Displaying 20 results from an estimated 31 matches for "model4".
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2006 Jun 27
8
Avaya 4610sw SIP setup problem
...t;
SET REGISTERWAIT 120
SET SIPDOMAIN "sip.mycompany.com"
SET SIPPROXYSRVR "204.140.111.219"
SET SIPPORT "5070" (this is not a typo)
SET SIPREGISTRAR "204.140.111.219"
SET SP_DIRSRVR 10.1.1.1
SET SP_DIRSRVRPORT 389
SET SP_DIRTOPDN ou=People,o=avaya.com
IF $MODEL4 SEQ 4602 goto SETTINGS4602
IF $MODEL4 SEQ 4610 goto SETTINGS4610
IF $MODEL4 SEQ 4620 goto SETTINGS4620
IF $MODEL4 SEQ 4621 goto SETTINGS4621
IF $MODEL4 SEQ 4622 goto SETTINGS4622
IF $MODEL4 SEQ 4625 goto SETTINGS4625
IF $MODEL4 SEQ 4630 goto SETTINGS4630
goto END
goto END
SET WMLHOME http://support...
2010 Jan 07
1
LD50 and SE in GLMM (lmer)
...3 different instars (non significant fixed effect) and several different blocks (I did 4 replicates at a time)= random effect).
Since there is no predict function for lmer, I used the following to get predicted values (thanks to a post by John Maindonald (I'll attach his post below)):
model4 <- lmer (y~time + (1|blc/instar),family=binomial(link="probit"))
summary(model4)
b <- fixef(model4)
X <- (model.matrix(terms(model4),zerotest))
hat <- X%*%b
pxal <- pnorm(hat) # probit link, for logit it would be: pval <- exp(hat)/(1+exp(hat))
pval
Once I...
2002 Jun 20
2
scatterplot3d
Hello,
I am trying to replicate example 4 in the package 'scatterplot3d':
s3d.dat_data.frame(cols=as.vector(col(my.model4)),
rows=as.vector(row(my.model4)),
value=as.vector(my.model4))
scatterplot3d(s3d.dat, type="h", lwd=5, pch=" ",
x.ticklabs=colnames(my.model4), y.ticklabs=rownames(my.model4),
main="Conditional probabilities, Model 3")
Nice! but,
1) Not clear how to attach a ve...
2009 Apr 27
2
refit with binomial model (lme4)
Dear R users,
I'm trying to use function 'refit' from lme4
and I get this error that I can't understand:
> refit(dolo4.model4,cbind(uu,50-uu))
Error in function (classes, fdef, mtable) :
unable to find an inherited method for function "refit", for signature
"mer", "matrix"
if I try:
> refit(dolo4.model4,uu)
Error in asMethod(object) : matrix is not symmetric [1,2]
I get this error m...
2012 Oct 26
2
Interpreting and visualising lme results
Dear R users,
I have used the following function (in blue) aiming to find the linear regression between MOE and XLA and nesting my data by Species. I have obtained the following results (in green).
model4<-lme(MOE~XLA, random = ~ XLA|Species, method="ML")summary(model4)
Linear mixed-effects model fit by maximum likelihood Data: NULL AIC BIC logLik -1.040187 8.78533 6.520094
Random effects: Formula: ~XLA | Species Structure: General positive-definite, Log-Cholesky parametr...
2006 Sep 12
4
variables in object names
Is there any way to put an argument into an object name. For example,
say I have 5 objects, model1, model2, model3, model4 and model5.
I would like to make a vector of the r.squares from each model by code
such as this:
rsq <- summary(model1)$r.squared
for(i in 2:5){
rsq <- c(rsq, summary(model%i%)$r.squared)
}
So I assign the first value to rsq then cycle through models 2 through 5
gathering there valu...
2004 Mar 03
7
Location of polr function
Hello
I am running R 1.8.1 on a Windows platform
I am attempting to fit an ordinal logistic regression model, using the
polr function, as described in Venables and Ripley. But when I try
model4 <- polr(ypsxcat~committed + as.factor(sex)
+ as.factor(drugusey) + anycsw + as.factor(sex)*committed
+ as.factor(sex)*as.factor(drugusey)+as.factor(sex)*anycsw, data =
duhray)
I get a message that the polr function was not found.
Any help appreciated
thanks
Peter
Peter L. Flom, PhD
Ass...
2005 Jul 15
1
nlme and spatially correlated errors
...st by
Crawley 2002 for linear models
e.g. (GUAMAC = a species, CRAS = soil water retention capacity, TRANSECTO =
transect)
> model1<-gls(GUAMAC ~ CRAS)
> model2<-lme(GUAMAC ~ CRAS, random = ~ 1 | TRANSECTO)
> model3<-lme(GUAMAC ~ CRAS, random = GUAMAC ~ CRAS | TRANSECTO)
> model4<-lme(GUAMAC ~ CRAS, random = GUAMAC ~ CRAS -1 | TRANSECTO)
> AIC(model1,model2,model3,model4)
df AIC
model1 3 3730.537
model2 4 3698.849
model3 6 3702.408
model4 4 3704.722
> plot(Variogram(model2, form = ~ X + Y))
> model5<-update(model2,corr=corSpher(c(30,0.8), form = ~ X + Y, n...
2012 Nov 08
2
Comparing nonlinear, non-nested models
...1.64, 1.89, 2.14, 2.47, 3.20, 4.47, 5.31, 6.48)
y <- c(0.00, 0.35, 0.41, 0.49, 0.58, 0.61, 0.71, 0.83, 0.98, 1.03, 1.06)
model1 <- function(x, xo, ym) ym * (x-xo)/x
model2 <- function(x, xo, ym, k) ym * (x-xo)/(k+x-xo)
model3 <- function(x, xo, ym) ym * (1-exp(-log(2)*(x-xo)/xo))
model4 <- function(x, xo, ym, k) ym * (1-exp(-log(2)*(x-xo)/k))
fit1 <- nls(y~model1(x, xo, ym), start=list(xo=0.5, ym=1))
fit2 <- nls(y~model2(x, xo, ym, k), start=list(xo=0.5, ym=1, k=1))
fit3 <- nls(y~model3(x, xo, ym), start=list(xo=0.5, ym=1))
fit4 <- nls(y~model4(x, xo, ym, k),...
2005 Mar 01
3
packages masking other objects
...th:
detach(package:lme4)
library(nlme)
but I still get an error message when I try to use getCovariateFormula. The line I use it in, and the error message is below:
rownames(table)<-c((getCovariateFormula(model1)),(getCovariateFormula(model2)),(getCovariateFormula(model3)),(getCovariateFormula(model4)),(getCovariateFormula(model5)),(getCovariateFormula(model6)),(getCovariateFormula(modelnull)))
Error in switch(mode(x), "NULL" = structure(NULL, class = "formula"), :
invalid formula
This line works fine when I run models with different structures that are not in th...
2010 Feb 09
1
Missing interaction effect in binomial GLMM with lmer
...ss ~ Origin + Treat + (1 | Female)
Df AIC BIC logLik Chisq Chi Df Pr(>Chisq)
model3 3 98.863 107.225 -46.431
model2 4 94.707 105.857 -43.353 6.1558 1 0.01310 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
model4<-update(model2,~.-Treat)
anova(model2,model4)
Data:
Models:
model4: success ~ Origin + (1 | Female)
model2: success ~ Origin + Treat + (1 | Female)
Df AIC BIC logLik Chisq Chi Df Pr(>Chisq)
model4 3 155.592 163.954 -74.796
model2 4 94.707 105.857 -43.353 62.885 1 2....
2009 Mar 31
1
using "substitute" inside a legend
...list("y"==slope%*%"x", R^2==rsq),
list(slope=round(mylm$coefficients[[1]],2),rsq=round(summary(mylm)$adj.r.squared,
2)))
I have four models and I need to list each equation in a legend. I'm using:
legend(20, 120, title = "fitted models", c(model1, model2, model3, model4),
lty=1, col=c("blue", "red", "black", "green"))
But this only works when I have up to two models. With four models the
formatting is lost. I suspect the concatenation converts the objects into
List and this destroys the formatting.
Does anybody have any su...
2009 Mar 09
1
lme anova() and model simplification
...h31 -0.609
awpm -0.360 0.345
oriensouth 0.379 -0.712 -0.225
Standardized Within-Group Residuals:
Min Q1 Med Q3 Max
-1.10623050 -0.20081291 -0.09441451 0.19507694 1.08369449
Number of Observations: 14
Number of Groups: 8
>
> model4<-lme(awsm~adh31+orien,random=~1|viney,method="ML")
> anova(model3,model4)
Model df AIC BIC logLik Test L.Ratio p-value
model3 1 6 41.47847 45.31281 -14.73924
model4 2 5 46.13353 49.32881 -18.06676 1 vs 2 6.655056 0.0099
>
>
> model5&...
2011 Apr 14
1
mixed model random interaction term log likelihood ratio test
...r as the
fixed effect, DrugPair as the random effect and the interaction between
these two as the random effect as well. Please confirm if that seems
correct.
2. Assuming the above code is correct, I have model2 in which I remove the
interaction term, model3 in which I remove the DrugPair term and model4 in
which I only keep the fixed effect of MatingPair.
3. I want to perform the log likelihood ratio test to compare these models
and that's why I have REML=F. However the code anova(model1, model2, model3,
model4) gives me a chisq estimate and a p-value, not the LRT values. How do
I get LRT (L....
2010 Sep 29
1
Understanding linear contrasts in Anova using R
...linear) = -2(1.805) -1(0.675) +0(-.912) +1(.544) +2(1.889) =
0.043
# SS(linear) = n*(Effect(linear)^2)/Sum((a(j)^2)) = 12(.043)/10 = .002
# F(linear) = SS(linear)/MS(error) = .002/2.263 = .001
# t(linear) = sqrt(.001) = .031
# To do this in R I would use
order.group <- ordered(group)
model4 <- lm(dv~order.group)
summary(model4)
# This gives:
#Coefficients:
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 0.79967 0.19420 4.118 0.000130 ***
# order.group.L 0.01344 0.43425 0.031 0.975422
# order.group.Q 2.13519 0.43425 4.917 8...
2010 Oct 03
5
How to iterate through different arguments?
If I have a model line = lm(y~x1) and I want to use a for loop to change the
number of explanatory variables, how would I do this?
So for example I want to store the model objects in a list.
model1 = lm(y~x1)
model2 = lm(y~x1+x2)
model3 = lm(y~x1+x2+x3)
model4 = lm(y~x1+x2+x3+x4)
model5 = lm(y~x1+x2+x3+x4+x5)...
model10.
model_function = function(x){
for(i in 1:x) {
}
If x =1, then the list will only add model1. If x =2, then the list will add
both model1 and model2. If x=3, then the list will add model1 model 2 and
model3 and so on. How do I translate...
2008 Nov 25
4
glm or transformation of the response?
...:40)
attach(poissondata)
However, I have run into a problem because it looks like the lm model (with sqrt-transformation)
fits the data best:
##
model1=lm(response~explanatory,poissondata)
model2=lm(sqrt(response+0.5)~explanatory,poissondata)
model3=lm(log(response+1)~explanatory,poissondata)
model4=glm(response~explanatory,poissondata,family=poisson)
model5=glm(response~explanatory,poissondata,family=quasipoisson)
model6=glm.nb(response~explanatory,poissondata)
model7=glm(response~explanatory,quasi(variance="mu",link="identity"))
plot(explanatory,response,pch=16)
lines(e...
2005 Oct 15
1
generating response curves
...ots. Having a large number of parameters the following formula is to time consuming. Perhaps a built in function exists to speed up the process.
Model1<-a~b
#Setting the scale extent
min(area)
max(area)
avals<-seq(0,10,.1)
# generating the plot
plot(area,incidence, las=1)
lines(avals,predict(model4,list(area=avals),type="response"))
Thanks in advance
Christian
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2008 Jan 30
0
95% confidence and prediction intervals for linear mixed models
Hi R-users,
>From the last week I've been working fitting a linear mixed model with
random intercept and fixed shape (model4) for a data set with 37 individuals
measured over time, using lme package. Results are at the end of this
message. The outcome is score and the covariate is age.
My question is: is possible (and how) to estimate both 95% confidence and
prediction intervals for the mixed model as in linear regressi...
2010 Mar 28
1
keeping track of who did what
hello everybody.
i need to log "who did what" on some models. i was reading recipe 59
of rails recipes and i created something like that:
class LogSweeper < ActionController::Caching::Sweeper
observe :model1, :model2, :model3, :model4, :model5, ...
after_save(model)
save_log(model, "save")
end
after_destroy(model)
save_log(model, "destroy)
end
private
save_log(model, event, user_id = controller.session[:account_id])
#create the log entry, i want to save the id of the model, the id
of th...