Hi,
I have a data set which is like this I write as the CSV I import:
Sample;Hole;Feature;Value
1;5;x;4,2334
1;5;y;3,3434
1;5;r;0,1080
1;10;x;5,2526
1;10;y;4,3434
1;10;r;0,1080
with 98 sample and 10 different holes. These are measured values.
Now I also have a list of nominel values:
Nathaniel,
If you are interested in the particular subject, you should consider them
as a fixed effect, which wil give you what you want.
If your subjects are really random, the only thing you could be interested
in, is whether considering the subjects as a grouping is helping you in
improving
Hello,
I have a problem with using Iwidgets in R while i'm able to use BWiget and
Tktable. If I let tcl list the names of
packages, Iwidgets seems available. I use R 2.6.0; in addition
addTclPath(C:/Tcl/lib) has been used (in which iwidets is present)
try(tcl(package, names))
Tcl http opt
Hi Klaus,
I am not exactly sure what you are asking for, but something like this? This
would be option (2) from your list - I don't know that it would be too
difficult in R that you would want to use another tool.
filt - function(x)
with(x,which(Hole 1))
normalize - function(x,y)
{
My previous suggestion was inconsistent with the Trellis/Lattice idea
of creating a trellis object without necessarily creating a plot. And
it also interfered with attempts to plot to a file device. So here is
a better solution, based on replacing `print.trellis`, though it is
still basically a
Hello Alexander
I doubt that such an analyis is very useful as the data is not sampled
synchronously (equity close in the US for ^gspc and even that is not
always at the same time, some average price from Oanda data). Also fx
data from others sources as suggested in another mail on this list
Are there parameters to set position and size of the windows created by
the commands in iplots? I could not find them.
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On 15/10/2007, Hans-Peter [EMAIL PROTECTED] wrote:
Is it possible to download somewhere the former 2.5.1 windows bin version
of
R? (Need it for testing).
Thanks,
Hans-Peter
I have the following code:
list1 - list()
for (i in list.files(pattern=filename1)){
x - read.table(i)
list1[[i]] - x
}
list2 - list()
for (i in list.files(pattern=filename2*)){
x - read.table(i)
list2[[i]] - x
}
anslist - vector('list', length(list1))
for(i in
Dear list,
I have a data frame with a number of events (factor) and the times at which
they occurred (continuous variable):
event time
A 10
A 12
B 15
A 17
C 13
...
Is it possible in R to make a plot against time of the cumulative frequency of
occurrence of each event? This would be, a raising
A new package 'nnls' is now available on CRAN.
The package provides an R interface to the Lawson-Hanson NNLS algorithm
for non-negative least squares that solves the least squares problem A x =
b with the constraint x = 0.
The Lawson-Hanson NNLS algorithm was published in
Lawson CL, Hanson RJ
Perhaps:
par(mfrow=c(3,1))
lapply(tapply(df$time, df$events, ecdf), plot)
On 15/10/2007, Dieter Vanderelst [EMAIL PROTECTED] wrote:
Dear list,
I have a data frame with a number of events (factor) and the times at
which they occurred (continuous variable):
event time
A 10
A 12
B 15
A
2007/10/15, Stephen Tucker [EMAIL PROTECTED]:
Hi Klaus,
I am not exactly sure what you are asking for, but something like this?
This
would be option (2) from your list - I don't know that it would be too
difficult in R that you would want to use another tool.
filt - function(x)
Dear All
I posted a similar question quite some time ago, but that was on an old
OS and an old version of R. This time I have RHEL 4, which is still
supported as an OS, and R 2.5.1 which is not *that* old.
My sessionInfo() gives:
sessionInfo()
R version 2.5.1 (2007-06-27)
This looks like what happens when you use a UTF-8 locale and don't have
Unicode X11 (meta-)fonts installed. Try running in LC_CTYPE=en_GB: if
that works you will both have a workaround and know where to look for a
solution.
On Mon, 15 Oct 2007, michael watson (IAH-C) wrote:
Dear All
I
Hi,
Suppose I have a data.frame like this
Lines - var1 var2 var3 var4 var5 var6
0 2 1 2 0 0
2 3 7 6 0 1
1.54 9 9 6 0
1.06 1022 3 3
DF - read.table(textConnection(Lines), skip=1)
names(DF) -
Perhaps,
names(which.max(sapply(DF, max)))
On 15/10/2007, Lauri Nikkinen [EMAIL PROTECTED] wrote:
Hi,
Suppose I have a data.frame like this
Lines - var1 var2 var3 var4 var5 var6
0 2 1 2 0 0
2 3 7 6 0 1
1.54 9 9 6 0
Dear All
Another one I have touched on before with a much older OS and version.
My sessionInfo() is:
sessionInfo()
R version 2.5.1 (2007-06-27)
i686-redhat-linux-gnu
locale:
LC_CTYPE=en_GB.UTF-8;LC_NUMERIC=C;LC_TIME=en_GB.UTF-8;LC_COLLATE=en_GB.U
Thank you Brian, setting the locale using.
Sys.setlocale(LC_CTYPE,en_GB)
Meant that my test plot command worked fine.
Will now install Unicode X11 fonts
-Original Message-
From: Prof Brian Ripley [mailto:[EMAIL PROTECTED]
Sent: 15 October 2007 14:12
To: michael watson (IAH-C)
Cc:
OK, tried google, got very, very lost. There are lots of different
packages out there.
Can anyone tell me where I can download the Unicode X11 (meta-)fonts
for Red Hat that R needs?
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of michael watson (IAH-C)
Hi
[EMAIL PROTECTED] napsal dne 15.10.2007 14:36:59:
2007/10/15, Stephen Tucker [EMAIL PROTECTED]:
Hi Klaus,
I am not exactly sure what you are asking for, but something like
this?
This
would be option (2) from your list - I don't know that it would be too
difficult in R that you
date - as.POSIXlt(Sys.time()) #present date
for (i in 1:difftime(as.POSIXlt(Sys.Date()),2007-10-01))
if (date$wday != 0 date$wday != 6) {print(date);assign(date,
(date-86400))} else (assign(date, (date-86400)))
I am trying to print dates from present day to a day in
Quite helpful indeed. Greatly appreciated.
Another problem I had was trying to simulate an example from my book.
Simulating 1000 coin tosses, and finding the frequency of sign changes. So
how will we plot this using R? (frequency of sign changes in Y axis)
Daniel Nordlund wrote:
See ?seq.Date, e.g.
now - Sys.Date()
dd - seq(now - 20, now, by = day)
dd[as.POSIXlt(dd)$wday %% 6 != 0]
and have a look at R News 4/1.
On 10/15/07, Vishal Belsare [EMAIL PROTECTED] wrote:
date - as.POSIXlt(Sys.time()) #present date
for (i in
On Sat, Oct 13, 2007 at 03:08:58PM +0300, Martin Ivanov wrote:
Dear R users,
I am using the vars package to calculate the impulse response functions and
the forecast error variance decomposition of a VAR model. Unfortunately I do
not know whether these functions assume unit or one
On Sun, 14 Oct 2007, coldeyes.Rhelp wrote:
Hi there:
i got a problem to get the prediction from a model recently. for
example if i use a survival analysis to predict the risk. i use the code
like below: i found the the prediction is not equal to (coef * x + coef
* sex) , could someone help
Henrique Dallazuanna wrote:
Perhaps,
names(which.max(sapply(DF, max)))
Nice.
I was thinking along the lines of
M - as.matrix(DF)
colnames(M)[col(M)[which.max(M)]]
On 15/10/2007, Lauri Nikkinen [EMAIL PROTECTED] wrote:
Hi,
Suppose I have a data.frame like this
Lines - var1 var2
Hello dear useRs,
I'm trying to export a barplot into an emf file. My problem is that
the plot is properly printed into the file, except the bars that do
not appear :(
I've experienced some problems also with simple points plots, in which
points did not appear (same problem).
Can you help
Your logic test is not correct. Here is what I think you want. See
if it makes sense.
x - list() # create some test data
x[[1]] - read.table(textConnection(1 2 3 4
+ 5 6 7 8
+ 4 3 2 1
+ 8 7 6 5))
# create another list element
x[[2]] - t(x[[1]])
str(x[[1]])
'data.frame': 4 obs. of 4
On Mon, 2007-10-15 at 17:36 +0200, Poirier Clement wrote:
Hello dear useRs,
I'm trying to export a barplot into an emf file. My problem is that
the plot is properly printed into the file, except the bars that do
not appear :(
I've experienced some problems also with simple points plots,
Group,
I have count data with one observation per subject. I would like to fit
a glmm to these data in order to account for overdispersion in the
outcome. The lmer() function does not appear to be able to handle data
that have only one observation per-cluster id, even though separate
Hi,
I have a vector of strings (class character) with 6 elements (length
6). I call it 'names'.
Graham Chapman
John Cleese
Terry Gilliam
Eric Idle
Terry Jones
Michael Palin
And I want to turn it into another vector of strings called
'shortnames' with the same length.
The new vector should
Lemon Curry ?
sapply( strsplit( monty, ), function(x) {
paste( substring(x,1,3), collapse = )
})
is a way to do it, ...
There is probably a better way to do that using the gsubfn package
Gonçalo Ferraz wrote:
Hi,
I have a vector of strings (class character) with 6 elements (length
On Mon, 2007-10-15 at 12:04 -0400, Gonçalo Ferraz wrote:
Hi,
I have a vector of strings (class character) with 6 elements (length
6). I call it 'names'.
Graham Chapman
John Cleese
Terry Gilliam
Eric Idle
Terry Jones
Michael Palin
And I want to turn it into another vector of
Gabor,
Thanks much. Your solution is elegant. My overall scheme is to take
present date, and check whether it is a weekend, if not, then create a
string based on the date, to concatenate into a url link for
download.file( ). The files I need to download have a part which is in
the format: mmddyy.
Gabor,
Thanks much. Your solution is elegant. My overall scheme is to take
present date, and check whether it is a weekend, if not, then create a
string based on the date, to concatenate into a url link for
download.file( ). The files I need to download have a part which is in
the format: mmddyy.
Hi everyone,
When I try to import data do R, the following message appears: \U
sequences are not supported on Windows.
I tried lots of methods to import (read.csv, read.table, RODBC, read.delim)
and the same message appears for all these methods. I think it is a bigger
problem.
Can
On Monday 15 October 2007 06:43:52 pm pintinho wrote:
pi I tried lots of methods to import (read.csv, read.table, RODBC,
read.delim)
pi and the same message appears for all these methods. I think
it is a bigger
pi problem.
pi
pi Can anyone help me solving this issue?
A little bit more
I really depends on what you want to plot. You can plot the
cumulative count of the sign changes with:
x - sample(c(0,1), 1000, TRUE)
plot(cumsum(x), type='l')
There are other things you can do. You can get a rough histogram of
the length of the run by:
stem(rle(x)$length)
The decimal
Looks fine by me. There are lots of other ways of doing it. Does
this look any nicer?
x - c('asdfghk', 'qwerrey')
gsub(.*(...)$, '\\1', x)
[1] ghk rey
On 10/15/07, Sergio Correia [EMAIL PROTECTED] wrote:
I want to extract the last 3 letters of a string.
So far, I've done this:
symbol
Hi,
I have a matrix that has a variable number of columns. I do not know, a
priori, the number of columns.
How can I get a sub matrix, for example, from row 10 to the end of the
columns?
In MatLab I would use something like this: SubMatrix = Matrix[10, 1:end]
Thanks a lot.
--
View this
I was hoping to avoid using regex except when necessary (u know what
they say), but I'm beginning to think that's the way things are done
in R.
Thanks
On 10/15/07, jim holtman [EMAIL PROTECTED] wrote:
Looks fine by me. There are lots of other ways of doing it. Does
this look any nicer?
x
Dear Sirs:
Is there a way to group multiple responses in variables sets. SPSS
have this feature: this is possible to group a set of variables by
their common categories. I would like to do the same in R.
Example:
my.df = data.frame(var1=c(1,2,2,1,2,2,1,2), var2=c(2,2,1,1,2,2,1,1),
?table to count the factors
x
[1] a b c d e
paste(head(x, -1), tail(x, -1), sep='')
[1] ab bc cd de
On 10/15/07, Tom Sgouros [EMAIL PROTECTED] wrote:
Hi All:
I feel like there must be a slick R-native no-loop way to get the counts
for the entries in a factor, but I'm unable to see how.
subura said the following on 10/15/2007 12:04 PM:
Care to explain how i can use a wildcard expression to source all files
ending with .R in a subdirectory ? I've tried something like this
'source(glob2rx(*.R))' without success.
Thank you
Try
R.files - list.files(my.path, pattern =
I knew it was simple. Thanks very much.
-tom
jim holtman [EMAIL PROTECTED] wrote:
?table to count the factors
x
[1] a b c d e
paste(head(x, -1), tail(x, -1), sep='')
[1] ab bc cd de
On 10/15/07, Tom Sgouros [EMAIL PROTECTED] wrote:
Hi All:
I feel like there must be a
Hi all!
How is it possible to estimate standard errors for coef obtained from lme?
Is there sth like se.coef() for lmer or what is the anaytical solution?
Thank you!
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Dirk Eddelbuettel wrote:
Sean,
On 15 October 2007 at 12:03, Sean Davis wrote:
| I am trying to build Rmpi on Suse 10.2 linux. While I would like to use
| the RPM version of liblam-7.1.2, I have not been able to do so. It
| seems that Rmpi makes some pretty strong assumptions when trying
Thanks for the response...
My confusion about plot stems from the fact I am plotting 82 points with 82
colours, so surely all colours get plotted?
As for updating X, I recently installed the latest version of XFree86 for my
version of linux, RHEL 4.
As for Brian's e-mail you quoted, I do try
Hello,
I've read the other posts with regard to chisq.test and goodness of fit
and am still missing something.
1. I create a simple vector of randomly generated lognormal values with
mean=0 and sd=1;
d1 - rlnorm(100,meanlog=0,sdlog=1);
2. I also create a vector of probabilities that are expected
For numerical accuracy, the coxph routine centers each covariate before doing
the computation. All of the downstream results (predict, survfit, etc) use
this
centered data.
Terry Therneau
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R-help@r-project.org mailing list
(This could be fixed, it has not happened to me in a long time, but I
will mention it mainly because it is not something you are likely to
think of.)
It used to be that the X colours might be defined by the first
application that needed them, so if the systems administrator happened
to start
Hello everybody,
I would like to use the SpDep-package (especially the Local Moran index
analysis and the Getis-Ord statistics) in R for analysing my data. However,
I don't have x-y coordinates, but my data is in a distance matrix format. Is
it possible to use the SpDep package with predefined
hi:
Yesterday I post a message about hoy to plot a time series, but someone told
me to post more information about the file so here it is:
the file was read using read.table and the name is list. When I use
str(list) it tells the following variables:
YEAR int: 2003,2003,20032004
MONTH
Thanks Jholtman.
However, the plot didnt come out the way I envisone dit to be. On the Y
axis, i should have sign changes in 1000 tosses, the range being from
negative to postitive, and a straight horizontal line across y=0. The
X-axis should have the toss number, range 0-1000
I'm glad u
Here are a couple of ways that you can do it:
x - expand.grid(YEAR=c(2003,2004), MONTH=1:12, DAY=1, STATE=LETTERS[1:4])
x$SALES - runif(nrow(x), 10, 100)
# add date for plotting
x$date - ISOdate(x$YEAR, x$MONTH, x$DAY)
# sort into date order for plotting
x - x[order(x$date),]
# you can use
Gad Abraham wrote:
Hi,
I'm using survreg() from the survival package for parametric survival
regression (modelling inter-arrival times of patients to a waiting list
as exponentially distributed, with various regressors such as queue size
and season).
Does anyone know which algorithm
Dan-157 wrote:
Dear R-Users,
I am new to R, so please excuse the ignorance.
I have data:
x (2.14, 2.41, 1.09, 0.17, 8.18)
y (3.81, 5.13, 0.63, 0.75, 6.35)
I would like to use simple linear regression and test 2 things:
1) slope of line of best fit is statistically different
I have executed a Repeated Measures ANOVA with one DV (latency) and
one within subject factor (acoustic condtion: 3 levels) by
bootstrapping my sampling distribution of F from the empirical sample
distribution. I chose to resample because the sample distribution
deviates from normality a lot.
The
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf
Of azzza
Sent: Monday, October 15, 2007 6:06 PM
To: r-help@r-project.org
Subject: Re: [R] Need some help
Thanks Jholtman.
However, the plot didnt come out the way I envisone dit to be. On the Y
Hi Stephen,
Check the help for predict.glm(). The argument for passing new data is
actually 'newdata', as in:
pred = predict(glm.model, newdata=form[150001:20,-1],
type=response)
Cheers Joe
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of
Did you look at the C source code? There are 4 different variants
(survregN.c, where N - 2:5) , depending on whether the distribution is
built-in or not, and penalized likelihood is being used or not. They all
look like NR to me, but I confess I haven't read the code in extreme
detail. It is well
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