases[[First@#2]]]) &, ...


42

Here is a solution that uses a BezierCurve to indicate a "snipped" axes. The function snip[x] places the mark on the axes at relative position x (0 and 1 being the ends). The function getMaxPadding gets the maximum padding on all sides for both plots (based on this answer). The two plots are then aligned one over the other, with the max padding applied for ...


42

I'd like to expand on Quantum_Oli's answer to give an intuitive explanation for what's happening, because there's a neat geometric interpretation. At one point in the animation it looks like there is a circle of colored dots moving about the center, this is a special case of so called hypocycloids known as Cardano circles. A hypocyloid is a curve generated ...


41

An easy way to add a vertical line is by using Epilog. Here is an example: f[x_] := (x^2 z)/((x^2 - y^2)^2 + 4 q^2 x^2) /. {y -> π/15, z -> 1, q -> π/600} Quiet[maxy = FindMaxValue[f[x], x]*1.1] lineStyle = {Thick, Red, Dashed}; line1 = Line[{{π/15 + 1/50, 0}, {π/15 + 1/50, maxy}}]; line2 = Line[{{π/15 - 1/50, 0}, {π/15 - 1/50, maxy}}]; ...


39

You can also use MeshFunctions: Plot[{Cos[x], x Sin[x]}, {x, -3 Pi, 3 Pi}, MeshFunctions -> {(Cos[#] - # Sin[#]) &}, Mesh -> {{0}}, MeshStyle -> Directive[Red, PointSize[Large]]] Update: Dealing with Tan[x] using Exclusions Plot[{Tan[x], x Sin[x]}, {x, -3 Pi, 3 Pi}, MeshFunctions -> {(Tan[#] - # Sin[#]) &}, Mesh ...



Only top voted, non community-wiki answers of a minimum length are eligible