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:

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;

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

   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["https://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]
  • $\begingroup$ Could you please explain in words what do you do with the points in the shell in {r , r+dr} $\endgroup$ Mar 22, 2016 at 20:32
  • $\begingroup$ @Dr. belisarius, Thank you for your comment ... I added some more information. $\endgroup$
    – mrz
    Mar 22, 2016 at 22:16

3 Answers 3


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.

  • 2
    $\begingroup$ Isn't this affected by a bias coming from the points near the border? (not sure, I'm just asking) $\endgroup$ Mar 23, 2016 at 2:14
  • 1
    $\begingroup$ @Dr.belisarius. Sorry, I meant I was binning distances that fall within a quarter of the box length not just binning a quarter of the points. Usually for molecular dynamics simulation where one bin distances for different time periods half the box length is the standard max r to consider. I chose a quarter of the box length since the local density dies out way before that distance. $\endgroup$
    – RunnyKine
    Mar 23, 2016 at 3:31
  • 1
    $\begingroup$ @Dr.belisarius. You're welcome. I also want to mention that the only v10 function here is DistanceMatrix. But there's a function HierarchicalClustering`DistanceMatrix in v9 that does the same thing. But I think EuclideanDistance is not the default distance function, so you'll have to specify that. $\endgroup$
    – RunnyKine
    Mar 23, 2016 at 3:49
  • 2
    $\begingroup$ @mrz. You're right it is easily adaptable to 3D systems, all you have to do is change the normalization from 2 Pi r dr to 4 Pi r^2 dr and of course use volume density instead of area density. With regards to your update see my comment to @Dr.belisarius above. $\endgroup$
    – RunnyKine
    Mar 23, 2016 at 9:18
  • 2
    $\begingroup$ @mrz. Not a physicist but a physical chemist, yeah I know a fair bit about correlation functions. If you ask a question, I will try to answer it. $\endgroup$
    – RunnyKine
    Mar 23, 2016 at 9:37

"Disclaimer": This is solely an attempt to clean/speed up OP's code as it was presented, as I have not tried to figure out an alternative algorithm to do the calculation.

First a few comments:

  • You might want to localize all the intermediate variables you use in a Block.
  • When the deepest indice(s) in Part is All, we don't need to specify them. That is, p[[All,2,All,All]] is simply equivalent to p[[All,2]].
  • Table is not a looping construct per se: Table produces a table of values, so unless we do something with the result (e.g. tab = Table[...]), we should use something else. In OP's case, they could be replaced with Do, but as we will see, we can do away with them entirely.

The bottleneck seems to be the calls to HistogramList, which happens nCentral times, 4397 in the example. There are two things we can do with this part of the code: turn the procedural g=0;Do[...;g=g+h] into a functional equivalent using Fold, and replacing HistogramList with a home-cooked, faster variant. We look at the last point first.

The bin specification {0, maxShell - 1, 1} gives unit bins $[0,1)$, $[1,2)$, etc., and since shell has integer values, HistogramList just tallies the values in shell that are less than maxShell - 1. Thus we can

  1. Pick the shell values that are less than maxShell - 1
  2. Tally a Sorted version of the result
  3. Turn it into a SparseArray with default element 0 and correct length

Here is the code:

histList[shell_, maxShell_] :=
  Rule @@ Transpose[Tally[1 +
    Sort[Pick[shell, Negative[shell - maxShell + 1]]]
, maxShell - 1]

The g=0;Do[...;g=g+h] part can now be made functional using

gAcc[g_, i_] := Block[{dist, shell},
  dist = {p[[1, i + 1 ;; All]] - p[[1, i]], p[[2, i + 1 ;; All]] - p[[2, i]]};
  shell = Floor[Sqrt[dist[[1]]^2 + dist[[2]]^2]/dr];
  g + histList[shell, maxShell]

and finally

g = Fold[gAcc, 0, Range[nCentral - 1]];

On my machine this is about 4 times faster than OP's code.

Finally, the last part that modifies g can be vectorized, so no loop is necessary. Simply:

shells = Range[maxShell - 1];
areaShells = Pi*(((shells + 1.0)*dr)^2 - (shells*dr)^2);
g = g/(nCentral*areaShells*pointDensity);
  • $\begingroup$ Thank you vey much for your help ... your explanations are great ... $\endgroup$
    – mrz
    Mar 22, 2016 at 23:41
image = Import@"https://i.stack.imgur.com/czhuI.png"; 
pts = ComponentMeasurements[Binarize@ImageSubtract[image, BilateralFilter[image, 4, 1]], 
                           "Centroid"][[All, 2]];

lpts = Length@pts;

(* the following is slow and not  really needed, we could take the center 
   to be at ImageDimensions/2 *)

nm = NMinimize[Tr[#.# & /@  Thread[Subtract[pts, {x0 + Cos[ArcTan[-y0 + #[[2]],
            -x0 + #[[1]]]], y0 + Sin[ArcTan[-y0 + #[[2]], -x0 + #[[1]]]]} & /@ 
                                     pts]]], {{x0, 400, 600}, {y0, 400, 600}}]

(*{9.29992*10^8, {x0 -> 504.879, y0 -> 505.181}}*)

ctr = {x0, y0} /. nm[[2]]
centeredPts = Subtract[#, ctr] & /@ pts;
xyLimits = Transpose[{-1/2 #, 1/2 #} &@ctr];
ptsInZone = Select[pts, xyLimits[[1, 1]] < #[[1]] < xyLimits[[1, 2]] &&
                        xyLimits[[2, 1]] < #[[2]] < xyLimits[[2, 2]] &];

f = Nearest[pts];

g[pt_, rad_] := Rest@f[pt, {lpts, rad}]
ran[n_] := Range[n, 250, n]
ptsAtAllRanges[pt_, n_] := g[pt, #] & /@ ran[n]

scale = 2;
pp = Differences[Length /@ ptsAtAllRanges[#, scale]] & /@ ptsInZone;
ListLinePlot[(Mean /@ Transpose@pp)/Rest@ran[scale], PlotRange -> All]

The horizontal scale is divided by 2:

Mathematica graphics

  • 1
    $\begingroup$ Your g(r) is not normalized to 1. $\endgroup$
    – RunnyKine
    Mar 22, 2016 at 23:35
  • $\begingroup$ @RunnyKine I didn't (and don't) understand the full requirements. $\endgroup$ Mar 22, 2016 at 23:38
  • $\begingroup$ @RunnyKine But in any case it seems just a multiplication factor for all the set $\endgroup$ Mar 22, 2016 at 23:50
  • $\begingroup$ Yes, I wonder where that factor got lost. $\endgroup$
    – RunnyKine
    Mar 22, 2016 at 23:55
  • $\begingroup$ This is again a wonderful code ... the curve profile is correct. I am sure it will take me one week to roughly understand it ;-). Concerning the normalization: see here physics.emory.edu/faculty/weeks//idl/gofr2.html $\endgroup$
    – mrz
    Mar 23, 2016 at 0:48

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