tag:blogger.com,1999:blog-8499895524521663926.post5379948537469447324..comments2023-09-17T04:53:31.019-04:00Comments on Phylogenetic Tools for Comparative Biology: New function: phyl.pca()Liam Revellhttp://www.blogger.com/profile/04314686830842384151noreply@blogger.comBlogger18125tag:blogger.com,1999:blog-8499895524521663926.post-32741815027540472382021-09-27T14:55:41.504-04:002021-09-27T14:55:41.504-04:00Dear Liam,
Thank you for the tool development and...Dear Liam,<br /><br />Thank you for the tool development and for keeping this blog so we can discuss the tools!<br /><br />I have run a phyl.pca on six tree traits using log-transformed and standardized variables and standard options (method="BM", mode="cov"). I got two variables that loaded on two different axes and I would like to try to rotate the axes in order to seek for a more easily interpretable solution. However the variamx basic function is not applicable to the phy.pca object it seems, since when I try to use<br /><br />rotacao = varimax(trait.pca3$rotation[, 1:3])<br /><br />I get<br /><br />Error in if (nc < 2) return(x) : argument is of length zero<br /><br />Is it possible to rotate the phyl.pca axes somehow? Thank you in advance for any comments.Anonymoushttps://www.blogger.com/profile/08514555587854985079noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-21623181420853292082015-06-04T10:01:03.254-04:002015-06-04T10:01:03.254-04:00Traits would just be the Y argument, a matrix with...Traits would just be the Y argument, a matrix with columns as your variables of interest with rows as your taxa.Raptor's Nesthttps://www.blogger.com/profile/01451618880276065935noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-1992421246349385572015-04-10T04:33:25.626-04:002015-04-10T04:33:25.626-04:00Hi Liam, I+ve trying to use this function, but I d...Hi Liam, I+ve trying to use this function, but I do not have a traits matrix (and I am actually not really sure about what it is). Could you give me some insight? ThanksSrita. Mhttps://www.blogger.com/profile/01003945670772796871noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-57856279338843654952014-01-16T12:29:04.815-05:002014-01-16T12:29:04.815-05:00Thank you both very much for your thoughts, I'...Thank you both very much for your thoughts, I'm looking into this.Anonymoushttps://www.blogger.com/profile/11530837713608220151noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-1223127467132435392013-12-26T16:54:50.041-05:002013-12-26T16:54:50.041-05:00Dear John,
Perhaps I can be of assistance to you. ...Dear John,<br />Perhaps I can be of assistance to you. The phylogenetic PCA is not yet implemented in the Geometric Morphometrics packages for R (such as Geomorph, Morpho or Shapes). Before proceeding, I suggest reading a little more on this ordination method in Polly et al. 2013 Hystrix and Monterio 2013 Hystrix (http://www.italian-journal-of-mammalogy.it/public/journals/3/issue_241_complete_100.pdf). Leandro Monterio has his own R code for including the phylogenetic covariance matrix during a PCA of landmark data, and David Polly has his Mathematica programs. Dean Adams, Erik Otárola-Castillo and I have no immediate plans to implement the phylo PCA into Geomorph (http://cran.r-project.org/web/packages/geomorph/), although you can put in a formal request to Dr. Adams. You can also contact the Morphmet list to see if anyone else has already worked up some R code.<br />Best wishes,<br />EmmaEmma Sherratthttps://www.blogger.com/profile/10593440736673735935noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-65951919036945728632013-12-23T16:38:24.063-05:002013-12-23T16:38:24.063-05:00Hi John. Although this seems reasonable, I'm a...Hi John. Although this seems reasonable, I'm afraid I don't know enough about geometric morphometrics to comment on this. There may be complicated issues that I don't appreciate. Why don't you try R-sig-phylo with this question? Good luck. - LiamLiam Revellhttps://www.blogger.com/profile/04314686830842384151noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-31382940007802509672013-12-21T17:17:34.892-05:002013-12-21T17:17:34.892-05:00Hello Liam. I'm currently using tpsRelw as I k...Hello Liam. I'm currently using tpsRelw as I know exactly how it handles sliding semi landmarks, yet still wish to perform your phyl.pca. I think the best way to do this would be to obtain instead my partial warps (as opposed to relative warps, the normal final step in tpsRelw) from tps Relw, and then subject this matrix (of taxa and scores) to your function. Can you see any issue with this?Anonymoushttps://www.blogger.com/profile/11530837713608220151noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-17998988350809915122011-07-14T18:59:27.733-04:002011-07-14T18:59:27.733-04:00Hmmm. This (in theory) should not be the problem,...Hmmm. This (in theory) should not be the problem, as the "change" to X is inside internal functions called by phyl.pca(). Still, that doesn't explain why changing it to be X throughout resolved the issue! I will investigate this.Liam Revellhttps://www.blogger.com/profile/04314686830842384151noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-53266449747476310362011-07-14T17:32:09.495-04:002011-07-14T17:32:09.495-04:00Hi Liam,
Thanks for the updated phyl.pca functio...Hi Liam, <br /><br />Thanks for the updated phyl.pca function. The code in the direct link doesn't run as written, but it looks like a small error. The trait matrix variable is initially named Y, but then switches to X after the line # analyze. I just changed Y to X and now it runs fine.zingiberhttps://www.blogger.com/profile/01500931939072109445noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-10957539910704118172011-04-13T14:07:46.525-04:002011-04-13T14:07:46.525-04:00Hi Scott.
I think in that case we would want to co...Hi Scott.<br />I think in that case we would want to compute MDS axes to minimize the influence of distances that are correlated due to the phylogeny. This should be possible so long as we have a way to get the expected distances between related species on a phylogeny given some model of evolution (say, BM).Liam Revellhttps://www.blogger.com/profile/04314686830842384151noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-53295658859567314442011-04-13T13:24:59.563-04:002011-04-13T13:24:59.563-04:00Hi Liam,
I have used ordination for species trai...Hi Liam, <br /><br />I have used ordination for species traits, and we used MDS (although we didn't compare to results of PCA). From my understanding, MDS avoids assuming linear relationships among variables (traits could be related in a nonlinear fashion), contrary to PCA I think. Might this be a good reason to use MDS instad of PCA for species traits? Of course transformation of traits could linearize them sufficiently. <br /><br />My thought was that you can use MDS axes just as you would use PCA axes. <br /><br />I haven't been able to find any attempts of phylogenetic MDS in the literature yet.Anonymoushttps://www.blogger.com/profile/01944150859555101169noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-34608519708226813872011-04-13T11:29:04.382-04:002011-04-13T11:29:04.382-04:00Hi Scott.
I don't know that this would be simp...Hi Scott.<br />I don't know that this would be simple.<br />What would be our goal in incorporating phylogenetic information into, say, classical MDS? I'd imagine it would be to minimize the influence of correlated distances between close relatives. Has this been attempted in any study, to your knowledge?<br />Thanks for the comment.Liam Revellhttps://www.blogger.com/profile/04314686830842384151noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-86650558350726605322011-04-13T00:28:12.228-04:002011-04-13T00:28:12.228-04:00Would it be simple enough to create a nonmetric mu...Would it be simple enough to create a nonmetric multidimensional scaling version of this function? Instead of using PCA that is.Anonymoushttps://www.blogger.com/profile/01944150859555101169noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-3146687606408101322011-04-12T14:32:13.207-04:002011-04-12T14:32:13.207-04:00Hi Vladimir.
You pre-empted my "improvements...Hi Vladimir.<br /><br />You pre-empted my "improvements" post <a href="http://phytools.blogspot.com/2011/04/fixes-and-improvements-to-phylpca.html" rel="nofollow">here</a>.<br /><br />The biggest issue is that the comparison is really:<br /><br />system.time(A<-solve(C); for(i in 1:1000) A%*%B)<br /># vs.<br />system.time(for(i in 1:1000) solve(C,B))<br /><br />That said, however, I did successfully swap solve(A,B) for solve(A)%*%B in one place in my code that it made a huge difference; that is for the likelihood calculation involving a n*m x n*m Kronecker product matrix. Here, on a sample 100 species tree with 10 traits, I saved about 0.75s per calculation by following your suggestion! This is fantastic since this calculation will be performed many times during ML optimization.<br /><br />Thanks!Liam Revellhttps://www.blogger.com/profile/04314686830842384151noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-28191767149881096412011-04-12T14:06:22.505-04:002011-04-12T14:06:22.505-04:00When you have two vectors that you use to multiply...When you have two vectors that you use to multiply C^-1 from the left, the speed up is not that impressive, only ~20%:<br /><br />system.time(for (i in 1:1000){A=solve(C); x%*%A%*%a; x%*%A%*%x; a%*%A%*%x; a%*%A%*%x;});<br />system.time(for (i in 1:1000){A=solve(t(C),x); A%*%a; A%*%x; A=solve(t(C),a); A%*%a; A%*%x});Vladimir Mininhttps://www.blogger.com/profile/00603257472903019482noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-7374341000752952082011-04-12T13:47:01.173-04:002011-04-12T13:47:01.173-04:00Liam,
It's probably a moot point anyway, sinc...Liam,<br /><br />It's probably a moot point anyway, since you found a more important inefficiency in your code, but from your original post it seems that it makes sense to cach 1'C^-1 rather than C^-1. This of course makes sense only if you are always multiplying C^-1 by the same vector from the left, but use different vectors for right multiplication. Demonstration is below:<br /><br />C = matrix(rnorm(10000),100,100);<br />a = rnorm(100);<br />x = rep(1,100);<br /><br />system.time(for (i in 1:1000){A=solve(C); x%*%A%*%a; x%*%A%*%x});<br />system.time(for (i in 1:1000){A=solve(C); colSums(A%*%a); sum(A)});<br />system.time(for (i in 1:1000){A=solve(t(C),x); A%*%a; A%*%x});<br /><br />On my machine, this gives more than 2 fold speed up.Vladimir Mininhttps://www.blogger.com/profile/00603257472903019482noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-84583038373339646712011-04-12T11:33:14.880-04:002011-04-12T11:33:14.880-04:00Thanks Vladimir.
I have basically been using solv...Thanks Vladimir.<br /><br />I have basically been using solve(...) to invert, rather than to "solve." Computing C^-1*Y is the same as solving Y=CX for X; except that solve(C,Y) is faster than solve(C)%*%Y, as you point out.<br /><br />I will note, though, that once I have computed invC<-solve(C), *then* invC%*%X becomes considerably faster than solve(C,X). So I guess it depends how many times I would normally invert C.<br /><br />After some reflection, though, I think the biggest inefficiency of my code is the repeated and unnecessary calls of vcv.phylo(). I will try and fix both of these issues and then I will post an updated version (along with the comparison).Liam Revellhttps://www.blogger.com/profile/04314686830842384151noreply@blogger.comtag:blogger.com,1999:blog-8499895524521663926.post-17449511529055684342011-04-12T00:38:36.096-04:002011-04-12T00:38:36.096-04:00Hi Liam,
First, I enjoy reading your posts!
I ju...Hi Liam,<br /><br />First, I enjoy reading your posts!<br /><br />I just wanted to point out that in most practical situations, one does not need to invert matrices. For example, in the above function, what you really want is matrix inverse times some vector. It is much faster to accomplish this task with solve() directly than first getting the inverse and then multiplying it by a vector.<br /><br />A simple demonstration is below:<br /><br />C = matrix(rnorm(10000),100,100);<br />a = rnorm(100);<br /><br />system.time(for (i in 1:1000) solve(C)%*%a);<br />system.time(for (i in 1:1000) solve(C,a));Vladimir Mininhttps://www.blogger.com/profile/00603257472903019482noreply@blogger.com