If A has more columns than in your example, you could always try to only
merge those columns of A with B that are relevant for the merging. You could
then cbind the result of the merging back together with the rest of A as
long as the merged data preserved the same order as in A.
Alternatively, yo
Hello!
In a grouped Dotplot, is there any way to set the color of error bars
to be the same as the corresponding symbols?
Example data:
require(lattice)
require(Hmisc)
data(barley)
Dotplot(variety~Cbind(yield, yield+2, yield-2)|year, groups=site,
data=barley)
I experimented with changing tre
Thanks Daniel, that helped me. Based on your suggestions I built this final
code:
library(foreign)
library(gdata)
AA = c(4,4,4,2,2,6,8,9)
A1 = c(3,3,11,5,5,7,11,12)
A2 = c(3,3,7,3,5,7,11,12)
A = cbind(AA, A1, A2)
BB = c(2,2,4,6,6)
B1 =c(5,11,7,13,NA)
B2 =c(4,12,11,NA,NA)
B3 =c(12,13,NA,NA
Hello list,
I am having trouble with "match" and "which" giving a NA result for some
values, but not others. Here is a simple example:
> aaa <- seq(0,1,by=0.05)
> aaa
[1] 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70
[16] 0.75 0.80 0.85 0.90 0.95 1.00
> match(0.5,
Dear all,
Does any one know if any R package or function can do Ordinary Least Products
regression? Many thanks!
Bill
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PLEASE do read the posting guide http://www.R-proje
Hi Jim,
Here is one way:
# data
x <- structure(list(category = structure(c(1L, 1L, 1L, 2L, 2L, 2L,
1L, 1L, 2L), .Label = c("case", "control"), class = "factor"),
SNP1 = c(1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L), SNP2 = c(0L,
1L, 2L, 1L, 2L, 0L, 0L, 1L, 0L), SNP3 = c(2L, 1L, 2L, 0L,
1L, 0L
Hello,
I have indicators for the present of absent of a snps in columns and the
categorey (case control column). I would like to extract ONLY the tables and
the indices (SNPS) that give me 2 x 3 tables. Some gives 2x 2 tables when
one of the allelle is missing. The data look like the matrix snpmat
On Wed, Jul 27, 2011 at 2:58 PM, Erica Crome wrote:
> Afternoon R help,
>
> I want to run Rasch/IRT analyses using the ltm package, however, I am
> using large scale survey data which requires weighting for accurate
> results. I attempted to create a weighted object to insert into the
> formulae o
Afternoon R help,
I want to run Rasch/IRT analyses using the ltm package, however, I am
using large scale survey data which requires weighting for accurate
results. I attempted to create a weighted object to insert into the
formulae of the ltm packages, however, the survey data only includes
30 re
Dear R User,
I am wondering if there is a way to generate correlated multivariate
non-normal distribution?
For example, I want to generate four correlated negative binomial
series with parameters r=10, p=0.2, based on the correlation
coefficient matrix
| 1 0.9 0.8 0.8 |
| 0.9 1 0.8 0.
Yes.
According to your suggestion,I modified my code.It works well.
Thanks you very much.
My best.
2011/7/26 Peter Ehlers
> On 2011-07-26 00:16, Lao Meng wrote:
>
>> Hi all:
>> There's a question about glht function.
>>
>> My data:data_ori,which inclue CD4, GROUP,time.
>>
>>
>> f_GROUP<-fac
Hi:
Here are a few options. The most important thing I'd recommend is to
rotate the bar plot so that the species can be read easily. In the
process, you also get the right justification you want in a readable
text size. Here's an example with the barplot() function in base R:
set.seed(103)
x <- r
This is clearly a message for the R-help mailing list, since it was
sent to the R help mailing list.
fisher.test(x)[1]
Jeremy
On 26 July 2011 14:51, Zmarz, Pawel wrote:
> Dear r-helpers,
>
> I would be very grateful if you could post the message below on the r-help
> discussion board. Than
Roland Sookias gmail.com> writes:
>
> Hi
>
> I'm trying to replicate Smith et al.'s
> (http://www.sciencemag.org/content/330/6008/1216.abstract) findings by
> fitting their Gompertz and logistic models to their data (given in
> their supplement). I'm doing this as I want to then apply the
> equ
On Tue, Jul 26, 2011 at 7:18 PM, Eduardo Mendes wrote:
> Hello again
>
> I do apologize for the previous email without any useful information that
> can lead to an answer. To those who felt offended by "to no avail", I do
> apologize again. I guess I will be always a newbie as far as R is
> con
Hello again
I do apologize for the previous email without any useful information that
can lead to an answer. To those who felt offended by "to no avail", I do
apologize again. I guess I will be always a newbie as far as R is
concerned.
The date format was wrong and although I have tried to us
Tia:
Please first read the Help file -- ?bwplot
and especially note the examples at bottom.
-- Bert
On Tue, Jul 26, 2011 at 12:39 PM, Tia Molte wrote:
> Hi guys,
> I need your help with the boxplot.
> I've to create a boxplot starting from a table (.csv) in which there are the
> result of a tes
?strptime
--
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology VOICE: (360) 407-6815
PO Box 47600FAX:(360) 407-7534
Olympia, WA 98504-7600
Dear r-helpers,
I would be very grateful if you could post the message below on the r-help
discussion board. Thank you very much!
Best Wishes,
Pawel
Hello R community,
I am generating lots of results using the fisher.test function, testing many
2x2 tables of SNPs for association with a parti
Hi guys,
I need your help with the boxplot.
I've to create a boxplot starting from a table (.csv) in which there are the
result of a test.
Each column is a question and the rows are the answer of the respondents
(from 0 to 5).
Some answer is missing and has been filled with NA.
the table has been l
I am a graduate student who's just starting to use more advanced
statistics and is completely new to R. I'm looking for whether
parasite prevalence varies with region. I'd like to try using just a
subset of the data defined by one these covariates, sex, looking at
data only from adults, without u
Making the second layer of legend have full transparency seems to have fixed
the issue. My first layer has to have a white background because I have a
colored grid behind (sorry that didn't make it into my toy example).
I did try to use type="Xlib", but I get the error "libpng warning:
Application
I am trying to tweak how my categorical x-axis labels are formatted in
my bar graph. Specifically, I would like to a) decrease the spacing
between lines (e.g. spacing between Dialium and guianensis) b) right
justify the text and c) have each species name align with the center
of the corres
Hello
I have a huge file (not an R-file) in which the first column is a string
with date, hour, minutes and seconds (For instance, "31-Jul-2010 23:59:00").
I tried as.Date but the error msg was "Error in charToDate(x) :
character string is not in a standard unambiguous format".
I have chec
If I understand you correctly, you are looking for partitions of an integer, so
look at Robin Hankin's package 'partitions'
David L. Reiner
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of David Pham
Sent: Tuesday, July 26, 2011 5:
This is much clearer. So here is what I think you want to do. In theory and
practice:
Theory:
Check if AA[i] is in BB
If AA[i] is in BB, then take the row where BB[j] == AA[i] and check whether
A1 and A2 are in B1 to B3. Is that right? Only if both are, you want the
indicator to take 1.
Here i
Hi,
I'm trying to get a plot that looks somewhat like the attached image
(sketched in word).
I think I need somthing called a rose diagram? but I can't get it to do what
I want. I'm happy to use any library.
Essentially, I want a circle with degree slices every 10 degrees with 0 at
the top repres
Hi
I'm trying to replicate Smith et al.'s
(http://www.sciencemag.org/content/330/6008/1216.abstract) findings by
fitting their Gompertz and logistic models to their data (given in
their supplement). I'm doing this as I want to then apply the
equations to my own data.
Try as a might, I can't quite
On Tue, 2011-07-26 at 13:42 -0700, Bert Gunter wrote:
> Not quite, Gavin. You have to assign the value of unname back:
Yes; thought that would be a given. The point was to give a simple
example to show that `unname()` removes names/dimnames.
G
> > z <- structure(2, names="a")
> > unname(z)
> [1]
Not quite, Gavin. You have to assign the value of unname back:
> z <- structure(2, names="a")
> unname(z)
[1] 2
> z
a
2
> zz <- unname(z)
> zz
[1] 2
> names(z) <- NULL
> z
[1] 2
Cheers,
Bert
On Tue, Jul 26, 2011 at 1:18 PM, Gavin Simpson wrote:
> On Tue, 2011-07-26 at 16:43 +0100, Barry Rowlin
On Tue, 2011-07-26 at 16:43 +0100, Barry Rowlingson wrote:
> On Tue, Jul 26, 2011 at 4:21 PM, ascoquel wrote:
> > Hi,
> >
> > I've done a linear fit on my data and I would like to get back the a (time)
> > coefficient ...
> >
> > mod<-lm(res_sql2$Lx0x~0+time)
> > result<-data.frame()
> > result<-c
OK, Ive done more research, and I think that what I am looking for is
"repeated cross section" or "pseudo-panel" estimators. Does anyone know if
these have been implimented inany r package?
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On Jul 26, 2011, at 12:28 PM, Anera Salucci wrote:
I am trying to used "ordgee" from "geepack" for an ordinal dataset.
When I write the code it returns
"Warning message:In binomial(link) : use of binomial(link=link) is
deprecated" ,
but the program runs.
fit <- ordgee(ordered(phy) ~ age+B
On 2011-07-26 08:47, marcel wrote:
I am making final adjustments to a multi-plot figure using basic and lattice.
In the lattice plot, I would like to include a legend that matches the
format of the legends in the other plots, which consist of appropriately
colored squares (pch 22) followed by tex
Please do *NOT* repeat post, nor send HTML (see the posting guide).
This is repeat of
https://stat.ethz.ch/pipermail/r-help/2011-July/284829.html
There is no error here. If there had been, the posting guide asks you
to contact the package maintainer, who will (I hope) be able to
explain the **
I am trying to used "ordgee" from "geepack" for an ordinal dataset.
When I write the code it returns
"Warning message:In binomial(link) : use of binomial(link=link) is deprecated" ,
but the program runs.
> fit <- ordgee(ordered(phy) ~ age+BMI++convert, id=id, data=newdata,
>int.const=FALSE)
W
I am making final adjustments to a multi-plot figure using basic and lattice.
In the lattice plot, I would like to include a legend that matches the
format of the legends in the other plots, which consist of appropriately
colored squares (pch 22) followed by text. In lattice, I was able to get a
le
It was
result[[1]]
I have no intercept
Thanks !!!
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Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-projec
Thank you, this works well - overall, I now have pretty good control over the
appearance of the L-hand side of the plot and relative distance between axis
line, ticks, tick labels and axis title by adjusting pad1, pad2 and the plot
position in the print statement. I am sure it is not the best code,
Will:
result$coef[[2]]
Give you want you want?
Jeremy
On 26 July 2011 08:21, ascoquel wrote:
> Hi,
>
> I've done a linear fit on my data and I would like to get back the a (time)
> coefficient ...
>
> mod<-lm(res_sql2$Lx0x~0+time)
> result<-data.frame()
> result<-coef(mod)
> print("result")
On Tue, Jul 26, 2011 at 4:21 PM, ascoquel wrote:
> Hi,
>
> I've done a linear fit on my data and I would like to get back the a (time)
> coefficient ...
>
> mod<-lm(res_sql2$Lx0x~0+time)
> result<-data.frame()
> result<-coef(mod)
> print("result")
> print(result)
> [1] "result"
> time
> 0.025
# for, e.g., four random individuals
x.sub<-data[data$ID %in% sample(1:16, 4),]
On Tue, Jul 26, 2011 at 10:10 AM, Soanes, Louise
wrote:
> Hello
> I have been trying to figure out how to randomly select a number of records.
> I have a column "ID" which contains 16 individuals (numbered 1-16), ea
Dear R-Gurus
I am a PhD student from South Africa working on chimpanzee behaviour.
I am looking at patterns of shade utilization and am using generalized
linear mixed models to examine the effects of various factors on
whether chimpanzees choose to spend time in the sun or shade. I
realise that th
Hello
I have been trying to figure out how to randomly select a number of records. I
have a column "ID" which contains 16 individuals (numbered 1-16), each
individual has a few hundred rows of GPS locations associated with it. I am
trying to select a random sample of these individuals (with all
Hi,
I've done a linear fit on my data and I would like to get back the a (time)
coefficient ...
mod<-lm(res_sql2$Lx0x~0+time)
result<-data.frame()
result<-coef(mod)
print("result")
print(result)
[1] "result"
time
0.02530191
But I would like just the value 0.02530191 ... I tried result$ti
The ls() function return the names of the objects not the objects.
So if you have :
a <- c(1,2,3)
b <- c('A','B','C')
ls() will return :
"a" "b" which only 2 letters.
-
Christophe Poulet
GIGA-Research.
Human Genetics Dept.
Liège, Belgium.
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On Tue, Jul 26, 2011 at 9:48 AM, thierrydb wrote:
> Hello,
>
> I have a population of 2000+ zoo time series (but my environment also
> contains objects that are not zoo time series). I'm trying to calculate the
> latest 90 days Z-Score of all zoo time series, using the following code:
>
>
> LZS<-f
Hi
> Re: [R] Big data and column correspondence problem
>
> Daniel, thanks for the answer.
> I will try to make myself i little bit clearer. Doing step by step I
would
> have (using a loop trough the lines of 'A'):
I am not sure if you are successful in your clarifying.
>
> 1. AA[1] is 4. As
On a point of information, the beta distribution is indeed
defined for x >= 0 and, respectively, for x <= 1 so long as
the parameters a="shape1" and b="shape2" are respectively
not less than 1:
dbeta(x,a,b) = (x^(a-1))*((1-x)^(b-1))/Beta(a,b)
When a=1 and b=1 we have the uniform distribution on
William Dunlap wrote:
>> $ cut(c(20.8, 21.3, 21.7, 23, 25), 2, dig.lab=1)
>> [1] (21,23] (21,23] (21,23] (23,25] (23,25]
>> Levels: (21,23] (23,25]
>>
>> So the first number, 20.8, get put in the interval (21,23], which seem
>> strange. I can see why this could happen, though, as perhaps the 20.8
Hi Merik,
Please keep the mailing list copied.
On Tue, Jul 26, 2011 at 6:44 AM, Merik Nanish wrote:
> You can convert my data into a dataframe simply by dat <- data.frame(id,
> month, value). That doesn't help though.
Can you be more specific? What is the problem you are having?
And no, that's
Hello,
I have a population of 2000+ zoo time series (but my environment also
contains objects that are not zoo time series). I'm trying to calculate the
latest 90 days Z-Score of all zoo time series, using the following code:
LZS<-function(ser) {
temp<-window(ser,start=Sys.Date()-90)
last((temp-
DODamen represents the dissolved oxygen values and is a valid predictor. I'm
not sure how the with(test1, table(Depart, DODamen)) helps. It does show that
the DODamen column is populated and in use since it lists out a table of
relation between the Depart and DODamen columns.
Thanks for the he
Daniel, thanks for the answer.
I will try to make myself i little bit clearer. Doing step by step I would
have (using a loop trough the lines of 'A'):
1. AA[1] is 4. As so, I would have to compare A1[1] = 20 and A2[1] =3 with
B1 B2 B3
B[3,2:4] 7 11 NA
beacause BB[3]=4. Since there is
Hi, but why we do the difference : ltemp <- 2 * diff(tfit$loglik[1:2]) ??
Where I can find information about Integrate Likelihooh and null like
lihood??
Thank you very much,
Roby
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I am trying to used "ordgee" from "geepack" for an ordinal dataset.
When I write the code it returns
"Warning message:In binomial(link) : use of binomial(link=link) is deprecated" ,
but the program runs.
Even when I run your example for "ohio" and "respdis", it returns the same
error.
Please
On Mon, Jul 25, 2011 at 8:03 PM, Jose Bustos Melo wrote:
> Who knows if there's people working in Life Cycle Assesment (carbon emition)
> with R? or If there's someone interested in doing a package about it, please
> let me know!
>
Some of the places to check would be:
rseek.org
http://cran.at.r
On 26/07/2011 8:48 AM, Rohini Sen wrote:
Hello,
I'm new to R and work with psychometrics (Item response theory). I was using
a package called "cirt" to model a data that contains only 0s, 1s and 9s. The
package is basically used for Monte Carlo Markov Chain estimation of
parameters and I speci
For the list to have any hope of helping, we need at the very least a small
reproducible example, one that comes with data and gives the same
error. We don't even know what commands you issued.
Sarah
On Tue, Jul 26, 2011 at 8:48 AM, Rohini Sen wrote:
> Hello,
>
>
>
> I'm new to R and work with p
1. This thread belongs on the r-sig-mixed-models list ,
2. I would bet that for some of the data sets, the model is
unbalanced/overspecified, causing te convergence issues (imagine
climbing a narrow ridge by zig-zagging back and forth across it).
Cheers,
Bert
On Tue, Jul 26, 2011 at 12:56 AM, ON
This is not very confusing. It is the exact same error in the sense that this
time the values of x1 are not only outside the interval (0-1) but within
[0-1] as in your first example, but this time they are also outside [0-1].
The reason is that you did not divide x1 by sum(x1) this time. In other
w
On Jul 26, 2011, at 8:43 AM, Terry Therneau wrote:
There is something very odd about your data to give se values that are
so very large. Usually, this means that the data is deterministic:
some
combination of predictors is able to separate a subset with no events
from all the others. This
Hi
this might be a little bit off topic, but here it goes: lets assume I have
the following:
set.seed(13)
dat1 <- rnorm(2000, mean=10, sd=10)
dat2 <- rnorm(100, mean=10, sd=20)
d.all <- density(dat, n=1024)
d.co <- density(x[[v]], , from=min(d.all$x), to=
Well, please install them for arch=i386 !
And do note what the rw-FAQ and the @ReadMe said about not reporting
here, and what you needed to do if a package was not available.
(FWIW, my machine correctly finds that arch for those packages. Are
you *sure* you installed them for R 2.13.1? Try
I don't know if it could help, but under Mac os, you have to re-install all
your packages after installing a new version of R.
just use the following command :
install.packages("package.name",dependencies = TRUE);
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Hello,
I'm new to R and work with psychometrics (Item response theory). I was using
a package called "cirt" to model a data that contains only 0s, 1s and 9s. The
package is basically used for Monte Carlo Markov Chain estimation of
parameters and I specified 5000 iterations. The model seems to r
[Follow-up -- see at end]
On 26-Jul-11 12:03:46, Ted Harding wrote:
> On 26-Jul-11 11:26:14, Jim Lemon wrote:
>> On 07/26/2011 02:40 AM, Naomi Robbins wrote:
>>> Hello!
>>> It's a shoot in the dark, but I'll try. If one has a total of 100
>>> (e.g., %), and three components of the total, e.g.,
>>>
There is something very odd about your data to give se values that are
so very large. Usually, this means that the data is deterministic: some
combination of predictors is able to separate a subset with no events
from all the others. This leads to a problem where the log-likelihood
is maximum at
You could doa manual uninstall. Read windows FAQ for Windows on the
entries the R installation program makes to the windows registry. It
also describes how to remove the registry entries. Once you have done
this you can simply delete the R directory. You will be missing a
very good piece of sof
I am using the latest version of R 2.13.1 and need to load the following
packages;
library(maps)
library(mapdata)
library(mapproj)
library(lattice)
library(tgp)
library(spatstat)
library(akima)
I get the follwing error when loading the packages, "maps", "mapdata" and
"mapproj"
Error: package 'ma
Hi,
I wanted to know if there existed an good implementation in R of the
following classical subset Integers :
$P_{n,m} = {k_1, \ldots, k_m \in \mathbb{N} : k_1 + \ldots + k_m = n }$
for any integers $m < n$. There is an obvious not optimal code which would
be to run through $m$ sums and put a
Surv(entry-time, last-follow-up-time, status-at-last-follow-up)
This works in survfit, coxph, and coxme. Left truncation is currently
not implemented for survreg.
In your case both times above would be "time since event".
Terry T.
--- begin included message --
I have a fairly simple que
Hello,
I'd like to compare the predictive power of two independent simple
regression models, which relate to data sets with different numbers of
points (n1 and n2). I think I could use a F-test to compare both
residual variances, but I don't know whether it's a good idea or not.
To my opinio
And at that point, one is essentially doing a meta-analysis. For example:
library(metafor)
ri <- c(.5, .4)
ni <- c(40, 25)
res <- rma(ri=ri, ni=ni, measure="ZCOR", method="FE")
predict(res, transf=transf.ztor, digits=2)
pred se ci.lb ci.ub
0.46 NA 0.24 0.64
You also get the CI (in additi
On 26-Jul-11 11:26:14, Jim Lemon wrote:
> On 07/26/2011 02:40 AM, Naomi Robbins wrote:
>> Hello!
>> It's a shoot in the dark, but I'll try. If one has a total of 100
>> (e.g., %), and three components of the total, e.g.,
>> mytotal=data.frame(x=50,y=30,z=20), - one could build a pie chart with
>> 3
On Tue, Jul 26, 2011 at 7:11 AM, Gabor Grothendieck
wrote:
> On Tue, Jul 26, 2011 at 4:48 AM, djdjoko wrote:
>> Hi,
>> I have an intraday timeseries of financial data (see below) which has gaps
>> due to market opening and closing hours. I am trying to plot it, but the
>> time gap is always visib
On 07/25/2011 05:49 PM, ATANU wrote:
i am trying to make 3-d barplots,pie-charts in R,just like Excel. i have used
rgl , but that does not produce beautiful graphs like excel(i dont need to
rotate the graph). can anyone help me to produce graphs,just like excel.
thanks in advance
Hi ATANU,
See
On 07/26/2011 02:40 AM, Naomi Robbins wrote:
Hello!
It's a shoot in the dark, but I'll try. If one has a total of 100
(e.g., %), and three components of the total, e.g.,
mytotal=data.frame(x=50,y=30,z=20), - one could build a pie chart with
3 sectors representing x, y, and z according to their pr
On Tue, Jul 26, 2011 at 4:48 AM, djdjoko wrote:
> Hi,
> I have an intraday timeseries of financial data (see below) which has gaps
> due to market opening and closing hours. I am trying to plot it, but the
> time gap is always visible in the plot. I tried converting data to xts, zoo,
> timeSeries
Try the logLik function with your model !
Arnaud
Date: Mon, 25 Jul 2011 12:16:37 +0100
From: Partha Pratim PATTNAIK
To: r-help@R-project.org
Subject: [R] How to find the likelihood of a null model in R
Message-ID: <20110725121637.jo1u2ctuxy8kw...@www.sms.ed.ac.uk>
Content-Type: text/plain;
Date: Mon, 25 Jul 2011 19:03:08 +0100
From: jbustosm...@yahoo.es
To: r-help@r-project.org
Subject: [R] Life Cycle Assessment with R.
Hello everyone,
There's something really important about climate change and how many
institutions around the globe a
At 15:47 25/07/2011, Joshua Wiley wrote:
Hi Martin,
Off hand I do not know of a pre-defined function to do it, but the
"z-transformation" is just the inverse hyperbolic function, the mean
is just the mean, and the back transformation is the hypoerbolic
function so...
x <- c(.5, .4)
> Date: Mon, 25 Jul 2011 11:39:22 -0700
> From: lukescore...@gmail.com
> To: r-help@r-project.org
> Subject: [R] Is there an R program that produces optimal solution/mix of
> multiple samples' varying volumes and values
>
> Sorry about the lengthy subj
Dear Joe,
You need to use offset()
lm(y ~ a + offset(b) + c)
Best regards,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie & Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium
Hi,
I am trying to do a linear regression but I want one of my variables to not
generate a coefficient. E.g. what I want to do is fit for y=a+b+c but
forcing the coefficient of b to be 1. Is this possible?
I have been fitting y-b=a+c but I have found that when I recalculate y it is
not close
On Tue, 26 Jul 2011, Peter Ehlers wrote:
On 2011-07-26 01:13, Achim Zeileis wrote:
On Tue, 26 Jul 2011, Lao Meng wrote:
Hi all:
There's a question about glht function.
My data:data_ori,which inclue CD4, GROUP,time.
f_GROUP<-factor(data_ori$GROUP)
f_GROUP is a factor of 3 levels(0,1,2,3)
On Tue, Jul 26, 2011 at 12:23 AM, Amelia McNamara
wrote:
> I am trying to create a plot that has multiple plot characters for
> each point (e.g. a point within a triangle, a triangle within a
> square, etc). The workaround I have found to do this is by plotting
> twice, as in this example:
>
> x <
On 2011-07-26 01:13, Achim Zeileis wrote:
On Tue, 26 Jul 2011, Lao Meng wrote:
Hi all:
There's a question about glht function.
My data:data_ori,which inclue CD4, GROUP,time.
f_GROUP<-factor(data_ori$GROUP)
f_GROUP is a factor of 3 levels(0,1,2,3)
result<- lme(sqrt(CD4) ~ f_GROUP*time ,r
Hi,
I have an intraday timeseries of financial data (see below) which has gaps
due to market opening and closing hours. I am trying to plot it, but the
time gap is always visible in the plot. I tried converting data to xts, zoo,
timeSeries and plotting it with different functions i.e. plot.xts, plo
This is a theoretical issue. It is impossible for beta-distributed values to
take the value of 0 or 1. Hence, an attempt to fit a beta distribution to a
vector containing these values fails.
HTH,
Daniel
baxy77 wrote:
>
> Hi,
>
> Well, i need some help, practical and theoretical. I am wonderi
On Tue, 26 Jul 2011, Lao Meng wrote:
Hi all:
There's a question about glht function.
My data:data_ori,which inclue CD4, GROUP,time.
f_GROUP<-factor(data_ori$GROUP)
f_GROUP is a factor of 3 levels(0,1,2,3)
result <- lme(sqrt(CD4) ~ f_GROUP*time ,random = ~time|ID,data=data_ori)
glht(resu
On Jul 26, 2011 Lao Meng wrote:
> glht(result, linfct = mcp(f_GROUP="Tukey") )
> Error in `[.data.frame`(mf, nhypo[checknm]) : undefined columns selected
It is almost certainly the underscore in the name ("_") that is causing the
problem. Try putting the term in quotes ("f_GROUP").
Regards, Mar
Hi,
Well, i need some help, practical and theoretical. I am wondering why the
fitdistplus (mle function) is returning an error for this code:
[code]
x1 <- c(100,200,140,98,97,56,42,10,2,2,1,4,3,2,12,3,1,1,1,1,0,0);
plotdist(x1);
descdist(x1, boot =1000);
y<- sum(x1);
d= as.vector(length(x1));
fo
Dear Ben,
Maybe the model converges more slowy than other models. Running more iterations
might solve the problem. Have a look at ?lme and ?lmeControl.
Best regards,
Thierry
PS R-sig-mixedmodels is a better list for questions on lme().
-
Steven's solution is great, but it will only work if the rows are really
duplicates. If the data frame contains another variable whose values vary,
it will not work because then the rows are obviously unique.
df<-data.frame(df,value=rnorm(11))
unique(df)
You would then have to make a decision, w
For question (a), do:
which(AA%in%BB)
Question (b) is very ambiguous to me. It makes little sense for your example
because all values of BB are in AA. Therefore I am wondering whether you
meant in question (a) that you want to find all values in BB that are in AA.
That's not the same thing. I am
On 2011-07-26 00:16, Lao Meng wrote:
Hi all:
There's a question about glht function.
My data:data_ori,which inclue CD4, GROUP,time.
f_GROUP<-factor(data_ori$GROUP)
f_GROUP is a factor of 3 levels(0,1,2,3)
result<- lme(sqrt(CD4) ~ f_GROUP*time ,random = ~time|ID,data=data_ori)
glht(result,
Hi all:
There's a question about glht function.
My data:data_ori,which inclue CD4, GROUP,time.
f_GROUP<-factor(data_ori$GROUP)
f_GROUP is a factor of 3 levels(0,1,2,3)
result <- lme(sqrt(CD4) ~ f_GROUP*time ,random = ~time|ID,data=data_ori)
glht(result, linfct = mcp(f_GROUP="Tukey") )
Error
On 2011-07-25 10:30, Manojit Roy wrote:
Dear all,
I am trying to create a 6-plot layout - 3 rows and 2 columns - so that
only the top two plots have variable widths, all else with their default
setting. Using
"layout(matrix(c(1,2,3,4,5,6),3,2,byrow=T),widths=c(5,2))" rescales
column #2 of all th
Naomi,
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