See under "Note" in ?strptime:
Remember that in most timezones some times do not occur and some
occur twice because of transitions to/from summer time.
âstrptimeâ does not validate such times (it does not assume a
specific timezone), but conversion by âas.POSIXctâ) will d
I have a problem with duplicated date_time stamps that I do not see as
duplicated.
I read a file with observations taken every 30 minutes:
> aur2009=read.csv(paste(datadir,"AUR_ECPP_2009.csv",sep="/"),sep=";",stringsAsFactors=F)
> aur2009[1:3,1:5]
Date.Time E_filled E_filled_flag LE_filled
Dear Lists,
What is the appropriate software package for dumping say 20 PDFS in a
folder, then creating data visualization with frequency counts of
certain words as well as measure correlation within each file for
certain key relationships or key words.
I am doing text analysis of biases in enter
Dear all,
I have made couple logistic regressions, what making a distribution of some
event.
Currently, i store it like this:
o1 <- lrm(...)
o2 <- lrm(...)
o3 <- lrm(...)
...
Then, i have made a function to peak required regression object from this
variables by it number:
get_object <- functio
#Hi all,
#Using the example data that follows, can someone please show me how to get
a scatterplot of points with #error bars in the Y direction. something like
this works for one Y:
xYplot(Cbind(y1, l1, u1) ~x1, data=y)
#but this:
xYplot(Cbind(y1, l1, u1) + Cbind(y2, l2, u2)~x1, data=y)
#
Hello,
I am attempting to utilize the 'alabama' package to solve a constrained
nonlinear optimization problem.
The problem has both equality and inequality constraints (heq and hin
functions are used). All constraints are smooth, i.e. I can differentiate
easily to produce heq.jac and hin.jac fun
I am using PAM with k = 10 clusters, but I only get one cluster ID for all my
observations. I couldn't find any discussion about this in the help file, or
mailing lists.
Is there a reasonable explanation for this result ?
cIDs <- pam(all, 10, cluster.only = TRUE, do.swap = FALSE)
> table(cIDs)
G'day Bill,
On 18/05/2011, at 10:36 AM, wrote:
> If you
>
> ?plot.stl
>
> you will see that that the second argument, set.pars, is a list of argument
> settings for par(), including a (variable) default setting for mfrow. I.e.
> plot.stl overrides your external setting (which will also ove
I converted code using the Design package to be what rms expects but hadn't
bothered to re-run the code. For the first note you sent, you can
attach(counties) (or better: use with(counties) inside the show.pts
function. But I'm getting a warning I'll need to track down before it works
100%:
Warn
If you
?plot.stl
you will see that that the second argument, set.pars, is a list of argument
settings for par(), including a (variable) default setting for mfrow. I.e.
plot.stl overrides your external setting (which will also override any layout()
setting).
It looks like to override it ba
Dear Community,
I try to compute the variance of a MC integral where I know the
analytical solution.
The function is exp(-x) integrated on the interval (2,4). (Rizzo example)
The true value is exp(-2)-exp(-4)~.1170196
I have the following short code:
iter <- 1
MCs <- numeric(iter)
n <- 50
G'day,
I am looking at monthly reports, and have three series of monthly data from
2007 to 2009. I would like to show the season decomposition of these two series
side by side on the one device, however using plot doesn't seem to respect any
use of layout(matrix(1:3, ncol=3)) or par(mfcol=c(1,3
The documentation for simprof says, with respect to the method.distance
argument, "This value can also be any function which returns a "dist" object."
So you should be able to use the Jaccard index by setting up your own function
to compute it. e.g.
Jaccard <- function(X) vegan::vegdist(X, me
Addendum: I tried a gamma fit in glmmPQL and got the same errors.
*Ben Caldwell*
PhD Candidate
University of California, Berkeley
On Tue, May 17, 2011 at 3:51 PM, Benjamin Caldwell
wrote:
> Hello
> After seeing this (
> https://stat.ethz.ch/pipermail/r-sig-mixed-models/2011q1/005213.html) ema
PS I should have followed the example with one using with() for something that
would often be done with attach(): Consider:
with(polyData, {
plot(x, y, pch=".")
o <- order(x)
lines(x[o], eta[o], col = "red")
})
I use this kind of dodge a lot, too, but now you can mostly use data= argument
Thank you all, this is exactly what I had in mind, except that I still have
to get my head around apply et al. Back to the books for me then!
Lara
On Tue, May 17, 2011 at 2:41 PM, Jannis wrote:
> Have a look at lapply(). Something like:
>
> entries.with.nrows=lapply(data,function(x)dim(x)[1]>1)
Amen to all of that, Bert. Nicely put. The google style guide (not perfect,
but a thoughtful contribution on these kinds of issues, has avoiding attach()
as its very first line. See
http://google-styleguide.googlecode.com/svn/trunk/google-r-style.html)
I would add, though, that not enough pe
Hi Charles,
It's not clear to me what you mean by "doesn't work".
> test <- "Interesting 1\nPoint\n"
> cat(test)
Interesting 1
Point
> test1 <- gsub("ing 1\nP","ing 3\nP", test)
> cat(test1)
Interesting 3
Point
>
Cheers
Andrew
On Tue, May 17, 2011 at 10:45:31AM +0200, Thibault Charles wrote:
On May 17, 2011, at 6:00 PM, Alaios wrote:
and what about the cdf of function?
Did you read the help page for dnorm?
--
David.
Regards
Alex
--- On Tue, 5/17/11, Downey, Patrick wrote:
From: Downey, Patrick
Subject: RE: [R] pdf (probability distribution function) and cdf
To: "Alaios"
and what about the cdf of function?
Regards
Alex
--- On Tue, 5/17/11, Downey, Patrick wrote:
> From: Downey, Patrick
> Subject: RE: [R] pdf (probability distribution function) and cdf
> To: "Alaios" , "Jonathan Daily"
> Cc: R-help@r-project.org
> Date: Tuesday, May 17, 2011, 6:17 PM
> ?dnorm
Hi:
Here's one way, using an abbreviated example:
du <- data.frame(v1 = factor(rep(1:10, each = 4)),
v2 = factor(rep(rep(1:2, each = 2), 10)),
v3 = factor(rep(1:2, 20)),
x1 = rnorm(40),
y1 = rnorm(40),
x2 = rnorm
Rolf,
Taking out the scalar multiples did it. Thanks for that.
*Ben Caldwell*
PhD Candidate
University of California, Berkeley
On Sun, May 15, 2011 at 10:30 PM, Rolf Turner wrote:
>
> In your example it appears that you are plotting a histogram (on the
> frequency
> scale) and then superim
Bumping this thread in order to see if there's any new info on this. :)
I'm interested in doing MaxDiff Scaling. Will Maximum Likelihood Difference
Scaling give me the same results as MaxDiff Scaling?
I found a paper, Discrete-Choice Logit Models with R, by Philip A. Viton.
It discusses using
I will take a look. In my real data I need to interpolate the 16 points into
64 points for each of the categories.
Thanks Marc
JR
-Original Message-
From: Marc Schwartz [mailto:marc_schwa...@me.com]
Sent: Tuesday, May 17, 2011 1:09 PM
To: ROLL Josh F
Cc: r-help@r-project.org
Subjec
Hi all,
I thought I would post code to send an email out of R. The code uses
Grothendieck and Bellosta's interface package rJython for executing Python
from R. The code itself provides basic email functionality for email servers
requiring authentication. It should be easy to extend it (e.g., for
Have a look at lapply(). Something like:
entries.with.nrows=lapply(data,function(x)dim(x)[1]>1)
should give you a vector with the elements of the list that you seek marked
with TRUE.
This vector can then be used to extract a subset from your list by:
data.reduced=data[entries.with.nrows]
Or s
On 18/05/11 08:24, Lara Poplarski wrote:
Hello All,
I have a list of dataframes, and I need to subset it by keeping only those
dataframes in the list that meet a certain criterion. Specifically, I need
to generate a second list which only includes those dataframes whose number
of rows is> 1.
C
shouldKeep <- sapply(listOfDataFrames, function(df)nrow(df)>1)
listOfDataFrames[shouldKeep]
or, compressed to get rid of the intermediate variable
listOfDataFrames[sapply(listOfDataFrames, function(df)nrow(df)>1)]
If you are writing production code and there is any chance that
listOfDataFra
Hi Lara,
You might try the following (untested):
yourlistofdataframes[sapply(yourlistofdataframes, function(d) nrow(d) > 1)]
HTH,
Jorge
On Tue, May 17, 2011 at 4:24 PM, Lara Poplarski <> wrote:
> Hello All,
>
> I have a list of dataframes, and I need to subset it by keeping only those
> dataf
Also I can not reproduce figure 7.11 by
f <- Newlabels(f, list(turnout='voter turnout (%)'))
windows()
nomogram(f, interact=list(income=incomes), turnout=seq(30,100,by=10),
lplabel='estimated % voting Democratic', cex.var=.8, cex.axis=.75)
On Tue, May 17, 2011 at 4:04 PM, John Smith w
Hello All,
I have a list of dataframes, and I need to subset it by keeping only those
dataframes in the list that meet a certain criterion. Specifically, I need
to generate a second list which only includes those dataframes whose number
of rows is > 1.
Could someone suggest how to do this? I have
Hi:
I'm assuming this is a snippet of a much larger data set, because one
certainly would not use box plots on singleton observations. You have
four grouping variables in your example, so the best you could likely
do on one graphics page would be to, for example, plot time vs.
distance, 'dodged' b
olafgy gmail.com> writes:
>
> Hi there,
>
> I am trying to import 100 files with the names of "vpn 1 .dat" to "vpn 100
> .dat"
> into a respective table calld vpn1 to vpn100.
>
> I therfore have created a variable X<-1:100
>
> I not want to use X as a subtitute for the number in my filename,
On May 17, 2011, at 2:55 PM, ROLL Josh F wrote:
> Marc,
> How could I also apply the spline function to each of the 'columns' found in
> the result from
>
> tapply(Df$Rate,list(Df$Bin,Df$Type),sum)
>
> ??
Something along the lines of the following:
apply(tapply(Df$Rate,list(Df$Bin,Df$
Dear R-users,
I am using R 2.13.0 and rms 3.3-0 , but can not reproduce figure 7.8 of the
handouts *Regression Modeling Strategies* (
http://biostat.mc.vanderbilt.edu/wiki/pub/Main/RmS/course2.pdf) by the
following code. Could any one help me figure out how to solve this?
setwd('C:/Rharrell')
re
Marc,
How could I also apply the spline function to each of the 'columns' found in
the result from
tapply(Df$Rate,list(Df$Bin,Df$Type),sum)
??
-Original Message-
From: Marc Schwartz [mailto:marc_schwa...@me.com]
Sent: Tuesday, May 17, 2011 12:42 PM
To: ROLL Josh F
Cc: r-help@r
On 17.05.2011 19:17, Sebastian Gibb wrote:
Hello Yan,
please use an empty password and write an e-mail to c...@r-project.org
afterwards.
Bye,
Sebastian
Dear all,
I'm just wondering how to submit a package to cran?
I followed the instruction, using anonymous as username and my email
addr
On May 17, 2011, at 12:53 PM, LCOG1 wrote:
> I will hit my own ball on this one
>
>
> tapply(Df$Rate,list(Df$Bin,Df$Type),sum)
>
Ahayou had mentioned creating a new column in your initial post, presumably
added to 'Df', as opposed to creating a new independent matrix of the results.
You
I also have a problem using odfWeave on Windows XP with R > R2.11.1. odfWeave
fails, giving mysterious error messages. (Not quite the same as yours, but
similar. I sent the info to Max Kuhn privately, but did not get a response
after two tries.) My odfWeave reporting system worked fine prior to
Brilliant!
Thanks
Martin
Quoting Henrique Dallazuanna :
Try this:
sapply(intvl.period.myrs, sum)
On Tue, May 17, 2011 at 1:13 PM, Martin Hughes wrote:
Dear R users
I have a list, as follows:
intvl.period.myrs
$Devonian
[1] 4.8 4.2 9.5 5.7
$Ordovician
[1] 7.2 5.1 10.2 1.9
$Silu
I will hit my own ball on this one
tapply(Df$Rate,list(Df$Bin,Df$Type),sum)
--
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Sent from the R help mailing list archive at Nabble.com.
___
Dear all,
it might not be a R-related question, however, I would appreciate if anyone
could suggest a mathematical/statistical framework that integrates graphs
and arrays, and potentially has a R implementation.
to give an example : given a graph of interactions (gene1 --- gene2, gene3
--- gene5,
On May 17, 2011, at 11:48 AM, LCOG1 wrote:
> Okay everyone heres a likely softball for someone.
>
> Consider the following data frame:
>
> #Create data
> x<-rep(c(1,15),10)
> y<-rnorm(20)
> z<-c(rep("auto",10),rep("bus",10))
> a<-rep(c(1,1,2,2,3,3,4,4,5,5),2)
> #Create Data frame
> Df<-data.fram
Like This?
x<-rep(c(1,15),10)
y<-rnorm(20)
z<-c(rep("auto",10),rep("bus",10))
a<-rep(c(1,1,2,2,3,3,4,4,5,5),2)
#Create Data frame
Df<-data.frame(Source=x,Rate=y,Bin=a,Type=z)
Df
ddply(Df,c('Type','Bin'),summarise,Summed=sum(Rate))
# Adding a column to Df
ddply(Df,c('Type','Bin'),mutate,Summed=s
Can you provide a brief reproducible example for us to play with?
Scott
On Tuesday, May 17, 2011 at 12:31 PM, Stijn Van Daele wrote:
> Dear R users,
>
> I have a problem with reshaping data. I know such questions have been asked
> before, but I can't get it right, neither with the reshape funct
One possibility is:
library(doBy)
summaryBy(Rate~Source+Bin, data=Df, FUN=sum)
On 5/17/2011 12:48 PM, LCOG1 wrote:
> Okay everyone heres a likely softball for someone.
>
> Consider the following data frame:
>
> #Create data
> x<-rep(c(1,15),10)
> y<-rnorm(20)
> z<-c(rep("auto",10),rep("bus",10))
Hglm does the work! Thanks!
Also, I find that the developing version of lme4, called lme4a, has the
capability to fit Gamma models. And both lme4a and hglm produce results
consistent with the published ones.
Problems solved!
Wayne (Yanwei) Zhang
Statistical Research
>CNA
-Original Me
It's worse if the model does converge because then you don't have a warning
about the result being nonsense.
Frank
Terry Therneau-2 wrote:
>
> -- begin included message ---
> I did realize that there are way more predictors in the model. My
> initial thinking was use that as an initial model fo
Hi:
This might be helpful:
http://glmm.wikidot.com/faq
It has some commentary and potential workarounds re gamma GLMMs toward
the bottom of the page. I might also suggest that r-sig-mixed-models
might be a better forum for these types of questions; you can
subscribe by going to
http://www.r-pro
Try this:
sapply(intvl.period.myrs, sum)
On Tue, May 17, 2011 at 1:13 PM, Martin Hughes wrote:
>
>
> Dear R users
>
> I have a list, as follows:
>
>> intvl.period.myrs
>
> $Devonian
> [1] 4.8 4.2 9.5 5.7
>
> $Ordovician
> [1] 7.2 5.1 10.2 1.9
>
> $Silurian
> [1] 4.7 3.0 7.8 2.0 3.3 1.6 2.6 2.
Dear R users,
I have a problem with reshaping data. I know such questions have been asked
before, but I can't get it right, neither with the reshape function nor with
the melt function.
My dataset has about 407 variables and about 48000 cases.
Each case looks as follows:
V1 v2 v3v4
Dear R users
I have a list, as follows:
intvl.period.myrs
$Devonian
[1] 4.8 4.2 9.5 5.7
$Ordovician
[1] 7.2 5.1 10.2 1.9
$Silurian
[1] 4.7 3.0 7.8 2.0 3.3 1.6 2.6 2.7
I want to write a loop that will sum up the values in each part, and give me a
vector containing the (in this case 3)
Okay everyone heres a likely softball for someone.
Consider the following data frame:
#Create data
x<-rep(c(1,15),10)
y<-rnorm(20)
z<-c(rep("auto",10),rep("bus",10))
a<-rep(c(1,1,2,2,3,3,4,4,5,5),2)
#Create Data frame
Df<-data.frame(Source=x,Rate=y,Bin=a,Type=z)
I want to create a new column th
Hi
When I call for the loadings of my plsr using the command,
x <- loadings(BHPLS1)
my loadings contain variable names rather than numbers.
>str(x)
loadings [1:94727, 1:10] -0.00113 -0.03001 -0.00059 -0.00734 -0.02969 ...
- attr(*, "dimnames")=List of 2
..$ : chr [1:94727] "PCIList1" "PCIL
?dnorm
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Alaios
Sent: Tuesday, May 17, 2011 12:50 PM
To: Jonathan Daily
Cc: R-help@r-project.org
Subject: Re: [R] pdf (probability distribution function) and cdf
I am assuming a Gaussian
Hello Yan,
please use an empty password and write an e-mail to c...@r-project.org
afterwards.
Bye,
Sebastian
> Dear all,
>
> I'm just wondering how to submit a package to cran?
> I followed the instruction, using anonymous as username and my email
> address as password, but it didn't connect.
I followed the directions and all was well. Sorry I have no insight.
On Tue, May 17, 2011 at 11:51 AM, Yan Jiao wrote:
> Dear all,
>
> I'm just wondering how to submit a package to cran?
> I followed the instruction, using anonymous as username and my email address
> as password, but it didn't
Dear all,
I'm just wondering how to submit a package to cran?
I followed the instruction, using anonymous as username and my email address as
password, but it didn't connect.
Any hints?
Yan
[[alternative HTML version deleted]]
__
R-help@r-pr
I am assuming a Gaussian distribution with specific mean and standard deviation
--- On Tue, 5/17/11, Jonathan Daily wrote:
> From: Jonathan Daily
> Subject: Re: [R] pdf (probability distribution function) and cdf
> To: "Alaios"
> Cc: R-help@r-project.org
> Date: Tuesday, May 17, 2011, 3:18 PM
Hi
I am analysing a dataset of 40 samples each with 90,000 intensity measures for
various peptides. I am trying to identify the Biomarkers (i.e. most significant
peptides). I beleive that PLS with jack knifing, or alternativeley
CMV(cross-model-validation) are multivariateThe 40 samples belong
Maas James Dr (MED uea.ac.uk> writes:
>
>
> What is the easiest way to extract a value from a 'mer'
> object from glmer? The first I need is the
> trtpair Std.Dev. which in this case is 0.17542?
>
> I've managed to get the fixed effects numbers from
> summary(fednmaout)@coeffs
You sho
I like the function Ecdf in the Hmisc package. For the pdf, you can also
use plot(density(x)), which is in the stats package.
??Ecdf
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Alaios
Sent: Tuesday, May 17, 2011 10:09 AM
To: R
sorry I solved it, was a typo on my part!!
--
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__
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https://stat
On Tue, May 17, 2011 at 2:15 PM, Jannis wrote:
> Otherwise I would stick to SPSS/Excel/Sigmaplot or browse for some R GUIs
> with some basics beeing implemented on a "click basis" like RStudio, Rkward
> or TinnR. All R GUI implementations are however limited to basic
> plotting/analysis routin
On 2011-05-17 06:50, Doran, Harold wrote:
Suppose I have data such as the following
set.seed(12345)
tmp<- data.frame(var1 = rnorm(100), var2 = rnorm(100), var3=rnorm(100, 10, 30))
tmp1<- data.frame(vars = with(tmp, c(var1, var2, var3)), type = gl(3, 100))
var3 is on a different scale, but I cr
- Reduce the model to a reasonable size with far less variables than
observations.
- Code factors as factors rather than numerics
- don't use variables with perfect correlation to other nor any duplicates
Best,
Uwe Ligges
On 17.05.2011 15:46, Songer, Katherine B - DNR wrote:
Uwe,
Thank you
The result is good, thanks a lot, but how can I with this method fill my raster
to color?
Le 17 mai 2011 à 15:43, Duncan Murdoch a écrit :
> I don't think filled.contour gives you access to the contour lines. If you
> use contourLines() to compute them, then you can draw them using code like
Thank you! I'm a bit of a newby but this has really helped. thank you
--
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__
R-help@r-project.org mailin
I work with large datasets (1 points) so I can't post them , but my
function is :
create_map <- function(grd, level ,map_output, format = c("jpeg"), width_map =
150, height_map = 150,...)
{
##sp <- spline(x = grd[,1], y = grd[,2])
grd2 <- matrix(grd[,
Thank you for your answer, but the function spline() (and a lot of other
function in R) can't take in its parameters the original contour which are
define by a vector, i.e. :
##creation of breaks for colors
i<-1
paliers <- c(-1.0E300)
while(i<=length(level[,1
Hi Dirk,
for me the following still hangs:
R --vanilla -e 'library(plyr); ddply(data.frame(a=1:101,b=1:101), .(a,b), nrow)'
but... when running the script via littler with "r -p bug.R" it returns.
Thanks,
Mikael
On 17 May 2011 13:50, Dirk Eddelbuettel wrote:
>
> On 17 May 2011 at 12:18, Mikae
Hi there,
I am trying to run the following code:
> dcOU<-function(x,t,x0,theta,log=FALSE){
+ Ex<-theta[1]/theta[2]+(x0-theta[1]/theta[2])*exp(-theta[2]*t)
+ Vx<-theta[3]^2*(1-exp(-2*theta[2]*t))/(2*theta[2])
+ dnorm(x,mean=Ex,sd=sqrt(Vx),log=log)
+ }
> OU.lik<-function(theta1,theta2,theta3){
+ n<
Dear list,
I have a data.frame with aggregated frequencies of occurrence of two-way
(and in one case three-way) features in my observations. Like this:
location condition transient fricationmrallvoiced correctstop
vpresent frequency
1 STNPOSTON TRUE FALSE FALSE
Two other options (which I prefer):
aggregate(x~id.var, data=dat, FUN=min)
or
library(doBy)
summaryBy(x~id.var, data=dat, FUN=min)
HTH,
Ivan
Le 5/17/2011 16:09, John Kane a écrit :
A bit convolulted but it works
library(reshape2)
yy<- melt(dat)
zz<- cast(yy, id.var ~ variable, min)
zz[,2]
--
Do you have a reproducible example? As posed I have no idea what this
vector contains.
Are you assuming a specific distribution type and using these vectors
to parameterize it?
On Tue, May 17, 2011 at 10:08 AM, Alaios wrote:
> Dear all,
> I would like for a given vector to calculate the pdf and
Hi Dennis,
Thanks for your input. I am quite shocked that the Gamma model is not
implemented in glmer yet, but do you happen to know if there is other packages
in R that could fit Gamma hierarchical models besides MASS? The glmmPQL results
are a bit off compared to what have been published. T
A bit convolulted but it works
library(reshape2)
yy <- melt(dat)
zz <- cast(yy, id.var ~ variable, min)
zz[,2]
--- On Tue, 5/17/11, Downey, Patrick wrote:
> From: Downey, Patrick
> Subject: [R] Minimum value by ID
> To: r-help@r-project.org
> Received: Tuesday, May 17, 2011, 9:44 AM
> Hello,
>
Perfect. Thank you.
-Mitch
-Original Message-
From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
Sent: Tuesday, May 17, 2011 9:51 AM
To: Downey, Patrick
Cc: r-help@r-project.org
Subject: Re: [R] Minimum value by ID
have a look at help file of function tapply(), and try this:
Dear all,
I would like for a given vector to calculate the pdf and the cdf and then plot
these two graphs.
Is there in R any toolkit that can help me do that fast?
Regards
Alex
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Folks:
> Only if the user hasn't yet been introduced to the with() function,
> which is linked to on the ?attach page.
>
> Note also this sentence from the ?attach page:
> " attach can lead to confusion."
>
> I can't remember the last time I needed attach().
>
> Peter Ehlers
Yes. But perhaps
In fact, my previous post is not correct:
you have to replace :
acfth <- ARMAacf(ar=numeric(0),ma=artest$ar)
by :
acfth <- ARMAacf(ar=numeric(0),ma=-artest$ar)
Then it coincides exactly the same correlations as SAS
The error comes from the equation estimated by the ar() function :
"For defini
have a look at help file of function tapply(), and try this:
with(dat, tapply(x, id.var, min))
I hope it helps.
Best,
Dimitris
On 5/17/2011 3:44 PM, Downey, Patrick wrote:
Hello,
I have a longitudinal dataset where each individual has a different number
of entries. Thus, it is of the follo
Suppose I have data such as the following
set.seed(12345)
tmp <- data.frame(var1 = rnorm(100), var2 = rnorm(100), var3=rnorm(100, 10, 30))
tmp1 <- data.frame(vars = with(tmp, c(var1, var2, var3)), type = gl(3, 100))
var3 is on a different scale, but I create the following plot, which looks
terr
Uwe,
Thank you very much for looking at this. I'm attaching the data, in case you
have any wisdom on why variables 10, 38, and 42 would appear constant.
Meanwhile, I'll remove colinear variables and read up a little more...
Thanks,
Katie
-Original Message-
From: Uwe Ligges [mailto:lig.
Hello,
I have a longitudinal dataset where each individual has a different number
of entries. Thus, it is of the following structure:
x <- runif(12)
id.var <- factor(c(rep("D1",4),rep("D2",2),rep("D3",3),rep("D4",3)))
dat <- as.data.frame(x)
dat$id.var <- id.var
dat
> dat
x id
I don't think filled.contour gives you access to the contour lines. If
you use contourLines() to compute them, then you can draw them using
code like this:
contours <- contourLines(V2b,levels=paliers)
for (i in seq_along(contours)) {
x <- contours[[i]]$x
y <- contours[[i]]$y
lines(
-- begin included message ---
I did realize that there are way more predictors in the model. My
initial thinking was use that as an initial model for stepwise model
selection. Now I wonder if the model selection result is still valid
if the initial model didn't even converge?
--- end inclusion -
On 17/05/2011 8:24 AM, Pierre Bruyer wrote:
Thank you for your answer, but the function spline() (and a lot of other
function in R) can't take in its parameters the original contour which are
define by a vector, i.e. :
If you post some reproducible code to generate the contours, someone
wi
On 2011-05-17 02:22, Timothy Bates wrote:
Dear Bryony: the suggestion was not to change the name of the data object, but
to explicitly tell glm.nb what dataset it should look in to find the variables
you mention in the formula.
so the salient difference is:
m1<- glm.nb(Cells ~ Cryogel*Day, da
On 2011-05-16 17:14, Sam Albers wrote:
Hello,
Apologies for a similar earlier post. I didn't include enough details in
that one.
I am having a little trouble subsetting some data based on a grouping
variable. I am using an instrument that does depth profiles of a water
column. The instrument re
What is the easiest way to extract a value from a 'mer' object from glmer? The
first I need is the
trtpair Std.Dev. which in this case is 0.17542?
I've managed to get the fixed effects numbers from
summary(fednmaout)@coeffs
but no luck with the Random effects stuff.
TIA
Jim
> fednmaout
The problem is the use of variable selection without simultaneous shrinkage.
It will result in an entirely unreliable model and essentially will choose a
random sample of the predictors. See
http://www.childrensmercy.org/stats/faq/faq12.asp
Frank
Shi, Tao wrote:
>
> Hi Frank,
>
> I know it's k
--- Vikas Garud schrieb am Di, 17.5.2011:
> Von: Vikas Garud
> Betreff: [R] Box Plot under GUI (R Commander/RKward)
> An: r-help@r-project.org
> Datum: Dienstag, 17. Mai, 2011 11:31 Uhr
> Hi,
>
> I tried to search the mailing list on box plot, but could
> not find
> anything relevant. I have
something like this will get you going, assuming your data are in a dataframe
called “qual”
# qual <- read.table(pipe("pbpaste"), header=T, sep='\t')
boxplot(formula=Time~Distance+Season, data=qual)
Followup question from me:
i can’t see why
boxplot(formula=Time, data=qual)
should return the
On May 17, 2011, at 12:57 , pcc wrote:
> This is a really simple question, I'm sure,but I can't make EpiR work!
> I keep getting the following:
>
>> epi.2by2(47, 263483, 282, 935028, method="cohort.time", conf.level=0.95)
> Error in epi.2by2(47, 263483, 282, 935028, method = "cohort.time",
> c
Furthermore, aggregate passes extra arguments to FUN, so explicit
declaration of na.rm = T can be done in the call to aggregate
directly:
aggregate(cbind(v1, v2) ~ v3, data = testDF, FUN = mean, na.rm = T)
HTH,
Jon
On Tue, May 17, 2011 at 5:22 AM, Jannis wrote:
> Andy,
>
>
> have a look at ?mea
On 17 May 2011 at 12:18, Mikael Högqvist wrote:
| Hi,
|
| I have run into a strange issue with Rscript and plyr. When running
| the following Rscript:
|
| #!/usr/bin/Rscript
| library(plyr)
| d = ddply(data.frame(a=1:101,b=1:101), .(a,b), nrow)
| head(d)
|
| the script does not terminate. It pr
Hi,
I am a consultant in Quality Management. I am exploring the use of R
with any GUI - R commander/Rkward for doing analytical work. Have
installed R, R Commander and Rkward.
I hope to learn by doing various exercises that I use for teaching
analytical techniques to my clients. I would be pos
This is a really simple question, I'm sure,but I can't make EpiR work!
I keep getting the following:
> epi.2by2(47, 263483, 282, 935028, method="cohort.time", conf.level=0.95)
Error in epi.2by2(47, 263483, 282, 935028, method = "cohort.time",
conf.level = 0.95) :
unused argument(s) (935028)
On May 17, 2011, at 4:09 AM, Bryony Tolhurst wrote:
Dear Bill
Many thanks. I will try this.
One question: why is the attach()function problematic? I have always
done it that way (well in my very limited R-using capacity!) as
dictated by 'Statistics, an Introduction using R' by Michael Cra
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