It is my understanding that the Hellinger transformation is basically the square root of relative abundance data (if rows are samples). However, my row sums do not represent the total community (I am only looking at a fraction of it), and I am interested in keeping it that way. Since my data is already divided by the total species number for each row, wouldn't it be enough to take the square root my data to simulate the Hellinger transformation?

Is Hellinger even required if I can use relative abundance data (having fully saturated samples)?

  • $\begingroup$ From your comment in your other related Q it makes me wonder if you have the real row totals or just the row totals of the subset of columns (species) that your are working with. I don't think it would change my answer much if you don't have the actual row totals, but just note that if this is the case, then you are not doing a Hellinger transformation on the full data set but relativising to the subset when doing the Hellinger. $\endgroup$ May 6, 2015 at 23:18
  • $\begingroup$ I only have a OTU classification for one bacterial phylum. I do know the total bacterial signal in all my samples (the real row totals), but i only have one partial matrix i can work with. So i am relativising to the real row totals before. $\endgroup$
    – nouse
    May 7, 2015 at 8:12

1 Answer 1


The Hellinger transformation is defined as

$$ y^{\prime}_{ij} = \sqrt{\frac{y_{ij}}{y_{i.}}} $$

Where $j$ indexes the species, $i$ the site/sample, and $i.$ is the row sum for the $i$th sample.

If your data are already of the form $\frac{y_{ij}}{y_{i.}}$, but you've only taken a subset of the species, then yes, you can just apply a square root transformation to the data you are using and it would have been the same if you'd done the entire Hellinger transformation on the entire data set and then thrown out some of the species.

If you have a large number of taxa, in my experience I have found applying the Hellinger transformation (or just the square root to already proportional abundance data) to be an improvement over and above just analysing the % (or proportional) abundance data.

  • $\begingroup$ Can you elaborate on how the Hellinger transform improved your analysis? $\endgroup$
    – Galen
    Jul 31 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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