XinMeng wrote:
> Hello sir:
> I use Rcmd to execute R code,such as :
>
> "C:\\Program Files\\R\\R-2.2.1\\bin\\Rcmd.exe" BATCH globalLowessRun.r
>
> globalLowessRun.r is a R function written by myself.
>
> But I wanna make the "globalLowessRun.r" changeable,such as globalLinearRun.r
> or gridba
?lines
On 20/09/06, Mauricio Cardeal <[EMAIL PROTECTED]> wrote:
> Hi. Please, how can I put together 2 or more lines at the same
> scatterplot ? Example: measures of protein intake (quantitative) of 4
> children over 30 days, by day. How to plot all children at same graphic:
> Protein X Time ? Is
Poizot Emmanuel wrote:
> Brian Edward a écrit :
>
>> Hello all,
>>
>> I have been a R user for about a year now, running on a MS Windows
>> machine.
>> I am in the process of making a complete switch to open-source. Linux
>> is a
>> new world to me. Ubuntu was my selection of the various distr
Darren Weber wrote:
>Hi Patrick,
>
>thanks for pointing me to your work and Rmetrics.
>
>I have a few questions on my mind right now. Do you have methods for
>automatic download of price quote histories?
>
Rmetrics and tseries
>I can use python to get
>XML data on FOREX price quotes from the N
On Wed, 20 Sep 2006, Murray Jorgensen wrote:
> I am in a discriminant analysis situation with a frame containing
> several variables and a grouping factor, if you like:
>
> set.seed(200906)
> exampledf <- as.data.frame(matrix(rnorm(50,5,2),nrow=10,ncol=5))
> exampledf$Group <- factor(rep(c(1,2,3),
Brian Edward a écrit :
Hello all,
I have been a R user for about a year now, running on a MS Windows machine.
I am in the process of making a complete switch to open-source. Linux is a
new world to me. Ubuntu was my selection of the various distributions.
Please pardon this very basic question
I am in a discriminant analysis situation with a frame containing
several variables and a grouping factor, if you like:
set.seed(200906)
exampledf <- as.data.frame(matrix(rnorm(50,5,2),nrow=10,ncol=5))
exampledf$Group <- factor(rep(c(1,2,3),c(3,3,4)))
exampledf
I'm sure there must be a simple wa
Hi
Gavin Simpson wrote:
> Dear list,
>
> Following advice posted to this list a while back by Prof Ripley [1], I
> have been trying to draw a per mille character [2] in an axis label.
>
> This should give the correct character:
>
> plot(1:10, ylab = "\u2030")
>
> but all I get is '"S'. I'm ru
Hello sir:
I use Rcmd to execute R code,such as :
"C:\\Program Files\\R\\R-2.2.1\\bin\\Rcmd.exe" BATCH globalLowessRun.r
globalLowessRun.r is a R function written by myself.
But I wanna make the "globalLowessRun.r" changeable,such as globalLinearRun.r
or gridbasedLowessRun.r,and so on.
How can
Hi. Please, how can I put together 2 or more lines at the same
scatterplot ? Example: measures of protein intake (quantitative) of 4
children over 30 days, by day. How to plot all children at same graphic:
Protein X Time ? Is there any command like "overlay" ?
Thank you,
Mauricio
_
Try either of these:
rbind(DF1, DF2[setdiff(rownames(DF2), rownames(DF1)),])
rbind(DF1, DF2[!(rownames(DF2) %in% rownames(DF1)),])
On 9/19/06, Kartik Pappu <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I have two data frames each with 5 columns and different number of
> rows. some of the row names i
How about,
uxy <- union(row.names(x), row.names(y))
ixy <- intersect(row.names(x), row.names(y))
rbind(x[is.element(row.names(x),uxy),],
y[!is.element(row.names(y),ixy),])
Note, simple rbind'ing of the two frames changes
common row.names.
Gamal
__
Andrew Zachary wrote:
>
> Here is an example (though the data are too large to send ). The dataset
> is (6530 x 15). Predictors are continuous N(0,1). Trying to build a
> regression tree.
>
> fit <- rpart( y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 +
> x11 + x12 + x13 + x14, data=my.
Tong Wang wrote:
> Hi, I have a data set , say, X : [1][2] mean 0 .8
> sd1 3
>
> I need to use it in do.call() , so when I refer to the first col of
> X, I need it to be a list with members mean, sd. thus I constructed X
> as a list of list : X[[1]]<- list(me
Brian Edward wrote:
> Hello all,
>
> I have been a R user for about a year now, running on a MS Windows machine.
> I am in the process of making a complete switch to open-source. Linux is a
> new world to me. Ubuntu was my selection of the various distributions.
> Please pardon this very basic
Michael Gormley wrote:
> I am trying to preprocess a large dataset of affymetrix data. Creating an
> affybatch is not possible with the computer I am running it on, so I have
> used the justRMA command to run RMA. I have read the affy document
> describing the justRMA command and the help doc
Hi,
I have a data set , say, X : [1][2]
mean 0 .8
sd1 3
I need to use it in do.call() , so when I refer to the first col of X, I
need it to be a list with member
Hi all,
I have two data frames each with 5 columns and different number of
rows. some of the row names in one data frame are the same as the row
names in the other. I want to be able to merge the two data frames to
get a new data frame in which the duplicated row names are only shown
once with t
Hello all,
I have been a R user for about a year now, running on a MS Windows machine.
I am in the process of making a complete switch to open-source. Linux is a
new world to me. Ubuntu was my selection of the various distributions.
Please pardon this very basic question (I was unable to locate
I am trying to preprocess a large dataset of affymetrix data. Creating an
affybatch is not possible with the computer I am running it on, so I have used
the justRMA command to run RMA. I have read the affy document describing the
justRMA command and the help documentation but I am unclear as t
Hi all, does somebody know how to store an array in MySQL with the
package RMySQL. Thanks in advance.
--
Web Page
http://www.geocities.com/ricardo_rios_sv/index.html
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
I am having a problem with RODBC's connections. It appears that
my connection to the database is closed by R automatically before
I am done with it.
Here is my code:
foo <- function(dsn) {
db <- odbcConnect(dsn)
odbcSetAutoCommit(db, FALSE)
data <- someDatabaseOperation(db)
data2 <- someL
You are referencing Ct[i+1] in the loop and it is not defined. From
then on everything is NA.
On 9/19/06, Guenther, Cameron <[EMAIL PROTECTED]> wrote:
> Hello everyone,
>
> I have a function here that I wrote but doesn't seem to work quite
> right. Attached is the code. In the calib funcion und
Dear R user,
Is there any well-developed R package for pre-processing MALDI-TOF or
SELDI-TOF data besides PROcess? The package should be able to perform
denoising, baseline correction, normalization, spectrum
alignment/calibration, peak identification and binning. Thank you.
Deming Mi
__
Hi,
I have two data frames each with 5 columns and different number of
rows. some of the row names in one data frame are the same as the row
names in the other. I want to be able to merge the two data frames to
get a new data frame in which the duplicated row names are only shown
once with the d
Hello everyone,
I have a function here that I wrote but doesn't seem to work quite
right. Attached is the code. In the calib funcion under the for loop
Bt[i+2]<-(1-m)*Bt[i+1]+Rt[i]*Rerr-Ct[i+1] returns NA's for everything
after years 1983 and 1984. However the code works when it reads
Bt[i+2]<-
Is this what you have in mind?
tmp <- data.frame(id=1:10, y1=sample(10), y2=sample(10))
tmp2 <- cbind(id=c(tmp$id, tmp$id), stack(tmp[,2:3]))
barchart(values ~ id | ind, data=tmp2)
If not, send an equally trivial example of what you want to the list
and someone will send you an optimized set
Peter Dalgaard:
> Ben Bolker <[EMAIL PROTECTED]> writes:
>> 1. compose your response
> I've always wondered why step 1. - often the time-consuming bit - is not
> listed last.
The advice applies to the situation when answering immediately would be
your knee-jerk reaction. It is assumed tha
I need to identify repeated items in p$a with
different s and d entries on the same row, given that
the "0" items should not be considered in the
comparison. Here is an example:
1. Items 3 and 5 in p$a are repeated with different
entries of s and d, should be removed.
2. Item 2 was repeated twi
Hello list,
I am interested in plotting a series of barcharts (package lattice) with
several metrics. Instead of calling barchart many times to call each
parameter, is there an approach to calling several parameters at once?
An alternative might be a barplot for a primary parameter and then addin
Have you read the books by Cleveland?
His experiments show that most people do better estimating things and
comparing things on a linear scale rather than looking at angles and
areas (also see
http://biostat.mc.vanderbilt.edu/twiki/pub/Main/StatGraphCourse/graphsco
urse.pdf)
With a dot chart you
You could use the weighted mean (?weighted.mean) or weighted median
(?wtd.quantile in package Hmisc).
On 9/19/06, Taka Matzmoto <[EMAIL PROTECTED]> wrote:
> Hi R-users
> I have a data set. There are 10 products and the numbers of people who
> ranked the products.
>
> The format of the data set is
Gavin,
the advice given here:
http://tolstoy.newcastle.edu.au/R/help/02b/4378.html
works on FreeBSD 6.1.
Try:
> pdf("~/tmp/test_per_mille.pdf", paper = "a4", family = "URWBookman",
+ encoding="WinAnsi.enc")
> plot(1:10, ylab="\211")
> dev.off()
Cheers
Andrew
On Tue, Sep 19, 2006 at 05:2
On Tue, 19 Sep 2006, Gavin Simpson wrote:
> Dear list,
>
> Following advice posted to this list a while back by Prof Ripley [1], I
> have been trying to draw a per mille character [2] in an axis label.
>
> This should give the correct character:
>
> plot(1:10, ylab = "\u2030")
>
> but all I get is
Hi R-users
I have a data set. There are 10 products and the numbers of people who
ranked the products.
The format of the data set is
productID rank1 rank2 rank3 rank4 rank5 rank6 rank7 rank8 rank9 rank10
--
On Tue, 2006-09-19 at 17:29 +0100, Gavin Simpson wrote:
> Dear list,
>
> Following advice posted to this list a while back by Prof Ripley [1], I
> have been trying to draw a per mille character [2] in an axis label.
Before I get asked for it, here's the sessionInfo() - apologies for not
supplyin
Dear list,
Following advice posted to this list a while back by Prof Ripley [1], I
have been trying to draw a per mille character [2] in an axis label.
This should give the correct character:
plot(1:10, ylab = "\u2030")
but all I get is '"S'. I'm running linux (FC5) and have fonts installed
tha
Here is an example (though the data are too large to send ). The dataset
is (6530 x 15). Predictors are continuous N(0,1). Trying to build a
regression tree.
fit <- rpart( y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 +
x11 + x12 + x13 + x14, data=my.data.set, weights=wts )
And the outp
Andrew,
Not sure what your problem is based on your email.
But data volume is not a problem if there is only 1400 obs and 15 predictors.
On 9/19/06, Andrew Zachary <[EMAIL PROTECTED]> wrote:
> Not sure if anyone has posted on this problem ... I want to use rpart to
> build a binary tree on a re
On Tue, 2006-09-19 at 22:16 +1000, Robert King wrote:
> Here is another thing that might help work out what is happening. If I
> use --no-install, ade4 actually fails as well, in the same way as zipfR.
>
> [Desktop]$ R CMD check --no-install ade4
> * checking for working latex ... OK
> * u
Pieter Provoost gmail.com> writes:
> I'm having some problems with a bubble plot (ps package). I don't want
> tick marks on all four sides (just two), I want to have a smaller font
> size, and I would like to be able to define bubble sizes shown in the
> legend (now it shows 0, 0, 0, 9.747 and
On Tue, 19 Sep 2006, Andrew Zachary wrote:
> Not sure if anyone has posted on this problem ... I want to use rpart to
> build a binary tree on a relatively large dataset with ~1400 data points
> and 15 predictors. But I've noticed that rpart fails almost immediately
> in the call to C_s_to_rp, as
Guoyan Zhao wrote:
> Dear everyone,
>
> I have only tpr (true positive rate) and fpr (false positive rate) which
> were used to plot the ROC curve. Is there any way I can calculate the AUC
> for this curve using just this information? Any help is highly
> appreciated.
>
> Isabel
To get the A
Gianna Monti unimib.it> writes:
>
> Dear Gregory R. Warnes,
> I'm a phd student in statistics, at the University of Milano Bicocca.
> I'm interested to the methods of estimate the parameters of the Dirichlet
distribution.
> Do you have implemented an algorithm in R?
> If so, can you give me th
zhaoshi u.washington.edu> writes:
>
> hi--
>
> I am new to R and try to use R cluster my binary data. I use
> hierarchical clustering
> plot (hclust (dist(x,method="binary"),method="average"),cex=0.1)
> I end up with a cluster Dendrogram. On the left of my dendrogram, there
> is scale called
Not sure if anyone has posted on this problem ... I want to use rpart to
build a binary tree on a relatively large dataset with ~1400 data points
and 15 predictors. But I've noticed that rpart fails almost immediately
in the call to C_s_to_rp, as that code returns nonsense. Looking at the
code itse
Jan Sabee gmail.com> writes:
> I know that is probability of predict for new dataset.
> My question is how can I know each probability according to class (sore).
> I mean that I need the result of predit something like (M=1, F=0):
> 1 2 3 4 5 6 7 8 9 10
> 1 0 0 0 1 0 1 1 0 1
Dear everyone,
I have only tpr (true positive rate) and fpr (false positive rate) which
were used to plot the ROC curve. Is there any way I can calculate the AUC
for this curve using just this information? Any help is highly
appreciated.
Isabel
__
On 9/17/06, Darren Weber <[EMAIL PROTECTED]> wrote:
> Hi,
>
> are there any good charting and analysis tools for use with
> currencies, stocks, etc. in R? I have some tools to download currency
> data from the NYFRB using python and XML. Can we get and parse an XML
> download using R? Can we hav
Would this do what you want:
> m=seq(10,1,length=10)
> v=seq(3,1,length=10)
> x <- seq(-5,20,length=500)
> plot(x,dnorm(x,m[1],v[1]))
> for (i in 1:10) plot(x,dnorm(x,m[i],v[i]),type="l",col=i)
On 19/09/06, Anupam Tyagi <[EMAIL PROTECTED]> wrote:
> march uniroma1.it> writes:
>
> >
> >
> > Hi e
march uniroma1.it> writes:
>
>
> Hi everybody
> I'm new in R so the question will be easy for you
> I'm running multiple density functions taking account of the following
> conditions:
> mean=seq(10,1,length=10)
> var=seq(3,1,length=10)
>
> How can I describe the density functions on t
Dear UseRs,
you can find on CRAN web site my last contribute about
R & regression techniques:
http://cran.r-project.org/doc/contrib/Ricci-regression-it.pdf
It's in Italian language.
Regards.
Vito Ricci
Se non ora, quando?
Se non qui, dove?
Se non tu, chi?
___
The legend can be controlled with the key.entries parameter. I don't
know about the tick marks, though note that bubble calls xyplot in the
lattice package, not the "standard" plot function.
On 19/09/06, Pieter Provoost <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I'm having some problems with a bubble p
[ rpy-help CC removed. Please don't blanket-cross post. ]
On 18 September 2006 at 11:52, Darren Weber wrote:
| Hi Patrick,
|
| thanks for pointing me to your work and Rmetrics.
|
| I have a few questions on my mind right now. Do you have methods for
| automatic download of price quote historie
In fact, in this example its not really even necessary to manipulate
the function since the new function will find z in its environment
(which is the body of make.function3):
make.function3 <- function() {
z <- 1
function(x, y) x*y+z
}
g <- make.function2()
g(3,2) # 7
On 9/19/06
Hi,
I'm having some problems with a bubble plot (ps package). I don't want
tick marks on all four sides (just two), I want to have a smaller font
size, and I would like to be able to define bubble sizes shown in the
legend (now it shows 0, 0, 0, 9.747 and 4265.757 which is not really
convenien
Alfonso,
It sounds like unzip and zip are not in your path (as opposed to odfWeave
needing additional configuration). The error message that you received is
generated when odfWeave is invoked. It checks to see if unzip can be used at
the command line.
After googling, I found a basic link for s
You could alternately use substitute:
make.function2 <- function() {
fun <- function(x, y) x*y+z
body(fun) <- do.call(substitute, list(body(fun), list(z = 1)))
fun
}
ff <- make.function2()
ff(3,2) # 7
On 9/19/06, Mstislav Elagin <[EMAIL PROTECTED]> wrote:
> Peter Dalgaa
Dear Gregory R. Warnes,
I'm a phd student in statistics, at the University of Milano Bicocca.
I'm interested to the methods of estimate the parameters of the Dirichlet
distribution.
Do you have implemented an algorithm in R?
If so, can you give me the script? or, in general, some helps?
Best Re
Berton,
Thanks for your inupt. The 'nist' link you mentioned was one of the reasons
for my confusion and how it is implemented in R. As for now I am assuming
predict function with 'prediction' option will provide me tolerance/prediction
interval. Is this a proper assumption?
TIA for
Hi R users
I haven't run odfWeave example, R give me:
Setting wd to
C:\DOCUME~1\MARIOM~1\CONFIG~1\Temp\Rtmph2Nzqb/odfWeave19070343633
Copying C:/ARCHIV~1/R/R-23~1.1/library/odfWeave/examples/simple.odt
Decompressing ODF file using unzip -o "simple.odt"
Erro en odfWeave(demoFile, outputFile)
Here is another thing that might help work out what is happening. If I
use --no-install, ade4 actually fails as well, in the same way as zipfR.
[Desktop]$ R CMD check --no-install ade4
* checking for working latex ... OK
* using log directory '/home/rak776/Desktop/ade4.Rcheck'
* using Ve
Dear R users,
I'm trying to fit a gamma-frailty model on a simulated dataset, with 6
covariates, and I'm running into some results I do not understand. I
constructed an example from my simulation code, where I fit a coxph model
without frailty (M1) and with frailty (M2) on a number of data sa
Mesomeris, Spyros [CIR] wrote:
> Thanks David,
>
> It has actually worked, the problem was the formatting of the N/A values
> in Excel. R apparently doesn't like to see #N/A that Excel produces if a
> formula cannot be returned. So, saving the file as csv (comma delimited)
> file and removing all
Thanks David,
It has actually worked, the problem was the formatting of the N/A values
in Excel. R apparently doesn't like to see #N/A that Excel produces if a
formula cannot be returned. So, saving the file as csv (comma delimited)
file and removing all the #N/A observations, leaving those cells
On 9/19/2006 4:41 AM, Prof Brian Ripley wrote:
> On Tue, 19 Sep 2006, Sean O'Riordain wrote:
>
>> Hi Duncan,
>>
>> Thanks for that. In the light of what you've suggested, I'm now using
>> the following:
>>
>> # generate a random integer from 0 to t (inclusive)
>> if (t < 1000) { # to avoid
On 9/19/06, Mesomeris, Spyros [CIR] <[EMAIL PROTECTED]> wrote:
> Dear R helpers,
>
> I am trying to read a CSV file in R called EUROPE (originally an Excel
> file which I have saved as a CSV file) using the command
>
> EUROPEDATA <- read.csv("EUROPE.csv")
>
> EUROPE.csv is basically a matrix of dim
I think that command should work (assuming that it is *comma* rather
than semi-colon delimited, which is used in countries where a comma is
used as a decimal point, in which case you should use read.csv2
instead). So, is your data definitely as clean as you think. Have
you looked at the data in a
No output from the tools command
Peter Dalgaard wrote:
> Robert King <[EMAIL PROTECTED]> writes:
>
>> I'm testing a FC5 machine for use in a student lab. R 2.3.1 is installed
>> and
>> seems to work fine. There is one peculiarity - the logins are
>> authenticating
>> to a server, and a "ver
Dear R helpers,
I am trying to read a CSV file in R called EUROPE (originally an Excel
file which I have saved as a CSV file) using the command
EUROPEDATA <- read.csv("EUROPE.csv")
EUROPE.csv is basically a matrix of dimension 440*44, and has a line of
headers, i.e. each column has a name.
Usin
Anupam Tyagi yahoo.com> writes:
> It seem the more complicated case is often of more substantive interest in
> many
> settings: is children's income more strongly correlated with parent's
> education
> than parent's income?
An even better example (same measurement scale)---Questions like this
On Tue, 19 Sep 2006, Sean O'Riordain wrote:
> Hi Duncan,
>
> Thanks for that. In the light of what you've suggested, I'm now using
> the following:
>
> # generate a random integer from 0 to t (inclusive)
> if (t < 1000) { # to avoid memory problems...
>M <- sample(t, 1)
> } else {
>
for lists look at ?lapply() and ?sapply(); for your 2nd question try
something like:
array(unlist(aa), dim = c(3, 3, 2))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer
On Mon, 18 Sep 2006, Ahn ChaeHyung wrote:
> Dear all,
>
> I have the following list, "aa", composed of two 3*3 tables. I would like
> to use "apply" function to summarize it, but "apply" cannot handle "list".
> I want to do it without using any interation.
>
> 1. Is there any function like "
Ahn ChaeHyung wrote:
> Dear all,
>
> I have the following list, "aa", composed of two 3*3 tables. I would like
> to use "apply" function to summarize it, but "apply" cannot handle "list".
> I want to do it without using any interation.
>
> 1. Is there any function like "apply" for list?
> 2.
Dear all,
I have the following list, "aa", composed of two 3*3 tables. I would like
to use "apply" function to summarize it, but "apply" cannot handle "list".
I want to do it without using any interation.
1. Is there any function like "apply" for list?
2. Is there any way to transform that "l
Greg Snow intermountainmail.org> writes:
>
> You may want to rethink your whole approach here:
>
> 1. Pie charts are usually a poor choice of graph, there are better
> choices.
> 2. Adding percentages to a pie chart is a way of admitting that the pie
> chart is not doing the job.
> 3. If you wa
You might want to read
Ligges (2003): R Help Desk: Getting Help - R's Help Facilities and
Manuals, R News 3(1), 26--28,
http://cran.r-project.org/doc/Rnews/Rnews_2003-1.pdf
Uwe Ligges
Lynda wrote:
> I have a few questions for R:
>
> 1. Other than using a built-in function such as mean(), ho
Hi
please try to consult some introduction manual and see some
suggestions below
On 18 Sep 2006 at 17:55, Lynda wrote:
Date sent: Mon, 18 Sep 2006 17:55:02 -0400
From: Lynda <[EMAIL PROTECTED]>
To: r-help@stat.math.ethz.ch
Subject:
Hi Duncan,
Thanks for that. In the light of what you've suggested, I'm now using
the following:
# generate a random integer from 0 to t (inclusive)
if (t < 1000) { # to avoid memory problems...
M <- sample(t, 1)
} else {
while (M > t) {
M <- as.integer(urand(1,min=0, max=
Hi
both lapply and mapply can give you what you want but you have to
select only desired part.
e.g.
spec.apply<- function(sl, x) {
d<-dim(x)[1]
d2<-dim(x)[2]*2*d
vec<-seq(d,d2,2)
Z<-as.numeric(mapply(predict,sl, x))[vec]
dim(Z) <-c(d,length(vec)/d)
Z
}
but it is probably slower than simple fo
Peter Dalgaard wrote:
> Mstislav Elagin <[EMAIL PROTECTED]> writes:
>
>> Dear All,
>>
>> the last expression in the following code snippet crashes R (version
>> 2.3.1 on Windows XP) when run interactively:
>>
>> make.bad.function <- function(kind)
>> {
>>zz <- switch(kind,
>>
Robert King <[EMAIL PROTECTED]> writes:
> I'm testing a FC5 machine for use in a student lab. R 2.3.1 is installed and
> seems to work fine. There is one peculiarity - the logins are authenticating
> to a server, and a "verbose" flag is set somewhere, leading to lots of
> spurious messages li
Peter Dalgaard biostat.ku.dk> writes:
> No, he wants to compare two correlation coefficients, not test that
> one is zero. That's usually a misguided question, but if need be, the
> Fisher z transform atanh(r) can be used to convert r to an
> approximately normal variate with a known variance 1/(
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