Which result for normalized mutual information is correct?

Question

I wanted to find the normalized mutual information to validate a clustering algorithm, but I've encountered two different values depending on the library I use.

In Python:

from sklearn import metrics
labels_true = [0, 0, 0, 1, 1, 1]
labels_pred = [1, 1, 0, 0, 3, 3]

nmi = metrics.normalized_mutual_info_score(labels_true, labels_pred)

This returns nmi = 0.52954.

In R:

library(aricode)
labels_true = c(0, 0, 0, 1, 1, 1)
labels_pred = c(1, 1, 0, 0, 3, 3)

nmi = NMI(labels_true,labels_pred)

This returns n = 0.42061.

Which one should I trust? I don't really know why they are returning different results if there is a closed formula for NMI...

Answer 1

It turns out that the R version uses by default the 'max' variant of the NMI, while Python's uses 'sqrt'.

Doing this in R will yield the same results:

NMI(labels_true, labels_pred, variant="sqrt")