Skip to main content

performance tuning - Radial distribution function


I have tried to rewrite my old IDL code to calculate the radial correlation function for a regular 2D crystal structure.


The theory behind this function is given here: https://en.wikipedia.org/wiki/Radial_distribution_function


Update 1:



The radial density distribution counts the number of points in a distance between $r$ and $r +\Delta r$ from each considered central point (below one marked as red). The area of such a "shell" is $2\pi r \Delta r$. The density distributions are averaged for all center points and then normalized by the total point density times the ring area for each radius.




enter image description here


Update 2:



It is important that the maximum radius (the maximum shell) for each center point does not cross the edges of the available point coordiantes (corresponding to the range defined by the smallest and largest x and y point value). That means that a point close to the edges has a maller maximum radius (smallest distance to the next edge) than a center point (for a square: maximum radius = half of the diagonal).



My question: is it possible to improve my "unreadable" and slow code, which does not use any special mathematica functions?.


Also I must have made a normalization error, since the function does not converge at g(r)=1 (see plot below).


As input I have taken a crystal image recorded with a high resolution camera:


enter image description here


Dr. belisarius has detected all coordinates by the following one-liner:



pts = ComponentMeasurements[Binarize@ImageSubtract
[image, BilateralFilter[image, 4, 1]], "Centroid"];

These points pts I have used to determine the radial correlation function.


The resulting plot is:


enter image description here


The full code is given here:


Clear[radialDensityDistribution];
radialDensityDistribution [listData_, mrr_: 0, mrc_: dDiag,
subdivision_: 50] :=

(
(*listData: list of 2D data points*)
(*mrr: distance from edge*)
(*mrc: calculation radius from central point*)
(*subdivision: number of in size's from mean point distance*)

n = Length[listData];

x = listData[[All, 1]];
y = listData[[All, 2]];


minCorner = {Min[x], Min[y]};
maxCorner = {Max[x], Max[y]};

diag = maxCorner - minCorner;
dDiag = Sqrt[diag.diag];

area = diag[[1]]*diag[[2]];

pointDensity = n/area;


deltaR = (area/n)^(1.0/2);
dr = deltaR/subdivision;

maxShell = Floor[mrc/dr];

g = Array[0 &, maxShell];
centralPoint = Array[0 &, n];

com = {Mean[x], Mean[y]};

radii = Sqrt[(x - com[[1]])^2 + (y - com[[2]])^2];
maxrad = Max[radii] // N;

centralIndex = Flatten@Position[radii + mrr, n_ /; n <= maxrad];
nCentral = Length[centralIndex];

p = {x[[centralIndex]], y[[centralIndex]]};

g = 0;


Table[
dist = {p[[1, 2 ;; All]] - p[[1, i]],
p[[2, 2 ;; All]] - p[[2, i]]}[[All, i ;; All]];
shell =
Floor[Sqrt[
dist[[1, All]]*dist[[1, All]] + dist[[2, All]]*dist[[2, All]]]/
dr];
h = HistogramList[shell, {0, maxShell - 1, 1}][[2, All]];
g = h + g,
{i, 1, nCentral - 1}

];

Table[
areaShell = Pi*(((shell + 1.0)*dr)^2 - (shell*dr)^2);
g[[shell]] = g[[shell]]/(1.0*nCentral*areaShell*pointDensity),
{shell, 1, maxShell - 1}
];

rn = (Range[maxShell - 1] - 0.5)*dr;
{g, rn, deltaR}

)

image = Import["http://i.stack.imgur.com/czhuI.png"];

pts = ComponentMeasurements[
Binarize@ImageSubtract[image, BilateralFilter[image, 4, 1]],
"Centroid"][[All, 2]];

extx = Max[pts[[All, 1]]] - Min[pts[[All, 1]]];
exty = Max[pts[[All, 2]]] - Min[pts[[All, 2]]];

ext = Min[extx, exty];

{g, rn, deltaR} = radialDensityDistribution [pts, ext/4, ext/4, 20];

ListLinePlot[Transpose[{rn, g}], PlotRange -> Full, Frame -> True,
FrameLabel -> {{"g(r)", ""}, {"r (pixels)", ""}}, ImageSize -> Large]

Answer



I will show a simple and fast approach to computing the pair correlation function (radial distribution function) for a 2D system of point particles.:


radialDistributionFunction2D[pts_?MatrixQ, boxLength_Real, nBins_: 350] :=
Module[{gr, r, binWidth = boxLength/(2 nBins), npts = Length@pts, rho},

rho = npts/boxLength^2; (* area number density *)
{r, gr} = HistogramList[(*compute and bin the distances between points of interest*)
Flatten @ DistanceMatrix @ pts, {0.005, boxLength/4., binWidth}];
r = MovingMedian[r, 2]; (* take center of each bin as r *)
gr = gr/(2 Pi r rho binWidth npts); (* normaliza g(r) *)
Transpose[{r, gr}] (* combine r and g(r) *)
]

Here is how you use it:


rdf = radialDistributionFunction2D[pts, 1023.];

ListLinePlot[rdf, PlotRange ->{{0, 150}, All}, Mesh -> 80]

Mathematica graphics


Notice that you get the correct normalization for free. This took about 1.2 seconds on my machine. I have restricted the plot range to show the interesting features.


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...