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. Any ideas as to what would cause this?
It does on a Linux platform ('install' has nothing to do with this):
> system.time(x <- lapply(test,function(x) loess.smooth(x
This does speed up on an OsX install, so something must be wacky with the
linux install. Any ideas as to what would cause this?
On Tuesday, August 27, 2013 4:19:31 PM UTC-7, joe meiring wrote:
>
> I can't seem to get mclapply to use more than a single core. I have a 64
> core server running Lin
Lianne,
"graph" can be found here:
http://bioconductor.org/packages/2.12/bioc/
You will need to read the bioconductor instructions. If you visit the
R Project CRAN site, you can find packages there. Each package will
have an associated description and a list of "depends" which are other
packag
In spite of your claim to be familiar with explanatory modeling, you are not
describing your questions about R in terms of theory you would like to apply
and assumptions you are willing to make. This list is about R, so don't go
fishing for statistical advice.
Have you read the help file for pr
Hello,
Use "zoo" instead.
Regards,
Pascal
2013/8/28 Sudheer Joseph
> Also, if that was the case, in below specification it should take the
> timeseries properly which so not takesplace.
> > 1825/365
> [1] 5
> > ts(ru, start = c(2003, 1, 1), end = c(2007, 12, 30), frequency = 365)
>
>
> On Wed
Also, if that was the case, in below specification it should take the
timeseries properly which so not takesplace.
> 1825/365
[1] 5
> ts(ru, start = c(2003, 1, 1), end = c(2007, 12, 30), frequency = 365)
On Wed, Aug 28, 2013 at 4:57 AM, Sudheer Joseph wrote:
> Hi Jim,
> This
Hi Jim,
This won't work as R so not allow frequency which is a
fraction. It accept only whole number frequencies. Do you have any example
where it allows to put fraction as freqency
> ts(ru, start = c(2003, 1, 1), end = c(2008, 1, 1), frequency = 365.33)
Error in ts(ru, start =
I'm trying to educate myself about predictive analytics and am using R to
generate a linear model with the following data.
age <- c(23, 19, 25, 10,9, 12, 11,8)
steroid <- c(27.1, 22.1, 21.9, 10.7, 7.4, 18.8, 14.7, 5.7)
gpa <- c( 2.1, 2.9, 2.8, 3.5, 3.2, 3.9, 2.8, 2.6)
sample
I can't seem to get mclapply to use more than a single core. I have a 64
core server running Linux.
Fore example:
library(parallel)
test <- lapply(1:100,function(x) rnorm(1))
system.time(x <- lapply(test,function(x) loess.smooth(x,x)))
system.time(x <- mclapply(test,function(x) loess.smooth(
Hello,
"graph" is a Bioconductor package.
Regards,
Pascal
2013/8/28 Lianne Schroeder
> I have tried to find pkg graph and R reports back that it doesn't exist for
> R 3.0.1, so how do I get ggm to work with out that? Or is there something
> that has replaced graph and ggm needs to be updated
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(timeseries,dataseries,type="n")
lines(timeseries,dataseries1,col="red",type="l")
lines(timeseries,dataseries2,col="blue",type="l")
c
I have tried to find pkg graph and R reports back that it doesn't exist for
R 3.0.1, so how do I get ggm to work with out that? Or is there something
that has replaced graph and ggm needs to be updated to work with it?
Thanks,
Lianne
On Tuesday, August 27, 2013, David Winsemius wrote:
> Not an
Hi,
set.seed(29)
myVector<- rnorm(100)
seq(1,100,by=2)
# [1] 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
#[26] 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99
myVector[seq(1,100,by=2)]
rev(myVector)
sum(myVector>0)
#[1] 46
#or
tabl
Not an error in installation of ggm , just a failure to install pkg graph
--
David
Sent from my iPhone
On Aug 27, 2013, at 3:33 PM, Lianne Schroeder wrote:
> I'm new to R and trying to use the ggm package and I get the following
> message:
>
>> library("ggm",
> lib.loc="/Library/Frameworks/R
On 26.08.2013 12:05, Tyagi, Shalabh wrote:
Hi,
As we are in thinking of installing and using R as a data analysis
software on our machines, we have few questions before going ahead with
the installation. Please find below the questions:
1) Does R allow data to be sent out or processed out
This is great!
Thank you so much.
On Tue, Aug 27, 2013 at 3:06 PM, arun wrote:
> Hi,
> May be this helps:
> dat1<- read.table(text="
> iduseraction
> 1 12 login
> 2 12 view
> 3 12 view
> 4 12 view
> 5 12 login
> 6 12 view
> 7
Hi,
May be this helps:
dat1<- read.table(text="
id user action
1 12 login
2 12 view
3 12 view
4 12 view
5 12 login
6 12 view
7 12 view
8 12 login
",sep="",header=TRUE,stringsAsFactors=FALSE)
split(dat1,cumsum(da
On 27/08/2013 4:38 PM, Rolf Turner wrote:
Actually what's needed is "deparse(substitute(...))":
file = paste(deparse(substitute(x)),"csv",sep="."))
I have no idea what "deparse" and "substitute" really mean; I just treat
the forgoing as a black box/magic incantation. It works.
In cas
Hi,
I have a data set with structure similar to this:
iduseraction
1 12 login
2 12 view
3 12 view
4 12 view
5 12 login
6 12 view
7 12 view
8 12 login
I want to create a list of sessions. That means to split ta
I'm new to R and trying to use the ggm package and I get the following
message:
> library("ggm",
lib.loc="/Library/Frameworks/R.framework/Versions/3.0/Resources/library")
Loading required package: graph
Error: package graph could not be loaded
In addition: Warning message:
In library(pkg, charac
Actually what's needed is "deparse(substitute(...))":
file = paste(deparse(substitute(x)),"csv",sep="."))
I have no idea what "deparse" and "substitute" really mean; I just treat
the forgoing as a black box/magic incantation. It works.
cheers,
Rolf Turner
On 28/08/13 09:19, Had
It's not really the inverse of assign (that's get), but I think you
want substitute.
See http://adv-r.had.co.nz/Computing-on-the-language.html for more details.
Hadley
On Tue, Aug 27, 2013 at 4:13 PM, Robert Lynch wrote:
> I am looking for a way to extract the name of a variable that has been
>
I am looking for a way to extract the name of a variable that has been
passed into a function
for example
foo <-function(x){
write.csv(x, file = paste(NAME(x), "csv", sep ="."))
}
is there a function "NAME" that would let the calls
foo(bar)
write the file bar.csv
and foo(stuff)
write the fi
Hello,
One more thing, if you do cbind(year = rownames(dat), dat), then the
year variable defaults to a factor. Maybe it's better if you transform
it into numeric:
cbind(year = as.numeric(rownames(dat)), dat)
Rui Barradas
Em 27-08-2013 21:09, catalin roibu escreveu:
Thank you very much!
I
Dear R users,
since I updated to R-3.0 I get some error messages while using a
parallelized process with the packages foreach and doMC.
The error message is:
Error: Error in sendMaster(try(lapply(X = S, FUN = FUN, ...), silent =
TRUE)) (from doMC.R#170) :
ignoring SIGPIPE signal
The st
If the goal is to change row names and the replacements are unique the it is
as easy as:
rownames(dfrm) <- newnames
Include example data for more complete answers.
--
David
Sent from my iPhone
On Aug 27, 2013, at 2:41 PM, Rui Barradas wrote:
> Hello,
>
> If your data.frame is named 'dat'
I'd suggest that r-sig-geo is a better choice for quick help in this case.
Depending on how your data is currently stored, and given your short time
frame, you might do better using software such as Quantum GIS, if you can
get it installed quickly enough.
In R, if you can get your data into an ob
Hello,
If your data.frame is named 'dat', try the following.
cbind(year = rownames(dat), dat)
Hope this helps,
Rui Barradas
Em 27-08-2013 20:34, catalin roibu escreveu:
hello all!
I have a problem! I don't know how to convert row names in first column
variables.
Thank you very much for your
hello all!
I have a problem! I don't know how to convert row names in first column
variables.
Thank you very much for your help!
CR
--
---
Catalin-Constantin ROIBU
Lecturer PhD, Forestry engineer
Forestry Faculty of Suceava
Str. Universitatii no. 13, Suceava, 720229, Romania
office phone +4
Well I don't see anything that looks like a year but go from this
dat1[match(2, dat1$samp.depth),] and pick off what you want I guess
John Kane
Kingston ON Canada
-Original Message-
From: catalinro...@gmail.com
Sent: Tue, 27 Aug 2013 20:46:27 +0300
To: jrkrid...@inbox.com
Subject: Re: [
Hi,
May be this helps:
#dat1 is dataset.
with(dat1,tapply(row.names(dat1),list(samp.depth),FUN=head,1))
# 1 2 3
#"1592" "1596" "1600"
row.names(dat1)[tapply(seq_len(nrow(dat1)),list(dat1$samp.depth),FUN=head,1)]
#[1] "1592" "1596" "1600"
#or
row.names(dat1)[!duplicated(dat1$sam
Hello,
Try the following.
dat <- read.table(text = "
HUMstdHUMres samp.depth
1592 0.5687519NA 1
1593 1.5251896NA 1
1594 1.8168727NA 1
1595 1.1265923 0.8253599 1
1596 0.8725422 0.7423323 2
1597 0.6744835 0.5443933
with data named dat1
match(2, dat1$samp.depth)
John Kane
Kingston ON Canada
> -Original Message-
> From: catalinro...@gmail.com
> Sent: Tue, 27 Aug 2013 20:24:04 +0300
> To: r-help@r-project.org
> Subject: [R] find row name with specific value
>
> Hello all!
> I want to find the first
Hello all!
I want to find the first year where is fixed value (eg 2).
My data is like this:
HUMstdHUMres samp.depth
1592 0.5687519NA 1
1593 1.5251896NA 1
1594 1.8168727NA 1
1595 1.1265923 0.8253599 1
1596 0.8725422 0.7423323
Hello,
I am struggling with a lattice graph.
I want to plot points and condition on a variable called 'Method'.
For each conditional plot I have two subgroups, say: 'first'and 'second'.
I want to plot the points in a conditional graph with a different color or
symbol for each subgroup.
More
Dear All,
I have installed rJava
as root
sudo -i
then
R CMD javareconf
Exit
R CMD INSTALL -l /home/itziar/R/x86_64-pc-linux-gnu-library/3.0
rJava_0.9-4.tar.gz
And it was correctly installed. But now when launching R I get a segmentation
fault (core dumped)
Can anyone help, please?
Regards,
Itziar
Henrik solution is probably just what you need, but just to mention a
similar solution, there is also:
if (!require('installr')) install.packages('installr'); require('installr')
install.packages.zip("zip_URL")
(the source code is here:
https://github.com/talgalili/installr/blob/master/R/install.
Hi Pascal,
Still... there is something wrong. I changed the two lines as you
suggested, and now I'm using the keys parameter. But the output is not as
expected. That's why I was controlling the points using points function,
because that works well.
Let me try to show you with the code:
# Example 1
If you look at:
str(lst1)
List of 10
$ 1.1 :'data.frame': 783 obs. of 6 variables:
..$ Time : num [1:783] 1.83e-05 2.37e-05 2.90e-05 5.06e-04 5.11e-04 ...
..$ ChannelA : num [1:783] 4.53 4.88 4.8 3.58 3.58 ...
..$ ChannelB : num [1:783] 6.87 13.6 21.09 22.09 21.55 ...
..$ FileNo
Thank you very much.
But why is this? The function applyfilter should return a numeric result,
shouldn't it?
Best regards,
Ingo
From: arun [smartpink...@yahoo.com]
Sent: Tuesday, August 27, 2013 4:06 PM
To: Ingo Reinhold
Cc: R help
Subject: Re: [R] [pl
#or
data2<-ddply(data,.(FileNo,ChannelNo),transform,filtered=as.numeric(applyfilter(ChannelB)))
A.K.
- Original Message -
From: arun
To: Ingo Reinhold
Cc: R help
Sent: Tuesday, August 27, 2013 10:03 AM
Subject: Re: [R] [plyr] Moving average filter with plyr
Hi,
May be this helps:
Hi,
May be this helps:
lst1<-dlply(data,.(FileNo,ChannelNo),transform,filtered=applyfilter(ChannelB))
data2<-do.call(rbind,lapply(lst1,function(x){x$filtered<-
as.numeric(x$filtered);x}))
row.names(data2)<- row.names(data)
head(data,3)
# Time ChannelA ChannelB FileNo ChannelNo
#116
On Sun, Aug 25, 2013 at 10:55 AM, Christofer Bogaso
wrote:
> Hello again,
>
> I need to install Rmpi package from this
> http://www.stats.uwo.ca/faculty/yu/Rmpi/download/windows/MPICH2
>
> I was wondering if there is any direct way to install this in R. The
> trivial method would obviously be down
Hi Earl
The cookies will only be written to the file specified by the cookiejar option
when the curl handle is garbage collected.
If you use
rm(ch)
gc()
the cookie.txt file should be created.
This is the way libcurl behaves rather than something RCurl introduces.
If you don't explic
Dear all,
I'm stuck with a problem using plyr to process a rather large junk of data.
What I'm trying to do is applying a moving average to all the subparts of the
dataframe (the example data can be found here
https://dl.dropboxusercontent.com/u/2414056/testData.Rdata).
require(plyr)
load("te
Dear all,
The question would be:
I would love to get information and match names of different chemistry
compounds. A compound looks like this.
Link: http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=2
The task is the following:
1. You input a name via R to the link, then a database
Way outside my area but I think, well "guess" that the two panel approach is
better since it should give you more discrimination or easily visualized detail
for each variable but clearly Jim's suggestion will work too.
Otherwise, get a sleeping bag and order in some food. It could be a long
Thanks. There is one more idea from Jim Lemon which can also help.
In my case I am dealing with a sys. admin who thinks the JVM is using all
the memory in the linux machine. I think it is
http://www.linuxatemyram.com/ syndrome.
So if the junior sys. admin loses perspective I will never be rele
Jim,
With those ranges isn't Mohan going to lose discrimination? That's one reason
I suggested a two-panel approach.
John Kane
Kingston ON Canada
> -Original Message-
> From: j...@bitwrit.com.au
> Sent: Tue, 27 Aug 2013 22:36:12 +1000
> To: mohan.radhakrish...@polarisft.com
> Subject:
Well I have not spent a lot of time looking at the data , especially with what
appeared to be a constant :) . It was not clear to me that they were the same
units for one thing.
From my point of view (essentially total ignorance) if you have a situation
where you have data plotted with the sa
On 08/27/2013 10:08 PM, Jim Lemon wrote:
On 08/27/2013 10:01 PM, mohan.radhakrish...@polarisft.com wrote:
Hi,
Can't one y-axis be used ? Both 'Kbytes' and 'RSS' use the
same unit.
Hi Mohan,
Yes, you can use a single y axis. What you probably want is something
like this:
ylim<-range(c(Kbytes,
Ok, the problem seems simpler than I thought. It works well if omit two
lines:
y_s[,1] <- V(y)$name[y_s[,1]]
y_s[,2] <- V(y)$name[y_s[,2]]
It seems melt creates a data.frame (unlike what it did two years ago?)
Joe
On Tue, Aug 27, 2013 at 12:35 AM, Joseph J. Bakker wrote:
> Two years ago, as sh
On 08/27/2013 10:01 PM, mohan.radhakrish...@polarisft.com wrote:
Hi,
Can't one y-axis be used ? Both 'Kbytes' and 'RSS' use the
same unit.
Hi Mohan,
Yes, you can use a single y axis. What you probably want is something
like this:
ylim<-range(c(Kbytes,RSS))
plot(rNo,Kbytes,col
Hi,
Can't one y-axis be used ? Both 'Kbytes' and 'RSS' use the
same unit.
Thanks,
Mohan
From: John Kane
To: mohan.radhakrish...@polarisft.com
Cc: r-help@r-project.org
Date: 08/27/2013 05:13 PM
Subject:RE: [R] Scale of axis for two data sets
Still looks l
Import data:
mydata <- read.table("filename.txt")
give following command for column name
colnames(mydata) <- c("V1","V2","V3",)
Gyanendra Pokharel
University of Guelph
Guelph, ON
On Mon, Aug 26, 2013 at 5:42 PM, Docbanks84 wrote:
> Hi ,
>
> I just imported a large data set fro
If I understand you correctly just import the data into R as a data.frame and
then use the unlist() command.
John Kane
Kingston ON Canada
> -Original Message-
> From: mban...@partners.org
> Sent: Mon, 26 Aug 2013 09:06:12 -0700 (PDT)
> To: r-help@r-project.org
> Subject: [R] Pasting exc
names(mydatat) <- c("V1","V2","V3")
For future reference
https://github.com/hadley/devtools/wiki/Reproducibility and/or
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
John Kane
Kingston ON Canada
> -Original Message-
> From: mban...@partners.or
Still looks like a constant to me :) I think I see what you mean though. It's
just that representation of the numbers in the email don't show any difference.
Personally I think I'd go with a two panel graph using par() rather than a
double-axis graph if you can. Everything I have read about d
It is not a constant but changes very slowly. Actually it should be like
this.
KbytesRSS rNo
2687188 1287532 1
2687188 1287532 2
2687188 1287536 3
2687188 1287540 4
2687188 1287640 5
2687188 1287640 6
Thanks,
Mohan
From: John Kane
To: mohan.radhakr
I am misreading something here I think. Why are you graphing against a constant?
John Kane
Kingston ON Canada
> -Original Message-
> From: mohan.radhakrish...@polarisft.com
> Sent: Tue, 27 Aug 2013 13:28:48 +0530
> To: r-help@r-project.org
> Subject: [R] Scale of axis for two data sets
>
In case the OP wanted to append columns, rather than rename existing
columns:
cbind(myData, var1=NA, var2=NA, var3=NA)
col1 col2 col3 var1 var2 var3
1123 NA NA NA
2234 NA NA NA
3345 NA NA NA
On 24/08/2013 17:22, arun wrote:
Hi,
You could
At 04:46 27/08/2013, Murat Tasan wrote:
hi all -- i'm running into a strange problem that i can't seem to
easily get around, but i'm probably just missing something obvious.
I think you just subtract it from y. But perhaps I too am missing the obvious.
i have a model to which some data is fi
On 08/27/2013 05:58 PM, mohan.radhakrish...@polarisft.com wrote:
Hi,
Kbytes RSS rNo
2689632 1450876 1433788
2689632 1450876 1433788
2689632 1450876 1433788
2689632 1450876 1433788
2689632 1450876 1433788
2689632 1450876 1433788
2689632 1450876 1433788
Dear Jari Oksanen,
Thanks a lot for the varpart R package help.
I preferred Adj. R square as mentioned by Peres-Neto (2006), but there
are two facts that let me struggling around with the adj. R square. a)
I received an Adj. R. square > 1 for an explanatory matrix (common and
individual f
Assuming you attribute the name "dataset" to your data. A way to name a
column using colnames and which.
Code to change V1 column name:
colnames(dataset)[which(colnames(dataset) == "V1")] <- "Toto"
You are asking to R in the column's names of dataset which one is "V1"
(TRUE) and attributing to
You have what is called a "wide" table and you want a "tall" table. In
this case the function melt from reshape2 package is suitable. Be
careful specifying properly the variables (columns) you want to use for
the reshape using correct syntax.
?melt {reshape2}
You can also use as already sugges
Hi,
Kbytes RSS rNo
2689632 1450876 1433788
2689632 1450876 1433788
2689632 1450876 1433788
2689632 1450876 1433788
2689632 1450876 1433788
2689632 1450876 1433788
2689632 1450876 1433788
2689632 1460168 1443084
plot(data$rNo,data$RSS,pch=0,type="b",c
Hello Sebastian,
I think you may need to use tensorA package. You can define different
indices and use Einstein or Reimann summation convention.
Grüße,
Mehmet
On 23 August 2013 15:14, Sebastian Hersberger
wrote:
> Hi all
>
> I have a short question relating to the usage of the summation sign in
I think you want the UsingR package
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54 43 61 85
thierry.onkel
Thanks
this does it:
script.name <- basename(sys.frame(1)$ofile)
On Tue, Aug 27, 2013 at 4:07 PM, Henrik Bengtsson wrote:
> See findSourceTraceback() of R.utils. /Henrik
>
> On Mon, Aug 26, 2013 at 10:58 PM, peter dalgaard wrote:
> >
> > On Aug 27, 2013, at 07:12 , nevil amos wrote:
> >
> >
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