Hot answers tagged

50

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 ...


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 ...


36

Manipulate[ ParametricPlot[({Sin[n t1], Sin[(n - 1) t1]}), {t1, 0, 2 Pi}, Epilog -> {Red, PointSize[Large], Table[If[OddQ[i + j], Point[{Cos[Pi i/(2 (n - 1))], Cos[Pi j/(2 (n))]}]], {i, 2 n - 3}, {j, 2 n - 1}]}], {{n, 5}, 2, 20, 1}] General Case I We can generalize to the Lissajous curve specified by the two non-negative ...


34

Using color functions efficiently in data visualizations is more of an art than a recipe, so don't worry if you're not "good" at it yet. It's only a matter of time :) Copying the color schemes from seaborn: The best way to mimic those color schemes in Mathematica would be to copy the RGB values from seaborn for your preferred color scheme. You can find ...


32

Based on Oleksandr's excellent design idea here is my reimplementation of his package which offers much richer set of shapes. How to install the package The most recent version of the package can be installed from GitHub by evaluating the following: (* Load the package code *) package = ...


31

Styling closer to your example, using The Toad's colors: colors = {RGBColor[{0.9312692223325372, 0.8201921796082118, 0.7971480974663592}], RGBColor[{0.8822898168737189, 0.695820866705742, 0.7065457119485431}], RGBColor[{0.8135380254700676, 0.5705055182357822, 0.639280859468155}], RGBColor[{0.7195800708349119, 0.45537982893127477`, ...


30

Fortunately, Wikipedia has the answer, as long as we are content to restrict ourselves to non-intersecting closed polygons. This will probably be an acceptable limitation, given that excessively complicated plot markers tend to look slightly distracting anyway. Because we seek an aesthetic rather than rigorously well defined result, we do not need to be ...


30

This can be done more-or-less easily with a combination of options for AxesOrigin, PlotRange, and PlotRangePadding and the CustomTicks package (for easy outward-facing ticks). Needs["CustomTicks`"]; GapAxes[plot_Graphics, ticks : {{x__}, {y__}}, scalefactor_: Automatic] := With[ {prange = ticks[[All, 1 ;; 2]], s = Flatten@{scalefactor /. Automatic ...


30

Scientific progress! In v10.3 with all the goodies in AnatomyData we can now use the simple code: Entity["AnatomicalStructure", "Skin"]["Graphics3D"] Zoom in on the appropriate part and you're done. pelvisLoc = AnatomyData[Entity["AnatomicalStructure", "Pelvis"], "RegionBounds"]; Show[ Entity["AnatomicalStructure", "Skin"]["Graphics3D"], ...


29

This is the Collatz function I know: Collatz[1] := {1} Collatz[n_Integer] := Prepend[Collatz[3 n + 1], n] /; OddQ[n] && n > 0 Collatz[n_Integer] := Prepend[Collatz[n/2], n] /; EvenQ[n] && n > 0 Generating a graph from this is easy: Graph[(DirectedEdge @@@ Partition[Collatz[#], 2, 1]) & /@ Range[500] // Flatten // Union, ...


25

Normally Plot uses machine precision numbers; your $x^x$ expression is hitting the limit of the numbers that can be represented in machine precision right about $x>143$. Note: Solve[$MaxMachineNumber == x^x, x] (* Out: {{x -> 143.016}} *) You can increase the WorkingPrecision setting for Plot adequately, and the plot will be complete: f[x_] = ...


24

maybe this will provide a little insight: first look at the evaluation points used by ContourPlot: f[x_?NumericQ, y_?NumericQ] := (Sow[{x, y}]; Sin[3.2 x]*Sin[1.3*y] - 2.1*Sin[1.3*x]*Sin[3.2*y]); {plot, dat} = Reap[ContourPlot[f[x, y] == 0, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, ContourStyle -> Red]]; Row[{ ...


23

This package provides couple of functions for plotting commits data from GiHub: Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/Misc/GitHubPlots.m"] GitHubDateListPlot["hadley", "plyr"] GitHubBarChart["hadley", "plyr"] I think these plots are similar enough to the image in the question. There are number of ...


23

Source of the problem (possibly) Here is a clear indication that your Fortran library and the Mathematica function are behaving in fundamentally different ways. I noticed the apparent high frequency oscillations in the difference functional, so I decided to see exactly how quickly they oscillate, Plot[funcfortran[w, 0.06, -1.0, 1.0, 1.0, N[π/2], 2.0, 3., ...


22

Update: I have realized since writing this that the below algorithm is not exactly what Plot uses anymore. Given the reference in my reference, a book written by someone who knew Mathematica very well, I guess that the algorithm has been changed in more recent versions of Mathematica. The algorithm that Plot uses is explained in the link provided by ...


21

Graphics[{Circle[{0, 0}, 1, {0, Pi}], Circle[{0, 0}, .03], Line[{{1, 0}, {1, -.1}, {-1, -.1}, {-1, 0}}], Rotate[ Line[{{.03, 0}, {.6, 0}}] , #, {0, 0}] & /@ {0, Pi/2, Pi}, GeometricTransformation[ Piecewise[{ {{Red, Line[{{.8, 0}, {1, 0}}], Black, Line[{{.2, 0}, {.5, 0}}], Rotate[{Red, Text[#, {.75, 0}, {0, ...


21

If you are serious about using this extensively, consider making a function based on CreateDocument... Here is one way to pursue Szabolcs's line of thought. What follows is a function based on CreateDocument[] that can be used in conjunction with the (now somewhat neglected) option DisplayFunction, which handles where the output of graphics functions ...


21

If I'm not mistaken, a complement is defined as the set of elements in one set that are not contained in a given other set. In your case, you have specified the 'other' set (the union of S1, S2 and S3), but not the 'one' set. As you phrased it, I guess that set must be $\mathbb R^3$. So, the complement is the difference between an infinite space and a finite ...


21

There are built-in magnifying glasses. However, spontaneously I don't know how to invoke one directly for a Plot. Therefore I'm going to demonstrate one way that converts the Plot Graphics object into an Image: Image@Plot[Sin[x], {x, 0, 4}] FrontEndExecute[FrontEnd`Select2DTool["GetRectangleImageSelection"]] The image ribbon itself is ...


20

Ok, here's a very brief toy example while I don't have access to my desktop computer at work. It's easy enough to figure out, that a LogPlot of f is basically the plot of Log[f[x]]. And A LogLinearPlot is the plot of f[Exp[x]]. But we can extend this to arbitrary scalings of the axes. I start with defining a piecewise function which maps x values between 0 ...


20

If you are comfortable using undocumented and unsupported functionality we can do this with a ScalingFunctions option as I did for ListLogLinearPlot for the whole real numbers. (* listability *) (self : fn[off_, scale_])[x_List] := self /@ x (self : invfn[off_, scale_])[x_List] := self /@ x fn[off_, scale_][x_?NumericQ] := If[x < off, Log[x], ...


20

Here is an alternative answer. Of course, since you answered your own question, you may not need this. But I think the following is a viable alternative that may end up looking comparable, and has additional dynamic features. Instead of ListPlot, just use BubbleChart. data = ConstantArray[Range[5], 4] + Range[0, 6, 2]; newData = Map[MapIndexed[Join[#2, ...


20

I think SmoothDensityHistogram (docs here) is what you are looking for: data1 = RandomVariate[BinormalDistribution[{0, 0}, {2, 3}, 0.5], 100000]; data2 = RandomVariate[BinormalDistribution[{3, 4}, {2, 2}, .1], 100000]; data = data1~Join~data2; This is just some random sample data. If you plot it using ListPlot, you obtain the "blob" you mentioned: ...


20

Equation taken form the wiki page x[u_, v_] := (1 + (v/2) Cos[u/2]) Cos[u] y[u_, v_] := (1 + (v/2) Cos[u/2]) Sin[u] z[u_, v_] := (v/2) Sin[u/2] plot = ParametricPlot3D[{x[u, v], y[u, v], z[u, v]}, {u, 0, 2 Pi}, {v, -1, 1}, Boxed -> False, Axes -> False]


19

$(x^2+y^2-1)^2+(y^2+z^2-1)^2+(x^2+z^2-1)^2=0$ is satisfied by a set of points. This can be established: f = (x^2 + y^2 - 1)^2 + (y^2 + z^2 - 1)^2 + (x^2 + z^2 - 1)^2; FullSimplify[Reduce[f == 0, {x, y, z}, Reals]] Reduce[x^2 + y^2 == 1 && z^2 + y^2 == 1 && x^2 + z^2 == 1, {x, y, z}] i.e. (x == -(1/Sqrt[2]) || x == 1/Sqrt[2]) && ...


19

Wanna listen to a story? :) It was around 2002 when I finally became fed up with ParametricPlot3D[] and its inability to adaptively plot space curves. Recall that this was the old Graphics[] system where all the pictures were effectively done in PostScript. Thus, I set out to look for a way to adaptively plot curves in general. I was at the time very ...


18

A crude attempt This is for Mathematica 10+ only. To construct each face, I use an intersection between a unit 3-ball centred at the origin and a pyramid whose base is at infinity and apex is at the origin. Each edge of the pyramid passes through each vertex of the spherical face. The pyramid is given by ConicHullRegion[{origin}, {vertices}]. The ...


18

You can use your ContourPlot, you just need to wrap coordinates with GeoPosition, note that you have to flip order. (if x is longitude and y is latitude, because GeoPosition assumes first is latitude and so on.) cp is Graphics[GraphicsComplex[coordinates, primitives]...], it is convenient to use this form. We can apply GeoPosition in one place and reduce ...


18

Thanks to the links and tips provided by Mr.Wizard I found the answer I was looking for, with a combination of EvaluationMonitor and the undocumented Bag functionality. Two problems with the code I found: For some reason updating the Bag object doesn't trigger the evaluation of a Dynamic expression containing it. I had to add another variable which ...


18

Quick&Dirty: pt = {0, 0}; full = MandelbrotSetPlot[]; r = 0.2; Column[{ Row[{"Zoom: ", Slider[Dynamic[r], {0.01, 1}]}], Row[ { LocatorPane[Dynamic[pt], Dynamic[Show[full, Graphics[{EdgeForm[Red], Transparent, Rectangle[pt + r, pt - r]}], ImageSize -> Scaled[.45]]]], Dynamic[ MandelbrotSetPlot[{pt + r, ...



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