From Cartesian Plot to Polar Histogram using Mathematica - wolfram-mathematica

Question

Please Consider:
dalist={{21, 22}, {26, 13}, {32, 17}, {31, 11}, {30, 9},
{25, 12}, {12, 16}, {18, 20}, {13, 23}, {19, 21},
{14, 16}, {14, 22}, {18,22}, {10, 22}, {17, 23}}
ScreenCenter = {20, 15}
FrameXYs = {{4.32, 3.23}, {35.68, 26.75}}
Graphics[{EdgeForm[Thick], White, Rectangle ## FrameXYs,
Black, Point#dalist, Red, Disk[ScreenCenter, .5]}]
What I would like to do is to compute, for each point, its angle in a coordinate system such as :
Above is the Deisred output, those are frequency count of point given a particular "Angle Bin".
Once I know how to compute the angle i should be able to do that.

Answer 1

Mathematica has a special plot function for this purpose: ListPolarPlot. You need to convert your x,y pairs to theta, r pairs, for instance as follows:
ListPolarPlot[{ArcTan[##], EuclideanDistance[##]} & ### (#-ScreenCenter & /# dalist),
PolarAxes -> True,
PolarGridLines -> Automatic,
Joined -> False,
PolarTicks -> {"Degrees", Automatic},
BaseStyle -> {FontFamily -> "Arial", FontWeight -> Bold,FontSize -> 12},
PlotStyle -> {Red, PointSize -> 0.02}
]
UPDATE
As requested per comment, polar histograms can be made as follows:
maxScale = 100;
angleDivisions = 20;
dAng = (2 \[Pi])/angleDivisions;
Some test data:
(counts = Table[RandomInteger[{0, 100}], {ang, angleDivisions}]) // BarChart
ListPolarPlot[{{0, maxScale}},
PolarAxes -> True, PolarGridLines -> Automatic,
PolarTicks -> {"Degrees", Automatic},
BaseStyle -> {FontFamily -> "Arial", FontWeight -> Bold, FontSize -> 12},
PlotStyle -> {None},
Epilog -> {Opacity[0.7], Blue,
Table[
Polygon#
{
{0, 0},
counts[[ang + 1]] {Cos[ang dAng - dAng/2],Sin[ang dAng- dAng/2]},
counts[[ang + 1]] {Cos[ang dAng + dAng/2],Sin[ang dAng+ dAng/2]}
},
{ang, 0, angleDivisions - 1}
]}
]
A small visual improvement using Disk sectors instead of Polygons:
ListPolarPlot[{{0, maxScale}},
PolarAxes -> True, PolarGridLines -> Automatic,
PolarTicks -> {"Degrees", Automatic},
BaseStyle -> {FontFamily -> "Arial", FontWeight -> Bold,
FontSize -> 12}, PlotStyle -> {None},
Epilog -> {Opacity[0.7], Blue,
Table[
Disk[{0,0},counts[[ang+1]],{ang dAng-dAng/2,ang dAng+dAng/2}],
{ang, 0, angleDivisions - 1}
]
}
]
A clearer separation of the 'bars' is obtained with the addition of EdgeForm[{Black, Thickness[0.005]}] in the Epilog. Now the numbers marking the rings still have the unnecessary decimal point trailing them. Following the plot with the replacement /. Style[num_?MachineNumberQ, List[]] -> Style[num // Round, List[]] removes those. The end result is:
The above plot can also be generated with SectorChart although this plot is primarily intended to show varying width and height of the data, and isn't fine-tuned for plots where you have fixed-width sectors and you want to highlight directions and data counts in those directions. But it can be done by using SectorOrigin. The problem is I take it that the midpoint of a sector codes for its direction so to have 0 deg in the mid of a sector I have to offset the origin by \[Pi]/angleDivisions and specify the ticks by hand as they get rotated too:
SectorChart[
{ConstantArray[1, Length[counts]], counts}\[Transpose],
SectorOrigin -> {-\[Pi]/angleDivisions, "Counterclockwise"},
PolarAxes -> True, PolarGridLines -> Automatic,
PolarTicks ->
{
Table[{i \[Degree] + \[Pi]/angleDivisions, i \[Degree]}, {i, 0, 345, 15}],
Automatic
},
ChartStyle -> {Directive[EdgeForm[{Black, Thickness[0.005]}], Blue]},
BaseStyle -> {FontFamily -> "Arial", FontWeight -> Bold,
FontSize -> 12}
]
The plot is almost the same, but it is more interactive (tooltips and so).

Answer 2

That seems to be the polar coordinate system. The Cartesian-to-polar conversion formulas are in that same article:
This returns the angle in radians.

Answer 3

This
N#ArcTan[#[[1]], #[[2]]] & /# (# - ScreenCenter & /# dalist)
returns the list of angles of the ray from ScreenCenter to each point, in radians and between -pi and pi.
That is, I assumed you want the angle between each point in your plot and the red dot.
Note the use of ArcTan[x,y] rather than ArcTan[y/x], which automatically chooses the appropriate sign (otherwise you'd have to do it by hand, as in #Blender's answer).