make specific cluster

I have set of coordinates. I want to make clusters in which every point is within 1.5 distance unit of it's neighbor.

ex of point coordinates:

{{-12.945, 20.6509, 12.5901}, {-13.4452, 20.307, 111.626},
{-12.9731, 22.8458, 12.4215}, {-13.2381, 24.8167, 10.7147},
{-11.3668, 23.3908,11.8499}, {-11.6828, 23.7311, 10.8839},
{-13.3929, 21.1835, 9.86324}, {-11.5016, 21.3324, 10.1392},
{-12.3079, 22.096, 8.57246}, {-12.5268, 20.9679, 10.5444},
{-12.1951, 24.5423, 10.1807}, {-11.8887, 22.3883, 10.0751},
{-14.2529, 20.4808, 9.81084}, {-11.9876, 21.8094, 11.0478},
{-12.3718, 23.6176, 11.8266}, {-11.6179, 20.8324, 11.2154},
{-12.5927, 21.7492, 12.5087}, {-12.1665, 24.6649, 11.2909},
{-12.3854, 21.5571, 9.51876}, {-12.2237, 23.4278, 9.9787}}


what is the quickest way in Mathematica for this (for large data sets).

I tried this to find all points that are within mentioned distance:

Table[Select[List, EuclideanDistance[List[[i]], #] < 1.5 &], {i, 1, Length[[List]]}]


but now I have troubles to join all sets that have common elements.

• I formatted your code. As you have been here a while, you should learn how to do so, yourself. To see what changes I made, click on the "Edited ..." link above my gravitar. Commented Aug 19, 2013 at 16:25
• @rcollyer: my apologize. I'll be more careful. Commented Aug 19, 2013 at 16:32
• how'd you cluster, if you have say points {0,0,0}, {0,0,1} and {0,0,2}? I.e. the 2nd point could be clustered with both other, but the 1st not with the 3rd Commented Aug 19, 2013 at 16:40
• 'two clusters' means there is no point in one being within 1.5 distance unit of any point from another cluster. so these three numbers are all in one cluster together. Commented Aug 19, 2013 at 16:43
• Something like Gather[data, EuclideanDistance[#1, #2] < 1.5 &] ? Commented Aug 19, 2013 at 17:53

3 Answers

Here is a possible alteernative, I was working on while Kuba posted his answer :-) I also started by using FixedPoint and the inner loop seems to work but the outer one is easier with While.

c = {{-12.945, 20.6509, 12.5901}, {-13.4452, 20.307,
111.626}, {-12.9731, 22.8458, 12.4215}, {-13.2381, 24.8167,
10.7147}, {-11.3668, 23.3908, 11.8499}, {-11.6828, 23.7311,
10.8839}, {-13.3929, 21.1835, 9.86324}, {-11.5016, 21.3324,
10.1392}, {-12.3079, 22.096, 8.57246}, {-12.5268, 20.9679,
10.5444}, {-12.1951, 24.5423, 10.1807}, {-11.8887, 22.3883,
10.0751}, {-14.2529, 20.4808, 9.81084}, {-11.9876, 21.8094,
11.0478}, {-12.3718, 23.6176, 11.8266}, {-11.6179, 20.8324,
11.2154}, {-12.5927, 21.7492, 12.5087}, {-12.1665, 24.6649,
11.2909}, {-12.3854, 21.5571, 9.51876}, {-12.2237, 23.4278,
9.9787}};
MyClustering[data_List, distance_?NumericQ] :=
Module[{dataoriginal = data, res = {}, temp = {}},
While[dataoriginal =!= {},
temp = {};
AppendTo[res,
FixedPoint[(
Map[
Function[p,
temp = Join[temp,
Select[dataoriginal, EuclideanDistance[#, p] < distance &]];
dataoriginal = Complement[dataoriginal, temp]], #];
temp) &, {dataoriginal[[1]]}]]];
Return[res]]


Just few notes: dataoriginal is needed because I modify the original list and the argument of a function (data in that case) cannot be modified inside the function's body. For huge lists AppendTo is generally slow, so a possible alternative is

res = Join[{res}, FixedPoint[...]]

• It seems approach is the same :) but yours is tidy and compact +1. I will probably delete mine then, but later, I don't have time now to parse this. :)
– Kuba
Commented Aug 19, 2013 at 19:20
• @Kuba: I was focused on a annoying job so when I saw this question I shifted my attention on Mathematica and spent a couple of hours just to have an alibi for the other job not completed ;-) Commented Aug 19, 2013 at 19:25
• haha, the same here, but now I have to focus back on work :P
– Kuba
Commented Aug 19, 2013 at 19:28

Here's a different approach, though I think it's quite inefficient.

I treat the points as vertices in a graph. I check each pair of points and if the distance between them is less than 1.5 I connect them with an edge. The clusters are just the ConnectedComponents of the graph.

v = Range @ Length @ data;
e = UndirectedEdge @@@ Select[Subsets[v, {2}], EuclideanDistance @@ data[[#]] < 1.5 &];

ConnectedComponents @ Graph[v, e]
(* {{14, 8, 10, 12, 16, 19, 7, 20, 9, 13, 6, 11, 5, 15, 18, 4, 3, 17, 1}, {2}} *)

• SparseArrayStronglyConnectedComponents@SparseArray@UnitStep[1.5^2 - DistanceMatrix@data] follows the same logic but uses an undocumented internal function first publicized by Carl Woll. He found it to be quite fast. Commented Aug 22, 2013 at 5:18
• @RayKoopman I highly encourage you to pose an answer with this method. Also, other mentioned in this link are wort recalling.
– Kuba
Commented Jan 19, 2014 at 22:27

This is my interpretation:

your cluster is a set of points that for each one there is at least one within 1.5 distance.

I will not be surprised if there is some kind of one-liner but I haven't played with Clusters etc much.

This is straightforward approach:

data = (* your data*)

SetAttributes[f, HoldAll];
f[cluster_] := Module[{n = Length@cluster},
Do[
cluster = Join[cluster,
Select[data, EuclideanDistance[#, cluster[[i]]] < 1.5 &]
];
data = Complement[data, cluster];
, {i, n}];
cluster
];


The the inner loop should be done with FixedPoint but I've failed in implementation so I used While:

i = 0;
While[Length[data] > 0,
i++;
clusters[i] = data[[{1}]];
data = Rest@data;
start = 0;
end = 1;
While[start != end,
With[{i = i}, start = Length@clusters[i];
f[clusters[i]];
end = Length@clusters[i]];
];
]

set = clusters /@ Range@i

ListPointPlot3D[set, PlotStyle -> {Red, Blue}, BaseStyle -> AbsolutePointSize@10,
PlotRange -> All]
`

Looks reasonable :)

I have to focus and it will take some time to write explanation, be patient. Or maybe it will be pointless if this is an overkill :)