ArrayPlot in polar coordinates?

Question

I have a 2-D array ($360 \times 512$). The 360 rows represents the azimuthal angle from (0° to 359°) and the 512 columns are radial segments.

This is the ArrayPlot of my data.

ArrayPlot[data, Frame -> True, FrameLabel -> {{Style["Range", 18], None},
          {Style["Azimuth", 18], None}}, ColorFunction -> "Rainbow",
          ImageSize -> 480, AspectRatio -> 1, PlotRangePadding -> None]

I would like to know how to plot the data in polar coordinates. I see there were already some discussions on this topic, e.g. How to make a ArrayPlot/MatrixPlot in polar coordinates?

I tried to follow the steps, but did not succeed. Not sure if my array is too large, the running time is extremely long. For example,

polararrayplot[array_, colourfunc_] :=
SectorChart[Map[Style[{1, 1}, colourfunc[#]] &, array, {2}], SectorSpacing -> None];

polararrayplot[data, ColorData["Rainbow", #] &]

May I know which way of plotting suggested in the above link is more efficient? Or there is any newly available function in Mathematica in recent years? I am using Version 11.1.

I have uploaded the data here or here if anyone would like to take a look.

Many thanks in advance.

Answer 1

Here is an alternative approach, which is a slightly simplified version of the "diskfun" methods available in Chebfun. One might wish to note the similarities and differences between this method and the method in this previous answer.

Briefly: I will construct an interpolaing function to the OP's data by using SVD to decompose the data into a "radial part" and "angular part".

Import the data first:

wkong = Import["https://pastebin.com/raw/rnig0VMq", "CSV"];

Then, prepare the data for the disk interpolant construction by padding and reflecting it appropriately:

rflct = Append[#, First[#]] &[Join[Reverse[Drop[wkong, None, 1], 2], wkong, 2]];

Now, perform the SVD:

{u, s, v} = SingularValueDecomposition[rflct, Min[Dimensions[rflct]]];

Construct the components of the interpolating function:

u[[{1, -1}]] = Mean[u[[{1, -1}]]];
uf = ListInterpolation[#, {{0, 2 π}}, PeriodicInterpolation -> True] &
     /@ Transpose[u];
vf = ListInterpolation[#, {{-1, 1}}] & /@ Transpose[v];
s = Diagonal[s];

Let us look at a log plot of the singular values:

ListLogPlot[s]

It looks like we should be fine with taking the first $100$ components for plotting; thus,

With[{p = 100}, 
     DensityPlot[Total[Through[Take[vf, p][Norm[{x, y}]]] Take[s, p]
                       Through[Take[uf, p][ArcTan[x, y]]], 
                       Method -> "CompensatedSummation"], {x, y} ∈ Disk[], 
                 ColorFunction -> "ThermometerColors", PlotPoints -> 95]]

It takes a bit of time to plot, but not too overly long. You can always increase or decrease p as needed.

Answer 2

For your type of data, you may want to consider a more continuous type of plot than the sector chart that the solution you reference uses. I would suggest DensityPlot. Here's an example:

data = Import["~/Downloads/data.csv"];
interp = Interpolation@Flatten[MapIndexed[{#2, #} &, data, {2}], 1];

DensityPlot[
 interp[(Arg[x + y I] + Pi)/Degree, Sqrt[x^2 + y^2]],
 {x, -500, 500},
 {y, -500, 500},
 RegionFunction -> Function[{xx, yy}, 0 < Sqrt[xx^2 + yy^2] < 500],
 PlotPoints -> 100,
 Exclusions -> None,
 Frame -> False
 ]

I got the idea of using Arg instead of other alternatives from this answer.