similar to: low-variance warning in lmer

Displaying 11 results from an estimated 11 matches similar to: "low-variance warning in lmer"

2003 Jun 14
2
A sapply() funny.
The sapply function is refusing to return a result for what seem to me to be mysterious reasons. Here is a toy example: set.seed(111) X <- list(x=runif(20),y=runif(20)) rvec <- seq(0.01,0.15,length=42) foo <- function(x,X,cc) { mean((X$x)^x + (X$y)^cc) } bar <- function(x,a,b){a+b*x} try.b <- sapply(rvec,bar,a=1,b=2) # This runs without a problem and
2005 Apr 29
0
handling of zero and negative indices in src/main/subscript.c:mat2indsub() (PR#7824)
This message contains a description of what looks like a bug, examples of the suspect behavior, a proposed change to the C code to change this behavior, example of behavior with the fix, and suggestions for 3 places to update the documentation to reflect the proposed behavior. It is submitted for consideration for inclusion in R. Comments are requested. Currently, the code for subscripting
2005 May 06
0
(PR#7824) handling of zero and negative indices in
I've put this in (with some different wording). Although S blithely accepts mis-dimensioned index matrices I agree this is wrong and have made it an error. On Fri, 29 Apr 2005 tplate@acm.org wrote: > This message contains a description of what looks like a bug, examples > of the suspect behavior, a proposed change to the C code to change this > behavior, example of behavior with
2008 Mar 21
1
(no subject)
Hi, I am fairly new to R, and am stuck. I want to write an R function with argument n that returns a vector of length n with n simulated observations from the double exponential distribution with density: ??g(y) = 1/2e^-y ? For the double exponential, I want to generate y~Exp(1) and then take ?y with probability 0.5 ? Does anyone know how I can do this in R? Thanks! Fran [[alternative
2011 Jul 02
1
Simulating inhomogeneous Poisson process without loop
Dear all I want to simulate a stochastic jump variance process where N is Bernoulli with intensity lambda0 + lambda1*Vt. lambda0 is constant and lambda1 can be interpreted as a regression coefficient on the current variance level Vt. J is a scaling factor How can I rewrite this avoiding the loop structure which is very time-consuming for long simulations? for (i in 1:N){ ... N <- rbinom(n=1,
2010 Jan 20
3
barchart with stacked and beside bars
Hi, Is there a way to stack bars in a barchart as well as "beside" bars for the same treatment? eg.... I have one barchart like this: bio<-matrix(c(10,23,9,25),nrow=2,byrow=T) ntreat<-c("n0","n96") colnames(bio)<-ntreat barplot(bio,beside=T) now i want a similar barchart but with stacked bars:
2003 Jan 16
2
polynomial contrasts in R
In S-Plus, I can obtain polynomial contrasts for an ordered factor with contr.poly(). The function also exists in R, however is limited to factors where the levels are equally spaced. In S-Plus, one can obtain the contrasts for a set of numeric values representing unequally spaced ordered factors. Has anyone implemented this in R? I see that the S-Plus function calls another function (poly.raw())
2011 Jul 29
2
Multifactor boxplots
Dear All I would like to produce interaction boxplots and this seems to work: par(mfrow=c(2,2)) A=sample(rnorm(50,50,10)) B=sample(rnorm(50,100,10)) Test=merge(A,B,by=0)#by=0 where 0 is the row.names TreatA=(gl(2,50,100,labels=c("High","Low"))) TreatB=rep(gl(2,25,50,labels=c("High","Low")),2) Newdata=data.frame(TreatA,TreatB,Test)
2008 Aug 15
2
Design-consistent variance estimate
Dear List: I am working to understand some differences between the results of the svymean() function in the survey package and from code I have written myself. The results from svymean() also agree with results I get from SAS proc surveymeans, so, this suggests I am misunderstanding something. I am never comfortable with "I did what the software" does mentality, so I am working to
2005 May 23
1
Can't reproduce clusplot princomp results.
Dear R folk: Perhaps I'm just dense today, but I am having trouble reproducing the principal components plotted and summarized by clusplot. Here is a brief example using the pluton dataset. clusplot reports that the first two principal components explain 99.7% of the variability. But this is not what princomp is reporting. I would greatly appreciate any advice. With best regards, -- Tom
2004 Sep 02
3
confidence intervals
Dear R users; Im working with lme and Id like to have an idea of how can I get CI for the predictions made with the model. Im not a stats guy but, if Im not wrong, the CIs should be different if Im predicting a new data point or a new group. Ive been searching through the web and in help-lists with no luck. I know this topic had been asked before but without replies. Can anyone give an idea of