Stress attribute -- sklearn.manifold.MDS / Python

Question

I'm using the scikit-learn method MDS to perform a dimensionality reduction in some data. I would like to check the stress value to access the quality of the reduction. I was expecting something between 0 - 1. However, I got values outside this range. Here's a minimal example:

%matplotlib inline

from sklearn.preprocessing import normalize
from sklearn import manifold
from matplotlib import pyplot as plt
from matplotlib.lines import Line2D

import numpy


def similarity_measure(vec1, vec2):
    vec1_x = numpy.arctan2(vec1[1], vec1[0])
    vec2_x = numpy.arctan2(vec2[1], vec2[0])
    vec1_y = numpy.sqrt(numpy.sum(vec1[0] * vec1[0] + vec1[1] * vec1[1]))
    vec2_y = numpy.sqrt(numpy.sum(vec2[0] * vec2[0] + vec2[1] * vec2[1]))

    dot  = numpy.sum(vec1_x * vec2_x + vec1_y * vec2_y)
    mag1 = numpy.sqrt(numpy.sum(vec1_x * vec1_x + vec1_y * vec1_y))
    mag2 = numpy.sqrt(numpy.sum(vec2_x * vec2_x + vec2_y * vec2_y))
    return dot / (mag1 * mag2)

plt.figure(figsize=(15, 15))

delta = numpy.zeros((100, 100))
data_x = numpy.random.randint(0, 100, (100, 100))
data_y = numpy.random.randint(0, 100, (100, 100))

for j in range(100):
    for k in range(100):
        if j <= k:
            dist = similarity_measure((data_x[j].flatten(), data_y[j].flatten()), (data_x[k].flatten(), data_y[k].flatten()))
            delta[j, k] = delta[k, j] = dist

delta = 1-((delta+1)/2)  
delta /= numpy.max(delta)

mds = manifold.MDS(n_components=2, max_iter=3000, eps=1e-9, random_state=0,
               dissimilarity="precomputed", n_jobs=1)
coords = mds.fit(delta).embedding_
print mds.stress_

plt.scatter(coords[:, 0], coords[:, 1], marker='x', s=50, edgecolor='None')
plt.tight_layout()

Which, in my test, printed the following:

263.412196461

And produced this image:

How can I analyze this value, without knowing the maximum value? Or how to normalize it, to have it between 0 and 1?

Thank you.

Answer 1

It is because current scikit-learn's implementation computes and returns raw Stress value (σ_r) while you are expecting Stress-1 (σ₁).

The former is not very informative (its high value does not necessarily indicate bad fit), and a better way of communicating reliability is to calculate a normed stress, eg. Stress-1 that according to Kruskal (1964, p. 3) has more or less the following interpretation: value 0 indicates perfect fit, 0.025 excellent, 0.05 good, 0.1 fair and 0.2 poor.

I just implemented calculation of Stress-1 and sent PR. In the meantime one can use version from this branch, where Stress-1 is used and returned instead of raw Stress when normalize parameter is set to True (False by default).

For more information cf. Kruskal (1964, p. 8-9) or Borg and Groenen (2005, p. 41-43).

Answer 2

While also searching for a Kruskal Stress, I found this french course of Ricco Rakotomalala. It contains an example of code that seems to calculate the correct Kruskal Stress :

import pandas
import numpy
from sklearn import manifold
from sklearn.metrics import euclidean_distances

## Input data format (file.csv) : dissimilarity matrix
#   ;  A  ;  B  ;  C  ;  D  ; E
# A ; 0   ; 0.9 ; 0.8 ; 0.5 ; 0.8
# B ; 0.9 ; 0   ; 0.7 ; 0   ; 1
# C ; 0.8 ; 0.7 ; 0   ; 0.2 ; 0.4
# D ; 0.5 ; 0   ; 0.2 ; 0   ; 0.8
# E ; 0.8 ; 1   ; 0.4 ; 0.8 ; 0


## Load data
data = pandas.read_table("file.csv", ";", header=0, index_col=0)

## MDS
mds = manifold.MDS(n_components=2, random_state=1, dissimilarity="precomputed")
mds.fit(data)
# Coordinates of points in the plan (n_components=2)
points = mds.embedding_

## sklearn Stress
print("sklearn stress :")
print(mds.stress_)
print("")

## Manual calculus of sklearn stress
DE = euclidean_distances(points)
stress = 0.5 * numpy.sum((DE - data.values)**2)
print("Manual calculus of sklearn stress :")
print(stress)
print("")

## Kruskal's stress (or stress formula 1)
stress1 = numpy.sqrt(stress / (0.5 * numpy.sum(data.values**2)))
print("Kruskal's Stress :")
print("[Poor > 0.2 > Fair > 0.1 > Good > 0.05 > Excellent > 0.025 > Perfect > 0.0]")
print(stress1)
print("")