Solved – Adding labels to points using mds and scatter3d package with R

multidimensional scalingr

I have a dataset forwhich i have performed an mds and visualized the results using scatterplot3d library. However i would like to see the names of the points on the 3d plot. How do i accomplish that? Each column belongs to a certain group i would like to see which points belong to which groups on the 3dplot.

#generate a distance matrix of the data
d <- dist(data)

#perform the MDS on  3 dimensions and include a Goodness-of-fit (GOF)

fit.mds <- cmdscale(d,eig=TRUE, k=3) # k is the number of dimensions; 3 in this case

#Assign names x,y,z to the result vectors (dimension numbers)
x <- fit.mds$points[,1]
y <- fit.mds$points[,2]
z <- fit.mds$points[,3]

plot3d <- scatterplot3d(x,y,z,highlight.3d=TRUE,xlab="",ylab="",pch=16,main="Multidimensional Scaling 3-D Plot",col.axis="blue")

Best Answer

Basically, what you need is to store your scatterplot3d in a variable and reuse it like this:

x <- replicate(10,rnorm(100))
x.mds <- cmdscale(dist(x), eig=TRUE, k=3)
s3d <- scatterplot3d(x.mds$points[,1:3])
text(s3d$xyz.convert(0,0,0), labels="Origin")

Replace the coordinates and text by whatever you want to draw. You can also use a color vector to highlight the groups of interest.

The R.basic package, from Henrik Bengtsson, seems to provide additional facilities to customize 3D plots, but I never tried it.

Related Solutions

Solved – Normalizing data before applying MDS with strain criterion

Suppose you don't normalize your data. You could have a situation like this:

    user1   user2   user3   user4   user5   user6   user7   user8   user9
1   0.00    0.00    0.00    0.00    0.00    0.00    0.00    0.00    0.00
2   18      96      21      45      93      19      21      19      90
3   0.08    0.12    0.19    0.70    0.12    0.01    0.00    0.09    0.04
4   1019    217     53      1082    1010    2       30      0       100

So it is clear that some features influence the results much more than others.

Remember that the MDS is a way to force differences between elements in n dimensions in differences in 2 dimensions, between all the couples of elements! I simply think that in the first case the strain function

is affected by the lack of normalization. When you normalize the data the algorithm is able to catch the real differences among the points considering in a proper way all the different features.

Solved – MDS and PCA eigenvalues and eigenvectors

I find it helpful to consider the singular value decomposition for questions like this with the assumption that $X$ is a real matrix. Writing $X = UDV^T$, we can see that $XX^T = UD^2U^T$ and $X^TX = VD^2V^T$. As we can see, the eigenvalues of both $XX^T$ and $X^TX$ are contained in the diagonal matrix $D^2$ and are indeed equal. Also, we see that the matrix of eigenvectors of $XX^T$ is $U$, while the matrix of eigenvectors of $X^TX$ is $V$.

Because $$ X^TU = VDU^TU = VD, $$ we have the relationship $X^TU = VD$, similar to that pointed out by whuber.

Assuming $D$ is nonsingular, two additional properties that prove to be quite useful are:

$$ \begin{align} U &= XVD^{-1}\\ V &= X^TUD^{-1}. \end{align} $$

Best Answer

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Solved – Normalizing data before applying MDS with strain criterion

Solved – MDS and PCA eigenvalues and eigenvectors

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