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

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plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.

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