I'm having a huge problem with a nested model I am trying to fit in R.
I have response time experiment with 2 conditions with 46 people each and 32 measures each. I would like measures to be nested within people and people nested within conditions, but I can't get it to work.
The code I thought should make sense was:
nestedmodel <- lmer(responsetime ~ 1 + condition +
(1|condition:person) + (1|person:measure), data=dat)
However, all I get is an error:
Error in checkNlevels(reTrms$flist, n = n, control) :
number of levels of each grouping factor must be < number of observations
Unfortunately, I do not even know where to start looking what the problem is here.
Any ideas? Please, please, please? =)
Cheers!
This might be more appropriate on CrossValidated, but: lme4 is trying to tell you that one or more of your random effects is confounded with the residual variance. As you've described your data, I don't quite see why: you should have 2*46*32=2944 total observations, 2*46=92 combinations of condition and person, and 46*32=1472 combinations of measure and person.
If you do
lf <- lFormula(responsetime ~ 1 + condition +
(1|condition:person) + (1|person:measure), data=dat)
and then
lapply(lf$reTrms$Ztlist,dim)
to look at the transposed random-effect design matrices for each term, what do you get? You should (based on your description of your data) see that these matrices are 1472 by 2944 and 92 by 2944, respectively.
As #MrFlick says, a reproducible example would be nice. Other things you could show us are:
fit the model anyway, using lmerControl(check.nobs.vs.nRE="ignore") to ignore the test, and show us the results (especially the random effects variances and the statement of the numbers of groups)
show us the results of with(dat,table(table(interaction(condition,person))) to give information on the number of replicates per combination (and similarly for measure)
Related
Background
I am trying to test for differences in wind speed data among different groups. For the purpose of my question, I am looking only on side wind (wind direction that is 90 deg from the individual), and I only care about the strength of the wind. Thus, I use absolute values. The range of wind speeds is 0.0004-6.8 m/sec and because I use absolute values, Gamma distribution describes it much better than normal distribution.
My data contains 734 samples from 68 individuals, with each individual having between 1-30 repeats. However, even if I reduce my samples to only include individuals with at least 10 repeats (which leaves me with 26 individuals and a total of 466 samples), I still get the problematic error.
The model
The full model is Wind ~ a*b + c*d + (1|individual), but for the purpose of this question, the simple model of Wind ~ 1 + (1|individual) gives the same singularity error, so I do not think that the explanatory variables are the problem.
The complete code line is glmer(Wind ~ 1 + (1|individual), data = X, family = Gamma(log))
The problem and the strange part
When running the model, I get the boundary (singular) fit: see ?isSingular error, although, as you can see, I use a very simple model and random structure. The strange part is that I can solve this by adding 0.1 to the Wind variable (i.e. glmer(Wind+0.1 ~ 1 + (1|Tag), data = X, family = Gamma(log)) does not give any error). I honestly do not remember why I added 0.1 the first time I did it, but I was surprised to see that it solved the error.
The question
Is this a problem with lme4? Am I missing something? Any ideas what might cause this and why does me adding 0.1 to the variable solve this problem?
Edit following questions
I am not sure what's the best way to add data, so here is a link to a csv file in Google drive
using glmmTMB does not produce any warnings with the basic formula glmmTMB(Wind ~ 1 + (1|Tag), data = X, family = Gamma(log)), but gives convergence problems warnings ('non-positive-definite Hessian matrix') when using the full model (i.e., Wind ~ a*b + c*d + (1|individual)), which are then solved if I scale the continuous variables
I will delete if this is too loosely programming but my search has turned up NULL so I'm hoping someone can help.
I have a design that has a case/control matched pairs design with repeated measurements. Looking for a model/function/package in R
I have 2 measures at time=1 and 2 measures at time=2. I have Case/Control status as Group (2 levels), and matched pairs id as match_id and want estimate the effect of Group, time and the interaction on speed, a continuous variable.
I wanted to do something like this:
(reg_id is the actual participant ID)
speed_model <- geese(speed ~ time*Group, id = c(reg_id,match_id),
data=dataforGEE, corstr="exchangeable", family=gaussian)
Where I want to model the autocorrelation within a person via reg_id, but also within the matched pairs via match_id
But I get:
Error in model.frame.default(formula = speed ~ time * Group, data = dataFullGEE, :
variable lengths differ (found for '(id)')
Can geese or GEE in general not handle clustering around 2 sets of id? Is there a way to even do this? I'm sure there is.
Thank you for any help you can provide.
This is definatly a better question for Cross Validated, but since you have exactly 2 observations per subject, I would consider the ANCOVA model:
geese(speed_at_time_2 ~ speed_at_time_1*Group, id = c(match_id),
data=dataforGEE, corstr="exchangeable", family=gaussian)
Regarding the use of ANCOVA, you might find this reference useful.
I have spent a lot of time on multiple posts and tutorials, but I still do not understand wich "rule" I have to apply to my current data, and why.
My experiment follows a within-subject design, as every subjects (n=17) performed a task in 2 conditions, accross 5 blocks of trials. VD is the mean RTs, fixed effects are condition and block, random effect is subject.
I would like to analyse the interaction between condition and block.
Using aov
I first aggregated my data:
ag<-aggregate(RT~condition+block+subject,data=d, FUN=mean)
But then I don't know if I have to include my within-subject factors into my Error term:
(1)aov<-aov(RT~ condition * block + Error(subject/(condition * block)), data=ag)
OR
(2)aov<-aov(RT~ condition * block + Error(subject), data=ag)
I have seen on several posts that the within factors have to be included in the error term, as in (1), but I do not understand how the dfs are calculated.
Using lmer
Additionally, I would like to attempt using lmer instead of aov.
I suspect that the equivalent of (2) would be:
lmer(RT ~ 1+(1|sujet)+condition*block, ag)
But if it is the (1) which is the correct one, I can not figure out how does it would have to be specified using lmer.
I'm using the following code to try to get at post-hoc comparisons for my cell means:
result.lme3<-lme(Response~Pressure*Treatment*Gender*Group, mydata, ~1|Subject/Pressure/Treatment)
aov.result<-aov(result.lme3, mydata)
TukeyHSD(aov.result, "Pressure:Treatment:Gender:Group")
This gives me a result, but most of the adjusted p-values are incredibly small - so I'm not convinced the result is correct.
Alternatively I'm trying this:
summary(glht(result.lme3,linfct=mcp(????="Tukey")
I don't know how to get the Pressure:Treatment:Gender:Group in the glht code.
Help is appreciated - even if it is just a link to a question I didn't find previously.
I have 504 observations, Pressure has 4 levels and is repeated in each subject, Treatment has 2 levels and is repeated in each subject, Group has 3 levels, and Gender is obvious.
Thanks
I solved a similar problem creating a interaction dummy variable using interaction() function which contains all combinations of the leves of your 4 variables.
I made many tests, the estimates shown for the various levels of this variable show the joint effect of the active levels plus the interaction effect.
For example if:
temperature ~ interaction(infection(y/n), acetaminophen(y/n))
(i put the possible leves in the parenthesis for clarity) the interaction var will have a level like "infection.y:acetaminophen.y" which show the effect on temperature of both infection, acetaminophen and the interaction of the two in comparison with the intercept (where both variables are n).
Instead if the model was:
temperature ~ infection(y/n) * acetaminophen(y/n)
to have the same coefficient for the case when both vars are y, you would have had to add the two simple effect plus the interaction effect. The result is the same but i prefer using interaction since is more clean and elegant.
The in glht you use:
summary(glht(model, linfct= mcp(interaction_var = 'Tukey'))
to achieve your post-hoc, where interaction_var <- interaction(infection, acetaminophen).
TO BE NOTED: i never tested this methodology with nested and mixed models so beware!
I have data of various companies' financial information organized by company ticker. I'd like to regress one of the columns' values against the others while keeping the company constant. Is there an easy way to write this out in lm() notation?
I've tried using:
reg <- lmList(lead2.dDA ~ paudit1 + abs.d.GINDEX + logcapx + logmkvalt +
logmkvalt2|pp, data=reg.df)
where pp is a vector of company names, but this returns coefficients as though I regressed all the data at once (and did not separate by company name).
A convenient and apparently little-known syntax for estimating separate regression coefficients by group in lm() involves using the nesting operator, /. In this case it would look like:
reg <- lm(lead2.dDA ~ 0 + pp/(paudit1 + abs.d.GINDEX + logcapx +
logmkvalt + logmkvalt2), data=reg.df)
Make sure that pp is a factor and not a numeric. Also notice that the overall intercept must be suppressed for this to work; in the new formulation, we have a different "intercept" for each group.
A couple comments:
Although the regression coefficients obtained this way will match those given by lmList(), it should be noted that with lm() we estimate only a single residual variance across all the groups, whereas lmList() would estimate separate residual variances for each group.
Like I mentioned in my earlier comment, the lmList() syntax that you gave looks like it should have worked. Since you say it didn't, this leads me to expect that really the problem is something else (although it's hard to tell what without a reproducible example), and so it seems likely that the solution I posted will fail for you as well, for the same unknown reasons. If you want more detailed guidance, please provide more information; help us help you.