similar to: multivariate analysis by using lme

Displaying 20 results from an estimated 3000 matches similar to: "multivariate analysis by using lme"

2004 Jun 01
1
swapping with data.frame
Hi there, I have some data which are convenient to enter as lists. For example: t1<-list(fname="animal1",testname="hyla",dspkr="left",res1=39.7,res2=15.0) t2<-list(fname="animal1",testname="bufo",dspkr="left",res1=14.4,res2=56.1)
2012 Jan 04
1
function in R for my exercise
Hi R helpers! I have a question. I'm trying to create a function for an exercise. Here are the arguments I should include: x and y are numeric z is a name ("plus","minus","multiply","divide") and swap is logical. Here is what the function should do: When z="plus", then x+y is performed and so on for the other z names. It should give a NA
2000 Jul 05
1
Tukey.aov with split-plot designs
I am using R 1.1 with Redhat 6.2 and RW 1.001 with Win98 (the upkey doesn't work on my IBM either as has been previously reported by others). The function aov doesn't return either the residuals or the residual degrees of freedom for split-plot designs. If you use the following code from Baron and Li's "Notes on the use of R for psycology experiments and questionnaires"
2013 Apr 29
1
how to add new rows in a dataframe?
Hi, dat1<- read.table(text=" id??????????????? t???????????????????? scores 2???????????????? 0??????????????????????? 1.2 2???????????????? 2???????????????????????? 2.3 2???????????????? 3??????????????????????? 3.6 2???????????????? 4??????????????????????? 5.6 2???????????????? 6??????????????????????? 7.8 3???????????????? 0??????????????????????? 1.6 3????????????????
2010 Jul 20
1
p-values pvclust maximum distance measure
Hi, I am new to clustering and was wondering why pvclust using "maximum" as distance measure nearly always results in p-values above 95%. I wrote an example programme which demonstrates this effect. I uploaded a PDF showing the results Here is the code which produces the PDF file: ------------------------------------------------------------------------------------- s <-
2011 Apr 12
1
question about optim
Dear R-users, I would like to use optim( ) to minimize a function which depends on 4 parameters: 2 vectors, a scalar, and a matrix. And I have a hard to define the parameters at the beginning of the function, and then to call optim. Indeed, all the examples I have seen dont treat cases where parameters are not all real. Here is my code, it doesnt work but its just to show you where is exactly my
2012 Aug 06
1
cannot find function "simpleRDA2"
Hi, I am trying to run the command "forward.sel.par," however I receive the error message: "Error: could not find function 'simpleRDA2'." I have the vegan library loaded. The documentation on "varpart" has not helped me to understand why I cannot call this function. Maybe I am missing something obvious because I am still an 'R' novice. Below is a
2003 Aug 21
4
anova(lme object)
Hi, I use lme to fit models like R> res1 <- lme(y~A+B, data=mydata, random=~1|subject) R> res2 <- lme(y~B+A, data=mydata, random=~1|subject) (only difference between these two models are the sequence in which the indep variables are written in formula) where y is continuous and A, B, and subject are factors. To get ANOVA table I used R> anova(res1) R> anova(res2) and found
2011 May 19
3
problem with optim()
Dear R-users, I would like to maximize the function g above which depends on 4 parameters (2 vectors, 1 real number, and 1 matrix) using optim() and BFGS method. Here is my code: # fonction to maximize g=function(x) { x1 = x[1:ncol(X)] x2 = x[(ncol(X)+1)] x3 = matrix(x[(ncol(X)+2):(ncol(X)+1+ncol(X)*ncol(Y))],nrow=ncol(X),ncol=ncol(Y)) x4 = x[(ncol(X)+1+ncol(X)*ncol(Y)+1):length(x)]
2005 Jul 19
1
initial points for arms in package HI
Dear R-users I have a problem choosing initial points for the function arms() in the package HI I intend to implement a Gibbs sampler and one of my conditional distributions is nonstandard and not logconcave. Therefore I'd like to use arms. But there seem to be a strong influence of the initial point y.start. To show the effect I constructed a demonstration example. It is reproducible
2007 Nov 12
2
strange `nls' behaviour
I initially thought, this should better be posted to r-devel but alas! no response. so I try it here. sory for the lengthy explanation but it seems unavoidable. to quickly see the problem simply copy the litte example below and execute f(n=5) which crashes. called with n != 5 (and of course n>3 since there are 3 parameters in the model...) everything is as it should be. in detail:
2013 Apr 14
1
possible loop problem
Hi, It would be better if you provided the output of dput(dataset).? I am not sure about the structure of your dataset. Just from reading the data as is shown. dat1<- read.table(text=" separator,tissID >,>,2 ,2,1 ,6,5 ,11,13 >,>,4 ,4,9 ,6,2 ,7,3 ,21,1 ,23,58 ,25,9 ,26,4 >,>,11 ,1,12 >,>,21 ,4,1 ,11,3
2011 Aug 04
1
How to get the test statistic corresponding to the p-value in mtable?
Dear R-Users, I want to use mtable from package "memisc" to produce Latex-style estimation output. However, mtable() only gives me a p-value and not the corresponding test-statistic. Does anyone know how to extract it, either from a glm/anova object or mtable? Here is a short example: # Run this #################### install.packages("memisc") library(memisc) set.seed(1)
2011 Feb 24
2
MCMCpack combining chains
Deal all, as MCMClogit does not allow for the specification of several chains, I have run my model 3 times with different random number seeds and differently dispersed multivariate normal priors. For example: res1 = MCMClogit(y~x,b0=0,B0=0.001,data=mydat, burnin=500, mcmc=5500, seed=1234, thin=5) res2 = MCMClogit(y~x,b0=1,B0=0.01,data=mydat, burnin=500, mcmc=5500, seed=5678, thin=5) res3 =
2007 Aug 15
3
Formula in lm inside lapply
I am trying to run separate regressions for different groups of observations using the lapply function. It works fine when I write the formula inside the lm() function. But I would like to pass formulae into lm(), so I can do multiple models more easily. I got an error message when I tried to do that. Here is my sample code: #generating data x1 <- rnorm(100,1) x2 <- rnorm(100,1) y <-
2023 Nov 14
1
data.frame weirdness
What is going on here? In the lines ending in #### the inputs and outputs are identical yet one gives a warning and the other does not. a1 <- `rownames<-`(anscombe[1:3, ], NULL) a2 <- anscombe[1:3, ] ix <- 5:8 # input arguments to #### are identical in both cases identical(stack(a1[ix]), stack(a2[ix])) ## [1] TRUE identical(a1[-ix], a2[-ix]) ## [1] TRUE res1 <-
2023 Nov 14
1
data.frame weirdness
They differ in whether the row names are "automatic": > .row_names_info(a1) [1] -3 > .row_names_info(a2) [1] 3 Best, -Deepayan On Tue, 14 Nov 2023 at 08:23, Gabor Grothendieck <ggrothendieck at gmail.com> wrote: > > What is going on here? In the lines ending in #### the inputs and outputs > are identical yet one gives a warning and the other does not. > >
2007 Oct 26
2
[LLVMdev] LLVM Newbie. Questions about backend.
Hello, I have been studying LLVM and started to create a new backend for a new RISC architecture. Now I need some help to get forward with my project. I'm quite new to compiling techniques so I'm sorry for the stupid questions. Question 1: My idea is to lower the select SDNode as follows: %res1 = %falseVal %res2 = setc %trueVal, %condition Where setc is conditional mov. The
2023 Nov 14
1
data.frame weirdness
In that case identical should be FALSE but it is TRUE identical(a1, a2) ## [1] TRUE On Tue, Nov 14, 2023 at 8:58?AM Deepayan Sarkar <deepayan.sarkar at gmail.com> wrote: > > They differ in whether the row names are "automatic": > > > .row_names_info(a1) > [1] -3 > > .row_names_info(a2) > [1] 3 > > Best, > -Deepayan > > On Tue, 14 Nov
2002 Mar 16
1
promax (PR#1389)
Full_Name: conor dolan Version: 1.4.0 OS: windows98 Submission from: (NULL) (146.50.170.247) On the basis of Lawley and Maxwell's explanation in Factor Analysis as a Statistical Method (London: Butterworth, 1971), the following line in the promax routine (library, mva): attr(z, "covariance") <- crossprod(U) should be attr(z, "covariance") <- solve(crossprod(U))