The mixed models list is r-sig-mixed-models .
nlme:lme is not really designed for crossed random effects. IIRC, it's
possible, but not easy. As Kevin said, lme4:lmer is really what you should
use.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking thi
You might get better answers on the r-sig-ME list.
The lmer() function from lme4 handles crossed and non-nested random effects
quite seamlessly. I cannot comment on whether or not lme() can as well.
--
Kevin E. Thorpe
Head of Biostatistics, Applied Health Research Centre (AHRC)
Li Ka Shing Kno
Dear R project,
I am a doctoral student in Zhejiang university in China, I am using lme
function in nlme package and learning the function by Package 'nlme' document.
I am writing this email for some help to build a lme model.
My goal was to include two non-nested random effects in the lme mode
My recommendation is:
Post on the BioConductor site, not here.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Thu, Nov 14, 2019 at 9:22 AM chziy429 wrote:
>
Dear Sir
I have downloaded the raw CEL data included in "GSE41418" from GEO and tried to
process the raw microarray data according to the following Rscripts
affydata <- ReadAffy(cdfname = "mouse4302mmentrezgcdf")
eset <- oligo::rma(affydata)
The raw data can be read by ReadAffy but fai
I prefer the idiom
c(TRUE, a[-1] != a[-length(x)])
because it works for character and other data types as well.
I also find that thinking in terms of runs instead of subscripting
tricks is easier.
__
R-help@r-project.org mailing list
https://stat.ethz
Better:
b <- c(a[1]-1,a[-length(a)])
On 07 Aug 2014, at 17:28, Bart Kastermans wrote:
> For readability I like:
>
>> b <- c(0,a[-length(a)])
>> which(a != b & a == 0)
> [1] 4 12 18
>> which(a != b & a == 1)
> [1] 1 6 16 23
>
>
> On 07 Aug 2014, at 17:23, William Dunlap wrote:
>
>> My sol
For readability I like:
> b <- c(0,a[-length(a)])
> which(a != b & a == 0)
[1] 4 12 18
> which(a != b & a == 1)
[1] 1 6 16 23
On 07 Aug 2014, at 17:23, William Dunlap wrote:
> My solution may be a bit clearer if you define the function isFirstInRun
> isFirstInRun <- function(x) {
> if (le
My solution may be a bit clearer if you define the function isFirstInRun
isFirstInRun <- function(x) {
if (length(x) == 0) {
logical(0)
} else {
c(TRUE, x[-1] != x[-length(x)])
}
}
Then that solution is equivalent to
which(isFirstInRun(a) & a==1)
If 'a' contains NA's then
On 07 Aug 2014, at 11:16 , jim holtman wrote:
> rle
...with a little tinkering, like
> m <- c(1,cumsum(rle(a)$lengths)+1)
> m
[1] 1 4 6 12 16 18 23 34
then look at every 2nd element, discarding the last.
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solb
> a<-c(1,1,1,0,0,1,1,1,1,1,1,0,0,0,0,1,1,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1)
> which( a==1 & c(TRUE, a[-length(a)]!=1) )
[1] 1 6 16 23
> which( a==0 & c(TRUE, a[-length(a)]!=0) )
[1] 4 12 18
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Aug 6, 2014 at 7:12 PM, Johnnycz wrote:
> Hello,eve
rle
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Wed, Aug 6, 2014 at 10:12 PM, Johnnycz wrote:
> Hello,everybody,
> I have a sequence,like
> a<-c(1,1,1,0,0,1,1,1,1,1,1,0,0,0,0,1,1,0,0,0,0,0,1
Hello,everybody,
I have a sequence,like
a<-c(1,1,1,0,0,1,1,1,1,1,1,0,0,0,0,1,1,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1),how to
get the position of each first 1 and 0, that's to say, how to get
b<-c(1,6,16,23) for first 1 and d<-c(4,12,18) for first 0.
Many thanks!
Johnny
[[alter
On 21-02-2013, at 15:39, JiangZhengyu wrote:
> Dear R experts,
>
> I have two matrix (seq & mat) & I want to retrieve in a new matrix all the
> numbers from mat that =1 (corresponding to the same row/ column position) in
> seq, or all the numbers in mat that =-1 in seq. - Replace all the numb
hengyu
Sent: 21 February 2013 14:39
To: r-help@r-project.org
Subject: [R] Ask for help: find corresponding elements between matrix
Dear R experts,
I have two matrix (seq & mat) & I want to retrieve in a new matrix all the
numbers from mat that =1 (corresponding to the same row/
[,4]
#[1,] 0.2588229 0.8971982 NA -1.0750013
#[2,] 1.8311207 0.4880163 NA 0.8550108
#[3,] NA -1.2553858 -0.1321224 NA
A.K.
- Original Message -
From: JiangZhengyu
To: r-help@r-project.org
Cc:
Sent: Thursday, February 21, 2013 9:39 AM
Subject: [R] Ask for h
Dear R experts,
I have two matrix (seq & mat) & I want to retrieve in a new matrix all the
numbers from mat that =1 (corresponding to the same row/ column position) in
seq, or all the numbers in mat that =-1 in seq. - Replace all the numbers with
NA if it's not 1/-1 in seq. There ar
Hello,
Maybe the code below answers to both your questions (Q1 and Q2)
cnames <- colnames(dat)
cnames <- sub("WHIRR\\.", "WHIRR_", cnames)
for(which_col in seq_len(ncol(dat)))
{
subset_data <- dat[which_col:ncol(dat)] # change 1, Q2
file_name <- sprintf('%s.csv', cnames[which_col]
Hi,
I have a script below.
dat <- read.table(file="pt.csv", header=T, sep=",", row.names=1,
col.names=1)
dat
for(which_col in seq_len(ncol(dat)))
{
subset_data <- dat[,which_col:ncol(dat)]
file_name <- sprintf('%s.csv', colnames(dat)[which_col])
write.csv(subset_data, file_name)
Hi there,
I have a experimental design as following:
P A B Y
1 1 1 1 -0.18524045
2 1 1 2 -1.64226232
3 2 2 1 -0.51342697
4 2 2 2 -0.29684874
5 3 3 1 0.71566733
6 3 3 2 -1.06097480
7 4 4 1 0.05772670
8 4 4 2 0.99316677
9 5 1 1 -0.61860414
10 5 1 2 0.50257548
1
On 20/09/2010 3:07 PM, Tianchan Niu wrote:
Dear All,
I am trying to use ode solver "rk4" to solve an ODE system, however, it
keeps saying: Error in eval(expr, envir, enclos) : object "dIN" not found.
The sample codes are enclosed as follows, please help me. Thank you very
much!
rm(list=ls(
Dear All,
I am trying to use ode solver "rk4" to solve an ODE system, however, it
keeps saying: Error in eval(expr, envir, enclos) : object "dIN" not found.
The sample codes are enclosed as follows, please help me. Thank you very
much!
rm(list=ls())
library(odesolve)
# The ODE syste
Please learn that capitalization is important both in R (sic) and in
effective communication.
There are CRAN packages RNetCDF, ncdf and ncdf4 on CRAN to read/write
NetCDF (sic) format files from R.
On Wed, 18 Aug 2010, mat chueng wrote:
hi,r project
I am a new user who has some questions ab
hi,r project
I am a new user who has some questions about r tool,I have some original data
file in netcdf format which is a famous geometry data file format, The file
contains very large weather information from every zone of the world,I want to
extract only a part of them to calcuate, like the
>
> question like this:
>
> I have data tag and phi, when tag are the same, then phi are the same. like
> below, tag have value 1 2 3 4 and phi have value .9 .3 1 0.
>
> my question is, how can I get the result like
> tag=c(1,2,3,4) and corresponding phi=c(.9,.3,1,0)
>
> tag phi
> 2.3
> 1
question like this:
I have data tag and phi, when tag are the same, then phi are the same. like
below, tag have value 1 2 3 4 and phi have value .9 .3 1 0.
my question is, how can I get the result like
tag=c(1,2,3,4) and corresponding phi=c(.9,.3,1,0)
tag phi
2.3
1.9
2.3
31
2
Try this:
unique(cbind(tag, phi))[order(unique(tag)),]
On Wed, Jan 27, 2010 at 4:13 PM, song song wrote:
>> question like this:
>>
>> I have data tag and phi, when tag are the same, then phi are the same. like
>> below, tag have value 1 2 3 4 and phi have value .9 .3 1 0.
>>
>> my question is, h
> question like this:
>
> I have data tag and phi, when tag are the same, then phi are the same. like
> below, tag have value 1 2 3 4 and phi have value .9 .3 1 0.
>
> my question is, how can I get the result like
> tag=c(1,2,3,4) and corresponding phi=c(.9,.3,1,0)
>
> tag phi
> 2.3
> 1.9
che Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von jlfmssm
Gesendet: Tuesday, June 23, 2009 6:07 PM
An: r-help@r-project.org
Betreff: [R] ask for help xyplot
I want to make such plot using the following data set
plot Mean with Sd(+/-) by
I want to make such plot using the following data set
plot Mean with Sd(+/-) by Dim group by Lac
here is data
Does anyone know how to plot using xyploy
Thanks,
Aimin
Mean Sd Var Min Max Dim Lac
704 44.0 NA NA 44.0 44.0 -30 3
703 45.9200
ject:
RE: [R] ask for help how to get accumulate contingency table?> To: [EMAIL
PROTECTED]> > I'm not at work so I'm not sure it will work but you can try
below.> > I am assuming the result of table(data1) is a list so> >
tablelist<-table(data1)> > cum
Hi, I post the question again.
x<-rep(c(2,3,4),20)
y<-sample(rep(c(0,1),30))
tr<-rep(c(1:5),6)
data1<-data.frame(x,y,tr)
data1<-data1[order(data1$tr),]
> data1
x y tr
2 1 1
4 0 1
3 1 1
2 1 1
4 1 1
3 1 1
2 0 1
4 0 1
3 0 1
2 0 1
4 0 1
3 1 1
3 0 2
2 0 2
>table(data1)
tr = 1
y
x
Ruqiang Liang gmail.com> writes:
>
> I have a table like the following. I want to fit Cm to Vm like this:
> Cm ~Cl+Q1*b1*38.67*exp(-b1*(Vm-Vp1)*0.03867)/(1+exp(-b1*(Vm-Vp1)*0.03867))^2+
Q2*b2*38.67*exp(-b2*(Vm-Vp2)*0.03867)/(1+exp(-b2*(Vm-Vp2)*0.03867))^2
>
> I use nls, with start=list(Q1=2e-3,
I have a table like the following. I want to fit Cm to Vm like this:
Cm ~
Cl+Q1*b1*38.67*exp(-b1*(Vm-Vp1)*0.03867)/(1+exp(-b1*(Vm-Vp1)*0.03867))^2+Q2*b2*38.67*exp(-b2*(Vm-Vp2)*0.03867)/(1+exp(-b2*(Vm-Vp2)*0.03867))^2
I use nls, with start=list(Q1=2e-3, b1=1, Vp1=-25, Q2=3e-3, b2=1,
Vp2=200). But
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