I'm doing linear regression on compositional data using log-ratio transformation with census data. The IVs are compositional (percents summing to 100). The DV is non-compositional and continuous.

The alr and clr results are more easily interpreted. They all produce the same measure of fit. I'm inclined to go with alr (or clr). Aitchison characterizes ilr as the "pure mathematics" approach, but my audience is not statisticians or mathematicians.

If my objective is only to communicate insight from the analysis, why should I go with the much more difficult to interpret ilr (with balances) approach?

I've read heaps of research by Aitchison, Juan Jose Egozcue and Vera Pawlosky-Glahn but not looking to debate.


3 Answers 3


Continuing off of marianess's answer, clr is really not suitable due to the colinearity issue. In words if you try to make inferences with clr transformed data, you may fall in the trap of trying to infer increase/decreases of variables, which you can never never do with proportions in the first place.

The ilr transformation attempts to resolve this by just sticking to ratios of partitions, since ratios are stable quantities. These partitions can be represented as trees, where internal nodes in the tree represents the log ratio of the geometric means of the subtrees. This log ratios of subtrees is known as balances.

I'd also recommend checking out these publications, since they all have nice explanations of how to interpret the ilr transform.




Here is an IPython notebook that goes in the details of how to calculate balances given a tree

I also gave a description how to this with the modules in scikit-bio here in case you curious.

  • $\begingroup$ Why is the size of the output m - 1? $\endgroup$
    – O.rka
    Jun 13, 2018 at 1:02
  • $\begingroup$ Are you able to directly associate a feature with its a value? $\endgroup$
    – O.rka
    Jun 13, 2018 at 1:18
  • $\begingroup$ its m-1 since it is an isomorphism - you can only have at most m-1 contrasts before you start hitting collinearity issues. And yes, you should be able to link a feature to a specific partition see this answer here: stats.stackexchange.com/a/270203/79569 $\endgroup$
    – mortonjt
    Jun 15, 2018 at 3:05

There is a problem with clr() transformation. It does preserve the same amount variables after you transform the data, but in case of clr() you get a singular data (actually you get a singular covariance matrix): y1 + ... yD = 0. And as you might know, some statistical analysis cannot be performed on a singular data. ilr() transformation will reduce the number of your variable, so let's say you had D-dimensional space, but after ilr() you will end up with D-1. As a result, your transformed data is nothing more, but ratios. I recommend to read this paper here: http://is.muni.cz/do/rect/habilitace/1431/Hron/habilitace/15_Filzmoser_et_al__2010_.pdf

  • 1
    $\begingroup$ a useful distinction for clr. alr reduces the vector to D-1 as well. ilr ratios are of groups of variables(one or more), whereas alr ratios are of each single variable with the last variable in the vector the common denominator. ilr could provide insight that alr does not, but with my data most of the ilr ratios do not make any intuitive sense, even with balances. My current thinking is that it depends on your data (i.e. sometimes you want to see ratios of groups of variables, sometimes not). $\endgroup$
    – M Kearny
    Nov 2, 2016 at 3:55

I would go with ALR as it makes more sense. You use one component as baseline, or reference and then see what the others do in relation to that one.

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    $\begingroup$ Please read the tour. This is not an answer, but a comment. As soon as you have enough reputation you can comment on any post. $\endgroup$
    – Ferdi
    Feb 27, 2017 at 16:55
  • 4
    $\begingroup$ The ALR transform is an oblique basis. It can be very difficult to analyze data with respect to an oblique basis. Beyond this the ALR transform does not preserve metric concepts such as distance or variance and is sensitive to which part is taken as the denominator. $\endgroup$
    – jds
    Mar 28, 2017 at 16:29
  • $\begingroup$ Just to follow up on my last comment: statsathome.com/2017/08/09/… $\endgroup$
    – jds
    Aug 11, 2017 at 1:46

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