Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with r
Search options not deleted
user 122077
Use this tag for any *on-topic* question that (a) involves `R` either as a critical part of the question or expected answer, & (b) is not *just* about how to use `R`.
5
votes
1
answer
2k
views
Which result for normalized mutual information is correct?
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? …
- 966
4
votes
Accepted
Which result for normalized mutual information is correct?
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") …
- 966