On Aug 11, 2012, at 07:53 , Paul Johnson wrote:
> Greetings.
>
> I'm trying to understand a problem on a Dell Laptop. Details below,
> also uploaded the R working example that I pasted below.
The difference is not in locator() but in the circles, which in turn is due to
different aspect ratio
Hi all,
I have a data frame that has the columns OFB1, OFB2, OFB3,... OFB10.
How do I select the first 8 columns efficiently without typing each and
every one of them. i.e. I want something like:
a<-data.frame(initial_data$OFB1-10) #i know this is wrong, what would be
the correct syntax?
Thanks
On 10.08.2012 22:59, sharx wrote:
Hi, I have researched batch mode for windows and could not find anything that
worked.
I know the code should be
$ R CMD BATCH inputfile.R outputfile.Rout
or
R outputfile.Rout
Almost right, see below.
I tried these without success. I need detailed, step b
mydata[ , 1:8]
Or let's say you only one the 4th, 6th and 8th columns
mydata[ , c(4,6,8)]
and so on.
There are several good intro's available on the R website that will walk you
through this type of thing.
If you're a recovering SAS or SPSS user this paper may be of real help
www.et.bs.eh
Hi Sachin,
There are at least two ways. The safer way is to use a regular
expression to find the matching columns, like this:
a <- initial_data[grep("^OFB[0-9]+", names(initial_data))]
Alternatively, if you know that the columns you want are the first 8
you can select them by position, like this
I should have mentioned that I do not know the number index of the columns,
but regardless, thanks for the responses
On Sat, Aug 11, 2012 at 10:46 PM, Ista Zahn wrote:
> Hi Sachin,
>
> There are at least two ways. The safer way is to use a regular
> expression to find the matching columns, like
On Sat, Aug 11, 2012 at 8:51 AM, Sachinthaka Abeywardana
wrote:
> I should have mentioned that I do not know the number index of the columns,
> but regardless, thanks for the responses
Right, so use my first method. This does not depend on the position of
the columns.
Best,
Ista
>
>
> On Sat, A
As David Carlson mentioned earlier your script is working well for me.
sessionInfo()
R version 2.15.1 (2012-06-22)
Platform: i686-pc-linux-gnu (32-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_CA.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=e
Hi all,
I often run a bunch of code with Run Selection, and, after it's done
running, I find there have been like 20 errors, all due to an error that
occurred early in my code, which caused problems from there on. (I must then
scroll up through lots of code to find it.)
What I would like is for R
[1] Oh, that kind of duplicate... I thought you meant duplicate in that
someone else had already asked how to use R to analyze FITS images. Well
every time I post, I get the "Mailing List Subscription Reminder." I suppose
you want me to just click on "I'm already a subscriber" and just post here
th
Hi,
I want to simulate a data set with similar covariance structure as my
observed data, and have calculated a covariance matrix (dimensions
8368*8368). So far I've tried two approaches to simulating data:
rmvnorm from the mvtnorm package, and by using the Cholesky
decomposition
(http://www.cereb
On 11.08.2012 06:19, eliza botto wrote:
Dear R users,
Is there a way to apply a “stopping
rule” in hierarchical Clustering? I have a data and I want to find the optimal
number of clusters while doing hierarchical clustering. I would be deeply
obliged
if someone, whoever applied stopping rule
1. You probably need to tell us on what platform (OS, version) you are
running, as requested by the posting guide.
2. To me, it sounds like your question is: How can I avoid using the
error trapping tools built into R but still trap errors. If so, pretty
nonsensical, no?
3. But maybe someone else
Easy: Wrap your code into a function. A function by default exits on the
first error.
Best,
Uwe Ligges
On 11.08.2012 10:07, enocko wrote:
Hi all,
I often run a bunch of code with Run Selection, and, after it's done
running, I find there have been like 20 errors, all due to an error that
o
Sampling error? Do you realize how large a sample size you would
need to precisely estimate an 8000 x 8000 covariance matrix? Probably
exceeds the number of stars in our galaxy...
Numerical issues may also play a role, but I am too ignorant on this
aspect to offer advice.
Finally, this is reall
On 11.08.2012 04:55, Manish Gupta wrote:
How to write values on bars using mtext?
Grouped Bar Plot
counts <- table(mtcars$vs, mtcars$gear) barplot(counts, main="Car
Distribution by Gears and VS",xlab="Number of Gears",
col=c("darkblue","red"),legend = rownames(counts), beside=TRUE, horiz=TRUE)
Hi, thanks for the reply.
I am not assuming that the supplied covariance vector in any way
captures the 'true' covariance matrix of the population, but thats not
what I am after either. I just want to simulate data that has a
similar covariance as that covariance matrix. And the numbers are so
huge
I have a batch file to run R commands in a text file and an output file that I
use.
"C:\Program Files\R\R-2.13.2\bin\x64\R.exe" CMD BATCH
"C:\Users\Frank\Documents\R\Scripts\TLT_2012.txt"
"C:\Users\Frank\Documents\R\Scripts\TLT_2012.out"
PAUSE
I worked in an environment that you had to
There is no command "For" in R. It is "for" and R is case sensitive.
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-
On Aug 11, 2012, at 6:34 AM, andrej wrote:
>
> [2] I obtained R (actually R64) from the following website:
> http://cran.cnr.berkeley.edu/bin/macosx/
>
> I guess it didn't come with an 'sos' package. I assume it's here:
> http://cran.r-project.org/web/packages/sos/index.html
>
> and when I d
Actually you could find it with RSiteSearch("FITS") which comes with base R
as the first hit:
1. R: Read a single data set from a FITS file (score: 51)
Author: unknown
Date: Wed, 22 Feb 2012 19:25:09 -0500
Read a single data set from a FITS file Description Usage Arguments
Details Valu
Am 11.08.2012, 01:18 Uhr, schrieb Rui Barradas :
[...] The solution is obviously to write 'for', with lowercase 'f'.
Hope this helps,
Dear Mr Barradas,
indeed it did help, thank you very much!
Best regards
Holger van Lishaut
__
R-help@r-project.o
Am 11.08.2012, 01:18 Uhr, schrieb Rui Barradas :
[...] The solution is obviously to write 'for', with lowercase 'f'.
Hope this helps,
Dear Mr Barradas,
indeed it did help,
thank you very much!
Best regards
Holger van Lishaut
__
R-help@r-project.o
Alternatively, xts has a convenience function for this
.indexwday(SPY)
will give weekdays as numbers with Sunday being 0 and Saturday being 6.
There are also several similar functions
.indexDate(x)
.indexday(x)
.indexmday(x)
.indexwday(x)
.indexweek(x)
.indexmon(
Hello,
The situation is a bit odd.
If I open an R session and I paste there the script, then it does work.
However, if I save it as a text file and then I run it with the
source("myfile.R") command, then I fall back to the case that started this
thread i.e. I see nothing.
Does anybody have a
I have no experience with spplot() or maps in general, but might this
be more or less tied to R FAQ 7.22?
Michael
On Sat, Aug 11, 2012 at 12:00 PM, Lorenzo Isella
wrote:
> Hello,
> The situation is a bit odd.
> If I open an R session and I paste there the script, then it does work.
> However, if
Hello,
I'm not sure it works, but try the following.
for(j in which(dtp)){
for (q in 1:N){
if(y[q, j] %in% c("d", "D")) break
[...etc...]
and in the other loop the same,
for (j in which(!dtp)) {
for (q in 1:N) {
if(y[q, j] %in% c("d", "D")) break
[...etc...]
Em 10-08-2012
That is it. If you are going to use source() change
spplot(gadm, "language", col.regions=col, main="Swiss Language
Regions")
To
print(spplot(gadm, "language", col.regions=col, main="Swiss Language
Regions"))
---
David
> -Original Message-
> From: R. Michael Weylandt [mailto:mi
Dear R users,
I keep getting the message "Error: cannot allocate a vector of size 543.2 Mb"
on my 64 bit Linux machine with 4 Gb of memory. I studied the problem and it
occurs at the last lines of my code:
..
a <- laply(.);
save(a, file= "file.RData");
rm(a);
"a" is a 4-d arra
Hi,
I'm doing cross-correlation correlograms with the ncf package. I have four
study sites ; four correlograms.
I'd like to get the same y scale for the four of them... only, using
"ylim=c(-1,1)" does not change the y scale never and I don't know why. I
tried with plot() too.
Is there another
Hi
>i have this data
> X
[1]5.79 1579.52 2323.70 68.85 426.07 110.29 108.29 1067.60 17.05
22.66
[11] 21.02 175.88 139.07 144.12 20.46 43.40 194.90 47.307.74
0.40
[21] 82.859.88 89.29 215.101.750.79 15.933.910.27
0.69
[31] 100.58 2
To someone who may concern,
I want to continue the submission. of r-help list.
Thanks a lot!
Chen
--
Best Regards,
Yours,
Chen He
PhD Candidate
Institute of Population Research
Peking University, Beijing, China
[[alternative HTML version deleted]]
_
> > sos::findFn('fits astronomical')
>found 4 matches; retrieving 1 page
Alright I installed SOS, but in the line below "found 4 matches; retrieving
1 page" R spit out the following:
"Error in print.findFn(list(Count = 1, MaxScore = 1, TotalScore = 1, Package
= "FITSio", :
could not find fun
Hi Frederic
You definitely want to be using xmlParse() (or equivalently
xmlTreeParse( , useInternalNodes = TRUE)).
This then allows use of getNodeSet()
I would suggest you use Rprof() to find out where the bottlenecks arise,
e.g. in the XML functions or in S4 code, or in your code th
HI,
Try this:
dat1<-as.data.frame(matrix(rnorm(50,5),ncol=10))
colnames(dat1)<-paste0("OFB",1:10)
#to select first 8 columns - easy method
dat1[,1:8]
#2nd method
wanted<-paste0("OFB",1:8)
dat1[,colnames(dat1)%in%wanted]
#3rd method
#regular expression to select 3rd, 5th columns
dat1[grep("[[:al
It is a complex function, functions are quoted frequently, you may read from
bottom up
The independent variable for final fit is q
%%Rg0 is a function of L and b
Rg0sq<-function(L,b)L*b/6*(1-3/2*b/L+3/2*(b/L)^2-3/4*(b/L)^3*(1-exp(-2*L/b)))
%%alpha is a defined function
alpha<-function(x)(1+(x/3.
oops, Line 5 of the code should read:
glm2<-glm(lot2 ~ log(u), data=clotting, family=Gamma)
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Hello everybody,
I would like to replace or recode a list of numbers between 1 and 12 (total
100). I have tried to make it with recode, but i have two types of
replacements. For 1,2,3,4,11,12 => invierno and for 5,6,7,8,9 and 10 =>
verano.
recode(datos.mx1[,7], "1='invierno'; 2='invierno';
3='in
Hello,
I'm trying to impute data below detection limit (with multiple detection
limits)
so i need just a method or a code for imputation and then extract the
complete dataset to do the analyses.
Is there any package which could do that simply as i'm a beginner in R
Thank you
--
View this messa
Hello,
Try the following, using index vectors, not recode().
inx <- datos.mx1[, 7] %in% 5:10
datos.mx1[inx, 7] <- "verano"
datos.mx1[!inx, 7] <- "invierno"
Hope this helps,
Rui Barradas
Em 11-08-2012 20:10, Dominic Roye escreveu:
Hello everybody,
I would like to replace or recode a list of
Hello,
Post your data using dput(). Copy the output of dput(X) and paste it in
a post.
Anyway, you can try the following.
sum( leftaj <= X & X <= rightaj )
Hope this helps,
Rui Barradas
Em 11-08-2012 14:59, hafida escreveu:
Hi
i have this data
X
[1]5.79 1579.52 2323.70 68.85 4
If this is an option for you: An xml database can handle (very) huge xml
files and let you query nodes very efficiently.
Then, you could query the xml databse from R (using REST) to do your
statistics.
There are some open source xquery/xml databases available.
2012/8/11 Frederic Fournier
> Hell
HI,
Not sure whether this is what you wanted.
dat3<-c(5.79, 1579.52, 2323.70, 68.85, 426.07, 110.29, 108.29, 1067.60,
17.05, 22.66,
21.02, 175.88, 139.07, 144.12, 20.46, 43.40, 194.90, 47.30, 7.74, 0.40,
82.85, 9.88, 89.29, 215.10, 1.75, 0.79, 15.93, 3.91, 0.27, 0.69,
100.58,
HI,
I guess this should also work.
set.seed(1)
df1<-data.frame(colA=sample(1:100,n,replace=TRUE))
numbers1<-c(1,2,3,4,11,12)
numbers2<-c(5,6,7,8,9,10)
df1[df1$colA%in%numbers1,]<-"invierno"
df1[df1$colA%in%numbers2,]<-"verano"
#or
set.seed(1)
df1<-data.frame(colA=sample(1:100,n,replace=TRUE))
use 'dput' to enclose your data. I think 'cut' is what you want:
> dput(x)
c(0, 1.179, 3.729, 9.418, 18.01, 29.746, 58.662, 131.566, 2323.7
)
> dput(y)
c(5.79, 1579.52, 2323.7, 68.85, 426.07, 110.29, 108.29, 1067.6,
17.05, 22.66, 21.02, 175.88, 139.07, 144.12, 20.46, 43.4, 194.9,
47.3, 7.74, 0.4,
HI,
Try this:
set.seed(1)
df1<-data.frame(colA=sample(1:100,n,replace=TRUE))
numbers1<-c(1,2,3,4,11,12)
numbers2<-c(5,6,7,8,9,10)
df1$colA[df1$colA%in%numbers1]<-"invierno"
df1$colA[df1$colA%in%numbers2]<-"verano"
head(df1,10)
colA
1 27
2 38
3 58
4 91
5 21
6
Or if you want to use recode (which you forgot to mention is in package
car):
set.seed(1)
df1<- c(sample(1:12, 50, replace=TRUE)) # sample data
library(car)
recode(df1, "c(1, 2, 3, 4, 11, 12)='invierno'; 5:10='verano'")
--
David L Carlson
Associate Pro
Hi, thanks for the ideas, folks.
I'm on Windows 7, R 2.15.0 x64, RStudio 0.97.71.
I do appreciate your time... I would like to say my goal of dealing with
errors without R's error trapping tools is not nonsensical given that those
tools are cumbersome and not well-suited to the development phase
Look at the leaps function in the leaps package. It will compute the
Cp statistic which is a function of AIC.
On Thu, Aug 9, 2012 at 7:28 AM, zel7223 wrote:
> Hi,
>
> I want to use four independent variables to predict the output of one
> dependent variable using a linear model lm. I want to com
hello everyone,
i am getting problems in graph plotting. When i attach file after adding
color attributes in my data set. i got problem of "GlobalEnv" and masked
the followings. Like this
>attach(machm)
The following object(s) are masked _by_ '.GlobalEnv':
coll, sp
The following object(s) ar
The fill patterns date back to when the main way to get quality graphs
was using a pen plotter. Filling a rectangle with color using a pen
plotter took a long time and often resulted a soggy hole in the paper,
so the fill lines were preferred back then. Now with high resolution
screens and printe
stephenxqy wrote
>
> It is a complex function, functions are quoted frequently, you may read
> from bottom up
> The independent variable for final fit is q
>
> %%Rg0 is a function of L and b
> Rg0sq<-function(L,b)L*b/6*(1-3/2*b/L+3/2*(b/L)^2-3/4*(b/L)^3*(1-exp(-2*L/b)))
>
> %%alpha is a defined
Dear R People:
I have a column in a data frame that has date and time (one per
minute), such as "2012-08-09 22:23:00 2012-08-09 22:24:00" etc.
What is the best way to change that into numeric dates to be used as
an x-axis, please?
I've been looking at as.Date and as.Date.numeric, but those are f
Here it is:
zz <- strptime(zip.df$Time,"%Y-%m-%d %H:%M:%S")
> plot(zz,zip.df$TempC,type="l")
>
Woo Hoo!
Thanks though!
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
__
Hi List,
I find the following .so file in the R folder of my Arch Linux machine:
,--
| usr/lib/R/lib/libR.so
`--
I copied it to my home folder and changed permissons from root to my
personal user and then tried to call R functions from a different
languag
I am using the propagate function of the qpcR package to estimate the
standard error for an expression. This expression is simple enough
that I am able to calculate the first-order propagation of error,
which is what the documentation on propagate states that it does.
However, the results are not
HI,
I am plotting one pie chart and need to write % inside it. How can i write %
inside it as show in figure below.
http://r.789695.n4.nabble.com/file/n4640078/pie_chart.png
Regards
--
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