Thank you for your kind help. Your script works very well. Would you please
show me how to change NaN to zero and column variables 1, 2, ..., 8 to var1,
var2, ..., var8? Thanks again.
Legen
jholtman wrote:
>
> Is this what you want:
>
>> x <- read.table(textConnection("idcode1code2
Henrik Wahren wrote:
>
> How can one or more levels be removed from a factor of a data frame. There
> was a similar post on how to do this when a factor meets some criterion
> (e.g. <= 2), but I can¹t seem to get that solution to work.
> Here, I simply want to drop some levels.
>
>
Simply ca
G'day fellow Pacific rim dwellers,
On Wed, 11-Nov-2009 at 11:13AM +0800, Berwin A Turlach wrote:
|> G'day Isabella,
|>
[...]
|> > However, when trying to convert a character date from the year 2000
|> > to a Julian date, I get an instead of the desired Julian date:
|> >
|> > > as.date("02-MA
Hi, your problem is that you run a regression on three observations with
independent variables. This is obviously nonsense as two independent
variables plus intercept MUST perfectly explain all variable. This you can
see from the fact that the r-squared in the regression is 1, and that the
standard
dvkirankumar wrote:
>
> Hi all,
> I am new to R-project
> I have to create custom package and some required functions in that
> package
> can any one help me how to create a custompackage and how to write my
> function in to that package
> These things I have to use in many places as per requir
You can find guidance from Writing R Extensions manual, which is
shipped with R and available on CRAN.
Besides, you can have a look at package.skeleton.
2009/11/11 venkata kirankumar
>
> Hi all,
> I am new to R-project
> I have to create custom package and some required functions in that packag
How can one or more levels be removed from a factor of a data frame. There
was a similar post on how to do this when a factor meets some criterion
(e.g. <= 2), but I can¹t seem to get that solution to work.
Here, I simply want to drop some levels.
[[alternative HTML version deleted]]
Dear R users,
For a uni-variable distribution represented in a numerical vector,
we can obtain a distribution function using 'ecdf', and then calculate
corresponding p-values. But if I have a 2-column dataframe representing
a bi-variable joint distribution, given a pair of values, how can
Hi all,
I am new to R-project
I have to create custom package and some required functions in that package
can any one help me how to create a custompackage and how to write my
function in to that package
These things I have to use in many places as per requirements in my solution
please help me to
G'day Isabella,
On Tue, 10 Nov 2009 20:11:31 -0800
"Isabella Ghement" wrote:
> I tried the solution you suggested and find that am having problems
> getting R to extract the year from an object created by as.date():
Perhaps it would be best to first clarify what you really need. :)
If you have
thanks a lot
--
Pleasure should be subordinate to duty.
[[alternative HTML version deleted]]
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-gu
Dear R People:
I am using qcc for a quality control class.
I have used qcc with type "xbar.one" for individuals but cannot
determine how to plot a moving range control chart.
Has anyone done that, please?
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematica
Consider the following example:
x <- c(2,4,3,6)
y <- c(4,9,5,10)
z <- factor(c(1,1,2,2))
summary(lm("y ~ x + z"))
The above works fine.
Suppose I change z so that
x <- c(2,4,3,6)
y <- c(4,9,5,10)
z <- factor(c(1,1,2,NA))
summary(lm("y ~ x + z"))
the last row/observation is not considered in the
David -- thank you for your response.
merge does work but it creates another dataframe. df1 is very large and I did
not want another copy created. What I ended up doing is:
df1 <- merge(df1, df2, by="week")
In terms of memory allocation, will memory for two dataframes be allocated or
will the
Hi David,
Thank you so much for your reply - the quick answer is that I inherited the
R code
I am working with from someone else so I am trying to use the same functions
they have used. In hindsight, this did not make my life any easier - on the
contrary.
Of course, little did I know that I would
On Nov 10, 2009, at 11:11 PM, Isabella Ghement wrote:
Hi Berwin,
Many thanks for your fast reply! It's evening time in Vancouver,
Canada but
it must be day time in Australia, so
good day to you.
Sorry about the confusion regarding the library name - I am using
the "date"
library, as yo
Hi Berwin,
Many thanks for your fast reply! It's evening time in Vancouver, Canada but
it must be day time in Australia, so
good day to you.
Sorry about the confusion regarding the library name - I am using the "date"
library, as you pointed out.
I tried the solution you suggested and find that
John Tillinghast wrote:
> Hi,
> I'm running some routines with standard matrix operations like solve() and
> diag().
> When I do a profile, the lead item under total time is standardGeneric().
> Furthermore, solve() and diag() have much greater total time than self time.
> ???
Hi John -- are you i
G'day Isabella,
On Tue, 10 Nov 2009 18:40:11 -0800
"Isabella Ghement" wrote:
> I am trying to use the function as.date() from the "dates" package
As far as I can tell, there is no package called "dates", did you mean
the package "date"?
> in R 2.10.0 to convert a character date to a Julian d
sj gmail.com> writes:
> I am using glmer() from lmer(lme4) to run generalized linear mixed models. I
> can't figure out how to extract the z values for the fixed effects that are
> reported using the summary function . Any help would be appreciated.
library(lme4)
example(glmer)
coef(summary(gm1
According to Amazon review, 'Statistical Models in S' is a key
reference for understanding the methods implemented in several of
S-PLUS' high-end statistical functions, including 'lm()', predict()',
'design()', 'aov()', 'glm()', 'gam()', 'loess()', 'tree()',
'burl.tree()', 'nls()' and 'ms()'.
But
Hi everyone,
I am trying to use the function as.date() from the "dates" package in R
2.10.0 to convert a character date to a Julian date, as follows:
> as.date("02-MAY-01", order="mdy") # convert May 2, 2001 to a Julian date
[1] 2May1
However, when trying to convert a character date from the yea
Hello,
I am using glmer() from lmer(lme4) to run generalized linear mixed models. I
can't figure out how to extract the z values for the fixed effects that are
reported using the summary function . Any help would be appreciated.
Thanks,
Spencer
[[alternative HTML version deleted]]
Hello R Users!
I have a question about the output of ARMAtoMA when used to calculate
the variance of
a model. I have a mixed model of the form ARMA(1,1). The actual
model takes the form:
X(t) = 0.75X(t-12) + a(t) - 0.4a(t-1)
Given that gamma(0) takes the form [(1 + theta^2 -
2*theta*phi)/(1-ph
Thank you. I think that this is what I was looking for. I am getting
the hang of it with my data. Thank you all.
Sincerely,
Sergios Charntikov (Sergey), MA
Behavioral Neuropharmacology Lab
Department of Psychology
University of Nebraska-Lincoln
Lincoln, NE 68588-0308 USA
On Tue, Nov 10,
Hello R Users!
I have a question about the output of ARMAtoMA when used to calculate
the variance of
a model. I have a mixed model of the form ARMA(1,1). The actual
model takes the form:
X(t) = 0.75X(t-12) + a(t) - 0.4a(t-1)
Given that gamma(0) takes the form [(1 + theta^2 -
2*theta*phi)/(1-ph
Is this what you want:
> x <- read.table(textConnection("idcode1code2 p
+ 148 0.1
+ 157 0.9
+ 218 0.4
+ 262 0.2
+ 243 0.6
+ 356
Cedrick is correct. I was referring to the command invoked by entering the
string "screen" at the shell prompt.
- Original message -
From: "Cedrick W. Johnson"
To: "Peng Yu"
Cc: r-help@r-project.org
Date: Tue, 10 Nov 2009 16:38:09 -0500
Subject: Re: [R] R process gets killed spontaneo
Dear R users,
I'm new in R and couldn't find the solution to this in the postings. I want
to be able to use the "leaps" package to perform an "exhaustive" regression.
Most of my variables are categorical with many levels. I'd like to restrict
the candidate subsets to either all levels included or
On Nov 10, 2009, at 2:36 PM, David Winsemius wrote:
On Nov 10, 2009, at 3:07 PM, soeren.vo...@eawag.ch wrote:
Quite often, I need those tables:
x <- sample(c("a", "b", "c"), 40, rep=T)
y <- sample(c("X", "Y"), 40, rep=T)
(tbl <- table(x, y))
(z <- as.factor(paste(as.vector(tbl), " (",
roun
On Tue, 10 Nov 2009, Sam Albers wrote:
Hello,
I am having a little trouble formatting my dates correctly. When I plot
something using the following commands, R plots the most recent date on the
left of the figure and then earlier date on the right of the figure. Given
that English is read from
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Sam Albers
> Sent: Tuesday, November 10, 2009 3:08 PM
> To: r-help@r-project.org
> Subject: [R] Dates plotting backwards
>
> Hello,
>
> I am having a little trouble formatting m
>
>
> >
> > Thanks in advance!
> >
> > Sam
> >
> > Date=as.Date(test$Date, format= "%d/%m/%Y")
>
> Change that to
> test$Data <- as.Date(...)
> or plot Date instead of test$Date.
>
Yes that worked. Silly mistake. Sometimes those are the hardest ones to
spot. Thanks!
>
> Bill Dunlap
> Spotfire,
Sam Albers wrote:
Hello,
I am having a little trouble formatting my dates correctly. When I plot
something using the following commands, R plots the most recent date on the
left of the figure and then earlier date on the right of the figure. Given
that English is read from left to right I would
Mark Connolly wrote:
Thanks. I guess it kind of works for me as well. The contents of a
dataframe will display in the specified width, but doing a
str(dataframe) will happily run past the width set in options and use up
the device width (command line). I guess this may be my issue.
Options
Hello,
I am having a little trouble formatting my dates correctly. When I plot
something using the following commands, R plots the most recent date on the
left of the figure and then earlier date on the right of the figure. Given
that English is read from left to right I would like to have the dat
Dear all,
I have a dataset as below:
idcode1code2 p
148 0.1
157 0.9
218 0.4
262 0.2
243 0.6
356 0.7
37
Hi Nana,
This is not a r-devel question.
I suspect this should be something like:
write.netcdf.time(paste(path,'cam_fore.nc',sep=""), fore[,,times]
,lon,lat,times)
Robert
On Tue, Nov 10, 2009 at 8:31 AM, nana wrote:
> Dear CSAG R users,
>
> I will be glad if someone can point out what I am do
Cool! Thanks for the sampling and ff tips! I think I've figured it out now
using sampling...
I'm getting a quad-core, 4GB RAM computer next week, will try it again using
a 64 bit version :)
Thanks for your time!!!
Maja
tlumley wrote:
>
> On Tue, 10 Nov 2009, maiya wrote:
>
>>
>> OK, it's
When I invoke qplot, I get the following error:
Error in rename.default(x, .base_to_ggplot) : object '.data' not found
I would appreciate any advice.
sessionInfo:
R version 2.10.0 (2009-10-26)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States
Please read the last line on every message to r-help and particularly
note the requirement to provide reproducible code. We don't have your
data so its not reproducible. Also please clarify what you want
where.
On Tue, Nov 10, 2009 at 10:46 AM, MichelleJ wrote:
>
> I have a plotted a stacked pa
Thanks. I guess it kind of works for me as well. The contents of a
dataframe will display in the specified width, but doing a
str(dataframe) will happily run past the width set in options and use up
the device width (command line). I guess this may be my issue.
Options(width) does work for
On Nov 10, 2009, at 5:00 PM, David Winsemius wrote:
On Nov 10, 2009, at 4:29 PM, Hongwei Dong wrote:
Sorry for the confusion.
Let me put it in this way. Here we have 2000 people and we want to
put them into 150 groups. The distribution of the group size follows
the Gamma distribution with s
Ista Zahn wrote:
>
> options(width=n) is supposed to work, and does for me. I don't use Lyx
> though...
>
> -Ista
> On Tue, Nov 10, 2009 at 4:27 PM, Mark Connolly
> wrote:
>> I am not really sure where in the interactions this is handled, but I
>> would
>> like to keep echo-ed R code chunks f
options(width=n) is supposed to work, and does for me. I don't use Lyx though...
-Ista
On Tue, Nov 10, 2009 at 4:27 PM, Mark Connolly wrote:
> I am not really sure where in the interactions this is handled, but I would
> like to keep echo-ed R code chunks from running past the right margin and
>
On Nov 10, 2009, at 4:29 PM, Hongwei Dong wrote:
> Sorry for the confusion.
>
> Let me put it in this way. Here we have 2000 people and we want to
> put them into 150 groups. The distribution of the group size follows
> the Gamma distribution with shape parameter 0.067 and scale
> parameter
If the number of groups can be set "endogenously" my previous email about
the smallest `n' would apply. You can view this as a "waiting time"
problem. Here is one approach:
x <- round(pmax(1, pmin(rgamma(500, shape=0.067, rate=0.008), 85)))
csx <- cumsum(x)
ind <- which(csx > 2000)[1]
xg <- c
By the way, maybe the number of groups can be determined endogenously. It
will be better if I do not have to set the total number of groups
exogenously.
Thanks
Garry
On Tue, Nov 10, 2009 at 1:29 PM, Hongwei Dong wrote:
> Sorry for the confusion.
>
> Let me put it in this way. Here we have 2000
I think he was referring to the actual 'screen' command..
but, I digress... Is there any way you can put your commands in a script
and execute them from the command line so that you see the actual memory
and GC output from R in realtime? I use the following (in WinXP) to
debug any faulty proce
Hi Jack,
try stepAIC with trace parameter: stepAIC(...,trace=FALSE)
Regards,
M
Jack Luo a écrit :
Hi,
I am now running a cross-validation using coxph coupled with stepAIC for
model selection, is there anyway to suppress the output? It's too much.
-Jack
[[alternative HTML version d
Sorry for the confusion.
Let me put it in this way. Here we have 2000 people and we want to put them
into 150 groups. The distribution of the group size follows the Gamma
distribution with shape parameter 0.067 and scale parameter 0.008. At the
same time, the minimum group size is 1, and the large
I am not really sure where in the interactions this is handled, but I
would like to keep echo-ed R code chunks from running past the right
margin and off the page. I started with R and options(width=n), but
this does not seem to do anything (in the context of a document -- line
command works j
Hi,
I am now running a cross-validation using coxph coupled with stepAIC for
model selection, is there anyway to suppress the output? It's too much.
-Jack
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.e
I run R in gnome-terminal. Is it what you referred as 'screen session'?
On Tue, Nov 10, 2009 at 3:01 PM, wrote:
> This was happening to me on Red Hat Linux when I was running huge jobs within
> a screen session. By any chance are your R processes running within a screen
> session? (screen is a
Hi,
I'm running some routines with standard matrix operations like solve() and
diag().
When I do a profile, the lead item under total time is standardGeneric().
Furthermore, solve() and diag() have much greater total time than self time.
???
I assume there is some time-consuming decision going on i
Thank you Baptiste and Jim- I look forward to trying these ideas out
when I have a chance.
Mike
baptiste auguie wrote:
Hi,
>From what I understand of your code, you might find the following
construct useful,
funs <- c("mean", "sum", "sd", "diff")
x <- 1:10
lapply(funs, do.call, args=list(x))
This was happening to me on Red Hat Linux when I was running huge jobs within a
screen session. By any chance are your R processes running within a screen
session? (screen is a very nice program that will keep your sessions alive
after you log out, but it was killing off my big memory jobs for u
Thank you for the responses. The Lattice library is indeed useful for
producing the graphs in which I am interested, and I appreciate the
clarification between a function and the result of a function.
Ideally, I would like to be able to page through the graphs rather than
(or in addition to) ha
Hello, can someone help? How come
> gsub("\bINDS\b","INDUSTRIES","ADVANCED ENERGY INDS")
[1] "ADVANCED ENERGY INDS"
not ADVANCED ENERGY INDUSTRIES
Thanks.
Richard
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing li
Ok, I figured it out. My stupid mistake, should be \\b instead of \b.
From: Tan, Richard
Sent: Tuesday, November 10, 2009 3:36 PM
To: 'r-help@r-project.org'
Subject: gsub does not support \b?
Hello, can someone help? How come
> gsub("\bINDS\b","INDUSTRIE
On Nov 10, 2009, at 3:07 PM, soeren.vo...@eawag.ch wrote:
Quite often, I need those tables:
x <- sample(c("a", "b", "c"), 40, rep=T)
y <- sample(c("X", "Y"), 40, rep=T)
(tbl <- table(x, y))
(z <- as.factor(paste(as.vector(tbl), " (",
round(prop.table(as.vector(tbl)) * 100, 1), "%)", sep=""))
On Nov 10, 2009, at 3:07 PM, soeren.vo...@eawag.ch wrote:
Quite often, I need those tables:
x <- sample(c("a", "b", "c"), 40, rep=T)
y <- sample(c("X", "Y"), 40, rep=T)
(tbl <- table(x, y))
(z <- as.factor(paste(as.vector(tbl), " (",
round(prop.table(as.vector(tbl)) * 100, 1), "%)", sep=""))
May be you are interested in the first `n' for which the sum of iid gamma
rvs exceeds 2000, subject to the min-max constraints on each rv.
If so, the following one-liner will give it to you:
which(cumsum(pmax(1, pmin(rgamma(500, shape=0.067, rate=0.008), 85))) >
2000)[1]
Note that I have used a
Why did you transpose the dataframe (TData <- t(Data))? Is your data
in the same structure as is expected. Provide either the data file
you are plotting, or at least an 'str' of the object.
On Tue, Nov 10, 2009 at 3:06 PM, Koraelus wrote:
>
> Hello,
>
> Thank you very much. Your string makes pe
I think he means "rate = 0.008", so he is looking for:
rgamma(n, shape=0.067, rate=0.008)
Even then his problem is not well-posed. You cannot have both "independent"
gamma rv's and have them sum to 2000.
Ravi.
---
Hello,
Thank you very much. Your string makes perfect sense to me, but I get an
error when I try this:
Data<-read.csv("Datacull.txt",header=T,row.names=1)
TData<-t(Data)
PlotFunction<-function (x) {
par(mfrow=c(3,6))
for (i in colnames(x)) {
plot(x[[i]],type="o",axes=F,xlab='',ylab='',ylim=c(0,2
Quite often, I need those tables:
x <- sample(c("a", "b", "c"), 40, rep=T)
y <- sample(c("X", "Y"), 40, rep=T)
(tbl <- table(x, y))
(z <- as.factor(paste(as.vector(tbl), " (",
round(prop.table(as.vector(tbl)) * 100, 1), "%)", sep="")))
matrix(as.factor(z), nrow=3, dimnames=dimnames(tbl))
But
Peter -
I am using 2.8.1 on linux. When I use 2.10.0 on the pc, it works, so it
must be a version issue.
> sessionInfo()
R version 2.8.1 (2008-12-22)
x86_64-pc-linux-gnu
locale:
LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.U
TF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF
Ana María Prieto wrote:
>
> Hello.
> My name is Ana. I´m doing an eology master, and I´m just learning how R
> works.
> I have a Mac OS X 10.5.6, and I´m tryng to run just a simple ANOVA
> nanalyses.
> I dowloaded R version 2.10.0, and it seems I have problems with the
> script.
> I don´t know
Thanks.
I tried the rgamma function too. But I'm still wondering how I can set the
min, max, and sum of the variates created by the random draws. Anyone has a
clue? Thanks.
Garry
On Tue, Nov 10, 2009 at 11:47 AM, David Winsemius wrote:
>
> On Nov 10, 2009, at 2:26 PM, Hongwei Dong wrote:
>
>
On Nov 10, 2009, at 2:26 PM, Hongwei Dong wrote:
Exactly! Thanks, Duncan.
Let me re-phrase me question like this:
1) X_i values are independent Gammas, with the shape 0.067 and scale
0.008
2) Min(X)=1 and Max(X)=85
You might want to check that your parameterization in in agreement
with
Bierbryer, Andrew wrote:
Does anyone know why the following code hangs on the do.call, but works
fine when I either comment out the require(timeSeries) or only do 2
levels of a for loop instead of 3?
Thanks,
Andrew Bierbryer
require(timeSeries)
num <- 1
x.list <- l
Exactly! Thanks, Duncan.
Let me re-phrase me question like this:
1) X_i values are independent Gammas, with the shape 0.067 and scale 0.008
2) Min(X)=1 and Max(X)=85
3) SUM(X)=2000
4) Do I also have to define the number of draws? if yes, it could be 250.
Based on these restrictions, I want to ge
Hi,
>From what I understand of your code, you might find the following
construct useful,
funs <- c("mean", "sum", "sd", "diff")
x <- 1:10
lapply(funs, do.call, args=list(x))
and then working with lists rather than naming every object
individually. You might find mapply useful too when you have t
Here is a hint of how you might want to do the first part. You might
want to study the 'lappy' function
vegList <- lapply(c('bray', ..., 'binomial'), function(.method){
vegdist(x, method=.method)
})
clustList <- lapply(vegList, function(.dist){
lapply(c('average', ..., 'centroid'), funct
On 11/10/2009 1:25 PM, Hongwei Dong wrote:
Hi, Dear R users,
I'm wondering if I can do Monte Carlo Simulation in R. My problem is like
this: I know variable X follows Gamma distribution with shape parameter
0.067 and scale parameter 0.008. The sum of the X is 2000. I need R help me
to simulate a
On Tue, 10 Nov 2009, Duncan Murdoch wrote:
--interactive tells R that there is a human producing the input stream, so it
can ask questions and expect them to be answered. In your experiments with
it, your input stream was the pipe holding the output of echo, and R got
confused because that p
Dear All,
I wrote a function for cluster analysis to compute cophenetic correlations
between dissimilarity matrices (using the VEGAN library) and cluster
analyses of every possible clustering algorithm (SEE ATTACHED)
http://old.nabble.com/file/p26288610/cor.coef.R cor.coef.R . As it is now,
it i
Hi, Dear R users,
I'm wondering if I can do Monte Carlo Simulation in R. My problem is like
this: I know variable X follows Gamma distribution with shape parameter
0.067 and scale parameter 0.008. The sum of the X is 2000. I need R help me
to simulate a vector of X that satisfies both the probabil
On Nov 10, 2009, at 11:24 AM, Ana María Prieto wrote:
Hello.
My name is Ana. I´m doing an eology master, and I´m just learning
how R
works.
I have a Mac OS X 10.5.6, and I´m tryng to run just a simple ANOVA
nanalyses.
I dowloaded R version 2.10.0, and it seems I have problems with the
scri
On 11/10/2009 12:35 PM, Nathan Harmston wrote:
Hi,
I m currently working on interfacing R and C++ (passing a matrix from
R to C++, doing some stuff, returning a vector of results). There seem
to be a number of ways of doing this, such as rcpp and swig. Is there
a recommended way of doing this? I
Hello
On 11/10/09, Ana María Prieto wrote:
> When copying the script in the R console, it seems that there is a problem
> with the
>
> ~ symbol. this symbol is not in the keyboard, so I select it from spetial
> characters,
For what language is your keyboard designed? If it is for English,
sim
Hi!
One of my time series happens to be a constant. When I call arima with c(0,
0, 0), it gives me error. Here is an example:
ts1 <- ts(rep(1, 29))
fit <- arima(ts1, order=c(0,0,0))
When I run the second line above, it gives me the following error
"Error in decompose(ts(x[1L:wind], start = sta
On Nov 10, 2009, at 12:36 PM, Chuck White wrote:
df1 -- dataframe with column date and several other columns. #rows
>40k Several of the dates are repeated.
df2 -- dataframe with two columns date and index. #rows ~130 This
is really a map from date to index.
I would like to create a colum
On 11/10/2009 11:14 AM, Stefan Zeugner wrote:
Hello,
After hours of googling I could not resolve the following (although it
seems simple):
I would like to put subfunctions in a separate .R file that is then
called with source() from inside several "main" functions. A crude
example would be a
Nice problem!
If I understand you correctly, here's how to do it (with list-based matrices):
set.seed(1)
(x <- matrix(lapply(rpois(10,2)+1, function(k) sample(letters[1:10], size=k)), ncol=2,
dimnames=list(1:5,c("A","B"
A B
1 Character,2 Character,5
2 Character,2 Char
Hello,
After hours of googling I could not resolve the following (although it
seems simple):
I would like to put subfunctions in a separate .R file that is then
called with source() from inside several "main" functions. A crude
example would be as follows:
file subtest.R **
sub
Hi guys,
I am a totally begginer with R. I am planning to use R to design and
optimize my experiment. The experiment includes 4 factors, and three of the
fators have 3 levels, and the last factor has 6 levels. I am having a really
hard time to learn this program on my own and didn't find enough
Hello.
My name is Ana. I´m doing an eology master, and I´m just learning how R
works.
I have a Mac OS X 10.5.6, and I´m tryng to run just a simple ANOVA
nanalyses.
I dowloaded R version 2.10.0, and it seems I have problems with the script.
I don´t know what to do, I´ve already change the languages,
I have a plotted a stacked panel graph (single x axis and multiple y axis)
using the package zoo and would like to add a title for each separate panel.
I am using the script:
z <- with(mydata,zoo(cbind(mydata$Water.level,mydata$Submerged.plants,
mydata$Crayfish.CPUE,mydata$Carp.CPUE),Year))
plo
Hi,
I m currently working on interfacing R and C++ (passing a matrix from
R to C++, doing some stuff, returning a vector of results). There seem
to be a number of ways of doing this, such as rcpp and swig. Is there
a recommended way of doing this? In the long run I would like what I m
working on t
df1 -- dataframe with column date and several other columns. #rows >40k
Several of the dates are repeated.
df2 -- dataframe with two columns date and index. #rows ~130 This is really a
map from date to index.
I would like to create a column called index in df1 which has the corresponding
inde
On Tue, 10 Nov 2009, nana wrote:
I will be glad if someone can point out what I am doing wrong or not doing at
all in this.
Sending the same message to both r-help and r-devel is one thing you are doing
wrong.
I am trying to write out netcdf file in R. I have 26 time step but only the
fir
Does anyone know why the following code hangs on the do.call, but works
fine when I either comment out the require(timeSeries) or only do 2
levels of a for loop instead of 3?
Thanks,
Andrew Bierbryer
require(timeSeries)
num <- 1
x.list <- list()
for ( i in 1:10 ) {
On 10 Nov 2009, at 17:16, tlum...@u.washington.edu wrote:
On Tue, 10 Nov 2009, jim holtman wrote:
Have you tried something like this:
my.results = apply(chr, 2, function(x){
result <- try(anova(lrm( cpstc.f ~ x + time.cpstc + age + sex +
mri))[1,3])
if (inherits(result, "try-error")) r
On Tue, 10 Nov 2009, jim holtman wrote:
Have you tried something like this:
my.results = apply(chr, 2, function(x){
result <- try(anova(lrm( cpstc.f ~ x + time.cpstc + age + sex + mri))[1,3])
if (inherits(result, "try-error")) return(NULL)
result
})
This should catch the error and hav
On Tue, 10 Nov 2009, maiya wrote:
OK, it's the simple math that's confusing me :)
So you're saying 2.4GB, while windows sees the data as 700KB. Why is that
different?
Your data are stored on disk as a text file (in CSV format, in fact), not as
numbers. This can take up less space.
And let
Dear R users,
I will be glad if someone can point out what I am doing wrong or not doing at
all in this.
I am trying to write out netcdf file in R. I have 26 time step but only the
first time step is written.
For example:
>library(ncdf)
>path <- '/home/work/'
>forecast <- open.ncdf(paste(path
Hi,
I'm trying to determine the phase of irregularly sampled data. Is there any
particular reason why both spec.pgram and spec.ls return phase<-NULL for
vectors?
Thank you.
Lisandro
x
Lisandro Bene
Hi Peng,
in a very simplistic manner, what happens is that the Operating System
"thinks" it is too dangerous to let the R process to use so much
memory. So, to protect the whole system, it kills R, before the system
becomes unstable.
I've been looking at the problem you observed last week
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