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Multivariate Wald-Wolfowitz test to compare the distributions of two samples
"""
Multivariate Wald-Wolfowitz test for two samples in separate CSV files.
See:
Friedman, Jerome H., and Lawrence C. Rafsky.
"Multivariate generalizations of the Wald-Wolfowitz and Smirnov two-sample tests."
The Annals of Statistics (1979): 697-717.
Given multivariate sample X of length m and sample Y of length n, test the null hypothesis:
H_0: The samples were generated by the same distribution
The algorithm uses a KD-tree to construct the minimum spanning tree, therefore is O(Nk log(Nk))
instead of O(N^3), where N = m + n is the total number of observations. Though approximate,
results are generally valid. See also:
Monaco, John V.
"Classification and authentication of one-dimensional behavioral biometrics."
Biometrics (IJCB), 2014 IEEE International Joint Conference on. IEEE, 2014.
The input files should be CSV files with no header and row observations, for example:
X.csv
-----
1,2
2,2
3,1
Y.csv
-----
1,1
2,4
3,2
4,2
Usage:
$ python X.csv Y.csv
> W = 0.485, 5 runs
> p = 0.6862
> Fail to reject H_0 at 0.05 significance level
> The samples appear to have similar distribution
"""
import sys
import numpy as np
import pandas as pd
import scipy.stats as stats
from sklearn.neighbors import kneighbors_graph
from scipy.sparse.csgraph import minimum_spanning_tree
SIGNIFICANCE = 0.05
def mst_edges(V, k):
"""
Construct the approximate minimum spanning tree from vectors V
:param: V: 2D array, sequence of vectors
:param: k: int the number of neighbor to consider for each vector
:return: V ndarray of edges forming the MST
"""
# k = len(X)-1 gives the exact MST
k = min(len(V) - 1, k)
# generate a sparse graph using the k nearest neighbors of each point
G = kneighbors_graph(V, n_neighbors=k, mode='distance')
# Compute the minimum spanning tree of this graph
full_tree = minimum_spanning_tree(G, overwrite=True)
return np.array(full_tree.nonzero()).T
def ww_test(X, Y, k=10):
"""
Multi-dimensional Wald-Wolfowitz test
:param X: multivariate sample X as a numpy ndarray
:param Y: multivariate sample Y as a numpy ndarray
:param k: number of neighbors to consider for each vector
:return: W the WW test statistic, R the number of runs
"""
m, n = len(X), len(Y)
N = m + n
XY = np.concatenate([X, Y]).astype(np.float)
# XY += np.random.normal(0, noise_scale, XY.shape)
edges = mst_edges(XY, k)
labels = np.array([0] * m + [1] * n)
c = labels[edges]
runs_edges = edges[c[:, 0] == c[:, 1]]
# number of runs is the total number of observations minus edges within each run
R = N - len(runs_edges)
# expected value of R
e_R = ((2.0 * m * n) / N) + 1
# variance of R is _numer/_denom
_numer = 2 * m * n * (2 * m * n - N)
_denom = N ** 2 * (N - 1)
# see Eq. 1 in Friedman 1979
# W approaches a standard normal distribution
W = (R - e_R) / np.sqrt(_numer/_denom)
return W, R
if __name__ == '__main__':
if len(sys.argv) != 3:
print('Usage: $ python runs_test.py [X] [Y]')
sys.exit(1)
X = pd.read_csv(sys.argv[1], header=None).values
Y = pd.read_csv(sys.argv[2], header=None).values
W, R = ww_test(X, Y)
pvalue = stats.norm.cdf(W) # one sided test
reject = pvalue <= SIGNIFICANCE
print('W = %.3f, %d runs' % (W, R))
print('p = %.4f' % pvalue)
print('%s H_0 at 0.05 significance level' % ('Reject' if reject else 'Fail to reject'))
print('The samples appear to have %s distribution' % ('different' if reject else 'similar'))
@CarboPablo
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CarboPablo commented Mar 29, 2022

Hi!

I'm trying to figure out what is the logic behind connecting the points for p=2.
I was thinking that when creating the MST and creating the edges to the nodes, you should take the minimal value of euclidean distance. But for example, in figure 1b, there is nodes that are closer to a node, but however, their edge goes to other point, for example:
In this case, we can connect the points withing this options;
logic_1
And then we will have this result:
logic_2
But instead of that, the figure show us that the correct way to do it is to do it this way:
logic_3
And I don't understand why. Maybe becuase you can't link twice nodes? That will make sense but I don't find where is that mentioned in the paper.

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