Zero cells in contrast matrix problem
at this point, if I were you I would set
options(error=recover)
and debug at the point at which the program stops: this will be at
this point in the code:
if (length(cov) == p) {
cov <- diag(cov, p)
} else if (length(cov) != p^2) {
stop("normal prior covariance of improper length")
}
so you should be able to print(p) to find out the appropriate size.
On Mon, Oct 26, 2015 at 7:37 AM, Francesco Romano
<francescobryanromano at gmail.com> wrote:
Nope.
revanaA<- bglmer(Correct~Syntax*Animacy*Prof.group.2+(1|Part.name)+(1|Item), data = revana, family = binomial, fixef.prior = normal(cov = diag(9,14)))
fixed-effect model matrix is rank deficient so dropping 2 columns / coefficients Error in normal(cov = cov, common.scale = FALSE) : normal prior covariance of improper length Could it be that I need three separate priors, one for each effect? On Mon, Oct 26, 2015 at 12:25 PM, Ben Bolker <bbolker at gmail.com> wrote:
Ah. So try normal(cov=diag(9,14)) ... On Mon, Oct 26, 2015 at 7:18 AM, Francesco Romano <francescobryanromano at gmail.com> wrote:
For some reason the silly bugger didn't past the full command:
revanaA<- bglmer(Correct~Syntax*Animacy*Prof.group.2+(1|Part.name)+(1|Item), data = revana, family = binomial, fixef.prior = normal(cov = diag(9,16)))
fixed-effect model matrix is rank deficient so dropping 2 columns / coefficients Error in normal(cov = cov, common.scale = FALSE) : normal prior covariance of improper length To give more info on this, it is the Animacy factor that is causing separation because two levels of it have zero counts in some cases. On Mon, Oct 26, 2015 at 12:13 PM, Ben Bolker <bbolker at gmail.com> wrote:
Well, that's a separate problem (and not necessarily a "problem"). R is telling you that you have 16 separate combinations of the factors, but only 14 unique combinations represented in your data set, so it can only estimate 14 parameters. Unless there is a weird interaction with blme I don't know about, this should still give you reasonable answers. On Mon, Oct 26, 2015 at 7:10 AM, Francesco Romano <francescobryanromano at gmail.com> wrote:
Many thanks Ben, but I tried that already:
revanaA<- bglmer(Correct~Syntax*Animacy*Prof.group.2+(1|Part.name)+(1|Item), data = revana, family = binomial, fixef.prior = normal(cov = diag(9,16)))
fixed-effect model matrix is rank deficient so dropping 2 columns / coefficients Error in normal(cov = cov, common.scale = FALSE) : normal prior covariance of improper length On Mon, Oct 26, 2015 at 12:06 PM, Ben Bolker <bbolker at gmail.com> wrote:
On 15-10-26 06:56 AM, Francesco Romano wrote:
I wonder if anyone can help with the separation problem originally solved by Ben Bolker (see thread). The model and fitting I used previously was trial<-bglmer(Correct ~ Syntax.Semantics, data = trialglm, family = binomial, fixef.prior = normal(cov = diag(9,4)) which now has to change because the Syntax.Semantcs factor needs to be split into separate within-subjects variables, Syntax, a factor with two levels, and Animacy, a factor with four levels. In addition a new between-subjects factor called Group with two levels (native vs non-native speaker) has to be added which determines the following model, fit by bglmer: trial<-bglmer(Correct ~ Syntax*Animacy*Group+ (1|Part.name)+(1|Item), data = trialglm, family = binomial, fixef.prior = normal(cov = diag???) What values should I use for the cov=diag portion in order to continue attempting convergence of a model that includes the random effects?
In general a reasonable form is normal(cov = diag(v,np)) where v
is
the prior variance (generally something reasonably
large/non-informative; 9 (=std dev of 3) is probably an OK default)
and
np is the number of fixed-effect parameters. You can figure this
out
via
ncol(model.matrix(~Syntax*Animacy*Group,data=trialglm)
or multiply 2*4*2 to get 16 ...
R returns the following error because I don't know how to establish the parameters when more than one fixed effect is involved: Error in normal(cov = cov, common.scale = FALSE) : normal prior covariance of improper length Many thanks in advance for any help! On Thu, May 28, 2015 at 10:46 PM, Ben Bolker <bbolker at gmail.com> wrote:
I don't see your data -- I see a little tiny subset, but that's
not
really enough for a reproducible example.
This is the example given in the URL I sent:
cmod_blme_L2 <- bglmer(predation~ttt+(1|block),data=newdat,
family=binomial,
fixef.prior = normal(cov = diag(9,4)))
trial<-bglmer(Correct ~ Syntax.Semantics+(1|Part.name),
data =trialglm,
family = binomial,
fixef.prior = normal(cov=diag(9,8)))
The last line specifies an 8x8 matrix (because you have 8 fixed
effect
parameters) with a value of 9 on the diagonal, meaning the priors
for
the fixed effects are independent and each is Normal with a sd of
sqrt(9)=3.
On Thu, May 28, 2015 at 3:25 PM, Francesco Romano
<francescobryanromano at gmail.com> wrote:
Yes but this seems a bit above my head without your help. The data are in the three variables at the bottom of my email but I forgot to mention the random participant effect (n = 17). Thanks! Il gioved? 28 maggio 2015, Ben Bolker <bbolker at gmail.com> ha scritto:
On 15-05-28 06:55 AM, Francesco Romano wrote:
Many thanks to both. The approaches you suggest (and others online) help one deal with the separation problem but don't offer any specific advice as to how getting a valid p coefficient when comparing two levels of the model vexed by separation. Ben, here's the output of the bglmer which by the way would be ideal since it allows me to retain the random effect so that all my pairwise comparisons are conducted using mixed effects.
trial<-bglmer(Correct ~ Syntax.Semantics+(1|Part.name), data = trialglm,
family = binomial) Warning message: package ?blme? was built under R version 3.1.2
summary(trial)
Cov prior : Part.name ~ wishart(df = 3.5, scale = Inf, posterior.scale = cov, common.scale = TRUE) Prior dev : 1.4371 Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) ['bglmerMod'] Family: binomial ( logit ) Formula: Correct ~ Syntax.Semantics + (1 | Part.name) Data: trialglm AIC BIC logLik deviance df.resid 269.9 305.5 -126.0 251.9 376 Scaled residuals: Min 1Q Median 3Q Max -0.9828 -0.4281 -0.2445 -0.0002 5.7872 Random effects: Groups Name Variance Std.Dev. Part.name (Intercept) 0.3836 0.6194 Number of obs: 385, groups: Part.name, 16 Fixed effects: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.8671 0.4538 -4.114 3.89e-05 *** Syntax.Semantics A 0.8121 0.5397 1.505 0.1324 Syntax.Semantics B -16.4391 1195.5031 -0.014 0.9890 Syntax.Semantics C -1.1323 0.7462 -1.517 0.1292 Syntax.Semantics D 0.1789 0.5853 0.306 0.7598 Syntax.Semantics E -0.8071 0.7500 -1.076 0.2819 Syntax.Semantics F -1.5051 0.8575 -1.755 0.0792 . Syntax.Semantics G 0.4395 0.5417 0.811 0.4171 --- Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 Unfortunately the separation problem is still there. Should I be constraining the parameter somehow? How would I do that? The data is below.
Did you read the section in the URL I suggested? Just using bglmer isn't enough; you also have to set a prior on the fixed effects. Your data don't seem to be attached (note that the mailing list strips most non-ASCII file types).
In passing I also tried brglm which solves the separation problem but tells me comparison is significant which I don't believe one bit (see the data below). I am pretty sure about this because when I reveled and look at the comparisons I was able to compute using glmer, these turn out to be non-significant, when glmer told me they were:
trial<-brglm(Correct ~ Syntax.Semantics, data = trialglm, family =
binomial) Warning messages: 1: package ?elrm? was built under R version 3.1.2 2: package ?coda? was built under R version 3.1.3
summary(trial)
Call: brglm(formula = Correct ~ Syntax.Semantics, family = binomial, data = trialglm) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.6358 0.4035 -4.053 5.05e-05 *** Syntax.Semantics A 0.6689 0.5169 1.294 0.1957 Syntax.Semantics B -3.0182 1.4902 -2.025 0.0428 * Syntax.Semantics C -1.0135 0.6889 -1.471 0.1413 Syntax.Semantics D 0.1515 0.5571 0.272 0.7857 Syntax.Semantics E -0.7878 0.6937 -1.136 0.2561 Syntax.Semantics F -1.2874 0.7702 -1.672 0.0946 . Syntax.Semantics G 0.4358 0.5186 0.840 0.4007 --- Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 262.51 on 384 degrees of freedom Residual deviance: 256.22 on 377 degrees of freedom Penalized deviance: 245.5554 AIC: 272.22 MCMCglmm is too complex for me. Wolfgang, I tried the penalized likelihood method (logistf function) but output is hard to read:
trial<-logistf(Correct ~ Syntax.Semantics, data = trialglm, family =
binomial) Warning messages: 1: package ?logistf? was built under R version 3.1.2 2: package ?mice? was built under R version 3.1.2
summary(trial)
logistf(formula = Correct ~ Syntax.Semantics, data = trialglm, family = binomial) Model fitted by Penalized ML Confidence intervals and p-values by Profile Likelihood Profile Likelihood Profile Likelihood Profile Likelihood Profile Likelihood Profile Likelihood Profile Likelihood Profile Likelihood coef se(coef) lower 0.95 upper 0.95 Chisq p (Intercept) 3.2094017 0.7724482 2.9648747 3.5127830 0.000000 1.000000e+00 Syntax.Semantics A 4.1767737 6.3254344 0.4224696 12.0673987 64.224452 1.110223e-15 Syntax.Semantics B -1.0583602 0.8959376 -1.3963977 -0.7625216 0.000000 1.000000e+00 Syntax.Semantics C -0.7299070 0.9308193 -1.0765598 -0.4180076 0.000000 1.000000e+00 Syntax.Semantics D 0.2314740 1.1563731 -0.1704535 0.6479908 1.156512 2.821901e-01 Syntax.Semantics E -0.6476907 0.9771824 -1.0076740 -0.3164066 0.000000 1.000000e+00 Syntax.Semantics F -0.8271499 0.9305931 -1.1743834 -0.5160799 0.000000 1.000000e+00 Syntax.Semantics G 0.9909046 1.3787175 0.5457741 1.5353981 0.000000 1.000000e+00 Likelihood ratio test=121.9841 on 7 df, p=0, n=385 Wald test = 5.334321 on 7 df, p = 0.6192356 In particular, what is this model telling me? That Z (my ref level) and B are significantly different? I'm happy to try the elrm function with exact logistic regression but I am not capable of programming it. Besides, would it give me valid estimates for the comparison between the Z and B levels? The data frame should look like this: Outcome variable (Correct, incorrect) Predictor variable (A, B, C, D, E, F, G, Z) Counts (E: 38,3; B: 51,0; Z: 37,7; G: 40,12; D: 36,8; C:45,3; A: 34,13; F:65,22). Thank you! F. On Thu, May 28, 2015 at 2:28 AM, Ben Bolker <bbolker at gmail.com> wrote:
And for what it's worth, you can do this in conjunction with lme4 by using the blme package instead (a thin Bayesian wrapper around lme4), or via the MCMCglmm package; see http://ms.mcmaster.ca/~bolker/R/misc/foxchapter/bolker_chap.html for an example (search for "complete separation"). On Wed, May 27, 2015 at 5:21 PM, Viechtbauer Wolfgang (STAT) <wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:
You may need to consider using an 'exact', Bayesian, or penalized
likelihood approach (along the lines proposed by Firth).
Maybe a place to start:
Best,
Wolfgang
-----Original Message----- From: R-sig-mixed-models [mailto:r-sig-mixed-models-bounces at r- project.org] On Behalf Of Francesco Romano Sent: Wednesday, May 27, 2015 23:00 To: r-sig-mixed-models at r-project.org Subject: [R-sig-ME] Zero cells in contrast matrix problem After giving up on a glmer for my data, I remembered a post by Roger
Levy
suggesting to try the use non mixed effects glm when one of the cells in a matrix is zero. To put this into perspective:
trial<-glmer(Correct ~ Syntax.Semantics + (1 | Part.name), data =
trialglm, family = binomial) Warning messages: 1: In checkConv(attr(opt, "derivs"), opt$par, ctrl = control$checkConv, : Model failed to converge with max|grad| = 0.053657 (tol = 0.001, component 4) 2: In checkConv(attr(opt, "derivs"), opt$par, ctrl = control$checkConv, : Model is nearly unidentifiable: large eigenvalue ratio - Rescale variables? My data has a binary outcome, correct or incorrect, a fixed effect predictor factor with 8 levels, and a random effect for participants. I believe the problem R is encountering is with one level of the factor (let us call it level B) which has no counts (no I won' t try to post the table from the paper with the counts because I know it will get garbled up!). I attempt a glm with the same data:
trial<-glm(Correct ~ Syntax.Semantics, data = trialglm, family =
binomial)
anova(trial)
Analysis of Deviance Table Model: binomial, link: logit Response: Correct Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev NULL 384 289.63 Syntax.Semantics 7 34.651 377 254.97
summary(trial)
Call: glm(formula = Correct ~ Syntax.Semantics, family = binomial, data = trialglm) Deviance Residuals: Min 1Q Median 3Q Max -0.79480 -0.62569 -0.34474 -0.00013 2.52113 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.6917 0.4113 -4.113 3.91e-05 *** Syntax.Semantics A 0.7013 0.5241 1.338 0.1809 Syntax.Semantics B -16.8744 904.5273 -0.019 0.9851 Syntax.Semantics C -1.1015 0.7231 -1.523 0.1277 Syntax.Semantics D 0.1602 0.5667 0.283 0.7774 Syntax.Semantics E -0.8733 0.7267 -1.202 0.2295 Syntax.Semantics F -1.4438 0.8312 -1.737 0.0824 . Syntax.Semantics G 0.4630 0.5262 0.880 0.3789 --- Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 289.63 on 384 degrees of freedom Residual deviance: 254.98 on 377 degrees of freedom AIC: 270.98 Number of Fisher Scoring iterations: 17 The comparison I'm interested in is between level B and the reference level but it cannot be estimated as shown by the ridiculously high estimate and SE value. Any suggestions on how to get a decent beta, SE, z, and p? It's the only comparison missing in the table for the levels I need so I think it
would
be a bit unacademic of me to close this deal saying 'the difference
could
not be estimated due to zero count'. And by the way I have seen this comparison being generated using other stats. Thanks in advance, Frank [[alternative HTML version deleted]]
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-- Frank Romano Ph.D. Tel. +39 3911639149 LinkedIn https://it.linkedin.com/pub/francesco-bryan-romano/33/1/162 Academia.edu https://sheffield.academia.edu/FrancescoRomano
-- Frank Romano Ph.D. Tel. +39 3911639149 LinkedIn https://it.linkedin.com/pub/francesco-bryan-romano/33/1/162 Academia.edu https://sheffield.academia.edu/FrancescoRomano