- Original Message
From: Shang Liu <[EMAIL PROTECTED]>
Subject: [R] how to replace NA values in a list
I have a matrix named "spec" (see below), it is a 6x3 matrix, and each element
of spec is a list. For example, spec[1,"wavenumber"] is a list, and it contains
1876 numeric numbers and
Hi gang,
I always open more than 1 R console in Windows. I can't figure out a way to
do this with OS X yet. I need that to utilize the duo core on my desktop.
How would I do that?
--
Regards,
Anh Tran
[[alternative HTML version deleted]]
__
R
Hi all,I know this topic has came up multiple times, but I've never fully
understand the apply() function.
Anyway, I'm here asking for your help again to convert this loop to apply().
I have 2 data frames with the following information: a1 is the fragment that
is need to be covered, a2 is the pro
See ?row ?col
On Fri, Aug 1, 2008 at 1:42 PM, Ralph S. <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> I am trying to get the positions in array coordinates (needed later) of
> certain elements in an array but I am not sure how to get them.
>
> My array is Q and the condition is dt>dV, where dt and dV are
Try view in the svViews package.
On Fri, Aug 1, 2008 at 1:29 PM, Rachel Schwartz <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I would like to view matrices I am working with in a clean, easy to read,
> separate window.
>
> A friend showed me how to do something like I want with edit(). I can view
> the m
From: Duncan Murdoch
Date: Tue, 29 Jul 2008 19:45:28 -0400
On 27/07/2008 3:10 PM, Stephen Tucker wrote:
> Hi list,
>
> I was using Sweave and was
wondering if anyone has had any luck changing the font colors of the
code chunks. For instance, in my .Rnw preample I tried including:
>
> ===
>
Is there a way to create a 'bubble plot' in R?
For example, if we define the following data frame containing the level of y
observed for 5 patients at three time points:
time<-c(rep('time 1',5),rep('time 2',5),rep('time 3',5))
y<-c('a','b','c','d','a','b','c','a','d','a','a','a','b','c','d')
D<-
Daren Tan <[EMAIL PROTECTED]> writes:
> For example, c("dog.is.an.animal", "cat.is.an.animal",
> "rat.is.an.animal"). How can I identify the common prefix is
> ".is.an.animal" and delete it to give c("dog", "cat", "rat") ?
The 'Rlibstree' package from omegahat is quite fun for this sort of
thing:
Thanks to Duncan Murdoch and Marc Schwartz for their
excellent help.
As we don't yet have the apache http: interface to svn
running yet, 'svn export' is the access mechanism.
As I don't want to leave stale files lying around,
or clobber a local file should there be one with
the same name, R's t
Rachel,
You may want to try JGR, http://jgr.markushelbig.org/JGR.html
which has, among many nice IDE features, an object browser that will
do what you want.
HTH,
Jim Porzak
Responsys, Inc.
San Francisco, CA
http://www.linkedin.com/in/jimporzak
useR Group SF: http://ia.meetup.com/67/
On Fri,
on 08/01/2008 06:49 PM Steven McKinney wrote:
Hi useRs
I'm trying to figure out how to source
an R script file straight from a subversion
repository, without having to put a copy
of the script into the local working directory.
Has anyone done this?
Something such as
source(file = paste(
On 01/08/2008 7:49 PM, Steven McKinney wrote:
Hi useRs
I'm trying to figure out how to source
an R script file straight from a subversion
repository, without having to put a copy
of the script into the local working directory.
Has anyone done this?
Something such as
source(file = paste(
Hi useRs
I'm trying to figure out how to source
an R script file straight from a subversion
repository, without having to put a copy
of the script into the local working directory.
Has anyone done this?
Something such as
source(file = paste("svn://myrepo.xxx.org/opt/svn/repos/",
On 01/08/2008 5:50 PM, Richard Chandler wrote:
After poking around some more I found the problem. The shortcut name
in PATH should be c:\progra~1\htmlhe~1 instead of c:\progra~1\htmhe~1.
Don't know how that happened, but thanks for pointing me in the right
direction.
The Windows path can handle
After poking around some more I found the problem. The shortcut name
in PATH should be c:\progra~1\htmlhe~1 instead of c:\progra~1\htmhe~1.
Don't know how that happened, but thanks for pointing me in the right
direction.
Richard
Quoting Duncan Murdoch <[EMAIL PROTECTED]>:
> On 8/1/2008 12:39 PM,
Thanks for the reply, but I'm still having trouble. The PATH appears
correct:
c:\Rtools\bin;c:\Rtools\perl\bin;c:\Rtools\MinGW\bin;c:\progra~1\R\R-2.7.1\bin;c:\progra~1\miktex~1.7\miktex\bin;c:\progra~1\htmhe~1;c:\windows;c:\windows\system32;C:\Program
Files\QuickTime\QTSystem;C:\ArcGIS\arcexe9x\b
On Fri, 1 Aug 2008, Peter Mueller wrote:
Hi dear R users,
i would like to have:
1. the legend outside of the plot and
2. add a grid to the plot.
For 1. I found a example in the R mailinglist archive [1].
1: http://tolstoy.newcastle.edu.au/R/help/06/01/19423.html
The total example code would be
Hi all... I¹m making a chart dealing with frequencies of isotopes of various
elements. For instance, I'd like the following text to appear on a chart
with the "35" and "37" as superscripts:
Based upon:
35Cl: 75%
37Cl: 25%
I am having problems properly parsing the superscript that preceeds the
"C
hi
I have list of matrix of lenggth 61 containg the mean values..I want to make
a boxplot for each of the matrix.
I used a for loop but i cant figure out the way to save in the boxplots
> all.the.mean
[[1]]
mean
0.5
o.6
0.8
[[2]]
0.6
0.6
0.9
now i want the boxplot for each of the matrix in a se
Hi dear R users,
i would like to have:
1. the legend outside of the plot and
2. add a grid to the plot.
For 1. I found a example in the R mailinglist archive [1].
1: http://tolstoy.newcastle.edu.au/R/help/06/01/19423.html
The total example code would be:
> ?par
> par(xpd=T, mar=par()$mar+c(0,0,
On Fri, Aug 1, 2008 at 10:56 AM, Doran, Harold <[EMAIL PROTECTED]> wrote:
> First off, Marc Schwartz posted this link earlier today, read it.
>
> http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-p_002dvalues-not-di
> splayed-when-using-lmer_0028_0029_003f
>
> Second, your email is not really de
Eleni,
A way to do this is to group the data first using 'split' and then sapply
the dist function to this list.
The slower step will be the split which took a couple of minutes on my
laptop but sapply should not take more than a minute or so.
size <- 1
df <- data.frame(
id=rep(sample
Here it is with the gc() in a print statement.
Tom
>
> library(boot)
> setwd("C:/Documents and Settings/Tom/Desktop")
>
> data.in <- read.csv("inputdata.csv",header=T,as.is=T)
>
> per95 <- function( annual.data, b.index) {
+ sample.data <- annual.data[b.index,]
+ return(quantile(sample
Hi,
yesterday i had the surprise not to be able to load the package "ca" on R 2.7.0
saying that cannot find required package rgl although it was there. So today
i've upgraded to 7.2.1. patched and i got the following error:
> local({pkg <- select.list(sort(.packages(all.available = TRUE)))
+
I'm interested in analysing some of my data using multinomial regression.
I have been using nnet's multinom so far.
However, I found that some of the data shows overdispersion and hence want to
change to robust multinomial regression, package: multinomRob
I have succesfully implemented Agresti's (2
It seems like the objects are reasonable size and the memory size also
seems reasonable. That is what I usually go by to see if there are
large objects in my memory. If it was showing that R had 1.2GB of
memory allocated to it, I wonder if there might be a memory leak
somewhere.
On Fri, Aug 1, 2
Rachel Schwartz wrote:
Thanks Erik, almost worked! I am a mac user and for some reason View
worked perfectly for my PC using friend, but
doesn't for me.
When I tried:
> mat=matrix(1:10,ncol=2)
> mat
[,1] [,2]
[1,]16
[2,]27
[3,]38
[4,]49
[5,]5 10
On 8/1/2008 12:39 PM, Richard Chandler wrote:
Hi,
I am getting the following error when using R CMD INSTALL ever since I
upgraded to R-2.7.1:
hhc: not found
CHM compile failed: HTML Help Workshop not intalled?
As indicated, the package is installed but without CHM help files. I
have downloaded
Dear R users,
I need to come up with an efficient method to compute the correlation (or at
least, the euclidean distance if that's easier) between specific rows in a data
frame (46,232 rows,29 columns). The pairs of rows between which I want to
find the correlation share a common value in one
Thanks Erik, almost worked! I am a mac user and for some reason View worked
perfectly for my PC using friend, but
doesn't for me.
When I tried:
> mat=matrix(1:10,ncol=2)
> mat
[,1] [,2]
[1,]16
[2,]27
[3,]38
[4,]49
[5,]5 10
> View(mat)
I get no error mes
Hello,
I am trying to compute a distance measure for the rows (var1) in a
sparse matrix:
require(Matrix)
dfx = xtabs(~ var1 + var2, data = df, sparse = T, drop.unused.levels = T
)
dm = dist(dfx)
on my data frame. Note that I am using the xtabs function from the
Matrix package, not
Hi,
to be honest, I never created a matrix of lists before, but hopefully
this code will help you?
set.seed(12345)
my.pool <- c(NA, 0:10)
n <- 25
alist <- list(sample(x=my.pool, size=n, replace=TRUE))
alist
mymatrix <- matrix(rep(alist, 6*3), nrow=6)
mymatrix2 <- lapply(X=mymatrix, FUN=funct
Dear Lotta,
Try
my.df <- data.frame(a <- c(1,6,10,11,3,5,10), group <- c(5,5,5,5,9,9,9))
tapply(my.df$a,my.df$group,median)
5 9
8 5
See ?tapply and/or ?aggregate for more information.
HTH,
Jorge
On Fri, Aug 1, 2008 at 1:30 PM, Lotta R <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I have data t
I have data that looks like
O.lengthO.age
176 1
179 1
182 1
...
493 5
494 5
514 5
606 5
462 6
491 6
537 6
553 6
432 7
522 7
625 8
661 8
687 10
704 10
615 12
(truncated)
with a simple VonB growth model from within nls():
plot
?aggregate
Something like this should do it.
aggregate(xx[,1], list(xx[,2], median)
--- On Fri, 8/1/08, stephen sefick <[EMAIL PROTECTED]> wrote:
> From: stephen sefick <[EMAIL PROTECTED]>
> Subject: Re: [R] simple help request
> To: "Lotta R" <[EMAIL PROTECTED]>
> Cc: r-help@r-project.org
>
I have a matrix named "spec" (see below), it is a 6x3 matrix, and each element
of spec is a list. For example, spec[1,"wavenumber"] is a list, and it contains
1876 numeric numbers and NAs. I want to replace the NAs to zero, but don't know
how to change it, the difficulty may be all the elements
I know the following is documented behaviour, in the sense that the
help page for as.character mentions that it truncates at about 500
characters... but wouldn't it be better if there was a warning of
some sort issued? Or am I misunderstanding what's happening here?
> str <- sample(LETTER
# I am sure there is a better way than this
a <- c(1,6,10,11,3,5,10)
group <- c(5,5,5,5,9,9,9)
my.df <- cbind(a,group)
a.5 <- subset(my.df, group==5)
median(a.5[,1])
a.9 <- subset(my.df, group==9)
median(a.9[,1])
# MASS 4 is a good book
On Fri, Aug 1, 2008 at 1:30 PM, Lotta R <[EMAIL PROTECTED]>
Apparently, the Fed changed the way they handled missing values in the
interest rates files; now they use a period instead of #N/A like they
did in my old files. When I do a global replace and replace the periods
with a blank prior to importing in R, I get what I used to get:
DTB3<-read.table("
My suspicion is that there is some value
that R does not think is numeric, so the
column becomes a factor, and you are
seeing the codes for the factor.
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")
Kerpel,
Hi Folks!
I used the code below previously with no problems, but now I get:
DTB3<-read.table("C:\\Program
Files\\R\\R-2.7.1\\DTB3.csv",header=TRUE,sep=",")
> tail(DTB3)
DATE VALUE
14233 2008-07-23 1.56
14234 2008-07-24 1.62
14235 2008-07-25 1.71
14236 2008-07-28 1.70
I meant 5a 5b 5c. Multiple-line commands are handled correctly.
What is is doing is looking for "> " and " +" prompts. Anything else
is removed.
Here is a selection from the *R* buffer and the result after cleaning.
It includes an example of par().
Rich
*R*
> options(chmhelp = FALSE)
> option
On 8/1/08 12:40 PM, "Richard M. Heiberger" <[EMAIL PROTECTED]> wrote:
>
>> 5a) save my entire history to a text file
>> 5b) open it up in Emacs
>> 5c) prune any lines that don't have assignment operators
>
> No one has yet mentioned the obvious. ESS does your 5a 5b 5c with
>M-x ess-transc
Dear all,
Your constant talking about what bootstrap is and is not suitable for
made me finally verify the findings in the Pawinski et al paper.
Here is the procedure and the findings:
- First of all I took the raw data (that was posted earlier on this
list) and estimated the AUC values usin
Hi Ronaldo,
... lmer p-values
There are two packages that may help you with this and that might work with
the current implementation of lmer(). They are languageR and RLRsim.
HTH, Mark.
Bugzilla from [EMAIL PROTECTED] wrote:
>
> Hi,
>
> I have a modelo like this:
>
> Yvar <- c(0, 0, 0, 0,
See ?View but I don't think it 'auto updates' per your last sentence.
Maybe there's a better option?
Rachel Schwartz wrote:
Hi,
I would like to view matrices I am working with in a clean, easy to read,
separate window.
A friend showed me how to do something like I want with edit(). I can vie
Hi,
I am trying to get the positions in array coordinates (needed later) of certain
elements in an array but I am not sure how to get them.
My array is Q and the condition is dt>dV, where dt and dV are arrays of exactly
the same dimensions as Q.
I know that I can extract the elements of Q by
>5a) save my entire history to a text file
>5b) open it up in Emacs
>5c) prune any lines that don't have assignment operators
>
>
>Ken Williams
>Research Scientist
>The Thomson Reuters Corporation
>Eagan, MN
No one has yet mentioned the obvious. ESS does your 5a 5b 5c with
M-x ess-transcript
Hi,
I would like to view matrices I am working with in a clean, easy to read,
separate window.
A friend showed me how to do something like I want with edit(). I can view
the matrix in the 'R Data Editor':
For a sample matrix:
> mat=matrix(1:15,ncol=3)
> mat
[,1] [,2] [,3]
[1,]16
Hi all,
I have data that looks like
number|grouping
I would like to preform stats for each grouping
so
1|5
6|5
10|5
11|5
3|9
5|9
10|9
Say I would like to take the median for above, I should be returned 2 lines,
one for group #5 and one for group #9
Does this make sense?
I am sorry for the
Same problem. The Windows Task Manager indicated that Rgui.exe was using
1,249,722 K of memory when the error occurred. This is R 2.7.1 by the way.
> library(boot)
> setwd("C:/Documents and Settings/Tom/Desktop")
>
> data.in <- read.csv("inputdata.csv",header=T,as.is=T)
>
> per95 <- function
Dear Bert:
Thank you for your reply. But the "eval.parent" function does not seem to
solve the problem. Below is the code. I have tried running it with
eval.parent(myknot)
as well as with integers from 0 through 4, as in
eval.parent(myknot, 4)
Each attempt produces the same error:
Error in ev
Em Sex 01 Ago 2008, Marc Schwartz escreveu:
> on 08/01/2008 10:11 AM Ronaldo Reis Junior wrote:
> > Hi,
> >
> > I have a modelo like this:
> >
> > Yvar <- c(0, 0, 0, 0, 1, 0, 0, 0, 1, 2, 1, 1, 2, 3, 6, 6, 3, 3, 4)
> > TIME <- 4:22
> > ID <- rep("PlotA",19)
> > m <- lmer(Yvar~TIME+(TIME|ID),family=p
try
str(myclara)
to see what you have - a data frame , matrix etc
Are you getting any error messages?
I tried your write.table commands and they work okay.
--- On Fri, 8/1/08, pacomet <[EMAIL PROTECTED]> wrote:
> From: pacomet <[EMAIL PROTECTED]>
> Subject: [R] Exporting data to a text file
>
Just recently the ability to handle POSIXt style formats in as.chron
was added:
> x <- c("07/01/2001 12:00:00","07/17/2001 15:00:00")
> y <- as.chron(x, "%m/%d/%Y %H:%M:%S")
> y
[1] (07/01/01 12:00:00) (07/17/01 15:00:00)
> # Here is a workaround
> format(as.POSIXct(y), tz = "GMT", format = "%m
I'm not really clear on what you want here. Are you talking about plotting
multiple data points for each value ? In that case something like boxplot
might be what you want. Otherwise if you just wish to plot a data point for
each occurance of Normal etc then this will work but I'm not sure h
Hi,
I am getting the following error when using R CMD INSTALL ever since I
upgraded to R-2.7.1:
hhc: not found
CHM compile failed: HTML Help Workshop not intalled?
As indicated, the package is installed but without CHM help files. I
have downloaded the latest version of Rtools and I have tried
u
Hello.
I don't know how to do to ouput segments between points like the chart attached.
Can you help me please?
Thanks.
.yahoo.fr__
R-help@r-project.org mailing list
h
Hello.
I don't know how to do to ouput segments between points like the chart attached.
Can you help me please?
Thanks.
_
Envo__
R-help@r-project.org mailing list
https:
Hi,
I'm trying to help a friend who is doing a thesis in a nurse college, to
evaluate medical question forms.
There are about 30 questions giving more than 110 parameters to describe
each responding person's (gender, health etc.) and there are about 120
question forms to evaluate.
I have basically
Use gc() in the loop to possibly free up any fragmented memory. You
might also print out the size of B (object.size(B)) since that appears
to be the only variable in your loop that might be growing.
On Fri, Aug 1, 2008 at 12:09 PM, Tom La Bone <[EMAIL PROTECTED]> wrote:
>
>
> I have a data file c
on 08/01/2008 10:21 AM Sandy Small wrote:
Hi
I'm sure this question has been asked before but I can't find it in the
archives.
I have a data frame which includes interval and ordered nominal results.
It looks something like
"Measured" "Eyeball"
46.5 Normal
43.5 Mild
56.2 Normal
41.
HI R users
With clara function I get a data frame (maybe this is not the exact word,
I'm new to R) with the following variables:
> names(myclara)
[1] "sample" "medoids""i.med" "clustering" "objective"
[6] "clusinfo" "diss" "call" "silinfo""data"
I want to export
Sandy,
You can re-order a factor with
df$Eyeball<-factor(df$Eyeball, levels=c("Normal", "Mild", "Moderate",
"Severe"), ordered=T)
(assuming df is your data frame and that you want an _ordered_ factor;
the latter is not essential to your plots)
Incidentally, "NULL" isn't a particularly friendly
I'm not quite sure if this is what you mean but have a look at ?lowess or
?smooth. I think you might get want you want if you play around with the
parameters in lowess
--- On Fri, 8/1/08, Ptit_Bleu <[EMAIL PROTECTED]> wrote:
> From: Ptit_Bleu <[EMAIL PROTECTED]>
> Subject: [R] Best way to s
Dear list,
I'm writing a long document (thesis) and as much as I would like to
use only ggplot2 for the graphics, some features are still a bit
undocumented so I often end up choosing either ggplot2, lattice, or
base plots (which i know better) depending on the particular graph to
produc
On Fri, 1 Aug 2008, Rainer M Krug wrote:
Hi
if I try to import a raster layer which consists only of NULL values
from grass by using the readRAST6, I get an error message:
readRAST6("HSericea_seedsDisperse_2007")
ERROR: Invalid value for null (integers only)
Error in readBinGrid(rtmpfl11,
I have a data file called inputdata.csv that looks something like this"
ID YearResult Month Date
1 71741954 103 540301
2 7174195443 540322
3 20924 1967 4 2 670223
4 20924
On Fri, Aug 1, 2008 at 6:03 PM, Roger Bivand <[EMAIL PROTECTED]> wrote:
> On Fri, 1 Aug 2008, Rainer M Krug wrote:
>
>> Hi
>>
>> if I try to import a raster layer which consists only of NULL values
>> from grass by using the readRAST6, I get an error message:
>>
>>
>>> readRAST6("HSericea_seedsDis
First off, Marc Schwartz posted this link earlier today, read it.
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-p_002dvalues-not-di
splayed-when-using-lmer_0028_0029_003f
Second, your email is not really descriptive enough. I have no idea what
OR is, so I have no reaction.
Third, you're
Thomas Chu wrote:
>
> Neither of those 3 lines of commands managed to drop x4 and its P value
> magically decreased from 0.94 to almost 0! I am also baffled by how R
> calculated those RSS.
>
Maybe it is using a different type of SS. If i have a lm() model, and i do:
options(contrasts=c("con
Hi Rostam,
did you check
?rgb
already?
Hope this helps,
Roland
rostam shahname wrote:
Hi R users,
I would like to create an image using three matrices which contain the
values of Red, Green, and Blue of each pixel, i.e. one matrix which has
values of red, one which has values of green, and on
Hi
if I try to import a raster layer which consists only of NULL values
from grass by using the readRAST6, I get an error message:
> readRAST6("HSericea_seedsDisperse_2007")
ERROR: Invalid value for null (integers only)
Error in readBinGrid(rtmpfl11, colname = vname[i], proj4string = p4,
intege
Hi R users,
I would like to create an image using three matrices which contain the
values of Red, Green, and Blue of each pixel, i.e. one matrix which has
values of red, one which has values of green, and one which has values of
blue.
The values are between 0 and 1 instead of 0-255.
I have obtained
The biggest problem is that SPSS cannot fit a generalized linear mixed
model but lmer does. So, why would you expect the GLM in SPSS and the
GLMM in lmer to match anyhow?
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Draga, R.
> Sent: Friday, Aug
Hi everyone,
I thought that for a selfStart function, these two should be exactly
equivalent
> nls(Aform, DF)
> nls(Aform, DF, start=getInitial(Aform, DF))
but in this example that is not the case in R (although it is in S-plus
V6.2)
--
SSbatch<-selfStart(
model=funct
Hi
I'm sure this question has been asked before but I can't find it in the
archives.
I have a data frame which includes interval and ordered nominal results.
It looks something like
"Measured" "Eyeball"
46.5 Normal
43.5 Mild
56.2 Normal
41.1 Mild
37.8 Moderate
12.6 Severe
17.3
on 08/01/2008 10:11 AM Ronaldo Reis Junior wrote:
Hi,
I have a modelo like this:
Yvar <- c(0, 0, 0, 0, 1, 0, 0, 0, 1, 2, 1, 1, 2, 3, 6, 6, 3, 3, 4)
TIME <- 4:22
ID <- rep("PlotA",19)
m <- lmer(Yvar~TIME+(TIME|ID),family=poisson)
anova(m)
summary(m)
How to get the p-value for this case?
Thanks
Hi,
I have a modelo like this:
Yvar <- c(0, 0, 0, 0, 1, 0, 0, 0, 1, 2, 1, 1, 2, 3, 6, 6, 3, 3, 4)
TIME <- 4:22
ID <- rep("PlotA",19)
m <- lmer(Yvar~TIME+(TIME|ID),family=poisson)
anova(m)
summary(m)
How to get the p-value for this case?
Thanks
Ronaldo
--
Just because you're paranoid doesn't me
> "VK" == Vadim Kutsyy <[EMAIL PROTECTED]>
> on Fri, 01 Aug 2008 07:35:01 -0700 writes:
VK> Martin Maechler wrote:
>>
VK> The problem is in array.c, where allocMatrix check for
VK> "if ((double)nrow * (double)ncol > INT_MAX)". But why
VK> itn is used and not long
On 8/1/2008 9:46 AM, Benjamin Otto wrote:
Hi,
I would like to convert a simple list into an environment object. It seems I
have to create an environment object with new.env() and assign the single
values afterwards. Now what I did not really understand from the guides
until now is, how the paren
I ended up finding one solution to my 2nd question.
Here it is :
http://tolstoy.newcastle.edu.au/R/help/04/01/0649.html
Sorry for the bother.
David
De: GOUACHE David
Date: ven. 01/08/2008 15:50
À: Deepayan Sarkar
Cc: [EMAIL PROTECTED]
Objet : RE : [R] bwplot
Hi,
I would like to convert a simple list into an environment object. It seems I
have to create an environment object with new.env() and assign the single
values afterwards. Now what I did not really understand from the guides
until now is, how the parent environment supplied to the new.env() func
Martin Maechler wrote:
VK> The problem is in array.c, where allocMatrix check for
VK> "if ((double)nrow * (double)ncol > INT_MAX)". But why
VK> itn is used and not long int for indexing? (max int is
VK> 2147483647, max long int is 9223372036854775807)
Well, Brian gave you all i
Hi,
I was just woundering if there is in R a stable package for:
- Diffusion maps [1]
- Isomaps [2]
1:
http://www.cs.tau.ac.il/~shekler/Seminar2007a/DM%20GH%20and%20App/dm_elsevier.pdf
2: http://isomap.stanford.edu/
Thanks
Peter
--
http://games.entertainment.gmx.net/de/entertainment/games/fr
Dear collegues,
I have used R statistical program, package 'lmer', several times
already.
I never encountered major differences in the outcome between SPSS and R.
...untill my last analyses.
Would some know were the huge differences come from.
Thanks in advance, Ronald
In SPSS the Pearson c
Yep you are totally right. I looked at the graphs to do the analysis
quickly, and sacrificed correctness.
z <- rnorm(5000)
z.ts <- ts(z)
f <- fft(z.ts)
d <- fft(f, inverse=T)
plot(z.ts, d/5000)
#this is how far off the algorithm was from recreating the series.
After it is divided by the signal le
Interestingly, if I fitted the model using glm() rather than lm(),
drop1() would behave as expected:
summary(model.glm <- glm(y ~ ., data = sim.set, family = 'gaussian'))
summary(model.lm <- lm(y ~ ., data = sim.set))
drop1(model.glm, test = 'F')
drop1(model.lm, test = 'F')
model.glm <- step(mod
Hi Jeff,
You are absolutely right. In all the mess, I forgot that my Rserve was
started by a different user! Sometimes it's the simplest solution, isn't
it? Thanks a lot.
-Prasad
-Original Message-
From: Jeffrey Horner [mailto:[EMAIL PROTECTED]
Sent: Thursday, July 31, 2008 5:14 PM
To:
Hello and thank you for your reply.
My dummy example was a bit too simple... I'm having difficulty correctly
specifying the 'at' component since my real situation concerns a multipanel
display with ' relation="free" '.
To illustrate :
dates<-as.Date(32768:32895,origin="1900-01-01")
plouf<-dat
Treat it as an over-determined linear system, that is:
A <- cbind(cos((2*pi/T)*(Times - Tau)), sin((2*pi/T)*(Times - Tau)))
qr.solve(A, Yis)
because 'solve' will only handle square matrices.
Hans W. Borchers
Josué Polanco wrote:
>
> Hi everybody,
>
> I am reading the Lomb paper
Hi list, I have some questions regarding
1) conversion of date + time characters to chron
2) formatting chron object printing
Regarding (1), Gabor's Rnews 2004 4/1 article has been indispensible,
but I often work with files where dates and times are contained in a
single field. In this case, I wo
> "VK" == Vadim Kutsyy <[EMAIL PROTECTED]>
> on Thu, 31 Jul 2008 15:43:56 -0700 writes:
>>> I am getting an error "allocMatrix: too many elements
>>> specified" when I am trying to create large matrix or
>>> vector (about 1 billion elements).
>>>
>>> How can I find
Another option is ?by
test <- as.data.frame(cbind(c(rep(1,5),rep(2,5)), rnorm(10), rnorm(10)))
names(test)[1] <- "groupID"
test$groupID <- factor(test$groupID)
by(test[, -1], test$groupID, mean)
HTH,
Thierry
ir. Thie
Hello,
When I plot y=f(x) from the file xy.txt (
http://www.nabble.com/file/p18773387/xy.txt xy.txt ), I can clearly see a
trend.
Is there a function or a package able to take the median value of y for an
interval of x (x +/- a defined value) to plot nice graph (at least a better
one) ?
Thanks
one option is aggregate(), e.g.,
test <- as.data.frame(cbind(c(rep(1,5),rep(2,5)), rnorm(10), rnorm(10)))
names(test)[1] <- "groupID"
aggregate(test[c("V2", "V3")], list(test$groupID), mean)
I hope it helps.
Best,
Dimitris
--
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
C
Hi Bertolt,
by(test,INDICES=test$groupID,FUN=mean)
And today's a holiday in Switzerland, so stop working already ;-)
HTH
Stephan
Bertolt Meyer schrieb:
Dear R users,
I have a newbie-question that I couldn't resolve after reading through
several pieces of documentation and searching the ar
something like that should work :
aggregate(test, list(test[,1]), mean)
2008/8/1 Bertolt Meyer <[EMAIL PROTECTED]>
> Dear R users,
>
> I have a newbie-question that I couldn't resolve after reading through
> several pieces of documentation and searching the archive.
>
> I have a data.frame cont
Dear R users,
I have a newbie-question that I couldn't resolve after reading through
several pieces of documentation and searching the archive.
I have a data.frame containing experimental data from a group experiment
in psychology. Each line represents a single participant, but
participants
On Fri, Aug 1, 2008 at 7:31 AM, Stephen Tucker <[EMAIL PROTECTED]> wrote:
> In the example below, a straight application of strsplit() is probably the
> simplest solution. In a more general case where it may be desirable to match
> patterns, a combination of sub() or gsub() with strsplit() might
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