Hi,
I would like to plot a bunch of tree ring width data (time series)
using ggplots, but I'm having trouble figuring out how to do it.
My data is in a data.frame, with years as rownames and a distinct tree
ring series in each column. So, something like this:
rwl<-matrix(rnorm(800), nrow =
Hi all-
this seems to be simple to figure out but since im new to writing functions
I dont know what is happening. Here is my code along with the error I am
receiving:
semivario=function(data,ids,times,resids){
id=unique(data$ids)
index=combinations(length(data$times[data$ids==id[1]]),2)
time=ga
I think you should have a look at the 'proto' package on CRAN.
Yvonnick Noel
University of Brittany, Rennes
France
__
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.or
you are welcomed.
Steve
On Wed, Sep 15, 2010 at 6:11 PM, Suphajak Ngamlak <
supha...@phatrasecurities.com> wrote:
> Thank you so much. It works well
>
>
>
> Best Regards,
>
> Suphajak Ngamlak
> Equity and Derivatives Trading
> Phatra Securities Public Company Limited
> Tel: +662-305-9179
> Emai
I will get R-dev tomorrow, and give it a try. Where do I check out the svn?
thanks,
Stephen
On Wed, Sep 15, 2010 at 10:00 PM, Duncan Murdoch
wrote:
> I think this is fixed now. There were actually two bugs:
>
> I fixed an old one a few days ago, but my fix didn't handle the case of
> unsorted
Hello Chuan,
If you just want a matrix with the numbers 1 to 64 arranged by row...
m <- matrix(1:64, ncol=8, byrow=TRUE)
But perhaps I don't understand your question properly ?
Michael
On 16 September 2010 12:46, chuan zun liang wrote:
> Dear Prof:
>
> My name is Chuan.I from Malaysia.I am a
Dear Prof:
My name is Chuan.I from Malaysia.I am a beginner for R.I need favor for Prof.
This is my data:
y<-c(52,55,61,66,70,61,64,73,63,59,55,90,109,85,69,72,62,59,68,113,144,104,66,73,63,58,71,122,154,106,70,69,67,61,68,104,126,88,68,70,79,65,60,70,77,68,58,75,85,71,64,59,55,61,65,83,87,79,69,
I am looking for a way to find the roots of a non-polynomial
expression. I know R has a few ways to deal with polynomials, but, I
could not find a method that deals with functions involving e^(x) type
arguments, that have complex roots as well as real roots.
Any ideas?
Thanks,
Toros
___
Hi
On 13/09/2010 8:03 p.m., Benoit Boulinguiez wrote:
Hi all,
I'm still seeking for tweaking the appearance of the "color" legend in a
bar goemetry with ggplot2.
I can't seem to control the filling of the colour legend squares
take this,
ggplot(diamonds, aes(clarity, fill=color,colour = cut)) +
Hi Jane,
Try this:
split.screen( rbind(c(0, .8,0,1), c(.8,1,0,1)))
# [1] 1 2
# first screen split into a (2, 1) configuration
ind <- split.screen(c(2,1), screen=1)
screen(ind[1])
image(x, y, z, col = cols, zlim = zr)
screen(ind[2])
image(x, y, z2, col = cols, zlim = zr)
# Place the common, comple
I think this is fixed now. There were actually two bugs:
I fixed an old one a few days ago, but my fix didn't handle the case of
unsorted x properly. (I haven't checked whether the old code handled
that properly; I'd guess not, but it might have.) Now I've fixed my new
bug.
*Please* test
Hi all,
I want to put several figures in a one figure for easy comparison, so i
need to use the same methods to plot these figures. The following is an
example. I also list my method, but it does not work.
#Example data
x<- 1:10; y<- 1:10; z<- outer( x,y,"+");z2<- outer( x,y,"-")
#Quick view them
Thanks for your help!
My computer died, and I set up Sweave on another computer, on which it is
working perfectly. So perhaps I'll never be sure what the problem was on
the other machine.
On Tue, Sep 14, 2010 at 12:39 PM, Duncan Murdoch
wrote:
> On 14/09/2010 1:18 PM, Katie Surrence wrote:
>
>
Hi:
Why is phase nested within sites? Shouldn't sites and phase be crossed?
After all, there is one intervention (works) that purportedly affects the
river both upstream and downstream. Moreover, why is log(flow) being treated
as both a fixed and random effect?
Just curious,
Dennis
On Wed, Sep 1
On Sep 15, 2010, at 6:10 PM, Magnus Thor Torfason wrote:
Hi all,
I ran into a small issue when converting a list of vectors to a data
frame. The Issue I'm having is described by the snippet below:
#
# Convert a list of vectors into a
I have a dataset relating the effects of engineering works on the level of
salinity in a river before and after the works. I have modelled this using
linear mixed effects models to determine if the significance and level of the
response to the works. I am wanting to calculate the se of differe
On Sep 15, 2010, at 7:40 PM, Dennis Murphy wrote:
Hi:
Try something like this:
ind <- as.matrix(expand.grid(rownames(U), colnames(U)))
Uempty[ind] <- U[ind]
ind should be a two column matrix of (row, col) indices. Then just
replace
the values of Uempty associated with those indices with th
Kevin,
On 2010-09-15 16:37, Phil Spector wrote:
Kevin -
Here's one way:
z = boxplot(mydata$score,outline=FALSE,ylim=range(mydata$score))
text(1,z$out,SubNo[which(score == z$out)])
- Phil Spector
Statistical
Hi:
Try something like this:
ind <- as.matrix(expand.grid(rownames(U), colnames(U)))
Uempty[ind] <- U[ind]
ind should be a two column matrix of (row, col) indices. Then just replace
the values of Uempty associated with those indices with the corresponding
entries of U.
HTH,
Dennis
On Wed, Sep
Hi Nicolas,
Try (untested):
with(pop, table(factor(xloc, levels = 1:20), factor(loc, levels = 1:25)))
HTH,
Jorge
On Wed, Sep 15, 2010 at 7:31 PM, Nicolas Gutierrez <> wrote:
> Hi All,
>
> Im trying to superimpose (or "add") two matrices:
>
> 1. resulting from a table function with frequenci
Hi All,
I’m trying to superimpose (or "add") two matrices:
1. resulting from a table function with frequencies:
> U=table(pop$xloc, pop$yloc))
14 15 16 17 18 19 20 21 22
5 0 0 0 0 0 0 0 0 0
7 0 0 0 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0
Hi Richard,
Something like workf[ , columns] <- data.frame(lapply(workf[,
columns], as.numeric)) should do what you're after. However, this is
really a bit of a work around to the real problem. Can you provide
more details on the csv file you are reading in? Perhaps the first
couple rows or som
I would approach this slightly differently. I would make func a
function of x and y.
func <- function(x,y){
m <- median(x)
return(m > 2 & m < y)
}
Now generate tmp just as you have. then:
require(plyr)
res <- daply(tmp, .(z), summarise, res=func(x,y))
I believe this does the trick
A
On Sep 15, 2010, at 5:45 PM, Mark Ebbert wrote:
Dear R gurus,
I regularly come across a situation where I would like to apply a
function to a subset of data in a dataframe, but I have not found an
R function to facilitate exactly what I need. More specifically, I'd
like my function to ha
Hi:
Try this:
library(plyr)
func <- function(x, y) {
m <- median(x)
if(m > 2 & m < mean(y)) ret <- TRUE else ret <- FALSE
ret
}
ddply(tmp, .(z), summarise, r = func(x, y))
z r
1 a FALSE
2 b TRUE
3 c TRUE
HTH,
Dennis
On Wed, Sep 15, 2010 at 2:45 PM, Mark Ebbert wrote:
>
I'm reading in some data from a csv file, and it's reading in some of the
columns as character variables instead of numeric. I know I can fix this by
doing as.numeric for each of the columns, but the problem is I have a LOT of
different quantitative variables that I would have to do this for.
I've
Dear R gurus,
I regularly come across a situation where I would like to apply a function to a
subset of data in a dataframe, but I have not found an R function to facilitate
exactly what I need. More specifically, I'd like my function to have a context
of where the data it's analyzing came from
Hi all:
I have a problem when I want to do operation a sequence or matrix created by
loop and list() or data.frame(). Here is the example.
> for(i in 1:n) (k[i]=list(c(0:max[i])))
> k[1]+1
Error in k[1] + 1 : non-numeric argument to binary operator
What should I do to correct this pro
Hi Cristofer,
Try
as.numeric(strsplit(x, ",")[[1]])
HTH,
Jorge
On Wed, Sep 15, 2010 at 4:42 PM, Christofer Bogaso <> wrote:
> Hi, I have a string like "1,2,5,7,8".
>
> From this I want to create a numeric vector of length 5 (total number
> of numbers in above string), with each element will b
Kevin -
Here's one way:
z = boxplot(mydata$score,outline=FALSE,ylim=range(mydata$score))
text(1,z$out,SubNo[which(score == z$out)])
- Phil Spector
Statistical Computing Facility
SubNo[identify(rep(1,8),mydata$score)]
--
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology VOICE: (360) 407-6815
PO Box 47600FAX:(360) 407-7
How can I get the outlier in this boxplot of "Score" to be represented by
the corresponding value in "SubNo"?
score=c(6,6,7,14,5,7,6,8)
SubNo=1:8
mydata=data.frame(SubNo, score)
boxplot(mydata$score)
Thanks!
Kevin
[[alternative HTML version deleted]]
___
Hi all,
I ran into a small issue when converting a list of vectors to a data
frame. The Issue I'm having is described by the snippet below:
#
# Convert a list of vectors into a data.frame
strlen = 256
s.long.a = paste( letters[1+(0:strle
Chien-Pang -
Here's a *reproducible* example that should answer your question:
k = list()
n = 10
max = 10:19
for(i in 1:n) (k[i]=list(c(0:max[i])))
k[[1]] + 1
[1] 1 2 3 4 5 6 7 8 9 10 11
- Phil Spector
Hi all:
I have a problem when I want to do operation a sequence or matrix created by
loop and list() or data.frame(). Here is the example.
> for(i in 1:n) (k[i]=list(c(0:max[i])))
> k[1]+1
Error in k[1] + 1 : non-numeric argument to binary operator
What should I do to correct this pro
Thanks to all that helped. I had not compiled R myself, so it was rather
puzzling. I finally decided to re-install R from scratch. First I verified I
had
the right repositories listed and ran:
sudo apt-get update
sudo apt-get remove r-base
sudo apt-get autoremove
sudo apt-get install r-base
On 15 September 2010 at 13:48, Matias Salibian-Barrera wrote:
| I'm trying to install the package RcppArmadillo in my R 2.11.1 which I
installed
| and regularly update via Ubuntu's repositories.
We welcome questions on Rcpp et al on the rcpp-devel list. There is also the
r-sig-debian list for
On Wed, Sep 15, 2010 at 3:48 PM, Matias Salibian-Barrera
wrote:
> Hi there,
>
> I'm trying to install the package RcppArmadillo in my R 2.11.1 which I
> installed
> and regularly update via Ubuntu's repositories.
>
>
> When I try to install RcppArmadillo from CRAN I get:
>
>> install.packages('Rc
> x <- "1,2,5,7,8"
> as.numeric(unlist(strsplit(x, ',')))
[1] 1 2 5 7 8
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On
I have a matrix:
C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12
R1 0 0 0 0 0 5 0 0 0 0 0 0
R2 0 0 0 0 0 3 0 0 0 0 0 0
R3 0 0 0 0 0 0 4 0 0 0 0 0
R4 0 0 0 0 0 0 2 0 0 0 0 0
R5 1 0 0 0 0 0 0 0 0 0 0 0
R6 8 0 0 0 0 0 0
Try this:
x <- scan(textConnection("1,2,5,7,8"), sep = ",")
On Wed, Sep 15, 2010 at 5:42 PM, Christofer Bogaso <
bogaso.christo...@gmail.com> wrote:
> Hi, I have a string like "1,2,5,7,8".
>
> From this I want to create a numeric vector of length 5 (total number
> of numbers in above string), wi
Hi, I have a string like "1,2,5,7,8".
>From this I want to create a numeric vector of length 5 (total number
of numbers in above string), with each element will be the number in
above string. Is there any possibility to doing this?
Thanks and regards,
The fact that you are asking what a p-value is indicates that you do not have
enough of a background for us to be able to help you. It is not that we do not
want to help, just that anything we could say in an e-mail would likely do more
harm than good at this point.
You should either consult w
Hi there,
I'm trying to install the package RcppArmadillo in my R 2.11.1 which I
installed
and regularly update via Ubuntu's repositories.
When I try to install RcppArmadillo from CRAN I get:
> install.packages('RcppArmadillo', lib='~/myRlibs')
[...]
g++ -shared -o RcppArmadillo.so RcppArmad
x <-
c(-0.48,-0.48,-0.42,-0.26,0.58,0.48,0.47,0.54,0.5,0.52,0.52,0.56,0.58,0.61,0.68)
y <- c(0,0.2,0.4,0.6,0.8,1,1.2,1.4,1.6,1.8,2,2.2,2.4,2.6,2.8)
s <- approxfun(x[4:5], y[4:5], ties=mean)
s(0)
#This is the value that I want. The first zero crossing in the order
of y. #In other words in betw
thanks. Ravi and Nash.
I will read the new package and may use it after I am familiar with it. I may
bother both of you when I have questions.thanks for that in advance.
Nan
from Montreal
Hi Nan,
You can take a look at the "optimx" package on CRAN. John Nash and I wrote
this
package t
Diogo B. Provete wrote:
I have a data set and I want to procedure to model fitting (e.g., Poisson,
Gausian, binomial, quasipoisson etc.). I'd like to know if there is an
easier way to do this in R.
Easier than what ?
There is no shortage of R functions and packages to fit almost any type
of
Thanks and I'll strip the code down even more in future posts.
Stephen
On Wed, Sep 15, 2010 at 3:05 PM, Duncan Murdoch
wrote:
> On 15/09/2010 3:48 PM, stephen sefick wrote:
>>
>> Below is the code that I am using in a much larger function. I would
>> expect a bankfull measure at zero to be bet
I have a data set and I want to procedure to model fitting (e.g., Poisson,
Gausian, binomial, quasipoisson etc.). I'd like to know if there is an
easier way to do this in R.
Thank you in advance.
--
Atenciosamente,
Diogo Borges Provete
==
Biólogo
Mestre em Biologia A
On 15/09/2010 3:48 PM, stephen sefick wrote:
Below is the code that I am using in a much larger function. I would
expect a bankfull measure at zero to be between 0.6 and 0.8 approxfun
is returning 0.8136986. I am sure that I am missing something.
measure_bkf<- (structure(list(measurment_num =
sorry, a typo. the following code is an example from my computer and the real
one just has more variables.
the w[11] is the w[5] in the wden equation. the value of wden is calculated
based on the ifelse condition. it may equal to 1 when one is working; if not,
it
will equal to the value from
Hi,
I have a dataset with a response variable and multiple factors with more
than two levels, which I have been fitting using lm() or glm(). In
these fits, I am generally more interested in deviations from the global
mean than I am in comparing to a "control" group, so I use contr.sum()
as the fa
Below is the code that I am using in a much larger function. I would
expect a bankfull measure at zero to be between 0.6 and 0.8 approxfun
is returning 0.8136986. I am sure that I am missing something.
measure_bkf <- (structure(list(measurment_num = c(0, 0.2, 0.4, 0.6,
0.8, 1, 1.2,
1.4, 1.6, 1.8
On 9/15/10 10:38 AM, dadrivr wrote:
Hi everyone,
I am trying to make some publication-quality plots for use in Microsoft
Word, but I am having trouble creating high-quality plots that are supported
by Microsoft Word.
If I use the R plot function to create the figure, the lines are jagged, and
Wongsang,
Just to be clear R.utils is different than "utils"
As Henrik notes gunzip has been in R.utils ( see http://cran.r-project.org/)
for some time. It works
like a champ. R.utils is a great package.
On Wed, Sep 15, 2010 at 9:30 AM, Wonsang You wrote:
> Dear Henrik,
>
> Thank you so mu
Dear R Users
on a self-written function for calculating maximum likelihood probability (plz
check function code at the bottom of this message), one value, wden, suddenly
jump to zero. detail info as following:
w[11]=2.14
lnw=2.37 2.90 3.76 ...
regw=1.96 1.77 1.82
w
Hi:
This isn't the most elegant way, I'm sure, but here's one approach.
library(reshape)
# read in your data...I had to surround the text strings with quotes because
of the spaces
df <- read.table(textConnection("
TractID StandID Species CruiseDate DBHClassTreesPerAcre
'Carbon Stand
Class("person",representation(age="numeric",weight="numeric"))
[1] "person"
> bob <- new("person",age=30)
> is.null(b...@weight)
[1] FALSE
> b...@weight
numeric(0)
> b...@weight == numeric(0)
logical(0)
How can i test to see if a s4 property is assigned?
--
View this message in context:
http:/
That's good insight, and gives me some good ideas for what direction
to this. Thanks everyone !
Doug
P.S. - I guess if you have a significant interaction, that implies the
slopes of the individual regression lines are significantly different
anyway, doesn't it...
On Tue, Sep 14, 2010 at 11:33
Thanks Josh and Dan!
I got it to work using the following code that Dan had suggested.
x """D:\Program Files\R\R-2.9.2\bin\R.exe"" --no-save --quiet <
""&rsource.\Rtest.R"" > ""&rsource.\Rtest.log""";
Thanks for your help!! :)
Sarah
On Wed, Sep 15, 2010 at 2:58 PM, Joshua Wiley wrote:
> Hi
Hi Sarah,
Just a couple additional notes to what's been said:
1) It seems like R CMD BATCH might be easier
2) If the space in Program Files is causing issues, you can use the
Windows environment variable %PROGRAMFILES% (which also means you do
not need the drive letter.
3) I find it handy to defi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Sarah Jilani
> Sent: Wednesday, September 15, 2010 10:44 AM
> To: r-help@r-project.org
> Subject: [R] Sas to R
>
> Hi,
>
> I need to call an R program from Sas. I have tried usi
David Winsemius wrote:
On Sep 15, 2010, at 1:44 PM, Sarah Jilani wrote:
Hi,
I need to call an R program from Sas. I have tried using the following
code
in Sas using the x command but it just calls up dos and says
I went searching for a worked example and found this:
http://www.nesug.or
On Sep 15, 2010, at 1:44 PM, Sarah Jilani wrote:
Hi,
I need to call an R program from Sas. I have tried using the
following code
in Sas using the x command but it just calls up dos and says
I went searching for a worked example and found this:
http://www.nesug.org/proceedings/nesug08/sa/
Oh - and also have a look at the R graph gallery for inspiration:
http://addictedtor.free.fr/graphiques/
cu
Philipp
--
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihenstephan
Maximus-von-Imhof-Forum 3
85354 Frei
Sarah,
This is a SAS question, not R. However, it seems clear that it
has something to do with the fact that there are spaces in the
command that you're sending to Windows.
Maybe try calling with the 'short directory name' notation,
I forget what that's called in Windows.
Or else follow-up on
On Wed, Sep 15, 2010 at 12:22:15PM -0400, Ayyappa Chaturvedula wrote:
> Dear all, I am new to R and this group. I have good experience in S
> scripts. I need some orientation on data imports, general plotting
> functions. Can you please direct me?
Welcome to R. Coming from an S background you sho
Hi.
On Wed, Sep 15, 2010 at 9:30 AM, Wonsang You wrote:
> Dear Henrik,
> Thank you so much for your kind help. Unfortunately, I could not find out
> any function such as 'gunzip' in R.utils package.
Hmm... it's been there since at least 2005.
library("R.utils");
help("gunzip")
See the example.
Hi,
I need to call an R program from Sas. I have tried using the following code
in Sas using the x command but it just calls up dos and says
'D:\Program' is not recognized as an internal or external command, operable
program or batch file.
]
SAS CODE:
options xwait xsync;
%let Rsource=S:\EPI_Dat
R Users,
I am new to R and have tried to figure out how to automate this
process instead of using excel. I have read in this dataframe into r
with read.table. I need to reshape the data from the first table into
the format of the second table.
TractID StandID Species CruiseDate DBHClass
>> I'd prefer to stick with JPEG, TIFF, PNG, or the like. I'm not sure EPS
would fly.
Preferring to stick with bitmap formats (like JPEG, TIFF, PNG) is likely to
give you the jagged lines and other distortions you profess to want to
avoid.
EPS (encapsulated postscript, which handles vector+bitm
You might want to check out the Reproducible Research task view:
http://cran.r-project.org/web/views/ReproducibleResearch.html
There is a section on Microsoft formats, as well as other formats that
can be converted.
Max
On Wed, Sep 15, 2010 at 11:49 AM, Thomas Lumley
wrote:
> On Wed, 15 S
Dear Henrik,
Thank you so much for your kind help. Unfortunately, I could not find out
any function such as 'gunzip' in R.utils package. Instead, I could be
successful by using the following command.
system("gunzip filename")
On the other hand, the function 'gzfile' supports the compression as g
On Sep 15, 2010, at 12:23 PM, David Winsemius wrote:
On Sep 15, 2010, at 12:05 PM, Sean Parks wrote:
Hi,
I'm attempting to add a "Julian Day" column to a data frame.
Here is my code and the resulting data frame:
vic.data <- read.table("C:/VIC/data/vic.data.csv", header=F)
name
In an intervention study with subjects randomly allocated to two treatments
(treat A and B) and three time points (time) plus an additional baseline
measurement (dv_base), I've set up the following model to test for differences
in temporal courses of treatments for the outcome (dv), thereby allo
great! Thanks
- Original Message -
From: David Winsemius
To: raje...@cse.iitm.ac.in
Cc: r-help
Sent: Wed, 15 Sep 2010 21:12:27 +0530 (IST)
Subject: Re: [R] characters in a string
On Sep 15, 2010, at 11:16 AM, raje...@cse.iitm.ac.in wrote:
>
> Hi,
>
> I need to check if a string "a,b,c
On Sep 15, 2010, at 12:05 PM, Sean Parks wrote:
Hi,
I'm attempting to add a "Julian Day" column to a data frame.
Here is my code and the resulting data frame:
vic.data <- read.table("C:/VIC/data/vic.data.csv", header=F)
names(vic.data) <- c("year", "month", "day", "precip", "ev
Dear all, I am new to R and this group. I have good experience in S
scripts. I need some orientation on data imports, general plotting
functions. Can you please direct me?
Regards,Ayyappa Chaturvedula
__
R-help@r-project.org mailing list
https://s
Hi,
I'm attempting to add a "Julian Day" column to a data frame.
Here is my code and the resulting data frame:
vic.data <- read.table("C:/VIC/data/vic.data.csv", header=F)
names(vic.data) <- c("year", "month", "day", "precip", "evap",
"runoff", "baseflow", "Tsup", "SM1", "SM2",
On Wed, 15 Sep 2010, Ravi Varadhan wrote:
Dear Thomas,
You said, "the log-binomial model is very non-robust when the fitted values
get close to 1, and there is some controversy over the best approach."
Could you please point me to a paper that discusses the issues?
I have written some code to
On Wed, 15 Sep 2010, dadrivr wrote:
Thanks for your help, guys. I'm looking to produce a high-quality plot (no
jagged lines or other distortions) with a filetype that is accepted by
Microsoft Word on a PC and that most journals will accept. That's why I'd
prefer to stick with JPEG, TIFF, PNG,
On Sep 15, 2010, at 10:25 AM, dadrivr wrote:
>
> Thanks for your help, guys. I'm looking to produce a high-quality plot (no
> jagged lines or other distortions) with a filetype that is accepted by
> Microsoft Word on a PC and that most journals will accept. That's why I'd
> prefer to stick with
On Sep 15, 2010, at 11:16 AM, raje...@cse.iitm.ac.in wrote:
Hi,
I need to check if a string "a,b,c,d" is delimited by two
"" 's as efficiently as possible(I need to do this a lot of
times) and return TRUE. Can someone suggest a good technique?
> txt <- "a,b,c,d"
>
> grep("^.+$", txt)
[1
On Wed, Sep 15, 2010 at 11:25 AM, dadrivr wrote:
>
> Thanks for your help, guys. I'm looking to produce a high-quality plot (no
> jagged lines or other distortions) with a filetype that is accepted by
> Microsoft Word on a PC and that most journals will accept. That's why I'd
> prefer to stick w
Le 15/09/10 17:16, raje...@cse.iitm.ac.in a écrit :
Hi,
I need to check if a string "a,b,c,d" is delimited by two"" 's as
efficiently as possible(I need to do this a lot of times) and return TRUE. Can someone suggest a good technique?
Hi Rajesh,
> f <- function( x ) grepl( "^.*$", x )
> f(
On Sep 15, 2010, at 10:16 AM, raje...@cse.iitm.ac.in wrote:
>
> Hi,
>
> I need to check if a string "a,b,c,d" is delimited by two "" 's
> as efficiently as possible(I need to do this a lot of times) and return TRUE.
> Can someone suggest a good technique?
See ?grep and ?regex
> grepl("^.*$
Hi,
That seems to be the case. Is there a way that I can put an object in some
global place where all the workers can access it?
Doing the following at the start (before I make multiple workers) does not work:
Assign('global.control', control, globalenv())
Thanks for your help!
Kind regards,
G
DeaR all,
The stripchart function (graphics) is provides jittered and stacked
univariate scatterplots, but I wonder if anyone has implemented a
*symmetric* version of this - as in the lower panel of Wilkinson's paper:
http://www.jstor.org/stable/2686111
I have looked through several functions i
Thanks for your help, guys. I'm looking to produce a high-quality plot (no
jagged lines or other distortions) with a filetype that is accepted by
Microsoft Word on a PC and that most journals will accept. That's why I'd
prefer to stick with JPEG, TIFF, PNG, or the like. I'm not sure EPS would
f
On 15/09/2010 11:13 AM, Tonja Krueger wrote:
Hi all,
I used contour() to add contour lines to a plot. Now I’m wondering if there is
a way to get an output of the calculated x- and y- coordinates of the contour
lines?
?contourLines
Duncan Murdoch
___
Hi,
I need to check if a string "a,b,c,d" is delimited by two "" 's as
efficiently as possible(I need to do this a lot of times) and return TRUE. Can
someone suggest a good technique?
[[alternative HTML version deleted]]
__
R-help@r-project.
On Sep 15, 2010, at 11:13 AM, Tonja Krueger wrote:
Hi all,
I used contour() to add contour lines to a plot. Now I’m wondering
if there is a way to get an output of the calculated x- and y-
coordinates of the contour lines?
?contourLines # as was suggested to be the first "See(n) Also"
Thanks. It seems that both Rtools210.exe and Rtools211.exe support R 1.10.x,
which one is better for R 2.10.1.
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
Hi Henrique,
Thanks for your advice which works for me.
B.R.
Stephen L
From: Henrique Dallazuanna
Cc: R-help@r-project.org
Sent: Wed, September 15, 2010 10:24:01 PM
Subject: Re: [R] About choosing file
You can try tcltk:
library(tcltk)
tk_choose.files()
Hi all,
I used contour() to add contour lines to a plot. Now I’m wondering if there is
a way to get an output of the calculated x- and y- coordinates of the contour
lines?
Tonja
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That approach will be unique to OSX, upon which PDF is a default format. You
can copy and paste from a PDF document using Preview into Office or iWork or
similar apps. However, when subsequently displaying that content on a non-OSX
system, the content will be shown as a bitmap not as the vector
On Wed, Sep 15, 2010 at 10:38 AM, dadrivr wrote:
>
> Hi everyone,
>
> I am trying to make some publication-quality plots for use in Microsoft
> Word, but I am having trouble creating high-quality plots that are supported
> by Microsoft Word.
>
> If I use the R plot function to create the figure, t
There's many ways to solve this, but you are close to one already: Make the
pdf, put the cursor where you want it in the document, then on the menu bar
Insert --> Picture --> From File... And navigate to the file. This works on
the Mac, and seems to store the picture internally in a different way
Hi everyone,
I am trying to make some publication-quality plots for use in Microsoft
Word, but I am having trouble creating high-quality plots that are supported
by Microsoft Word.
If I use the R plot function to create the figure, the lines are jagged, and
the picture is not of high quality (sa
Hi
I do not want to go too much deep to internals of your function. What do
you suppose to get as the result.
If you want to get results of your function for a vector of reynolds and
dk you can use function outer and probably get rid of for cycle in the
function.
outer(c(100, 530,2410), c(10,
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