Skip to main content

programming - K-means clustering


In MATLAB, there is a command kmeans() that divides an array into $k$ clusters and calculates the centroid of each cluster. Is there any command in Mathematica to perform the same action?


For example:


x = {{1,2,3,4,5}, {6,7,8,9,10}, {11,12,13,14,15}, {16,17,18,19,20}, {21,22,23,24,25}}


The MATLAB kmeans command does this:


[b, c] = kmeans(x, 2)

It divides x into two clusters and calculates the centroids of these two clusters, and indicates for each element in the array which cluster it is a member of


b = {1,1,2,2,2}

and


c={{3.5000,4.5000,5.5000,6.5000,7.5000},{16.0000,17.0000,18.0000,19.0000,20.0000}}

Answer



ClusteringComponents is indeed the function to go for. To get the same results as MATLAB you need to do the following:



x = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 
15}, {16, 17, 18, 19, 20}, {21, 22, 23, 24, 25}};

cc = ClusteringComponents[x, 2, 1, Method -> "KMeans",
"DistanceFunction" -> SquaredEuclideanDistance, "RandomSeed" -> 1]


{1, 1, 2, 2, 2}



The arguments x and 2 are the same as with MATLAB. The '1' is used to indicate the level of the nested array to consider the data points. In this case we're looking at the top level, so we're considering 5D points.



Mathematica can use various methods. "KMeans" is default, so it isn't necessary to provide it here. The default "DistanceFunction" is EuclideanDistance. MATLAB's is SquaredEuclideanDistance, so we have to explicitly use that.


Since clustering uses a process with random initializations the results may differ depending on the RNG state. I used "RandomSeed" -> 1 to initialize the RNG to a state that yields the results you showed. {1,1,1,2,2} is a possible output too.


Now to the centroids:


Mean /@ {Pick[x, cc, 1], Pick[x, cc, 2]} // N


{{3.5, 4.5, 5.5, 6.5, 7.5}, {16., 17., 18., 19., 20.}}



I added //N since you seemed to want machine precision results. Leave it away for exact results.





On a side note: KMeans may sometimes yield disastrous results. It's a well-known property of this algorithm.


x1 = RandomVariate[MultinormalDistribution[{0, 0}, {{1, 0}, {0, 20}}],500];
x2 = RandomVariate[MultinormalDistribution[{6, 0}, {{1, 0}, {0, 20}}],500];
Graphics@{Red, Point@x1, Green, Point@x2}

Mathematica graphics


xx = Join[x1, x2];
cc = ClusteringComponents[xx, 2, 1,
Method -> "KMeans",
"DistanceFunction" -> SquaredEuclideanDistance, "RandomSeed" -> 1];

{c1, c2} = {Pick[xx, cc, 1], Pick[xx, cc, 2]};
Graphics@{Red, Point@c1, Green, Point@c2}

Mathematica graphics


In this case, one of the additional three clustering methods that Mathematica knows (Method -> "PAM") works wonders.


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...