Hello, Could someone please tell me how to find the estimate of inertia for the second axis of an RLQ analysis using ade4? Using the example from the ade4 package, I *suspect* that the inertia for the 2nd axis of the R table would be 4.139332 (see below results summary from rlq1). However, the row is labeled "12" not "2" which suggests this is not correct. In addition, this would mean that axis 2 of the RLQ would explain more than 100% of the inertia of the separate ordination for dudimil ((100*4.139)/dudimil$eig[2]=176%) which I believe is not supposed to occur. Could anyone set me straight by either letting me know how to get the inertia for axis 2 or that my thinking is perhaps wrong? Thanks, John data(aviurba) coa1 <- dudi.coa(aviurba$fau, scannf = FALSE, nf = 2) dudimil <- dudi.hillsmith(aviurba$mil, scannf = FALSE, nf = 2, row.w = coa1$lw) duditrait <- dudi.hillsmith(aviurba$traits, scannf = FALSE, nf = 2, row.w = coa1$cw) rlq1 <- rlq(dudimil, coa1, duditrait, scannf = FALSE, nf = 2) plot(rlq1) summary(rlq1) randtest.rlq(rlq1)> summary(rlq1)Eigenvalues decomposition: eig covar sdR sdQ corr 1 0.4782826 0.6915798 1.558312 1.158357 0.3831293 2 0.1418508 0.3766308 1.308050 1.219367 0.2361331 Inertia & coinertia R: inertia max ratio 1 2.428337 2.996911 0.8102800 12 4.139332 5.345110 0.7744148 Inertia & coinertia Q: inertia max ratio 1 1.341791 2.603139 0.5154512 12 2.828648 4.202981 0.6730098 Correlation L: corr max ratio 1 0.3831293 0.6435487 0.5953384 2 0.2361331 0.5220054 0.4523576
Hello, Could someone please tell me how to find the estimate of inertia for the second axis of an RLQ analysis using ade4? Using the example from the ade4 package, I *suspect* that the inertia for the 2nd axis of the R table would be 4.139332 (see below results summary from rlq1). However, the row is labeled "12" not "2" which suggests this is not correct. In addition, this would mean that axis 2 of the RLQ would explain more than 100% of the inertia of the separate ordination for dudimil ((100*4.139)/dudimil$eig[2]=176%) which I believe is not supposed to occur. Could anyone set me straight by either letting me know how to get the inertia for axis 2 or that my thinking is perhaps wrong? Thanks, John data(aviurba) coa1 <- dudi.coa(aviurba$fau, scannf = FALSE, nf = 2) dudimil <- dudi.hillsmith(aviurba$mil, scannf = FALSE, nf = 2, row.w = coa1$lw) duditrait <- dudi.hillsmith(aviurba$traits, scannf = FALSE, nf = 2, row.w = coa1$cw) rlq1 <- rlq(dudimil, coa1, duditrait, scannf = FALSE, nf = 2) plot(rlq1) summary(rlq1) randtest.rlq(rlq1)> summary(rlq1)Eigenvalues decomposition: eig covar sdR sdQ corr 1 0.4782826 0.6915798 1.558312 1.158357 0.3831293 2 0.1418508 0.3766308 1.308050 1.219367 0.2361331 Inertia & coinertia R: inertia max ratio 1 2.428337 2.996911 0.8102800 12 4.139332 5.345110 0.7744148 Inertia & coinertia Q: inertia max ratio 1 1.341791 2.603139 0.5154512 12 2.828648 4.202981 0.6730098 Correlation L: corr max ratio 1 0.3831293 0.6435487 0.5953384 2 0.2361331 0.5220054 0.4523576
Dear John, In RLQ (similarly to other methods in ade4), inertia are given by eigenvalues. Thus, the percentage of inertia are obtained simply by: rlq1$eig/sum(rlq1$eig)*100 For more details on outputs or RLQ, you can have a look at http://listes.univ-lyon1.fr/wws/arc/adelist/2009-03/msg00001.html Lastly, if you have questions about ade4, please subscribe and send an email to adelist (http://listes.univ-lyon1.fr/wws/info/adelist) Cheers. John Poulsen wrote:> > Hello, > > Could someone please tell me how to find the estimate of inertia for > the second axis of an RLQ analysis using ade4? Using the example from > the ade4 package, I *suspect* that the inertia for the 2nd axis of the > R table would be 4.139332 (see below results summary from rlq1). > However, the row is labeled "12" not "2" which suggests this is not > correct. In addition, this would mean that axis 2 of the RLQ would > explain more than 100% of the inertia of the separate ordination for > dudimil ((100*4.139)/dudimil$eig[2]=176%) which I believe is not > supposed to occur. > > Could anyone set me straight by either letting me know how to get the > inertia for axis 2 or that my thinking is perhaps wrong? > > Thanks, > John > > data(aviurba) > coa1 <- dudi.coa(aviurba$fau, scannf = FALSE, nf = 2) > dudimil <- dudi.hillsmith(aviurba$mil, scannf = FALSE, nf = 2, row.w > = coa1$lw) > duditrait <- dudi.hillsmith(aviurba$traits, scannf = FALSE, nf = 2, > row.w = coa1$cw) > rlq1 <- rlq(dudimil, coa1, duditrait, scannf = FALSE, nf = 2) > plot(rlq1) > summary(rlq1) > randtest.rlq(rlq1) > >> summary(rlq1) > > Eigenvalues decomposition: > eig covar sdR sdQ corr > 1 0.4782826 0.6915798 1.558312 1.158357 0.3831293 > 2 0.1418508 0.3766308 1.308050 1.219367 0.2361331 > > Inertia & coinertia R: > inertia max ratio > 1 2.428337 2.996911 0.8102800 > 12 4.139332 5.345110 0.7744148 > > Inertia & coinertia Q: > inertia max ratio > 1 1.341791 2.603139 0.5154512 > 12 2.828648 4.202981 0.6730098 > > Correlation L: > corr max ratio > 1 0.3831293 0.6435487 0.5953384 > 2 0.2361331 0.5220054 0.4523576 > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > >-- St?phane DRAY (dray at biomserv.univ-lyon1.fr ) Laboratoire BBE-CNRS-UMR-5558, Univ. C. Bernard - Lyon I 43, Bd du 11 Novembre 1918, 69622 Villeurbanne Cedex, France Tel: 33 4 72 43 27 57 Fax: 33 4 72 43 13 88 http://pbil.univ-lyon1.fr/members/dray/