18
$\begingroup$

I come across three different statistical measures to compare two sets, in particular to segmentation on images (e.g., comparing the similarity between the ground truth and the segmented result).

What are the differences between these measurements (they are quite similar mathematically):

I see papers using Dice more often, but others also suggest using Jaccard and overlap coefficients. What are their differences?

$\endgroup$
2
  • 2
    $\begingroup$ If you don't know about these measures already, you have to click on each Wikipedia page and then hold together in your head the key points from each. This isn't a self-contained question that will be predictably useful to anyone else. References are fine, but a question mustn't depend on people reading them. $\endgroup$
    – Nick Cox
    Oct 12, 2017 at 14:33
  • 2
    $\begingroup$ And also search this site for them; there is a lot of threads about, already. $\endgroup$
    – ttnphns
    Oct 12, 2017 at 14:35

1 Answer 1

13
$\begingroup$

From the wikipedia page: $$J=\frac{D}{2-D} \;\; \text{and}\;\; D=\frac{2J}{J+1}$$ where $D$ is the Dice Coefficient and $J$ is the Jacard Index. In my opinion, the Dice Coefficient is more intuitive because it can be seen as the percentage of overlap between the two sets, that is a number between 0 and 1.

As for the Overlap it represents the percentage of overlap as it relates only to the smallest volume: $$overlap = \frac{|X\cap Y|}{min(|X|, |Y|)}$$ The relation between it and the other two measures is not direct, but one can be get from one the others and vice-versa with information of the area/volume of $X$ and $Y$

$\endgroup$
1
  • 5
    $\begingroup$ I'm not sure if Dice is really more intuitive than Jaccard, however there are two distinct differences between the two. Dice does not satisfy the triangle inequality, hence $1-D$ can not properly be used as a metric, while $1-J$ can. The second is that $D\geq J$ and I have the suspicion that Dice is used more often because of this property - although I do not have any proof for that ;) $\endgroup$
    – reox
    Feb 11, 2021 at 6:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.