Dear Mike, You are absolutely right, and it the amount of covariance is another other element that is of interest to me. In broad terms I was thinking of relating covariation back to inital shape variation to gain a magnitude measure through its predictive% on shape variation, but I want and need to think about it some more on how to go about it (and its validity!).
Thank you for helping me, and your example is very useful Have a lovely day, Katrien > On 14 May 2020, at 13:09, Mike Collyer <[email protected]> wrote: > > Katrien, > > Even though one can calculate the fractional covariance as Philipp described, > one has to consider whether there is much overall covariance between > matrices. The following is an example of how the fractional covariances > based on squared singular values can be deceiving, in R. > > > set.seed(22) > > n <- 20 > > p <- 3 > > X <- matrix(rnorm(n * p), n, p) > > Y <- matrix(rnorm(n * p), n, p) > > > > X <- scale(X, scale = FALSE) > > Y <- scale(Y, scale = FALSE) > > > > sxy <- svd(crossprod(X, Y)/(n-1)) > > > > sxy$d^2 # squared singular values > [1] 0.2689045372 0.0543512900 0.0001299134 > > sum(sxy$d^2) # their sum > [1] 0.3233857 > > sxy$d^2/sum(sxy$d^2) # their fraction > [1] 0.8315287394 0.1680695319 0.0004017288 > > > > sx <- svd(crossprod(X)/(n-1)) > > sy <- svd(crossprod(Y)/(n-1)) > > > > sx$d # eigenvalues for X > [1] 1.6143390 1.1115365 0.2711724 > > sum(sx$d) # their sum > [1] 2.997048 > > > > sy$d # eigenvalues for Y > [1] 1.5372356 0.9224574 0.4432287 > > sum(sy$d) # their sum > [1] 2.902922 > > > So, by generating two matrices of random data, and any covariation between > them is incidental, we find that 83.15% of the overall covariance is > explained in the first PLS vectors. However, the summed squared singular > values (overall covariance) is 0.32, which is quite small compared to the > 2.99 or 2.90 summed eigenvalues (total variances) of either matrix. It is > possible to have much of the very little covariance explained by the first > axis. Focusing on the 83.15%, alone, could be misleading about the amount of > covariance between X and Y. > > Cheers! > Mike > >> On May 14, 2020, at 2:59 AM, [email protected] >> <mailto:[email protected]> wrote: >> >> Dear Katrien, >> >> >> >> The sum of the squared singular values in PLS equals the sum of all the >> squared pairwise covariances between the two blocks of variables. Therefore, >> what you can compute is the fraction of "squared" total covariance that each >> PLS dimension accounts for. You can do this by dividing each squared >> singular value by the summed squared singular values. >> >> Also, if one insists on a significance test, it should be based on the >> distribution of these singular values. >> >> >> >> Best, >> >> >> >> Philipp >> >> >> >> Am Donnerstag, 14. Mai 2020 03:31:12 UTC+2 schrieb katrien.janin: >> Hello everybody, >> >> I am hoping you can help me figure out the following: after having used the >> integration.test function (geomorph), I would like to know the % each PLS >> captures of the total covariance but I am kinda stumped on what might be a >> good way how to go about it. >> >> I was initially thinking along the lines of generating the covariance matrix >> of the two block (e.g. cov.il.is <http://cov.il.is/> <-cov(il.is.int >> <http://il.is.int/>$A1.matrix,il.is.int <http://il.is.int/>$A2.matrix) and >> computing the eigenvalue for each PLS , and then regress these. But of >> course the cov matrix has a very different structure, so that will not work. >> I am clearly thinking in the wrong direction, but for now I can’t seem to >> see the tree in the forest. Any ideas? >> >> Best wishes, >> Katrien >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Morphmet" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to [email protected] >> <mailto:[email protected]>. >> To view this discussion on the web visit >> https://groups.google.com/d/msgid/morphmet2/76c0b670-b6ed-448a-b80d-4d662263aa83%40googlegroups.com >> >> <https://groups.google.com/d/msgid/morphmet2/76c0b670-b6ed-448a-b80d-4d662263aa83%40googlegroups.com?utm_medium=email&utm_source=footer>. > > > -- > You received this message because you are subscribed to the Google Groups > "Morphmet" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected] > <mailto:[email protected]>. > To view this discussion on the web visit > https://groups.google.com/d/msgid/morphmet2/0F0BC6C1-D02A-46FE-847F-C8F6BE069DD4%40gmail.com > > <https://groups.google.com/d/msgid/morphmet2/0F0BC6C1-D02A-46FE-847F-C8F6BE069DD4%40gmail.com?utm_medium=email&utm_source=footer>. -- You received this message because you are subscribed to the Google Groups "Morphmet" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/morphmet2/8933EC72-B6C4-4D75-AF00-218CB3C8F2A3%40gmail.com.
