问题
I am interested in taking a look at the Eigenvalues after performing Multidimensional scaling. What function can do that ? I looked at the documentation, but it does not mention Eigenvalues at all.
Here is a code sample:
mds = manifold.MDS(n_components=100, max_iter=3000, eps=1e-9, random_state=seed, dissimilarity="precomputed", n_jobs=1) results = mds.fit(wordDissimilarityMatrix) # need a way to get the Eigenvalues
回答1:
I also couldn't find it from reading the documentation. I suspect they aren't performing classical MDS, but something more sophisticated:
“Modern Multidimensional Scaling - Theory and Applications” Borg, I.; Groenen P. Springer Series in Statistics (1997)
“Nonmetric multidimensional scaling: a numerical method” Kruskal, J. Psychometrika, 29 (1964)
“Multidimensional scaling by optimizing goodness of fit to a nonmetric hypothesis” Kruskal, J. Psychometrika, 29, (1964)
If you're looking for eigenvalues per classical MDS then it's not hard to get them yourself. The steps are:
- Get your distance matrix. Then square it.
- Perform double-centering.
- Find eigenvalues and eigenvectors
- Select top k eigenvalues.
- Your ith principle component is sqrt(eigenvalue_i)*eigenvector_i
See below for code example:
import numpy.linalg as la
import pandas as pd
# get some distance matrix
df = pd.read_csv("http://rosetta.reltech.org/TC/v15/Mapping/data/dist-Aus.csv")
A = df.values.T[1:].astype(float)
# square it
A = A**2
# centering matrix
n = A.shape[0]
J_c = 1./n*(np.eye(n) - 1 + (n-1)*np.eye(n))
# perform double centering
B = -0.5*(J_c.dot(A)).dot(J_c)
# find eigenvalues and eigenvectors
eigen_val = la.eig(B)[0]
eigen_vec = la.eig(B)[1].T
# select top 2 dimensions (for example)
PC1 = np.sqrt(eigen_val[0])*eigen_vec[0]
PC2 = np.sqrt(eigen_val[1])*eigen_vec[1]
来源:https://stackoverflow.com/questions/38112853/how-to-obtain-the-eigenvalues-after-performing-multidimensional-scaling