Zero cells in contrast matrix problem

options(error=recover)
if (length(cov) == p) {cov <- diag(cov, p)} else if (length(cov) != p^2)

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:

http://sas-and-r.blogspot.nl/2010/11/example-815-firth-logistic-regression.html

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

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