How to create a wind rose with Mathematica?

Question

I am trying to plot wind rose in Mathematica, but have no idea how to do this.

Any suggestions ?

Answer 1

Edit:

PolarTicks uses the built-in option for "Direction". An earlier version of this answer shows how to manually add PolarTicks.

---

The following displays the wind rose on the right (with different data points). As rcollyer notes, the data points and joining lines can be both achieved in a single use of ListPolarPlot through PlotMarkers->Automatic. The PlotLabel is from Kuba.

r = Table[{2 t Pi/16, RandomReal[{1, 6}]}, {t, 0, 15}];
ListPolarPlot[Append[r, r[[1]]],
  PlotLabel -> "风向图", BaseStyle -> 14,
  PolarTicks -> {"Direction", Automatic},
  Joined -> True, PlotMarkers -> Automatic, 
  PlotStyle -> {PointSize[Large]}, PolarAxes -> True, 
  PolarGridLines -> {Table[2 k Pi/16, {k, 0, 15}], Automatic}, 
  PolarAxesOrigin -> {Pi/2, 6}]

Answer 2

Edit note: I want to thank to all upvoters, this is really shocking and motivating :). Just to make this answer covering both graphs I've added right graph made with SectorChart like I suggested in comments and to not clone David's solution.

data = RandomReal[{1, 5}, 16];

Left graph: For equally spaced (in angle) measurements it is easier to use Mesh for ParametricPlot:

data2 = ({Cos@#, Sin@#} & /@ Range[0, 15 Pi/8, Pi/8]) data; AppendTo[data2, data2[[1]]];
f = BSplineFunction[data2, SplineDegree -> 1];

g2 = ParametricPlot[(r f[t]), {t, 0, 1}, {r, 0, 1}, BoundaryStyle -> {Thick, Black},
                                  Frame -> False, PlotRangePadding -> .2, PlotRange -> 6
                                  MeshFunctions -> (#3 &), Mesh -> (Length[data2] - 2), 
                                  MeshShading -> {White, Black},  Axes -> True, 
                                  Ticks -> None, AxesStyle -> Arrowheads@.07]

Right graph:

data2 = Transpose[{ConstantArray[1, 16], data}];

g1 = SectorChart[data2, PolarTicks -> {"Direction", Automatic}, PlotLabel -> "风向图", 
           BaseStyle -> {15, Bold},  SectorOrigin -> -Pi/16, PolarAxes -> True,
           PolarGridLines -> {Range[Pi/16, 2 Pi, Pi/8], Range[0, Ceiling@Max[data]]}, 
           PolarAxesOrigin -> {(Pi/8 (Position[#, Min@#] &@data-1)])[[1, 1]], 
                               Ceiling@Max[data]},
           ColorFunction -> (Blend["Rainbow", #2] &)]

Row[{g2, g1}]

---

Old one: works also for not equally spaced points (in angle). Why does ListPolarPlot has no Filling->{0,0} option? :(

Graphics[{EdgeForm@Thick,
          Polygon[Riffle[data2, f[0, 0], {2, -1, 3}] /. f -> List],
          White, 
          Polygon[Riffle[data2, f[0, 0], {3, -1, 3}] /. f -> List]
         },
         Axes -> True, PlotRangePadding -> .2, AxesStyle -> Arrowheads@.07,
         Ticks -> None, PlotRange -> 5]

Answer 3

Using WeatherData and Kuba's code we can use Mathematica to produce an actual wind rose with real data. This is the function I came up with:

windRose[city_] := Module[{data, total},
  (* Base the wind rose on thirty years of data, as seems to be customary *)
  data = WeatherData[city, "WindDirection", {{1983, 1}, {2013, 1}}];
  data = Select[
    DeleteCases[data /. {{__}, x_} :> x, _Missing], # < 360 &];
  data = HistogramList[data, {22.5}];
  data = data[[2]];
  total = Total[data];
  data = Transpose[{ConstantArray[1, 16], data/total}];

 SectorChart[data, PolarTicks -> {"Direction", Automatic}, 
 PlotLabel -> city, BaseStyle -> {15, Bold}, SectorOrigin -> -Pi/16, 
 PolarGridLines -> {Range[Pi/16, 2 Pi, Pi/8], Automatic}, 
 PolarAxesOrigin -> {(Pi/8 Position[#, Min@#] &@data[[;; , 2]])[[1, 1]], Max@data[[;; , 2]]}, 
 ColorFunction -> (Blend["Rainbow", #2] &), PolarAxes -> True]
  ]

The argument could be a pair of coordinates, a name of a city or a weather station ID. In the end the wind rose will represent the weather station. If we enter a city, Mathematica will choose a weather station in/by that city. So for example:

windRose["Chicago"] (* Chicago, Illinois *)

windRose["Gothenburg"] (* Gothenburg, Sweden *)

Answer 4

Nothing fancy here, just showing that you can fake Filling in ListPolarPlot by creating a Polygon (with a vertex at the origin) from each line segment:

windrose[data_] := Module[{dat = Append[data, First@data]},
  ListPolarPlot[dat, DataRange -> {0, 2 Pi}, Joined -> True, 
    Ticks -> None, Mesh -> Length[dat] - 2, 
    MeshShading -> {White, Black}, MeshStyle -> None] /. 
   Line[{x__}] :> {Polygon[{x, {0, 0}}], Black, Line[{x}]}]

windrose[RandomReal[{0.3, 1.0}, 16]]