Hi
How does one tell R that one is using an augmented matrix as appose to an
non-augmented matrix?
For instance, when I want to solve two equations in two unknowns, I will
have to solve a 2 by 3 augmented matrix, however all I know to type into R
is something like this;
weights = c(1,2,3,4,5,6
Hi:
For a normal probability plot, use qqnorm().
Example:
x <- rnorm(50)
qqnorm(x)
qqline(x)
See the help page ?qqnorm for customization options.
HTH,
Dennis
On Tue, Jan 19, 2010 at 11:09 PM, Natalia Slobodina wrote:
> I am hoping to create a graph that will look like it is plotted on
>
I am hoping to create a graph that will look like it is plotted on
arithmetic probability plot. Something similar to how this is done in Excel:
http://peltiertech.com/Excel/Charts/ProbabilityChart.html.
I can't quite find a code that will transform the linear axis into a
cumulative percent axis wi
Dear R helpers
(I have already written the required R code which is giving me correct results
for a given single set of data. I just wish to wish to use it for multiple
data.)
I have defined a function (as given below) which helps me calculate Macaulay
Duration and Modified Duration and its
I am having the exact same error, and the suggested work-around does not
help.
My env is:
64-bit SUSE Linux v 10.1 on Xeon processors. gcc and gfortran version is
4.1.2
64-bit R 2.10.1 compiled from soure using CC="gcc -m64" etc.
64-bit unixODBC driver manager 2.2.14 compiled from source
Orac
On Jan 19, 2010, at 11:17 PM, Ron_M wrote:
Hi all, I have a expression like that :
rep(2,4)
[1] 2 2 2 2
> paste(rep(2,4), sep="", collapse="")
[1] ""
Now I want to write it as "" through some automated way. Is
there is
function for doing that? I have tested with paste(), but
Hi:
The : operator is meant for numeric sequences; see ?':'
On Tue, Jan 19, 2010 at 7:57 PM, rusers.sh wrote:
> Hi,
> I know we can use 1:10 to represent the 1,2,3,...,10 numbers, but the
> following conditions are except.
> Anybody knows how to represent the following two cases with similar
On Jan 19, 2010, at 10:57 PM, rusers.sh wrote:
Hi,
I know we can use 1:10 to represent the 1,2,3,...,10 numbers, but the
following conditions are except.
Anybody knows how to represent the following two cases with similar
usage
of ":" or others? Usually, i will get several hundred names fo
Hi all, I have a expression like that :
> rep(2,4)
[1] 2 2 2 2
Now I want to write it as "" through some automated way. Is there is
function for doing that? I have tested with paste(), but count not get any
desired result.
Thanks,
--
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Hi,
I know we can use 1:10 to represent the 1,2,3,...,10 numbers, but the
following conditions are except.
Anybody knows how to represent the following two cases with similar usage
of ":" or others? Usually, i will get several hundred names for them, such
as a1,a2,... or f[[1]],f[[2]],...
#Exa
plot(..., xaxs='i', yaxs='i')
On Tue, Jan 19, 2010 at 7:42 PM, Douglas M. Hultstrand
wrote:
> Hello,
>
> I am creating plots of hourly precipitation and accumulated precipitation
> (on different axis, see attached image). I was wondering how can I have the
> plot frame (black border) start at zer
I did not post this question to the ecology list, even though it is about an
analysis using Vegan, since my question is based more on basic data
manipulation in R. I think this is a fairly simple question but for the life
of me I cannot find a reference for in the archives.
I have an analysis
Thanks for the swift replies.
I have found this to work for my purpose:
data <- subset(X, !duplicated(X[,3])
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Sent from the R help mailing list
See:
?duplicated
On Tue, Jan 19, 2010 at 9:47 PM, Tuatara wrote:
>
> Hi everybody,
>
> I would like to delete rows based on duplicate entries in column 3 in the
> data matrix X (size 6 x 57). I have tried the unique(x) command as
>
>> data <- X[unique(X[,3]),]
>
> however, for some reason th
Use duplicated indeed of unique.
On Wed, Jan 20, 2010 at 12:47 AM, Tuatara wrote:
>
> Hi everybody,
>
> I would like to delete rows based on duplicate entries in column 3 in the
> data matrix X (size 6 x 57). I have tried the unique(x) command as
>
>> data <- X[unique(X[,3]),]
>
> however, fo
Hi everybody,
I would like to delete rows based on duplicate entries in column 3 in the
data matrix X (size 6 x 57). I have tried the unique(x) command as
> data <- X[unique(X[,3]),]
however, for some reason the command introduces a lot of NA's into the
dataset.
So, now I'm looking for a
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Douglas M. Hultstrand
> Sent: Tuesday, January 19, 2010 4:42 PM
> To: R mailing list
> Subject: [R] Plot frame border to start at zero?
>
> Hello,
>
> I am creating plots of h
Dear Seth,
> You might be able to reproduce this and get some more information like
> this:
>
>
>R -d gdb
>run
>source("src/tests/Examples/base-Ex.R")
>
> Assuming you get a crash, type bt in the gdb console and send output.
I think it may have ended up being related to different Fo
Thanks. I got it right now.
The reason is that I am using a 2.8.1 version of R, which seems causing
trouble with the installation.
After I updated my R to 2.10.x, everything works great now.
best,
senlin
2010/1/19 Uwe Ligges
>
>
> On 19.01.2010 03:42, Rolf Turner wrote:
>
>>
>> On 19/01/2010,
On Wed, Jan 20, 2010 at 8:18 AM, Douglas Bates wrote:
> On Tue, Jan 19, 2010 at 9:26 AM, Martin Maechler
> wrote:
>> Scanning for 'Matrix' in old R-help e-mail, I found
>>
>>> "GA" == Gad Abraham
>>> on Fri, 27 Nov 2009 13:45:00 +1100 writes:
>>
>> GA> Hi,
>> GA> I'd like to st
On Wed, Jan 20, 2010 at 2:26 AM, Martin Maechler
wrote:
> Scanning for 'Matrix' in old R-help e-mail, I found
>
>> "GA" == Gad Abraham
>> on Fri, 27 Nov 2009 13:45:00 +1100 writes:
>
> GA> Hi,
> GA> I'd like to store large covariance matrices using Matrix classes.
>
> GA> dsy
Hi:
Per Baptiste's suggestion (not the iapply one, the plyr one :)
> aaply(b, 1, function(x) sort(x, na.last = FALSE))
Var1 1 2 3 4 5 6
1 1 2 4 5 7 9
2 NA 2 3 3 3 5
HTH,
Dennis
On Tue, Jan 19, 2010 at 2:27 AM, marco salvini wrote:
> Can you please help on the issue?
> I using the appl
Hello R users,
I think, I have an easy problem, but I can't solve it. I have defined home
range size of my study animal by using kernelbb, but I couldn't find how to
see the data of home range polygons like size and used smoothing parameter.
Could anybody help me?
Thank you,
Mustafa
[
Hello,
I am creating plots of hourly precipitation and accumulated
precipitation (on different axis, see attached image). I was wondering
how can I have the plot frame (black border) start at zero, it looks
like it is plotted less than zero?
The code I use to create the png files is below:
On Tue, Jan 19, 2010 at 9:51 PM, Liviu Andronic wrote:
> Why would this be evil? For R, for example? I've already read some
> objections to this on r-help, but I'm not sure I understand the
> reasons. As long as the 'ping' happens once, at first start,
> anonymously, and requires confirmation fro
On Jan 19, 2010, at 5:06 PM, Henrik wrote:
Hi,
I try to do statistics on small samples (32 and 16 data points)
using the
bootstrap method. The results from the ordinary non-parametric
bootstrap
show significant impact from bias, which suggests the use of a
bias-corrected bootstrap method
Maybe bc(..., args = "") or perhaps install a more standard version of
bc. See links at http://r-bc.googlecode.com
On Tue, Jan 19, 2010 at 5:43 PM, kayj wrote:
>
> Hi ,
>
> Thanks for your help, I have been trying to use the bc package but I am
> getting the following error
>
> "Error in system(c
Yes. I am looking for them. Thanks.
2010/1/19 jim holtman
> ?cut
> ?split
>
> On Tue, Jan 19, 2010 at 3:18 PM, rusers.sh wrote:
> > Hi,
> > I just cannot remember the R function to divide a dataset into equal
> > parts. And i searched the "divide dataset into equal parts" in R site,
> but
> >
Hi ,
Thanks for your help, I have been trying to use the bc package but I am
getting the following error
"Error in system(cmd, input = input, intern = TRUE) : -l not found"
any idea??
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Hi,
I try to do statistics on small samples (32 and 16 data points) using the
bootstrap method. The results from the ordinary non-parametric bootstrap
show significant impact from bias, which suggests the use of a
bias-corrected bootstrap method. Is there any built in function for this in
R? If s
Héctor Villalobos ipn.mx> writes:
>
> Hi,
>
> I'm sure this one is very easy
>
> I am trying to write a function where one of its arguments has two posible
> (strings) values,
> defaulting to one of them if none is specified. My problem is that when
> evaluating the function
> the foll
Hi,
On 1/19/10 12:05 PM, Karl-Dieter Crisman wrote:
I work with the Sage project, and we are trying to improve the ability
to use R through Sage. Most things work, but make check seems to
cause problems on certain platforms, and now that we want to upgrade
to 2.10.1 I thought we should ask for
Tena koe Hector
In addition to Jim's comment, you might like to define your function as:
fun <- function(n, result = "simple")
This will set the default to 'simple' but one can still call it as
fun(result='complete').
HTH ...
Peter Alspach
> -Original Message-
> From: r-help-boun...
?cut
?split
On Tue, Jan 19, 2010 at 3:18 PM, rusers.sh wrote:
> Hi,
> I just cannot remember the R function to divide a dataset into equal
> parts. And i searched the "divide dataset into equal parts" in R site, but
> cannot find it.
> Anybody can tell me that function. I am just blocked by it
Look at what happens when you don't pass in 'result':
> c('simple', 'complete') %in% c("simple", "s", "complete", "c")
[1] TRUE TRUE
result is a vector of length two and provides two results and the 'if'
was only expecting one. You might have to debug your code some more.
You probably want to u
On 1/19/10, Barry Rowlingson wrote:
> In a similar vein, has anyone ever put any 'phone home' code in a
> package, so that authors can track usage? Something in the package
> startup code that pings a logging server, for example?
>
> Yes I know doing such a thing without telling the user and
Using data frame, a, from the post below this is how it would be done
in SQL using sqldf. We join together the original table, a, with a
table of minimums (computed by the nested select) and then choose only
the rows where dt - mindt < 7 (in the where clause).
> library(sqldf)
> sqldf("select va
Hi,
I'm sure this one is very easy
I am trying to write a function where one of its arguments has two posible
(strings) values,
defaulting to one of them if none is specified. My problem is that when
evaluating the function
the following warning is produced:
"the condition has length > 1
On Mon, Jan 18, 2010 at 1:54 PM, Bert Gunter wrote:
> One way to do it:
>
> 1. Convert your date column to the Date class using the as.Date() function.
> This allows you to do the necessary arithmetic on the dates below.
> dt <- as.Date(a[,4],"%d/%m/%Y")
>
> 2. Create a factor out of your first th
On Tue, Jan 19, 2010 at 9:26 AM, Martin Maechler
wrote:
> Scanning for 'Matrix' in old R-help e-mail, I found
>
>> "GA" == Gad Abraham
>> on Fri, 27 Nov 2009 13:45:00 +1100 writes:
>
> GA> Hi,
> GA> I'd like to store large covariance matrices using Matrix classes.
>
> GA> dsy
manchester.ac.uk> writes:
> [...]
>
> I suspect this is an invented computation -- the "3456" strikes
> me as "unlikely" (it reminds me of my habitual illustrative use
> of set.seed(54321)).
>
> There is a definite problem with the development given by kayj.
> When k=2000 and i=k, the formula
Thanks a lot for the answers!
Somehow I thought there will be a facility to modify the formula. But
setting unwanted variables to zero will of course work and maybe is simple
enough. Thanks again!
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On Tue, Jan 19, 2010 at 8:23 PM, David Winsemius wrote:
>
> On Jan 19, 2010, at 2:51 PM, Christophe Genolini wrote:
>
>> Hi the list
>>
>> Is there a way to know how many times an R package (on CRAN) has been
>> download ?
>
> No, or at least not a comprehensive number. The question came up and wa
On Jan 19, 2010, at 2:51 PM, Christophe Genolini wrote:
Hi the list
Is there a way to know how many times an R package (on CRAN) has
been download ?
No, or at least not a comprehensive number. The question came up and
was discussed in March last year. Search term "popular".:
http://fin
Hi,
I just cannot remember the R function to divide a dataset into equal
parts. And i searched the "divide dataset into equal parts" in R site, but
cannot find it.
Anybody can tell me that function. I am just blocked by it.
Thanks a lot.
--
-
Jane Chang
Queen's
[[al
Hello
On Tue, Jan 19, 2010 at 7:51 PM, Christophe Genolini
wrote:
> Is there a way to know how many times an R package (on CRAN) has been
> download ?
>
Recently this page [1] was set up, but it doesn't seem updated for
some time now.
Liviu
[1] http://neolab.stat.ucla.edu/cranstats/
___
Dear R Help,
I work with the Sage project, and we are trying to improve the ability
to use R through Sage. Most things work, but make check seems to
cause problems on certain platforms, and now that we want to upgrade
to 2.10.1 I thought we should ask for help!
R builds just fine on both Mac and
Hi the list
Is there a way to know how many times an R package (on CRAN) has been
download ?
Christophe
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posti
You could also try read.csv.sql in sqldf. See examples on sqldf home page:
http://code.google.com/p/sqldf/#Example_13._read.csv.sql_and_read.csv2.sql
On Tue, Jan 19, 2010 at 9:25 AM, wrote:
> I'm sure this has gotten some attention before, but I have two CSV
> files generated from vmstat and f
On 19-Jan-10 18:48:47, Gabor Grothendieck wrote:
> On Tue, Jan 19, 2010 at 1:41 PM, Ted Harding
> wrote:
>> On 19-Jan-10 17:55:43, Ben Bolker wrote:
>>> kayj yahoo.com> writes:
Hi All,
I was wodering if it is possible to increase the precision using R.
I ran the script below i
'x' is a matrix and not a dataframe. You should be doing
colnames(x) <- c("Date", "quarter")
x[,"Date"] <- as.Date(x[,"Date"])
It would help if you took a look at the structure you were using to
understand how to access. 'names' applied to a vector would give you
the output for 13000 more entr
This suggestion does not work. x seems to have twice the number of
entries as spl.
> x <- do.call(rbind, spl)
> names(x) <- c('Date', 'quarter')
> x$Date <- as.Date(x$Date)
Error in x$Date : $ operator is invalid for atomic vectors
> x$quarter <- as.numeric(x$quarter)
Error in x$quarter : $ operato
i'll post in r-forge vegan help forum and appreciate your help very much.
greetings,
kay
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___
I'm trying to compile R 2.10.1 on AIX 5.3, and am getting the following
error:
Error in read.dcf(file = descfile) :
Line starting 'Package: tools ...' is malformed!
Calls: makeLazyLoading ... code2LazyLoadDB -> loadNamespace ->
parseNamespaceFile -> read.dcf
Execution halted
make[3]: *** [all]
Charles C. Berry wrote:
On Tue, 19 Jan 2010, Charles C. Berry wrote:
and the values in those places are different:
On Tue, 19 Jan 2010, Barry Rowlingson wrote:
On Tue, Jan 19, 2010 at 1:36 AM, Charles C. Berry
wrote:
> Its the environment thing.
> > I think you want something like t
I read vmstat data in just fine without any problems. Here is an
example of how I do it:
VMstat <- read.table('vmstat.txt', header=TRUE, as.is=TRUE)
vmstat.txt looks like this:
date time r b w swap free re mf pi po fr de sr intr syscalls cs user sys id
07/27/05 00:13:06 0 0 0 27755440 13051648
On 19/01/2010, at 11:40 PM, Dieter Menne wrote:
next time you post homework here, please make sure that you modify the
language of the task a bit so that the discrepancy between the task
and your
helplessness is less evident.
This sounds like a fortune to me! How about it Prof.
On Tue, Jan 19, 2010 at 1:41 PM, Ted Harding
wrote:
> On 19-Jan-10 17:55:43, Ben Bolker wrote:
>> kayj yahoo.com> writes:
>>> Hi All,
>>>
>>> I was wodering if it is possible to increase the precision using R.
>>> I ran the script below in R and MAPLE and I got different results
>>> when k is lar
On 19-Jan-10 17:55:43, Ben Bolker wrote:
> kayj yahoo.com> writes:
>> Hi All,
>>
>> I was wodering if it is possible to increase the precision using R.
>> I ran the script below in R and MAPLE and I got different results
>> when k is large.
>> Any idea how to fix this problem? thanks for your hel
Mihai.Mirauta wrote:
>
>
> Could someone tell me, how can I select from a dataframe only those
> columns whose names contain a certain text?
>
> For example, if the column names are
> "Bond1.Creditclass","Bond1.Price","Bond2.Creditclass","Bond2.Price", how
> do I select only the columns corres
On Tue, 19 Jan 2010, Christian Hennig wrote:
are there any R-packages for computations required in sampling theury (such
as confidence intervals under random, stratified, cluster sampling; I'd be
partoculary interested in confidence intervals for the population variance,
which is difficult eno
Barry Rowlingson wrote:
On Tue, Jan 19, 2010 at 5:37 PM, Charles C. Berry wrote:
Note:
i <- 20
bquote(y ~ poly(x,.(i)))
y ~ poly(x, 20)
I see it now. bquote(y~poly(x,.(i))) gets it's 'i' there and then, sticks
it in the returned expression as the value '20', so any further evaluati
Whats wrong with Power Point or anyone of its equivalents?
Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034
steve_fried...@nps.gov
Office (305) 224 - 4282
Fax (305) 224 - 4147
On Tue, Jan 19, 2010 at 6:56 PM, Ista Zahn wrote:
> You might be able to do that with Rgraphviz or another R package, but
> if I was doing it I would probably use PGF/tikZ. The homepage is here:
> http://pgf.sourceforge.net/
I second that - gives you really good results.
Cheers,
Rainer
>
>
On Tue, 19 Jan 2010, Barry Rowlingson wrote:
On Tue, Jan 19, 2010 at 5:37 PM, Charles C. Berry wrote:
Note:
i <- 20
bquote(y ~ poly(x,.(i)))
y ~ poly(x, 20)
I see it now. bquote(y~poly(x,.(i))) gets it's 'i' there and then, sticks
it in the returned expression as the value '20', so
If you want to keep the same coefficients but ignore certain ones just
use 0 for the ones you don't want:
mod <- lm(Sepal.Length ~ Petal.Length + Petal.Width, iris)
predict(mod, list(Petal.Length = 3, Petal.Width = 0))
Regarding the problem with the example, the example data was not the
best for
kayj yahoo.com> writes:
>
>
> Hi All,
>
> I was wodering if it is possible to increase the precision using R. I ran
> the script below in R and MAPLE and I got different results when k is large.
> Any idea how to fix this problem? thanks for your help
>
> for (k in 0:2000){
> s=0
> for(i in
Hi there,
are there any R-packages for computations required in sampling theury
(such as confidence intervals under random, stratified, cluster sampling;
I'd be partoculary interested in confidence intervals for the population
variance, which is difficult enough to find even in books)?
Thank
Stats Wolf gmail.com> writes:
>
> Dear all,
>
> Consider a completely randomized block design (let's use data(Oats)
> irrespoctive of the split-plot design it was arranged in). Look:
>
> library(nlme)
> fit <- lme(yield ~ nitro, Oats, random = ~1|Block, method="ML")
> fit2 <- lm(yield ~ nitro
On Tue, Jan 19, 2010 at 5:37 PM, Charles C. Berry wrote:
>
>> Note:
>>
>> i <- 20
>>> bquote(y ~ poly(x,.(i)))
>>>
>> y ~ poly(x, 20)
>>
>>
I see it now. bquote(y~poly(x,.(i))) gets it's 'i' there and then, sticks
it in the returned expression as the value '20', so any further evaluations
get
See the Rgraphviz package in bioconductor.
On Tue, Jan 19, 2010 at 11:27 AM, Gavin Simpson wrote:
> Dear List,
>
> A student in the Department where I work would like to produce a graphic
> similar to this one:
>
> http://image.guardian.co.uk/sys-files/Guardian/documents/2009/09/16/Public_spendin
and the values in those
places are different:
On Tue, 19 Jan 2010, Barry Rowlingson wrote:
On Tue, Jan 19, 2010 at 1:36 AM, Charles C. Berry wrote:
Its the environment thing.
I think you want something like this:
models[[i]]=lm( bquote( y ~ poly(x,.(i)) ), data=d)
Use
terms
On Tue, 19 Jan 2010, Charles C. Berry wrote:
and the values in those places are different:
On Tue, 19 Jan 2010, Barry Rowlingson wrote:
On Tue, Jan 19, 2010 at 1:36 AM, Charles C. Berry
wrote:
> Its the environment thing.
>
> I think you want something like this:
>
> ? ? ? ? ? ? ? ?mo
Jim,
Did you read the posting guide?
Did you do a google search, for example, with terms like "[R] generalized
linear models", "[R] count models", "[R] poisson regression"?
I think you should do.
Walmes.
-
..oooO
..
Try this:
f <- tempfile()
download.file("http://nseindia.com/content/historical/EQUITIES/2010/JAN/cm15JAN2010bhav.csv.zip";,
f)
myData <- read.csv(unzip(f))
On Tue, Jan 19, 2010 at 2:56 PM, Velappan Periasamy wrote:
> How to unzip this file?.
>
>> mydata <-
>> unzip("http://nseindia.com/content
On Jan 19, 2010, at 12:05 PM, werner w wrote:
Thanks Gabor and Henrique!
Sorry for the imprecise question. I want predict() to use the
coefficients
estimated by the original regression but to exclude terms from the
prediction formula. If I originally estimated y ~ x1 + x2 and got
coefficie
Thanks Gabor and Henrique!
Sorry for the imprecise question. I want predict() to use the coefficients
estimated by the original regression but to exclude terms from the
prediction formula. If I originally estimated y ~ x1 + x2 and got
coefficients b0, b1, b2, I would like to remove x2 and predict
Werner,
You could set 0 to that regressors you don't want to consider for
prediction.
da <- expand.grid(A=1:20, B=rnorm(20, 4, 0.2), C=10^seq(1,2,l=20))
da$y <- rnorm(da$A, 0, 0.3)
m0 <- lm(y~A+B+C, data=da)
new <- da
new$C <- 0
predict(m0)[1:5]
predict(m0, newdata=new)[1:5]
At your disposal
Hi All,
I was wodering if it is possible to increase the precision using R. I ran
the script below in R and MAPLE and I got different results when k is large.
Any idea how to fix this problem? thanks for your help
for (k in 0:2000){
s=0
for(i in 0:k){
s=s+((-1)^i)*3456*(1+i*1/2000)^3000
}
}
Super, thanks a lot!! I didn't think about using names()
Ivan
Le 1/19/2010 17:35, Carlos Ortega a écrit :
> OK.
> For the names of the variables you can include this code in the loop
> (variable nv):
>
>
> seq.dat<-c(seq(7,10,1), seq(12,17,1))
> for( i in 1:length(seq.dat) ) {
>
> j<-seq.dat[i]
Dears useRs,
I have 2 factors, (for the sake of explanation - A and B), with 4 levels each.
I've already fitted a negative binomial generalized linear model to my data,
and now I need to split the factors in two distinct analysis of deviance table:
- A within B1, A within B2, A within B3 and A
MS PowerPoint (version 2007 or beta 2010) although difficult for so dense
graphic.
Prefearable: MindManager although is $$. Use the Trial.
Regards,
Carlos.
On Tue, Jan 19, 2010 at 5:27 PM, Gavin Simpson wrote:
> Dear List,
>
> A student in the Department where I work would like to produce a gra
You might be able to do that with Rgraphviz or another R package, but
if I was doing it I would probably use PGF/tikZ. The homepage is here:
http://pgf.sourceforge.net/
-Ista
On Tue, Jan 19, 2010 at 4:27 PM, Gavin Simpson wrote:
> Dear List,
>
> A student in the Department where I work would lik
How to unzip this file?.
> mydata <-
> unzip("http://nseindia.com/content/historical/EQUITIES/2010/JAN/cm15JAN2010bhav.csv.zip";)
Warning message:
In
unzip("http://nseindia.com/content/historical/EQUITIES/2010/JAN/cm15JAN2010bhav.csv.zip";)
:
error 1 in extracting from zip file
>
OK.
For the names of the variables you can include this code in the loop
(variable nv):
seq.dat<-c(seq(7,10,1), seq(12,17,1))
for( i in 1:length(seq.dat) ) {
j<-seq.dat[i]
nv<-names(ssfa)[j]
with( ssfa, twoplots(TO_POS, ssfa[[j]], nv) )
}
And this modification in the function (nm):
#defines
Hello,
I have a barchart. The y-axis represents counts and thex-axis is divided
into 10 equal intervals ranging fronm 0 to 0.1, 0.1 to 0.2, ..0.9 to
1.0.
Is there a way to model the counts in R?
thanks,
Jim
[[alternative HTML version deleted]]
_
Dear List,
A student in the Department where I work would like to produce a graphic
similar to this one:
http://image.guardian.co.uk/sys-files/Guardian/documents/2009/09/16/Public_spending_160909.pdf
Does anyone know if the figure in the pdf can be generated in a specific
software application fo
On Jan 19, 2010, at 10:21 AM, khaz...@ceremade.dauphine.fr wrote:
--
Hello all
My computer is MacBook and I want to draw a plot in R, for example for
x <- c(1,3,6,9,12)
y <- c(1.5,2,7,8,15)
I use this command plot(x,y).
Y
This recomputes the lm but if that is ok then:
mod <- lm(y1 ~ x1 + x2 + x3 + x4, anscombe)
mod$call$formula <- update(as.formula(mod$call$formula), ~ . - x1 - x2)
predict(eval(mod$call), list(x3 = 1, x4 = 1))
On Tue, Jan 19, 2010 at 11:10 AM, Werner W. wrote:
> Hi,
>
> probably just a quick qu
Thank you for your answer, I got the second part!
Ivan
Le 1/19/2010 17:03, Carlos Ortega a écrit :
> Hello,
>
> You can loop in the subset you need by storing in a variable and
> looping on that variable with indexes:
>
> seq.dat<-c(seq(7,10,1), seq(12,17,1))
> for( i in 1:length(seq.dat) ) {
>
Try this;
mod1 <- lm(y ~ u + v + w, data = d)
update(mod1, . ~ . -v)
On Tue, Jan 19, 2010 at 2:10 PM, Werner W. wrote:
> Hi,
>
> probably just a quick question: can I somehow change the formula used with
> predict? E.g., the regression was run on "y ~ u + v + w" but for the
> prediction the te
You should be able to set limits on memory use for a process in the operating
system, eg with limits or ulimits under Unix-alike shells.
-thomas
On Tue, 19 Jan 2010, Nathan Stephens wrote:
My group is working with datasets between 100 Mb and 1 GB in size, using
multiple log ins. From
Hi,
probably just a quick question: can I somehow change the formula used with
predict? E.g., the regression was run on "y ~ u + v + w" but for the prediction
the term v should be removed from the formula contained in the regression
object and only "y ~ u + w" be used.
I could use model.matrix
Hello,
You can loop in the subset you need by storing in a variable and looping on
that variable with indexes:
seq.dat<-c(seq(7,10,1), seq(12,17,1))
for( i in 1:length(seq.dat) ) {
j<-seq.dat[i]
with(ssfa, twoplots(TO_POS, ssfa[[j]]))
}
Regards,
Carlos.
On Tue, Jan 19, 2010 at 4:53 PM, Ivan C
Hi again!
I feel like I cannot do anything by myself but I would now like to plot
for all numeric variables I have (14 of them). I wanted to add a loop then.
The code is:
--
#defines the function for the plots (as written by Duncan Murdoch)
twoplots <- function(x, y) {
ylab <- deparse(subs
On Tue, 2010-01-19 at 06:19 -0800, Kay Cichini wrote:
> hello gavin,
>
> you are right, i didn't get into the documentation to deep and i'm also a
> beginner, that's why i'm just about to get into the logical part of the
> syntax.
> now, the output from perm.disp() says:
As I said, you *can't* d
I am running into a separate, but related issue. On Linux, one may impose
memory limits via the --max-vsize, --max-nsize, and --max-ppsize arguments
upon starting R. I do not know if similar arguments are available on
Windows. HTH
--
View this message in context:
http://n4.nabble.com/Server-hang
Scanning for 'Matrix' in old R-help e-mail, I found
> "GA" == Gad Abraham
> on Fri, 27 Nov 2009 13:45:00 +1100 writes:
GA> Hi,
GA> I'd like to store large covariance matrices using Matrix classes.
GA> dsyMatrix seems like the right one, but I want to specify just the
--
Hello all
My computer is MacBook and I want to draw a plot in R, for example for
x <- c(1,3,6,9,12)
y <- c(1.5,2,7,8,15)
I use this command plot(x,y).
but it dosn't work.
Could you please help me?
thank you
khazaei
My group is working with datasets between 100 Mb and 1 GB in size, using
multiple log ins. From the documentation, it appears that vsize is limited
to 2^30-1, which tends to prove too restrictive for our use. When we drop
that restriction (set vsize = NA) we end up hanging the server, which
requi
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