Well, sticking to your loop that you seem comfortable with try this:
## list to hold dataframes
extracList <- vector("list", length(1981:2010))
cnt <- 0
for(i in 1981:2010){
cnt <- cnt + 1
rasterObj <-get( c(paste("RR_", i, "_1", sep="")))
extractList[[cnt]]<- extract(ras
On 04/17/2014 01:01 PM, Mark Manger wrote:
Hi,
I’m trying to figure out how to label points in a contour plot produced from
the output of MclustDR, the dimension reduction function in the Mclust package.
The original data frame has row names
RRcoarse govtDeficitres_gdp Gro
On Tue, 22 Apr 2014, William Dunlap wrote:
For me that other software would probably be Octave. I'm interested if
anyone here has read in these files using Octave, or a C program or
anything else.
I typed 'octave read binary file' into google.com and the first hit was
the Octave help file f
also
do.call(rbind, result)
Sent from my iPad
On Apr 21, 2014, at 16:47, "Timothy W. Cook" wrote:
> Okay, all day on this and I send the msg. and almost immediately discover
> that:
>
> dat <- ldply(result)
>
> solves the problem.
>
>
>
>
> On Mon, Apr 21, 2014 at 5:40 PM, Timothy W. Coo
> For me that other software would probably be Octave. I'm interested if
> anyone here has read in these files using Octave, or a C program or
> anything else.
I typed 'octave read binary file' into google.com and the first hit was
the Octave help file for its fread function. In C fread is also
Hi,
#or you could use:
do.call(rbind,result)
A.K.
On Monday, April 21, 2014 4:48 PM, Timothy W. Cook wrote:
Okay, all day on this and I send the msg. and almost immediately discover
that:
dat <- ldply(result)
solves the problem.
On Mon, Apr 21, 2014 at 5:40 PM, Timothy W. Cook wrote:
>
After saving a file like so...
con <- gzcon("file.gz", "wb"))
writeBin(vector, con, size=2)
close(con)
I can read it back into R like so...
con <- gzcon("file.gz", "rb"))
vector <- readBin(con, integer(), 4800, size=2, signed=FALSE)
close(con)
...and I'm wondering what other programs might
On 04/22/2014 06:32 AM, VG wrote:
Hello Everyone,
I have a text file which has data in this format.
genestartendfreq
Slc19a1811441
Slc19a11922556
Slc19a12733367
Slc19a13634264
Slc19a14655283
Slc19a15406035
Slc19a
Also, you might want to check out the gamm4 package.
Ken
On 04/21/14, Bert Gunter wrote:
> See gam() and gamm() in the mgcv package.
>
> Cheers,
> Bert
>
> Bert Gunter
> Genentech Nonclinical Biostatistics
> (650) 467-7374
>
> "Data is not information. Information is not knowledge. And knowled
Thank you Arun for your quick answer.
Ill try it and ill let you know.
Cheers
On 21/04/2014, at 11:30 PM, "arun kirshna [via R]"
wrote:
> Hi,
>
> Using the example data from library(gvlma)
>
> library(gvlma)
> data(CarMileageData)
> CarMileageNew <- CarMileageData[,c(5,6,3)]
> lst1 <-
See gam() and gamm() in the mgcv package.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
H. Gilbert Welch
On Mon, Apr 21, 2014 at 3:25 PM, John Sorkin
wrote:
> Collea
Colleagues,
Is there any R program that will allow me to run a localized regression (e.g.
smoothing spline) in the context of a random effects model? I have data on the
growth of animals and want to create growth curves. I am not certain what shape
the growth curve would take, so I would like to
Well, yes and no, David.
Yours is one possible interpretation -- just work verbatim with the
finite sample and, as you say, hope it's well behaved.
But if the data are a noisy *sample* from a bimodal distribution and
the purpose is to estimate the *population* modes, then, of course,
it's a diff
Hello Everyone,
I have a text file which has data in this format.
genestartendfreq
Slc19a1811441
Slc19a11922556
Slc19a12733367
Slc19a13634264
Slc19a14655283
Slc19a15406035
Slc19a18138765
Slc19a1912
Okay, all day on this and I send the msg. and almost immediately discover
that:
dat <- ldply(result)
solves the problem.
On Mon, Apr 21, 2014 at 5:40 PM, Timothy W. Cook wrote:
> I am processing an arbitrary number of XML files and extracting specific
> nodes. I then create a dataframe for
Look below at your question. R-help does not support html email
so you must use plain text if you want to get an answer.
-
David L Carlson
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Marta Tob
I am processing an arbitrary number of XML files and extracting specific
nodes. I then create a dataframe for each nodeset.
I return a list containing these dataframes.
Example:
> str(result)List of 2
$ :'data.frame': 1 obs. of 5 variables:
..$ data-name : chr "Etiologic diagnosi
This will work, as long as there are exactly 2 distinct modes:
> runs <- rle(sign(diff(d.rv$y)))
> length(runs$lengths) # There should be 4 runs if there are
exactly 2 modes
[1] 4
> mode1 <- runs$lengths[1]+1
> mode2 <- length(d.rv$x)- runs$lengths[4]
> d.rv$y[c(mode1, mode2)] # The 2 modes:
[1]
@Duncan I understand. But it seems to be working. At least I could use
basic functions like htmlParse.
On Mon, Apr 21, 2014 at 10:37 AM, Duncan Murdoch
wrote:
> On 21/04/2014 10:30 AM, Alpesh Pandya wrote:
>
>> Frede's suggestion was very helpful to me and I could resolve the issue
>> finally.
>
Hi , Is it possible select a different number of rows by each group with R
I must to select different number (specific quantity in field2:Table1) of rows
in each group(field1:Table2). I have these 2
tables:Table1Table2field1field2field1field3field4field51310.375SpRm12610.416667SpRm23910.45833
*dear all,I have a bimodal distribution and i would like to identify the
two most frequent values. Using unimodal values i found from the R archive
that is possible to identify the most frequent value, which corresponds to
the peak of the density function:from Bert Gunter, Fri Mar 13 04rv <-
rbinom
You seem overly intent on getting a for loop into your code. Jorge's solution
has the same effect as your for loop.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#.
I was able to grab the logos by bringing the .svg file into
Inkscape (open source .svg editor). The three logos are grouped
so you can select them and then paste them into another
document. Then you'll have to ungroup the three logos and select
each logo separately.
---
Google for png vs. svg.
png is a raster graphics format, svg is a vector graphics format.
Which one is the right one to used depends totally on the purpose.
B.
On 2014-04-21, at 11:49 AM, Spencer Graves wrote:
> Awesome.
>
>
> Are there things one might like to do in R reading them
Awesome.
Are there things one might like to do in R reading them directly
as svg that aren't so easy after converting to another format like png?
Thanks again,
Spencer
p.s. To confirm, I received the three files you sent offline, converted
then to png using GIMP (
Indeed you don't need R. This is accomplished in two minutes with the
excellent, free, and open software inkscape. But before I type more and since I
have it open anyway, I'll just send you the three images off-list.
B.
On 2014-04-21, at 10:43 AM, Spencer Graves wrote:
> p.s. I don't need
Frede's suggestion was very helpful to me and I could resolve the issue
finally.
I created new archive manually using source code for XML package (only R
code) and winzip utility. Then ran this command for installation and that
worked fine.
install.packages("C:/Course/WorkingData/XML_3.98-1.1.zip
On Apr 21, 2014, at 10:55, Erin Hodgess wrote:
> Never mindsolved it
xkcd.com/979/
>
>
> On Mon, Apr 21, 2014 at 9:48 AM, Erin Hodgess wrote:
>
>> Hello!
>>
>> I have the following:
>>
>>> time1a <- as.POSIXct(paste(2008,08,01,sep="-"))
>>> time1b <- as.POSIXct(paste(2008,08,11,sep=
Never mindsolved it
On Mon, Apr 21, 2014 at 9:48 AM, Erin Hodgess wrote:
> Hello!
>
> I have the following:
>
> > time1a <- as.POSIXct(paste(2008,08,01,sep="-"))
> > time1b <- as.POSIXct(paste(2008,08,11,sep="-"))
> > time1a
> [1] "2008-08-01 CDT"
> > time1b
> [1] "2008-08-11 CDT"
> > time1c
Hello!
I have the following:
> time1a <- as.POSIXct(paste(2008,08,01,sep="-"))
> time1b <- as.POSIXct(paste(2008,08,11,sep="-"))
> time1a
[1] "2008-08-01 CDT"
> time1b
[1] "2008-08-11 CDT"
> time1c <- seq(from=as.POSIXct(time1a),to=as.POSIXct(time1b),by="days")
> class(time1c)
[1] "POSIXct" "POSI
p.s. I don't need to use R for this, though I'd prefer to stay with
something that's free, open-source software. I tried GIMP, but couldn't
see how to access the layers.
Hi, Peter:
On 4/21/2014 3:34 AM, Peter Crowther wrote:
> Spencer, what "components" do you want to extract? Do the SVG f
Hi, Peter:
On 4/21/2014 3:34 AM, Peter Crowther wrote:
> Spencer, what "components" do you want to extract? Do the SVG files
> have a reasonably well-defined structure? I've done a fair amount
> with SVG and may be able to help.
Thanks. I should have been more specific: I'd like to
On 21/04/2014 10:30 AM, Alpesh Pandya wrote:
Frede's suggestion was very helpful to me and I could resolve the
issue finally.
I created new archive manually using source code for XML package (only
R code) and winzip utility. Then ran this command for installation and
that worked fine.
insta
Hi Frede,
Many thanks for your reply.
1. The first argument in extract is a Formal class RasterLayer in the
Workspace (e.g RR_1981_1 ).
2. I created an intermediate name to hold the result fromthe extract
function because I'd like to create several dataframes with the output
of the iterative
On 21/04/2014 9:19 AM, Milan Bouchet-Valat wrote:
Le lundi 21 avril 2014 à 07:21 -0400, Duncan Murdoch a écrit :
> On 21/04/2014, 7:10 AM, Milan Bouchet-Valat wrote:
> > Le lundi 21 avril 2014 à 11:38 +0200, Frede Aakmann Tøgersen a écrit :
> >> Hi
> >>
> >> Couldn't find a zip file in the archiv
Hi,
Using the example data from library(gvlma)
library(gvlma)
data(CarMileageData)
CarMileageNew <- CarMileageData[,c(5,6,3)]
lst1 <- list()
y <- c("NumGallons", "NumDaysBetw")
for(i in seq_along(y)){
lst1[[i]] <- gvlma(lm(get(y[i])~MilesLastFill,data=CarMileageNew))
lst1}
pdf("gvlmaplot.pdf
Le lundi 21 avril 2014 à 07:21 -0400, Duncan Murdoch a écrit :
> On 21/04/2014, 7:10 AM, Milan Bouchet-Valat wrote:
> > Le lundi 21 avril 2014 à 11:38 +0200, Frede Aakmann Tøgersen a écrit :
> >> Hi
> >>
> >> Couldn't find a zip file in the archives.
> >>
> >> Why didn't 'R CMD build' (in a command
Hi Beatriz
Did you read the help for extract{raster} carefully?
Several things can be wrong.
1) First argument to extract is not a file name but a raster object.
2) In the loop you name an object extract as an intermediate name to hold the
result from the extract function. Do you think there co
Sorry guys
I didn't do my homework, sorry for cluttering the list. I did know about the
source argument.
This worked like a charm:
> install.packages(choose.files(), type="source", repos=NULL)
* installing *source* package 'climatol' ...
** R
** data
** inst
** preparing package for lazy loadin
Dear all,
I'm starting to work with loops and I'm stucked on something.
I've been searching and trying different possibilities but I don't get
to the solution.
I'd be very grateful if you could share any ideas that you think may help.
library("raster")
# All my variables are in the workspace
Hello Uwe
I tried both the options you mentioned but all i get as output then is the
header line and the headers as "row.names" and "X."
--
View this message in context:
http://r.789695.n4.nabble.com/Read-table-mucks-up-headers-tp4688742p4689185.html
Sent from the R help mailing list archive a
On 21/04/2014, 7:10 AM, Milan Bouchet-Valat wrote:
Le lundi 21 avril 2014 à 11:38 +0200, Frede Aakmann Tøgersen a écrit :
Hi
Couldn't find a zip file in the archives.
Why didn't 'R CMD build' (in a command shell on Windows) not make a
zip file? I did try to build from the climatol library that
Le lundi 21 avril 2014 à 11:38 +0200, Frede Aakmann Tøgersen a écrit :
> Hi
>
> Couldn't find a zip file in the archives.
>
> Why didn't 'R CMD build' (in a command shell on Windows) not make a
> zip file? I did try to build from the climatol library that holds the
> typical files as DESCRIPTION,
Le lundi 21 avril 2014 à 11:38 +0200, Frede Aakmann Tøgersen a écrit :
> Hi
>
> Couldn't find a zip file in the archives.
>
> Why didn't 'R CMD build' (in a command shell on Windows) not make a
> zip file? I did try to build from the climatol library that holds the
> typical files as DESCRIPTION,
Spencer, what "components" do you want to extract? Do the SVG files have a
reasonably well-defined structure? I've done a fair amount with SVG and
may be able to help.
- Peter
On 21 April 2014 08:30, Spencer Graves <
spencer.gra...@structuremonitoring.com> wrote:
> Hello:
>
>
> What wou
At 08:43 21/04/2014, Jeff Newmiller wrote:
Frede... Windows uses zip files (binary, aka
pre-compiled format) for packages by default,
because most installations don't have the
development tools installed. You need to setup
RTools and use the "source" option to
install_package in order to hand
Hi
Couldn't find a zip file in the archives.
Why didn't 'R CMD build' (in a command shell on Windows) not make a zip file? I
did try to build from the climatol library that holds the typical files as
DESCRIPTION, LICENSE, etc. as well as folders like R, etc.
Did you see the debug information o
Hi Jorge,
Thanks so much! Exactly what I wanted. Finally I wrote:
for(i in 1976:1981){
PE.files_01_7681 <- paste("val_mapped_petpe_", 1976:i, "01.txt", sep="")
}
Cheers,
Bea
On 21/04/2014 10:46, Jorge I Velez wrote:
> Hi Beatriz,
>
> Try
>
> paste("val_mapped_petpe_", 1976:1981, "01.txt", s
Hi Beatriz,
Try
paste("val_mapped_petpe_", 1976:1981, "01.txt", sep="")
Best,
Jorge.-
On Mon, Apr 21, 2014 at 6:43 PM, Beatriz R. Gonzalez Dominguez <
aguitatie...@hotmail.com> wrote:
> Dear all,
>
> I'm trying to create a loop to select a series of files into my computer
> but I haven't been
Dear all,
I'm trying to create a loop to select a series of files into my computer
but I haven't been successful until now. I've looked into different
possibilities but none has worked. I'd appretiate if you could help me
by providing me with some ideas.
Basically what I'd like to do is to
Frede... Windows uses zip files (binary, aka pre-compiled format) for packages
by default, because most installations don't have the development tools
installed. You need to setup RTools and use the "source" option to
install_package in order to handle the tar.gz package file, or download and
i
Hello:
What would you suggest I use to read and manipulate an SVG file?
I'd like to extract components of an svg file. I see it's XML,
but I have very little experience with either SVG or XML. I've tried
GIMP and findFn{sos} without finding a clear suggestion of where to start.
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