Yimeng Lu yl2058 at columbia.edu writes:
Could you give me some advice in
solving the problem of such error message from gnls and nls?
## begin error message
Problem in gnls(y1 ~ glogit4(b, c, m, t, x), data.frame(x..: Step halving
factor reduced below minimum in NLS step
Try to set
Baoqiang == Baoqiang Cao [EMAIL PROTECTED]
on Mon, 18 Jul 2005 15:02:05 -0400 writes:
Baoqiang Hello All, I'm learning R. Just wonder, any
Baoqiang package or function that I can use to get the
Baoqiang dissimilarity matrix? Thanks.
Yes,
learn to use help.search() {also read
JarekT == Tuszynski, Jaroslaw W [EMAIL PROTECTED]
on Mon, 18 Jul 2005 16:00:43 -0400 writes:
JarekT Hi, A minor announcement. I just added two functions
JarekT for reading and writing GIF files to my caTools
JarekT package. Input and output is in the form of standard
JarekT
I did manage eventually to reproduce this via
data(package=makecdfenv)
which is a BioC package with a non-standard use of the data directory.
(There are no objects in the single R file in the data directory, which
from its name I suggest is a dummy and the data directory should be
removed
Or read the *sources* for package boot, and find the original code with
all the comments.
On Mon, 18 Jul 2005, Spencer Graves wrote:
Excellent question. Try 'getAnywhere(index.array)'. It's hidden
in namespace:boot. Ditto for boot.return.
spencer graves
Obrien, Josh
Just change the names! E.g.
names(DF)[c(4,6)] - names(DF)[c(6,4)]
Strictly a data frame has names, not column names, hence the use the
names() and names-() functions here.
On Tue, 19 Jul 2005, wu sz wrote:
I have a data frame with 15 variables, and want to exchange the data
of 4th column
My thanks to Achim Zeileis and Prof Ripley for their responses. In a very
short time I not only had an answer and solved my problem, but also learned
something about R that I can employ in other situations.
Much appreciated,
MT
__
Hi Adai,
Many Thanks for the examples.
I work for a financial institution. We are exploring R as a tool to implement
our portfolio optimization strategies. Hence, R is still a new language to us.
The script I wrote tried to make a returns matrix from the daily return indices
extracted from a
On Tue, 19 Jul 2005, Prof Brian Ripley wrote:
Just change the names! E.g.
names(DF)[c(4,6)] - names(DF)[c(6,4)]
Strictly a data frame has names, not column names, hence the use the
names() and names-() functions here.
That answered the subject line. If you want to exchange the columns
Dear R users:
I encountered difficulties in michaelis-menten equation. I found
that when I use right model definiens, I got wrong Km vlaue,
and I got right Km value when i use wrong model definiens.
The value of Vd and Vmax are correct in these two models.
#-right model
Chun-Ying Lee [EMAIL PROTECTED] writes:
Dear R users:
I encountered difficulties in michaelis-menten equation. I found
that when I use right model definiens, I got wrong Km vlaue,
and I got right Km value when i use wrong model definiens.
The value of Vd and Vmax are correct in these
Thanks a lot! But DF[c(4,6)] - DF[c(6,4)] seems to just exchange
the data, not the names. If exchanging the columns(both data and
names) needs two steps?
DF[c(4,6)] - DF[c(6,4)]
names(DF)[c(4,6)] - names(DF)[c(6,4)]
Shengzhe
__
On Tue, 19 Jul 2005, wu sz wrote:
Thanks a lot! But DF[c(4,6)] - DF[c(6,4)] seems to just exchange
the data, not the names. If exchanging the columns(both data and
names) needs two steps?
DF[c(4,6)] - DF[c(6,4)]
names(DF)[c(4,6)] - names(DF)[c(6,4)]
Yes, it does. You did however say in
Chun-Ying Lee wrote:
Dear R users:
I encountered difficulties in michaelis-menten equation. I found
that when I use right model definiens, I got wrong Km vlaue,
and I got right Km value when i use wrong model definiens.
The value of Vd and Vmax are correct in these two models.
First, your problem could be boiled down to the following example. See
how the colnames of the two outputs vary.
df - cbind.data.frame( 100=1:2, 200=3:4 )
df/df
X100 X200
111
211
m - as.matrix( df ) # coerce to matrix class
m/m
100 200
1 1 1
2 1 1
It appears that
When I callculate a linear model, then I can compute via confint the
confidencial intervals. the interval level can be chosen. as result, I get
the parameter of the model according to the interval level.
On the other hand, I can compute the prediction-values for my model as well
with
Stephen,
This has nothing to do with your R but to do with your email settings.
Are you sure you are sending mails in plain text ? Your email on the
R-help mailing archive (see link below) appears to be unreadable
https://stat.ethz.ch/pipermail/r-help/2005-July/074210.html
Please try to use
hi all
i am having a problem with the expression/paste command
say we estimate a variable, named PHI
it contains the value of say 2
and we want to display this value as hat(phi) = PHI onto a graphic
i.e. hat(phi)=2
how does one do this?
i've tried the following:
1.
Clark Allan wrote:
hi all
i am having a problem with the expression/paste command
say we estimate a variable, named PHI
it contains the value of say 2
and we want to display this value as hat(phi) = PHI onto a graphic
i.e.hat(phi)=2
how does one do this?
i've tried the
Something like this :
x - 0.5
plot( 1:10, main=substitute( hat(Phi) ~ = ~ x, list(x=x) ) )
Also see http://tolstoy.newcastle.edu.au/R/help/04/09/3371.html
Regards, Adai
On Tue, 2005-07-19 at 13:35 +0200, Clark Allan wrote:
hi all
i am having a problem with the expression/paste command
Seems like no one had responded to this one yet, so I'll take a stab:
## Generate some bogus data:
set.seed(45)
dat - cbind(expand.grid(LETTERS[1:2], 1:3), round(runif(6), 2))
names(dat) - c(state, psu, weight)
dat2 - data.frame(state=sample(c(A, B), 100, replace=TRUE),
Dear all,
I'm looking for some material on data mining with R. I have something
from Luis Torgo but I'd like to see something else.
If anybody could help me I'll be thankful
Adrián
__
R-help@stat.math.ethz.ch mailing list
Dear all,
How can I read data from posgresql?
Thanks
Adrián
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Dear R users
a question about outer explanatory variables in lme:
I have measured size of a population of insects in fields.
These fields were spread out over a large region.
The fields are grouped (spatially) in pairs: one with fertiliser high,
the other one low.
I want to test effect of mean
On Tue, 19 Jul 2005 10:12:48 -0300 secretario academico FACEA wrote:
Dear all,
I'm looking for some material on data mining with R. I have something
from Luis Torgo but I'd like to see something else.
There have been several discussions on the list, so browsing the
archives will probably
for the stats gurus
does anyone know if there exists a general formula relating the median
of a continuous distribution to its moments. the distribution could be
skewed or symmetric and is definitely not normal.
the reason for asking is since the median of the particular distribution
that i am
On 7/19/2005 9:28 AM, Clark Allan wrote:
for the stats gurus
does anyone know if there exists a general formula relating the median
of a continuous distribution to its moments. the distribution could be
skewed or symmetric and is definitely not normal.
Not in general, and probably not any
On Tue, 19 Jul 2005, Uwe Ligges wrote:
Clark Allan wrote:
hi all
i am having a problem with the expression/paste command
say we estimate a variable, named PHI
it contains the value of say 2
and we want to display this value as hat(phi) = PHI onto a graphic
i.e. hat(phi)=2
how does
The search site at
http://finzi.psych.upenn.edu/
is back up. This allows you to search mailing list archives,
R functions from most packages, and documents.
You can also use the R function RSiteSearch() in R itself, which
opens the results in your browser.
I'm sorry the site was down for so
On Tue, 19 Jul 2005, Adaikalavan Ramasamy wrote:
First, your problem could be boiled down to the following example. See
how the colnames of the two outputs vary.
df - cbind.data.frame( 100=1:2, 200=3:4 )
df/df
X100 X200
111
211
That one is probably unintentional.
m -
Hi Adai,
Thank you very much for your suggestions. Your optimized function would come in
very handy cause I will need to generate a matrix of size around 2250 * 1000.
Regards,
Gilbert
-Original Message-
From: Adaikalavan Ramasamy [mailto:[EMAIL PROTECTED]
Sent: 19 July 2005 12:20
To:
Thanks For pointing that out. S - Original Message - From:
Adaikalavan Ramasamy To: Stephen Cc: Frank E Harrell Jr ; Prof
Brian Ripley ; Sent: Tuesday, July 19, 2005 1:30 PM Subject: Re: [R]
Survival dummy variables and some questions Stephen, This has
nothing to do with your R but to
Hello: I'm reading in a series of text files (100 files that are each 2000
rows by 6 columns). I wish to combine the columns (6) of each file (100) and
get the row mean. I'd like to end up with a data.frame of 2000 rows by 6
columns.
foo - list()
for(i in 1:10){
# The real data are read in
Although I tried to find an answer in the manuals and archives, I cannot
solve this (please excuse that my English and/or R programming skills
are not good enough to state my problem more clearly):
I want to write a function with an indeterminate (not pre-defined)
number of arguments and think
Hi Adai,
When I tried the optimized routine, I got the following error message:
r1
899188 902232 901714 28176U 15322M
20050713 7.595 10.97 17.96999 5.1925 11.44
20050714 7.605 10.94 18.00999 5.2500 11.50
20050715 7.480 10.99 17.64999 5.2500 11.33
20050718 7.415 11.05
Andy Bunn wrote:
Hello: I'm reading in a series of text files (100 files that are each 2000
rows by 6 columns). I wish to combine the columns (6) of each file (100) and
get the row mean. I'd like to end up with a data.frame of 2000 rows by 6
columns.
foo - list()
for(i in 1:10){
#
On Tue, 2005-07-19 at 17:29 +0200, Dirk Enzmann wrote:
Although I tried to find an answer in the manuals and archives, I cannot
solve this (please excuse that my English and/or R programming skills
are not good enough to state my problem more clearly):
I want to write a function with an
Hi everyone,
I am doing 5 years mortality predictive index score with survival analysis
using a Cox proportional hazard model where I have a continous predictive
variable and a right censored response which is the mortality, and the
individuals were followed a maximum of 7 years.
I'd like to
I have two files to compare, one is a regular txt file that I can read
in no prob.
The other is a .gct file (How do I read in this one?)
I tried a simple
read.table(data.gct, header = T)
How do you suggest reading in this file??
thank you.
__
On 7/19/2005 12:10 PM, mark salsburg wrote:
I have two files to compare, one is a regular txt file that I can read
in no prob.
The other is a .gct file (How do I read in this one?)
I tried a simple
read.table(data.gct, header = T)
How do you suggest reading in this file??
.gct is
Dear Matthias
Can you provide an example to demonstrate what you did? Two
remarks to your email. Maybe that answers already your question.
1) Using predict() you will get the estimated value for each
observation or for new data. You can reproduce this value by
using the coefficients from
On 7/16/05, Federico Gherardini [EMAIL PROTECTED] wrote:
On Friday 15 July 2005 17:00, Deepayan Sarkar wrote:
On 7/15/05, Federico Gherardini [EMAIL PROTECTED] wrote:
On Friday 15 July 2005 14:42, you wrote:
Hi all,
I've used the split argument to print four lattice plots on a single
Dear R-users
I have a problem choosing initial points for the function arms()
in the package HI
I intend to implement a Gibbs sampler and one of my conditional
distributions is nonstandard and not logconcave.
Therefore I'd like to use arms.
But there seem to be a strong influence of the initial
On Tue, 2005-07-19 at 12:28 -0400, Duncan Murdoch wrote:
On 7/19/2005 12:10 PM, mark salsburg wrote:
I have two files to compare, one is a regular txt file that I can read
in no prob.
The other is a .gct file (How do I read in this one?)
I tried a simple
read.table(data.gct,
ok so the gct file looks like this:
#1.2 (version number)
7283 19 (matrix size)
Name Description Values
... ..
How can I tell R to disregard the first two lines and start reading
the 3rd line in this gct file. I would just delete them, but I do not
know how to open
If it is a text file ?read.table should provide enough details to read the
file into R. Based on the file format referenced below it shouldn't be too
hard to get at the parts you want.
Randy
On 7/19/05 1:06 PM, Marc Schwartz (via MN) [EMAIL PROTECTED] wrote:
On Tue, 2005-07-19 at 12:28 -0400,
On 7/19/2005 1:16 PM, mark salsburg wrote:
ok so the gct file looks like this:
#1.2 (version number)
7283 19 (matrix size)
Name Description Values
... ..
How can I tell R to disregard the first two lines and start reading
the 3rd line in this gct file. I
Hi, All;
I tried to use library mclust in 64-bit compiled R 2.0.1 but failed.
Installation went smoothly without any warning or error. However, when I
tried to use them with the following simple code, it crashed.
Library(mclust)
Dat - c(rnorm(20, mean=0, sd=0.2), rnorm(30, mean=1, sd=0.2))
Ind -
Try ?read.table or args(read.table). Might skip=2 do what you want?
spencer graves
p.s. I routinely readLines(File, n=11) to see how many headers there
are AND identify the sep character. Then I
quantile(count.fields(File, ...)) to see if all records have the same
number
Hi R Users
I have a code which I am running for my thesis work. Just want to make sure that
its ok. Its a t test I am conducting between two gamma distributions with
different shape parameters.
the code looks like:
sink(a1.txt);
for (i in 1:1000)
{
x-rgamma(40, 2.5, 10) # n = 40, shape = 2.5,
Hi all,
i'm new to R,
I need to modelize in R a statistic algorithm,
This algo use Weibull, normal law, linear regression, normalisation, root mean
square, to find eta and beta fitting the weibull model (to analyse few results)
and further when we will get more information apply bayes model .
On Tue, 2005-07-19 at 13:16 -0400, mark salsburg wrote:
ok so the gct file looks like this:
#1.2 (version number)
7283 19 (matrix size)
Name Description Values
... ..
How can I tell R to disregard the first two lines and start reading
the 3rd line in this
This is all extremely helpful.
The data turns out is a little atypical, the columns are tab-delemited
except for the description columns
DATA1.gct looks like this
#1.2
23 3423
NAME DESCRIPTION VALUE
gene1 a protein inducer 1123
. . ..
How do I get R to
On 7/19/05, Dirk Enzmann [EMAIL PROTECTED] wrote:
Although I tried to find an answer in the manuals and archives, I cannot
solve this (please excuse that my English and/or R programming skills
are not good enough to state my problem more clearly):
I want to write a function with an
For the TAB delimited columns, adjust the 'sep' argument to:
read.table(data.gct, skip = 2, header = TRUE, sep = \t)
The 'quote' argument is by default:
quote = \'
which should take care of the quoted strings and bring them in as a
single value.
The above presumes that the header row is also
Peter Dalgaard [EMAIL PROTECTED] writes:
Chun-Ying Lee [EMAIL PROTECTED] writes:
Dear R users:
I encountered difficulties in michaelis-menten equation. I found
that when I use right model definiens, I got wrong Km vlaue,
and I got right Km value when i use wrong model definiens.
On Tue, 2005-07-19 at 13:08 -0500, Marc Schwartz (via MN) wrote:
For the TAB delimited columns, adjust the 'sep' argument to:
read.table(data.gct, skip = 2, header = TRUE, sep = \t)
The 'quote' argument is by default:
quote = \'
which should take care of the quoted strings and bring
On Tue, 2005-07-19 at 13:08 -0500, Marc Schwartz (via MN) wrote:
For the TAB delimited columns, adjust the 'sep' argument to:
read.table(data.gct, skip = 2, header = TRUE, sep = \t)
The 'quote' argument is by default:
quote = \'
which should take care of the quoted strings and bring
Hi,
I'm using a fairly simple HP Compaq desktop PC running Windows 2K. When
running a large process in R, the process RGUI.exe will never exceed 50%
of the CPU usage.
The program used to be able to use more of the computer, but does not now.
I don't believe this is a multiple processor
SUBIRANA CACHINERO, ISAAC wrote:
Hi everyone,
I am doing 5 years mortality predictive index score with survival analysis
using a Cox proportional hazard model where I have a continous predictive
variable and a right censored response which is the mortality, and the
individuals were
Hello,
I have been trying to take the derivative of a quadratic B-spline
obtained by using the COBS library. What I would like to do is
similar to what one can do by using
fit-smooth.spline(cdf)
xx-seq(-10,10,.1)
predict(fit, xx, deriv = 1)
The goal is to fit the spline to data that is
Dear Michael:
Why is it a problem that R is not using more CPU space than it seems to
need?
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Greene, Michael
Sent: Tuesday, July 19, 2005 2:29 PM
To: '[EMAIL PROTECTED]'
Subject: [R] CPU Usage with R 2.1.0
On 7/19/2005 2:53 PM, James McDermott wrote:
Hello,
I have been trying to take the derivative of a quadratic B-spline
obtained by using the COBS library. What I would like to do is
similar to what one can do by using
fit-smooth.spline(cdf)
xx-seq(-10,10,.1)
predict(fit, xx, deriv = 1)
Hi guys,
I ran into a problem of not being able to create unique labels when creating a
factor. Consider an example below:
hb - factor(c(1,1,1,2,2,2,3,3,3), levels=c(1,2,3),labels=c(1,1,2))
hb
[1] 1 1 1 1 1 1 2 2 2
Levels: 1 1 2
unique(hb)
[1] 1 1 2
Levels: 1 1 2
How come there are three
Dnia 2005-07-19 20:28, Użytkownik Greene, Michael napisał:
Hi,
I'm using a fairly simple HP Compaq desktop PC running Windows 2K. When
running a large process in R, the process RGUI.exe will never exceed 50%
of the CPU usage.
If you have hyperthreading, R catches only one virtual
I wish it were that simple (perhaps it is and I am just not seeing
it). The output from cobs( ) includes the B-spline coefficients and
the knots. These coefficients are not the same as the a, b, and c
coefficients in a quadratic polynomial. Rather, they are the
coefficients of the quadratic
The derivative of a quadratic B-spline is the centered finite difference of
a linear B-spline, so if you have access to the underlying coefficients of
the B-spline expansion you can do this easily. I believe the coefficients
are passed as the $coef component of the return value.
Reid Huntsinger
On 7/19/2005 3:34 PM, James McDermott wrote:
I wish it were that simple (perhaps it is and I am just not seeing
it). The output from cobs( ) includes the B-spline coefficients and
the knots. These coefficients are not the same as the a, b, and c
coefficients in a quadratic polynomial.
I think about half of my question in R can be solved with a judicious
do.call.
Thanks, Andy
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
To All,
I am using the BRugs package. In running the meta file BRugsFit() with a
syntactically correct model .txt file, I see the message:
Error in FUN(X[[1]], ...) : .C(..): 'type' must be real for this format
I haven't found information on this kind of error either here or on the
Dear all,
I just start using package mix in R and am inexperienced with it. When I
ran my program several times, I sometimes got the warning message, and
sometimes not :
Warning message:
Loglik converged before variable 2 ; beta may be infinite. in: fitter(X,
Y, strats, offset, init, control,
Hello
1. a few months ago, I had sciviews working fine with R (rw2001) under
windows XP
2. now, upgrading to rw2011, the stuff seems fine (every package
installed),but I find a conflict when launching sciviews:
- it runs, apparently
- but when I try to work (import/export In: text for
Quoting Christoph Buser [EMAIL PROTECTED]:
Dear R-users
I have a problem choosing initial points for the function arms()
in the package HI
I intend to implement a Gibbs sampler and one of my conditional
distributions is nonstandard and not logconcave.
Therefore I'd like to use arms.
But
I remember in one slide of Prof. Ripley's presentation overhead, he
said the most popular data analysis software is excel.
So is there any resource or tutorial on this topic?
Thank you so much!
__
R-help@stat.math.ethz.ch mailing list
Hello,
today I've updated on the newest R-Version. But sadly a function I needed
didnt want to work:
The input is e.g.
days(as.Date(21-07-2005,%d-%m-%y))
the error is: Fehler in Math.Date(dts): floor nicht definiert für Date
Objekte
(Error in Math.Date(dts): floor not defined for date
Dear R-help subscribers,
Can some one please help me figure out how to write code that will allow me to
use a for loop to scan a number of files one by one, and then save a summary of
each file as the for loop progresses. For example I have 24 files named a1
through a24, and I want to do
Justin Rhodes [EMAIL PROTECTED] writes:
Dear R-help subscribers,
Can some one please help me figure out how to write code that will allow me
to use a for loop to scan a number of files one by one, and then save a
summary of each file as the for loop progresses. For example I have 24
Justone more comment in addition to Sundar's solution:
If these are all numeric matrices, I would read them into R
as such, instead of data frames. Actually, I would read them
all into a 3-dimensional array (2000 x 6 x # of files).
Assuming you have such an array, then you can do something
like:
Hi,
I am interested in using the numeric output from the gradient attribute of
deriv's output in subsequent analyses.
But, I have so far been unable to determine how to do so.
I will use the example from the deriv help to illustrate.
## function with defaulted arguments:
(fx - deriv(y ~ b0
Hi,
I am interested in using the numeric output from the gradient attribute of
deriv's output in subsequent analyses.
But, I have so far been unable to determine how to do so.
I will use the example from the deriv help to illustrate.
## function with defaulted arguments:
(fx - deriv(y ~ b0
Hi there,
I've looked through the very helpful advice about adding fitted lines to
plots in the r-help archive, and can't find a post where someone has offered
a solution for my specific problem. I need to plot logistic regression fits
from three differently-sized data subsets on a plot of
Jay Rotella wrote:
Hi,
I am interested in using the numeric output from the gradient attribute of
deriv's output in subsequent analyses.
But, I have so far been unable to determine how to do so.
I will use the example from the deriv help to illustrate.
## function with defaulted
Hello everyone,
Would you please somebody explain me what my sin is (please see the code
and timing bellow)? And how to improve myself and the following piece of
R code? BTW, the code works.
This is R 64-bit built by myself on Sun SPARC Solaris 9 with gcc-4.0.1
(64-bit) also built by yours
Really sorry for the wrong addressing. It was intended to the list only.
I apologize.
Latchezar
-Original Message-
From: Latchezar Dimitrov
Sent: Tuesday, July 19, 2005 9:47 PM
To: Latchezar Dimitrov; 'Prof Brian Ripley'
Cc: 'r-help@stat.math.ethz.ch'
Subject: RE: [R] Memory
I think its likely that you are using different versions of chron.
I noticed that version 2.2-33 of chron had the statement:
tms - dts - trunc(dts)
but version 2.2-35 seems to have replaced it with:
tms - dts - floor(dts)
and that seems to be causing the problem.
As a workaround:
Would the unique quadratic defined by the three points be the same
curve as the curve predicted by a quadratic B-spline (fit to all of
the data) through those same three points?
Jim
On 7/19/05, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 7/19/2005 3:34 PM, James McDermott wrote:
I wish it were
Dear R-helpers,
I am trying to estimate a model that I am proposing, which consists of putting
an extra hidden layer in the Markov switching models. In the simplest case the
S(t) - Markov states - and w(t) - the extra hidden variables - are independent,
and w(t) is constant. Formally the model
On 7/19/05, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Dear R-helpers,
I am trying to estimate a model that I am proposing, which consists of putting
an extra hidden layer in the Markov switching models. In the simplest case the
S(t) - Markov states - and w(t) - the extra hidden variables -
Hi,
We used known Vm and Km to simulate the data set (time, Cp) without
adding random error in there. Yes, the line looks like very close
to a straight line. But why can't we obtain the correct values with
fitting process? We used optim first and then followed by using nls
to fit the model.
On 7/19/05, Gabor Grothendieck [EMAIL PROTECTED] wrote:
I think its likely that you are using different versions of chron.
I noticed that version 2.2-33 of chron had the statement:
tms - dts - trunc(dts)
but version 2.2-35 seems to have replaced it with:
tms - dts - floor(dts)
and
Laura M Marx wrote:
Hi there,
I've looked through the very helpful advice about adding fitted lines to
plots in the r-help archive, and can't find a post where someone has offered
a solution for my specific problem. I need to plot logistic regression fits
from three differently-sized
Dear All,
Is there a way I can turn off the following warning message for using
multi-argument returns?
multi-argument returns are deprecated in: return(p1, p2, p3, p4)
Steve.
**
Steve Su ([EMAIL PROTECTED])
Dear all,
I have an error message installing SJava package.
So I searched web site(google) and R-mailing list to find a similar error
message.
But I couldn't find it.
I installed R-2.1.1 like this on Fedora Core4
1) /configure --enable-R-shlib --with-libpng --with-jpeglib
2) make - make check
Numeric data that is part of a mixed type data frame is converted into
character. How can I tell apply to maintain the original class of a
column and not convert it into character. I would like to do this of
the vector and not inside the apply function individually over each
element. Consider the
On Wed, 20 Jul 2005, Steve Su wrote:
Is there a way I can turn off the following warning message
for using multi-argument returns?
multi-argument returns are deprecated in: return(p1, p2, p3, p4)
doubleEm - function(p1, p2, p3, p4) {
return(list(p1 = p1*p1,
p2 = p2*p2,
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