Thank you it works,
With best regards
Sudheer
On Wednesday, August 28, 2013, Pascal Oettli wrote:
> Hello,
>
> Use "zoo" instead.
>
> Regards,
> Pascal
>
>
> 2013/8/28 Sudheer Joseph 'sjo.in...@gmail.com');>>
>
>> Also, if that was the case, in below specification it should take the
>> timeserie
So am I crazy or does read.csv block concurrent operations?
On Wed, Aug 28, 2013 at 11:28 AM, joe meiring wrote:
> Yeah it seems like read.csv is blocking concurrent reads, can anyone
> verify if that is true? If so, is there a better way to read in csv files
> into a data frame?
>
>
> On Wed, A
Hi,
plot(strptime(data$Time,"%H:%M:%S"),data$Kbytes,pch=0,type="b",col =
"red", col.axis="red", ylab="", xlab="",las=2,lwd=2.5,cex.axis=1.5)
title("",cex.main=3,xlab="Seconds", line=5.2,ylab="Kbytes", cex.lab=2,1)
Hope I am not simplifying this in a bad way. These lines plot everything
properl
Hello again,
I was trying to install the rCharts package using following rule, however I
am getting error while doing that
> require(devtools)
Loading required package: devtools
WARNING: Rtools is required to build R packages, but no version of Rtools
compatible with R 3.0.1 was found. (Only the
On 08/29/2013 02:52 AM, Shane Carey wrote:
Hi,
Has anyone ever created scale breaks in R something like what is shown here
in the section,
Use a Scale Break
http://www.r-bloggers.com/graphing-highly-skewed-data/
Thanks
Hi Shane,
As Sarah answered, axis.break in the plotrix package is a start
Thanks for all the help! I was able to get all the packages to run now.
On Wed, Aug 28, 2013 at 6:25 AM, Ista Zahn wrote:
> On Tue, Aug 27, 2013 at 8:07 PM, Pascal Oettli wrote:
> > Hello,
> >
> > "graph" is a Bioconductor package.
>
> you can install it from
> http://www.bioconductor.org/pack
Hello,
In the absence of a commented, minimal, self-contained, reproducible code
(as kindly requested), it is hard to help you.
Regards,
Pascal
2013/8/29 Wildgruber, Christoph U.
> Hi,
>
> I started evaluating the 'Peaks' package a couple of months ago and found
> it to be quite
> useful. Get
On Wed, Aug 28, 2013 at 7:44 PM, Steve Lianoglou
wrote:
> Hi,
>
> On Wed, Aug 28, 2013 at 3:58 PM, Ista Zahn wrote:
>> Or go all the way and put
>>
>> options(stringsAsFactors = FALSE)
>>
>> at the top your script or in your .Rprofile. This will prevent this
>> kind of annoyance in the future wit
Hi,
On Wed, Aug 28, 2013 at 3:58 PM, Ista Zahn wrote:
> Or go all the way and put
>
> options(stringsAsFactors = FALSE)
>
> at the top your script or in your .Rprofile. This will prevent this
> kind of annoyance in the future without having to say stringsAsFactors
> = FALSE all the time.
I go ba
Or go all the way and put
options(stringsAsFactors = FALSE)
at the top your script or in your .Rprofile. This will prevent this
kind of annoyance in the future without having to say stringsAsFactors
= FALSE all the time.
Best,
Ista
On Wed, Aug 28, 2013 at 5:19 PM, arun wrote:
> Hi,
> Try:
> i
Hi Robert,
Your legend is for fill, not color, so you need
guides(fill = guide_legend(nrow = 3))
instead of
guides(colour = guide_legend(nrow = 3))
Best,
Ista
On Wed, Aug 28, 2013 at 5:09 PM, Robert Lynch wrote:
> I am having trouble getting my legend to format the way I want it to. I
> sus
duh!
Ivo Welch (ivo.we...@gmail.com)
http://www.ivo-welch.info/
J. Fred Weston Professor of Finance
Anderson School at UCLA, C519
Director, UCLA Anderson Fink Center for Finance and Investments
Free Finance Textbook, http://book.ivo-welch.info/
Editor, Critical Finance Review, http://www.cri
On Wed, Aug 28, 2013 at 4:32 PM, ivo welch wrote:
> is it possible to temporarily change the destination environment where
> objects are written to? I am thinking
>
> a <- new.env()
> attach(a)
> ### run some code, such as...
> b <- function(x) x
> detach(a)
> a$b
>
> obviously, this
Hello,
If you want to perform a chi suqare test, you don't need to make a
contingency table. From the help page for ?chisq.test, section Details:
"If x is a matrix with at least two rows and columns, it is taken as a
two-dimensional contingency table"
So all you need is to pass a two column
Good Afternoon,
My name is Gabriel, I'm doing an analysis if there is increase or decrease in
dependence on the mutated genes, using 3 or more genes using the fisher exact
test.I performed with success an analysis for two genes using fisher.test( ).
example of the 2x2 contigency table:
Good Afternoon,
My name is Gabriel, I'm doing an analysis if there is increase or decrease in
dependence on the mutated genes, using 3 or more genes using the fisher exact
test.I performed with success an analysis for two genes using fisher.test( ).
example of the 2x2 contigency table:
A relatively concise, commented, working solution to the problem
originally motivating
this thread was found (below). I suspect the approach I've taken has a
major inefficiency through the use of the "scan" statement appearing inside
the function "g". The way the code works right now, it has to r
is it possible to temporarily change the destination environment where
objects are written to? I am thinking
a <- new.env()
attach(a)
### run some code, such as...
b <- function(x) x
detach(a)
a$b
obviously, this is wrong. attach() only attaches for read access. I could
copy the gl
Yeah it seems like read.csv is blocking concurrent reads, can anyone verify
if that is true? If so, is there a better way to read in csv files into a
data frame?
On Wed, Aug 28, 2013 at 8:50 AM, joe meiring wrote:
> I'll be that one does give me 1 second. What I'm really trying to do is
> proces
Hi,
I started evaluating the 'Peaks' package a couple of months ago and found it to
be quite
useful. Getting back to it last week I had to set up my R environment due to
hardware
changes again. The Peaks package loads with no problem.
After successfully reinstalling all packages (RedHat 4.4.7-
On 29/08/13 05:03, Sarah Goslee wrote:
Yes. Using my rudimentary telepathic powers, I suppose that you also
want to know how to do it, not just whether it has been done. In that
case, perhaps you should look at axis.break() from the plotrix package.
(In response to the question, from S
Hi,
I am a beginner at R and do not have a strong background in statistics. I hope
somebody can help me with this. I have a data frame which looks like this
Facial.Type North.Indians South.Indians
1 Leptoprosopic96 115
2 Hyperleptoprosopic 189
Thanks a lot. That worked perfectly.
Best,Farnoosh Sheikhi
Sent: Wednesday, August 28, 2013 10:37 AM
Subject: Re: list to data frame
If the expected result is data.frame()
as.data.frame(t(as.data.frame(lapply(dat1,sum# would be data.frame. But,
not
On 28/08/13 20:17, Prof Brian Ripley wrote:
[That is Microsoft, whose messages are infamously accurate but
maximally uninformative.]
Fortune nomination.
cheers,
Rolf
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mail
Hi,
Try:
iris1<-data.frame(iris,annot=c(""),stringsAsFactors=FALSE)
str(iris1)
#'data.frame': 150 obs. of 6 variables:
# $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
# $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
# $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7
Hi Ed,
Because that's the default for data.frame(), and a lot of people trip
over it just as you did. It's an easy fix:
R> data(iris)
R> iris <- data.frame(iris, annot=c(""), stringsAsFactors=FALSE)
R> iris[1,"annot"]<-"annotation"
Thanks for the very useful reproducible example.
Sarah
On Wed,
I am having trouble getting my legend to format the way I want it to. I
suspect it is something simple.
>
> the code I have is
> library(ggplot2)
> ggplot(Chem.comp, aes(Course, GRADE.)) + geom_boxplot(notch =
> TRUE,aes(fill = COHORT))+
> labs(title = "Comparison between ISE cohorts and Peers
See
>?data.frame
In particular, the stringsAsFactors parameter which defaults to TRUE.
Hope this is helpful,
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
I have a table, and I want a new column to add some annotations to.
But it ends up as a factor instead of characters, and won't let me add
arbitrary text.
> data(iris)
> iris<-data.frame(iris,annot=c(""))
> iris[1,"annot"]<-"annotation"
Warning message:
In `[<-.factor`(`*tmp*`, iseq, value = "anno
I have constructed a GAM using the package mgcv to test whether the lengths of
an emerging insect (Length) varies with day of the year (DOY) and between two
sites (SiteCode). The data are collected at irregular time steps ranging from 2
days to 20 days between samples. The GAM takes the form
M3
thanks
On Wed, Aug 28, 2013 at 11:58 AM, Greenberg, Jonathan wrote:
> Hi Saptarshi:
>
> There are quite a few parallel mapply's out there -- my recommendation is
> to use the foreach package, since it allows you to be flexible in the
> parallel backend, and you don't have to write two statements
What you need to do is to create the plot without an x-axis (xaxt =
'n') and then add your own values on the axis with 'axis'
x <- read.table(text = " Time Kbytes RSS Dirty_Mode
1 11:42:02 2691296 15997961582736
2 11:43:42 2691396 15998041582744
3 11:45:22 2691496 1599804158
Hi Saptarshi:
There are quite a few parallel mapply's out there -- my recommendation is to
use the foreach package, since it allows you to be flexible in the parallel
backend, and you don't have to write two statements (a sequential and a
parallel statement) -- if a parallel backend is running,
Hello,
I find Map to be nice interface to mapply. However
Map calls mapply which in turn calls mapply via .Internal.
Is there a parallel version of mapply (like mcapply) or do I need to write
this myself?
Regards
Saptarshi
[[alternative HTML version deleted]]
___
On Aug 28, 2013, at 7:35 AM, Donald Catanzaro wrote:
> Hi All,
>
> Since R is used primarily as a statistical package by individuals using
> statistics to answer questions it makes sense to ask a question such as
> this. Oftentimes individuals have encountered similar issues as the poster
> and
On Wed, Aug 28, 2013 at 1:35 PM, Bert Gunter wrote:
> Sarah et. al.:
>
> Heh heh.
>
> However .. my always fallible judgment says, don't do it. Axis scale breaks
> invite misreading, Consider alternatives:
> http://stats.stackexchange.com/questions/1764/what-are-alternatives-to-broken-axes
Well,
Sarah et. al.:
Heh heh.
However .. my always fallible judgment says, don't do it. Axis scale breaks
invite misreading, Consider alternatives:
http://stats.stackexchange.com/questions/1764/what-are-alternatives-to-broken-axes
Cheers,
Bert
On Wed, Aug 28, 2013 at 10:03 AM, Sarah Goslee wrote:
>
Homework? We don't do homework here...
-- Bert
On Wed, Aug 28, 2013 at 10:02 AM, Edoardo Baldoni wrote:
> Dear R-help,
>
> I would like to ask you how to find the critical points of a function of
> several variables such as the following one:
>
> x = seq(-10,10,0.2)
> y = seq(-10,10,0.2)
> z
Hi,
set.seed(249)
dat1<- as.data.frame(matrix(sample(1:20,40*20,replace=TRUE),ncol=20))
#Based on your code:
vecNew<-sort(t(as.data.frame(lapply(dat1,sum))),decreasing=TRUE)[1:10]
#or
vec1<-sort(unlist(lapply(dat1,sum),use.names=FALSE),decreasing=TRUE)[1:10]
identical(vec1,vecNew)
#[1] TRUE
#
On Wed, Aug 28, 2013 at 12:52 PM, Shane Carey wrote:
> Hi,
>
> Has anyone ever created scale breaks in R something like what is shown here
> in the section, Use a Scale Break,
> http://www.r-bloggers.com/graphing-highly-skewed-data/
Yes.
Using my rudimentary telepathic powers, I suppose that yo
Dear R-help,
I would like to ask you how to find the critical points of a function of
several variables such as the following one:
x = seq(-10,10,0.2)
y = seq(-10,10,0.2)
zfunc = function(x,y) x^2 + y^2
z = outer(x,y,zfunc)
persp(x,y,z,theta = 30,phi=1,ticktype='detailed')
Thanks
Edoardo
Hi,
Has anyone ever created scale breaks in R something like what is shown here
in the section,
Use a Scale Break
http://www.r-bloggers.com/graphing-highly-skewed-data/
Thanks
--
Shane
[[alternative HTML version deleted]]
__
R-help@r-projec
On Aug 28, 2013, at 8:14 AM, Bala Chand wrote:
> Hi
> can you please give guidance for generating future prediction values from
> existing heart disease data and we want to know the which type of models
> are using for generating the future prediction. can you please tell me how
> can i write th
Sorry, I found what I was doing wrong!!
Thanks
On Wed, Aug 28, 2013 at 5:09 PM, Shane Carey wrote:
> Hi,
>
> I am trying to create a break on the y-axis of a boxplot using axis.break.
> My data range from 4 to 12 with a break at 11.
>
> I have tried the following code:
>
> A<-c(4,5,6,7,8,9,10,
Phil, sorry; I didn't see your response. You are right; the "IS" is
superfluous
On Wednesday, August 28, 2013 8:56:19 AM UTC-7, Alex Gilgur wrote:
>
> "having" is right; use "HAVING Premie IS NOT NULL" instead. The sqldf
> package has a SQLite database running behind it. All NA get internally
Hello R-Users,
I am using glmmPQL (library MASS) on time-series count data that are aggregated
by zipcode. My model includes natural cubic splines for season and day-of-week
as fixed effects and random intercept term for zipcode. I need to extract the
fitted values AND the standard error of the
"having" is right; use "HAVING Premie IS NOT NULL" instead. The sqldf
package has a SQLite database running behind it. All NA get internally
converted to NULL, which is the standard representation for N/A in SQL, and
then they become in the data.frame that is returned by the sqldf
command
C
Hi
can you please give guidance for generating future prediction values from
existing heart disease data and we want to know the which type of models
are using for generating the future prediction. can you please tell me how
can i write the formula for future predictive values.
can you see th
I'll be that one does give me 1 second. What I'm really trying to do is
process a bunch of (~300) relatively large CSV files simultaneously.
Something like this:
testFunc <- function(fileName) {
print(fileName)
df <- process_raw_data(fname)
# process_raw_data just reads in CSV file and
Hi,
I am trying to create a break on the y-axis of a boxplot using axis.break.
My data range from 4 to 12 with a break at 11.
I have tried the following code:
A<-c(4,5,6,7,8,9,10,12)
B<-c(4,5,6,7,8,9,10,12)
axis(2,A,B)
Where A and B are the tick marks and labels respectively?
However the plot w
Thank you so much, Pascal. Works perfectly! :)
On Tue, Aug 27, 2013 at 8:14 PM, Pascal Oettli wrote:
> Hello,
>
> The following works for me:
>
> library(Hmisc)
> timeseries<-c(1950,2000,2050,2100)
> dataseries<-seq(1:4)
> dataseries1<-c(1,2,3,4)
> dataseries2<-c(1.5,2.5,3.5,4)
> plot(timeserie
Dear list,
I am currently working with presence/absence GLM. Therefore I am using
binomial family and selection my models this way :
null <- glm(respvarPAT ~ 1 , family = binomial, data = datafit)
full <- glm(respvarPAT ~ CSpp + FSpp + Gpp + Mpp + Ppp + Lpp + TempPoly2
+ DepthPoly2 + DepthPo
I would tend to go for very basic approach.
Since there is no column of data named 'year' I will assume that the row
names contain the year.
min( row.names(mydata)[mydata$samp.depth==2] )
Although it's not pretty, I think it comes close to representing in R the
language the request as presented i
R users,
The deadline for poster abstracts is tomorrow, August 29, for the 2014 ASA
Conference on Statistical Practice, February 20-24, Tampa, Florida, USA.
Based on a survey of the 2013 CSP attendees there was particular interest
in R. It would be nice to see contributions from the R community
You appear to have no idea what you are doing. I suggest you post on
the Bioconductor list, not here, for guidance. Better yet, get help
locally from someone who does know what they are doing.
Cheers,
Bert
On Wed, Aug 28, 2013 at 6:36 AM, Robin Mjelle wrote:
> Hi,
>
> I have two data frames with
Hi All,
Since R is used primarily as a statistical package by individuals using
statistics to answer questions it makes sense to ask a question such as
this. Oftentimes individuals have encountered similar issues as the poster
and can quickly point them in the correct direction - that is all I am
SO what could possibly be causing this? Has anyone encountered this? This
is RedHat Enterprise 5.7.
On Tue, Aug 27, 2013 at 11:14 PM, Prof Brian Ripley
wrote:
> On 28/08/2013 06:54, joe meiring wrote:
>
>> This does speed up on an OsX install, so something must be wacky with the
>> linux install
Hi,
Does anyone why the following code produces different results?
a <- cbind(1:10,1:10)
b <- a
colnames(a) <- c("a","b")
colnames(b) <- c("c","d")
colnames(cbind(a,b))
> [1] "a" "b" "c" "d"
colnames(cbind(ts(a),ts(b)))
> [1] "ts(a).a" "ts(a).b" "ts(b).c" "ts(b).d"
Is this or compatibility r
Hi,
I have two data frames with time serie datamatrix. I want to pick a row X
from the first matrix and see if it correlates with row Y in the second
matrix. These are gene expression values and I probaly need to do some
scaling first, but I wonder if you have any suggestions on how to do the
corr
This is not an R question. Post elsewhere. Better yet, contact a local
statistical consultant.
-- Bert
On Wed, Aug 28, 2013 at 7:18 AM, Donald Catanzaro wrote:
> Good Day All,
>
> I am working with a diagnostic test and comparing the new test to an old
> test. Normally I would be able to calcul
Good Day All,
I am working with a diagnostic test and comparing the new test to an old
test. Normally I would be able to calculate sensitivity and specificity
quite easily.
However, the 'gold standard' that I am comparing my new diagnostic with is
really 'gold-plated' in that sometimes the 'gold
Toth, Denes ttk.mta.hu> writes:
>
>
> Sorry, I missed to attach the sessionInfo, here it goes:
>
> > sessionInfo()
> R version 3.0.1 (2013-05-16)
> Platform: x86_64-pc-linux-gnu (64-bit)
>
[snip]
> other attached packages:
> [1] lme4_1.1-0 Matrix_1.0-12 lattice_0.20-15
[snip]
> > Con
Hi,
The question is not clear.
May be this helps:
set.seed(29)
dat1<- as.data.frame(matrix(sample(c(0:7,NA),10*10,replace=TRUE),ncol=10))
which(is.na(dat1))
#[1] 11 23 24 28 37 43 47 72 77 87
#missing value index for each column
sapply(dat1,function(x) which(is.na(x)))
A.K.
how to Keep a
Hi Mohan,
as I said, it is difficult to help you without beeing able to reproduce
your plots. Could you please send to the list:
1. the exact command you use to create the plot
2. The output of dput(data)
Thanks
Jannis
On 28.08.2013 14:35, mohan.radhakrish...@polarisft.com wrote:
Hi,
plo
Hi,
I was trying to get calender axis for a arima time series
forecast.I could not succeed even though I used zoo time series object.
Please suggest if there is a way to plot with calender axis. Forgot to
attach the link! The working code is available at below link.
--
https://docs.goog
Hey
Sure sorry for the mistake, That was i meant to be also. I this what you
were searching?
-
- László-András Zsurzsa,-
- Msc. Infromatics, Technical University Mun
Sorry, I missed to attach the sessionInfo, here it goes:
> sessionInfo()
R version 3.0.1 (2013-05-16)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=en_US.UTF-8LC_MES
Hi,
For some reason do.call on anova fails if the models are named lmer objects.
Consider the following example:
library(lme4)
models <- list(
lmer(Reaction ~ Days + (1| Subject), sleepstudy),
lmer(Reaction ~ Days + (Days | Subject), sleepstudy))
#
# models is an unnamed list, do.call wor
Hi,
I was trying to get calender axis for a arima time series
forecast.I could not succeed even though I used zoo time series object.
Please suggest if there is a way to plot with calender axis.
--
with best regards
Sudheer
*
HI,
I was talking about another method to get the same result:
myVector[1:100%%2==1]
identical(myVector[1:100%%2==1],myVector[seq(1,100,by=2)])
#[1] TRUE
Hope it helps.
A.K.
From: Zsurzsa Laszlo
To: arun
Cc: R help
Sent: Wednesday, August 28, 2013 5:21
Hi,
Did you mean this?
dat3$year<-1901:2009
dat4<-dat3[,c(13,1:12)]
head(dat4,2)
# year Jan Feb Mar Apr May Jun Jul
#1 1901 NA NA NA NA 1.0309351 1.76204866 1.87906713
#2 1902 -1.195531 -0.8941016 -0.4601555 0.15921 0.3515
HI,
'odd' and 'oddVector' are the same.
identical(odd,oddVector)
#[1] TRUE
identical(odd,myVector[seq(1,100,by=2)]) myVector#
#[1] TRUE
A.K.
From: Zsurzsa Laszlo
To: arun
Cc: R help
Sent: Wednesday, August 28, 2013 8:58 AM
Subject: Re: [R] R Language
HI Catalin,
This seems to work:
spi(6,"spi.txt",1963,2010,"Standardized Precipitation Index - Rio Grande do
Norte State",2,"years","months")
>From the source code:
spi #on your R console
---
if (nargs < 3) {
return("Error: very small number of arguments")
Hello,
#3
myVector <- c(1:100)
oddVector <- myVector[c(1:100) %% 2 == 1]
odd <- myVector[c(1:100) %% 2 == 1]
identical( odd, seq(1,100,by=2)) gives me FALSE
#3
Is this what you sugg
Hi,
str(dat2)
# ts [1:1308, 1] NA NA NA NA 1.03 ...
#- attr(*, "tsp")= num [1:3] 1 110 12
# - attr(*, "dimnames")=List of 2
# ..$ : NULL
# ..$ : chr "Series 1"
If you want it as a data.frame,
dat3<-as.data.frame(matrix(as.numeric(dat2),ncol=12,byrow=TRUE))
colnames(dat3)<- month.abb
head(dat3,
Hi,
plot(strptime(data$Time,"%H:%M:%S"),data$Kbytes,type="l",col = "blue",
ylab="", xlab="",las=2,lwd=2.5, lty=1,cex.axis=1.5)
strptime functions draws a proper graph but now all the time values are
not in the x-axis.
> 1 11:42:02 2691296
> 2 11:43:42 2691396
> 3 11:45:22 2691496
> 4 11:47:02
Hi,
If you wanted the 'fitted' values:
source("cat1.txt")
#dat1 is the dataset
dat2<-dat1$fitted
A.K.
From: catalin roibu
To: arun
Sent: Wednesday, August 28, 2013 8:26 AM
Subject: Re: [R] SPI package error
I have another question, how can extract 12 mo
Hi Jannis,
I have tried that. It doesn't work. Jumps are not there in my other graphs
using numbers. Does this anything to do with time series ?
Can I just convert this time representation into milliseconds and plot the
graph ? The x-axis should still show this time format though(names.arg ?
)
Dear all!
I find a problem from SPI package. When I try a 6 months scale I have this
error:
write.table(test,file="spi.txt",quote=FALSE,row.names=TRUE)
spi3<-spi(6,"spi.txt",1902,2009)
Error in paste(title, sep = "") :
argument "title" is missing, with no default
If I use 3 months scale everythi
Hi Mohan,
i am not sure whether I understand your question correctly. Without
beeing able to easily reproduce your plot, I would guess that the
"breaks" come from the type='b' option you choose. When you use type
='l', the line would be continuous (though the jumps would still be
there). If y
Hi,
The plot function draws a broken line. The graph breaks when it jumps
from a lower value to a higher value with a big break in between. Why does
this type of data not use 'pch' or 'type'.
When I plot I don't indicate anywhere it is time.
> head(data)
Time Kbytes RSS Dirty
> -Original Message-
> From: rip...@stats.ox.ac.uk
> Sent: Wed, 28 Aug 2013 09:17:01 +0100
> To: r-help@r-project.org
> [That is Microsoft, whose messages are infamously accurate but >maximally
> uninformative.]
I don't think it's on-topic enough to be a fortune but it's going into my
c
On Tue, Aug 27, 2013 at 8:07 PM, Pascal Oettli wrote:
> Hello,
>
> "graph" is a Bioconductor package.
you can install it from
http://www.bioconductor.org/packages/release/bioc/html/graph.html
>
> Regards,
> Pascal
>
>
> 2013/8/28 Lianne Schroeder
>
>> I have tried to find pkg graph and R report
Can you specify you're task? I don't understand totally what you need to
do? You're already getting odd numbers with : seq(1,100,by=2)
-
- László-András Zsurzsa,-
-
This is not an R question, and you have not provided the 'at a minimum
information' the posting guide asked of you.
But this seems to be Windows, and
>LoadLibrary failure: The specified module could not be found.
is a Windows error message. You seem to misunderstand what it means:
check
Thank you Ligges for your valuable inputs. This will definitely be of
great help.
Best Regards,
Shalabh Tyagi
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: 28 August 2013 03:40
To: Tyagi, Shalabh
Cc: r-help@r-project.org; Tewari, Vishesh
Subject: Re:
Trying to access R from Netlogo5 (using the NetLogo R-Extension),
running the configuration validation tests in
NetLogo5/extensions/r/Systemcheck.nlogo, I get several loadlibrary()
errors ...
in rJava Check2,
> library(rJava); .path.package('rJava')
Error : .onLoad failed in loadNamespace() fo
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