Hi, I have a dataset like this,> testdataGrouped Data: expr ~ visit | subject expr visit subject 1 6.502782 V1 A 2 6.354506 V1 B 3 6.349184 V1 C 4 6.386301 V2 A 5 6.376405 V2 B 6 6.758640 V2 C 7 6.414142 V3 A 8 6.354521 V3 B 9 6.396636 V3 C I tried the command> testresult=lme(expr~visit,data=testdata,random=~visit|subject)I got the following error message. Error in MEestimate(lmeSt, grps) : Singularity in backsolve at level 0, block 1 In addition: There were 50 or more warnings (use warnings() to see the first 50) Could someone give me some hint on what went wrong? Thanks. Donglei Hu
Hi Donglei, what is your goal in fitting this model? The statement that you've used is creating an overparameterized model, as it fits a random intercept for each visit within each subject. I wonder if you might prefer testresult <- lme(expr~visit, data=testdata, random=~1|subject) which will fit a random intercept for each subject? Andrew On Monday 22 March 2004 14:29, donghu at itsa.ucsf.edu wrote:> Hi, > > I have a dataset like this, > > > testdata > > Grouped Data: expr ~ visit | subject > expr visit subject > 1 6.502782 V1 A > 2 6.354506 V1 B > 3 6.349184 V1 C > 4 6.386301 V2 A > 5 6.376405 V2 B > 6 6.758640 V2 C > 7 6.414142 V3 A > 8 6.354521 V3 B > 9 6.396636 V3 C > > I tried the command > > > testresult=lme(expr~visit,data=testdata,random=~visit|subject) > > I got the following error message. > Error in MEestimate(lmeSt, grps) : Singularity in backsolve at level 0, > block 1 In addition: There were 50 or more warnings (use warnings() to see > the first 50) > > Could someone give me some hint on what went wrong? Thanks. > > Donglei Hu > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > stat.math.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > R-project.org/posting-guide.html-- Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885 6226 University of Idaho E : andrewr at uidaho.edu PO Box 441133 W : uidaho.edu/~andrewr Moscow ID 83843 Or: biometrics.uidaho.edu No statement above necessarily represents my employer's opinion.
Dear All, I need help both with analytical and computational aspects. My problem: I have counted and measured increments on otoliths (inner-ear bony structures) of fish larvae collected at different times of day. This was repeated three times for each fish larvae (in random order and no particular dates). Then, I calculated the ratio of marginal-to-previous increment for each fish and reading. I wish to study if differences exist in ratios between hours of day (accounting for differences in readings within each fish) - in data.frame format... code hour read.no ratio 33001 5 1 0.56 33003 7 1 0.93 37123 19 1 0.23 ... 33001 5 2 0.62 33003 7 2 0.87 37123 19 2 0.22 ... 37123 19 3 0.26 ... My solution: Use function lme, with hr<-factor(hour)...> hr.lme<-lme(sqrt(ratio)~hr,random=~1|code/read.no) > anova(hr.lme) ; summary (hr.lme)Does this commands correspond to the initial problem statement? Thanks in advance. Regards, Eduardo Esteves