plotting - Turning a set of random walk data 45 degrees

I have a set of data that is just a "random" (generated by me, not by computer) sequence of length 2000 of 1's and (-1)'s. I used it to plot a 1-D random walk where +1 is step up, (-1) is step down, so my graph looked like this: enter image description here

I was asked to break the sequence in half and plot the first half against the second to create a 2-D random walk graph. This was easy enough, I just did it in excel. enter image description here

But I was then asked to turn the graph 45 degrees CCW so it looks like it's laid over a grid. So i guess if the first point is (1,1) it would go to (0,1), if it's (-1,1) it would go to (-1,0) and so on. I'm not sure how to do that. Please let me know if you know of a way to do it in Excel or Mathematica.

Answer

ListLinePlot[Accumulate @ Prepend[RandomChoice[{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}, 1000], 
  {1, 1}], AspectRatio -> Automatic]

Starting with some random data

rd = RandomChoice[{1, -1}, 2000];

ListLinePlot[Accumulate@rd]

Creating a random walk similar to the one shown in your question

rw = Accumulate@Transpose[{rd[[;; 1000]], rd[[1001 ;;]]}];

ListLinePlot[rw, AspectRatio -> Automatic]

and rotating it by 45°

rwr = rw.(RotationMatrix[45 Degree]/Sqrt[2]);


ListLinePlot[rwr, AspectRatio -> Automatic]

Comments

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Adding a thick curve to a regionplot

Suppose we have the following simple RegionPlot: f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}] Now I'm trying to change the curve defined by $y=g[x]$ into a thick black curve, while leaving all other boundaries in the plot unchanged. I've tried adding the region $y=g[x]$ and playing with the plotstyle, which didn't work, and I've tried BoundaryStyle, which changed all the boundaries in the plot. Now I'm kinda out of ideas... Any help would be appreciated! Answer With f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 You can use Epilog to add the thick line: RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, Epilog -> (Plot[g[x], {x, 0, 2}, PlotStyle -> {Black, Thick}][[1]]), PlotStyle -> {Directive[Yellow, Opacity[0.4]], Directive[Pink, Opacity[0.4]],

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