If you just want the value which is the most frequent try
as.numeric(names(which.max(table(c( -295.8045 ,-295.8045, -295.8045,
-295.8045 ,-325.4754 ,-295.8045, -295.8045, -295.8045, -413.2099,
-295.8045)
[1] -295.8045
There may be easier ways
Regards
Duncan Mackay
Department of Agronomy
Hi Gundala,
This would be a way to do it:
names(which.max(table(myvector)))
For example:
> set.seed(99)
> y <- sample(letters[1:3], 40, replace=TRUE)
[1] "b" "a" "c" "c" "b" "c" "c" "a" "b" "a" "b" "b" "a" "b" "c" "b"
[17] "b" "a" "a" "a" "a" "a" "c" "b" "c" "b" "a" "c" "a" "a" "b"
Hi,
Is there a way to do it?
For example I have the following vector:
> print(myvector)
> [1] -295.8045 -295.8045 -295.8045 -295.8045 -325.4754 -295.8045 -295.8045
[8] -295.8045 -413.2099 -295.8045
I want it to return -295.8045, which occur most often.
- Gundala Viswanath
Jakarta - Indonesia
Just out of curiosity, what system do you have?
These are the results in my machine:
> system.time(exp(m), gcFirst=TRUE)
user system elapsed
0.520.040.56
> library(pnmath)
> system.time(exp(m), gcFirst=TRUE)
user system elapsed
0.660 0.016 0.175
Juan Pablo
>
>> syst
Mike
see ?strip.custom and strip.default
Basically you need to create a string of factor levels equal to the number
of panels.
Below is an example for a plot of 24 panels 4 columns and six rows.
so that row 1 is:
"2000 Summer" | "2000 Autumn" | "2000 Winter" | "2000 Spring"
and so on for the
Thanks! I'll use the which.min function. I've been learning R from the pdfs
on the CRAN website, but I find the texts to be generally introductory, not
comprehensive, and focused on one niche purpose. Can anyone suggest a good
self-teach guide?
Thanks again!
R_Learner wrote:
>
> I have a long
Hi all,
By default a call to xyplot from the Lattice package when using 2
factors [eg xyplot( dv~iv | XY * AB ) ] yields the following shingle
structure:
|_A_|_A_|_B_|_B_|
|_X_|_Y_|_X_|_Y_|
However, I'm wondering if it is possible to merge the upper shingle
within levels of that factor,
Hi
Tobias Verbeke wrote:
> Dear list,
>
> Can someone explain why the childNames below
> gives
>
> character(0)
>
> instead of the (canonical) names of the children grobs
> of the xaxis gTree ?
>
> [1] "major" "ticks" "labels"
The problem is that you xaxis has an 'at' component of NULL, w
Le mer. 09 juil. à 06:20, Rainer M Krug a écrit :
Hi
I tried to install rkward under ubuntu hardy heron, but it tried to
use the one from the cran repository which was newer, but it did not
install. To be able to install rkward, I had to disable the cran
repository, install rkward, lock it's ve
Dear all:
I ‘m a newer. I have some problem when I check my package.
The error messages were as follows. It shows that the problem is about Rd
file. How can I find where the error is? How can I do??
c:\temp>Rcmd check bear
* checking for working pdflatex ... OK
* using log directory 'c:/tem
Indeed.
Presumably R was installed using an RPM from the Fedora repository and
the 'devel' RPM, which contains the requisite header files was not.
Using:
# yum install R-devel
should get you moving forward.
BTW, there is a new Fedora specific e-mail list for R. More info here:
https://
Is this what you want:
> x <- read.table(textConnection("Date Apples Oranges Pears
+ 1/7 2 35
+ 2/7 1 47
+ 3/7 3 810
+ 4/7 5 72
+ 5/7 6 35"), header=TRUE)
> closeAllCo
If its a zoo or ts time series you can use the lag function.
On Wed, Jul 9, 2008 at 2:57 PM, rcoder <[EMAIL PROTECTED]> wrote:
>
> Hi everyone,
>
> I have some data in a matrix, and I want to shift it down by one row. The
> matrix in question has a date column. Does anyone know of a way to shift t
Maybe I am wrong about this, but I think your problem is actually:
Mutils.h:9:41: error: R.h: No such file or directory
Mutils.h:10:22: error: Rversion.h: No such file or directory
Mutils.h:11:56: error: Rdefines.h: No such file or directory
I think these are important files for compiling additi
Suggest you read the relevant article in R News 4/1.
Also the zoo package and read.zoo, in particular, might help.
On Wed, Jul 9, 2008 at 6:09 AM, Caroline Keef
<[EMAIL PROTECTED]> wrote:
> Dear all,
>
> I've come across a problem using strptime, can anyone explain what's
> going on? I'm using ve
Right click the R icon and choose Run As Administrator and then
install the package. This is only needed for package installation
if you don't have have a local library in your home tree
and also is not needed for just everyday running of R.
On Wed, Jul 9, 2008 at 9:15 PM, Luciano La Sala
<[EMAIL
OK, never-mind. I found length(x). It's been a long day.
Gabor Csardi wrote:
>
> Why don't you write it for yourself, it takes less time than writing
> an email:
>
> mysummary <- function(x) {
> require(plotrix)
> require(e1071)
> c(Mean=mean(x), Std.Error=std.error(x), Std.Deviation=sd
The two describe functions have it. Maybe the one in psych too?
Also see valid.n in the prettyR package and of course there is length,
nrows and dim.
On Wed, Jul 9, 2008 at 8:53 PM, nmarti <[EMAIL PROTECTED]> wrote:
>
> Thanks Gabor.
> I was able to create a function that works really well. One
Hello all,
New user here. I have R 2.7.1. installed, but don’t have internet access in my
PC. Therefore, I downloaded the ISwR package (Introductory Statistics with R)
from CRAN and then moved it into my HD.
I can’t seem to install the local package under Windows Vista system. I've
tried using
Original Table:
Date Apples Oranges Pears
1/7 2 35
2/7 1 47
3/7 3 810
4/7 5 72
5/7 6 35
What I want after shift (data shifted, dates left unchanged)
Date Apples O
Hi,
My question is how do I gain control over what values the X and Y axis show.
Below is a sample plot I have made and want the X axis to represent a time
vector with values taking the form Q1.60, Q2.60, Q3.60...Q1.90..etc...Currently
the X axis starts with value 0 and increases by 1 through t
Yes I do want a random assignment, instead of rounding. (From what I
understand of the rbinom command, it will randomly assign 1 or 0, and the
higher the given probability, the higher the likelihood of a 1... Feel free
to correct me if I'm wrong!)
Ben Bolker wrote:
>
> -BEGIN PGP SIGNED ME
Ben:
Thanks for the reply. One further question, and this is where my novice
status at R shows through. The code makes sense, but what would I put it for
"m"? Is it the same number for all three (that was my first thought since it
was the same placeholder for all three). Number of rows in the matr
Thanks Gabor.
I was able to create a function that works really well. One more quick
question if you don't mind.
I want to report the number of observations also (i.e. n = 5420). What
function would do this? I really can't find anything that would just simply
count the number of x's.
Thanks aga
Hi,
I have been trying to install the Matrix package on R running in Linux
(Fedora). But, I get the following error message (at the bottom). I am logged
in as the root user. Any help would be appreciated! Thanks!
-Hyunseung Kang
* Installing *source* package 'Matrix' ...
Using GNU make for b
Georg Otto wrote:
Hi,
I have a problem using figures in Sweave:
To save my figures, I use
\SweaveOpts{prefix.string=figures/figure}
I adjust the figure size for my pdf document using
<>=
this works fine. The file
figures/figure-graphicsFun.pdf
has the right size, and so has the figur
On Wednesday 09 July 2008, hippie dream wrote:
> This might not possible in R but I thought I would give it shot. I am have
> to set up a 40 x 40 cm grid of 181 points equidistant from each other. Is
> there any way to produce a graph with R that can do this for me? Actual
> sizes are unimportant a
Ahhh. That worked perfectly. Thank you very much.
On Wed, Jul 9, 2008 at 4:19 PM, Dylan Beaudette <[EMAIL PROTECTED]>
wrote:
> On Wednesday 09 July 2008, hippie dream wrote:
> > This might not possible in R but I thought I would give it shot. I am
> have
> > to set up a 40 x 40 cm grid of 181 poi
I am still trying to get used to R, and apparently haven't found the right
place in the documentation to see how to do what I want in R. If I were to do
this in C++, there'd be no problem: I'd write it all myself, but I want to
learn R well enough it can save me a lot of coding time.
Here is t
Basically, I want 181 points equally spaced over a 40 x 40 cm area. I want
to be able to specify the number of points and the area to which they are
plotted on. I think you are right that grid is what I am looking for but I
was the grid to have axes which your code below, although appreciated, did
Still not sure exactly what you want, but it sounds like the 'grid'
package may be of some help.
It has very flexible ways partitioning regions for plotting. Is this
anything like you're after?
library(grid)
for(i in 0:10)
for(j in 0:10)
grid.points(i / 10, j / 10, default.unit = "npc
a <- "apple"
b <- "pear"
c <- "banana"
result <- c(2, 3, 5)
names(result) <- c(a, b, c)
Qian R wrote:
Here is my problem
a <- data$name[1]
b <- data$name[2]
c <- data$name[3]
a
"apple"
b
"pear"
c
"banana"
result <- c( a = 2 , b=3, c=5)
output:
a b c
2 3 5
But I want my output
Here is my problem
a <- data$name[1]
b <- data$name[2]
c <- data$name[3]
a
"apple"
b
"pear"
c
"banana"
result <- c( a = 2 , b=3, c=5)
output:
a b c
2 3 5
But I want my output
apple pear banana
2 3 5
[[alternative HTML version deleted]]
__
Right equidistant was clearly the wrong word. Sorry. I just meant that any
given point should have an equal distance from the four points immediately
surrounding it (x,-x,y-y) aside from those on the edge which will obviously
only have two or three points surrounding.
On Wed, Jul 9, 2008 at 3:12 P
What do you mean by equidistant? You can have three points that are
equidistant on the plane, but there's no way to add another point and
have it be the same distance from all of the existing points. (Unless
all the points are in the same place)
Hadley
On Wed, Jul 9, 2008 at 5:02 PM, hippie dre
I've determined that the problem with getting the subsetting to work with
the error bars was related to not dropping the unused levels of the Grouping
factor. The simplest solution was to follow a combination of Deepayan's
suggestions 1 and 2 below, i.e., supply one's own colors and subset
before
This might not possible in R but I thought I would give it shot. I am have to
set up a 40 x 40 cm grid of 181 points equidistant from each other. Is there
any way to produce a graph with R that can do this for me? Actual sizes are
unimportant as long it is to scale. Thanks
--
View this message in
on 07/09/2008 04:38 PM stephen sefick wrote:
I am plotting a twelve panel plot of a zoo object. I have tried to raise
the cex from 0.6 to 1 to 2 and it does not seem to do anything. I am trying
to output this to a tiff file
tiff()
how do I get the labels of a larger font size
Stephen
I had t
base , if you're not seeing it, upgrade your version of R to the latest.
Carl Witthoft wrote:
Query: what package is Reduce in?
(per GG's suggestion to use it)
thanks
Carl
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-h
Query: what package is Reduce in?
(per GG's suggestion to use it)
thanks
Carl
__
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.org/posting-guide.html
and provide com
I am plotting a twelve panel plot of a zoo object. I have tried to raise
the cex from 0.6 to 1 to 2 and it does not seem to do anything. I am trying
to output this to a tiff file
tiff()
how do I get the labels of a larger font size
Stephen
--
Let's not spend our time and resources thinking abo
On 7/9/08, David Afshartous <[EMAIL PROTECTED]> wrote:
>
>
>
> On 7/9/08 1:07 PM, "Deepayan Sarkar" <[EMAIL PROTECTED]> wrote:
>
> > On 7/9/08, David Afshartous <[EMAIL PROTECTED]> wrote:
> >>
> >>
> >> All,
> >>
> >> I'm plotting points and lines for various groups.
> >> I'd like subseq
on 07/09/2008 02:17 PM Juliet Hannah wrote:
Using this data as an example
dat <- read.table(textConnection("Id myvar
12 1
12 2
12 6
34 9
34 4
34 8
65 15
65 23"), header = TRUE)
closeAllConnections()
how can I create another data set that does not have duplicate entries
for 'Id', but the
How about this:
> dat <- read.table(textConnection("Id myvar
+ 12 1
+ 12 2
+ 12 6
+ 34 9
+ 34 4
+ 34 8
+ 65 15
+ 65 23"), header = TRUE)
> closeAllConnections()
> # split by the id and then choose one
> x <- lapply(split(dat, dat$Id), function(.grp){
+ .grp[sample(seq(length(.grp)), 1)
> I built a 1,273,230 by 6 data set named "mydata2", it was saved in the
> following command,
>
> write.table(mydata2, "mydata2.txt", row.name=F,col.name=T,quote=F,sep="\t")
>
> The next day I read in above saved text file into R,
>
> temp<-read.table("mydata2.txt",header=T,sep="\t",na.strings=
Try this:
do.call(rbind, lapply(split(dat, dat$Id),
function(x)x[sample(1:nrow(x), 1),]))
On 7/9/08, Juliet Hannah <[EMAIL PROTECTED]> wrote:
> Using this data as an example
>
> dat <- read.table(textConnection("Id myvar
> 12 1
> 12 2
> 12 6
> 34 9
> 34 4
> 34
Can you provide commented, minimal, self-contained, reproducible code.
If you don't have code, at least provide a before/after version of
the matrix that you would like. It is easy to use indexing to move
stuff around, we just have to know what is it that you want to move.
On Wed, Jul 9, 2008 at
On Wednesday 09 July 2008, Ben Bolker wrote:
> Dylan Beaudette wrote:
> | On Wednesday 09 July 2008, Ben Bolker wrote:
> |> ACroske yahoo.com> writes:
> |>> I have a large matrix full of probabilities; I would like to convert
>
> each
>
> |>> probability to a 1 or a 0 using rbinom.
> |>> How can I
Dear all,
I have problem when reading a table into R. The total row of read in table
has is much less than the original saved table.
I built a 1,273,230 by 6 data set named "mydata2", it was saved in the
following command,
write.table(mydata2, "mydata2.txt", row.name=F,col.name=T,quote=F,sep="\t
On Wed, Jul 9, 2008 at 3:09 PM, Ben Bolker <[EMAIL PROTECTED]> wrote:
> Katharine Mullen few.vu.nl> writes:
>
>>
>> It is not an R package, but rather a collection of Fortran functions
>> that R uses from netlib:
>> http://www.netlib.org/port/
>>
>
> Where is Martin Maechler when we need him?
> "
Using this data as an example
dat <- read.table(textConnection("Id myvar
12 1
12 2
12 6
34 9
34 4
34 8
65 15
65 23"), header = TRUE)
closeAllConnections()
how can I create another data set that does not have duplicate entries
for 'Id', but the included values
are randomly selected from th
Hi everyone,
I have some data in a matrix, and I want to shift it down by one row. The
matrix in question has a date column. Does anyone know of a way to shift the
data by one row, whilst preserving the date column in the matrix - i.e.
shift the data and leave the date column in the current locat
phoebe kong gmail.com> writes:
> I built a 1,273,230 by 6 data set named "mydata2", it was saved in the
> following command,
>
> write.table(mydata2, "mydata2.txt", row.name=F,col.name=T,quote=F,sep="\t")
>
> The next day I read in above saved text file into R,
>
> temp<-read.table("mydata2.tx
Hi,
is it possible to get content of ftp directory similar to for
local files?
Directory is password protected.
Thanks
--
View this message in context:
http://www.nabble.com/ftp-directory-tp18368309p18368309.html
Sent from the R help mailing list archive at Nabble.com.
_
Katharine Mullen few.vu.nl> writes:
>
> It is not an R package, but rather a collection of Fortran functions
> that R uses from netlib:
> http://www.netlib.org/port/
>
Where is Martin Maechler when we need him?
"That's not a package, that's a library!" :-)
Ben
___
Hi there,
Is this what you want?
your.number=5.43
# For a vector
x=c(1,2,4,3,2,5,6,7,5.42,6)
which.min(abs(x-your.number))
[1] 9
# For a matrix
set.seed(123)
X=matrix(rpois(100,4.5),ncol=10)
apply(X,2,function(x) which.min(abs(x-your.number)))
[1] 9 3 2 3 2 5 1 2 2 2
HTH,
Jorge
On Wed, Ju
No, not at all, I'm glad that I do get the decomposition for non-square
matrices. My problem is not with elu$U but with as.matrix(elu$U), which is
not an upper diagonal matrix. Can I do something to fix this ?
Regards, Ulrike
Douglas Bates-2 wrote:
>
> On Wed, Jul 9, 2008 at 12:01 PM, Ulrike G
On 7/9/08 1:07 PM, "Deepayan Sarkar" <[EMAIL PROTECTED]> wrote:
> On 7/9/08, David Afshartous <[EMAIL PROTECTED]> wrote:
>>
>>
>> All,
>>
>> I'm plotting points and lines for various groups.
>> I'd like subsequent plots done on subsets to maintain the color assignments
>> from the origin
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Dylan Beaudette wrote:
| On Wednesday 09 July 2008, Ben Bolker wrote:
|> ACroske yahoo.com> writes:
|>> I have a large matrix full of probabilities; I would like to convert
each
|>> probability to a 1 or a 0 using rbinom.
|>> How can I do this on the
the function is there in grDevices
http://stat.ethz.ch/R-manual/R-patched/library/grDevices/html/png.html
type ? png
if tiff() is not listed, you need to update R to get the new base with new
grDevices.
thanks
y
Daniel Steinberg wrote:
>
> Greetings R users!
>
> I am working with the ENSEM
Daniel Malter wrote:
x=c(1:100)
your.number=5.43
which(abs(x-your.number)==min(abs(x-your.number)))
or [depending on the problem]:
which.min(abs(x-your.number))
HTH,
Tobias
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Vo
On Wednesday 09 July 2008, Ben Bolker wrote:
> ACroske yahoo.com> writes:
> > I have a large matrix full of probabilities; I would like to convert each
> > probability to a 1 or a 0 using rbinom.
> > How can I do this on the entire matrix? The matrix was converted from a
> > raster ArcMap dataset,
Even using POSIXlt it seems to work fine when you are looking for NAs
in the dates (ones that did not convert correctly. So you must be
doing something different or your data is different from the example
you have in the mail. You are always requested to provide commented,
minimal, self-contained
Is this what you're looking for?
test <- matrix(runif(100, 0, 1), nrow = 20)
nr <- nrow(test)
matrix(sapply(test, rbinom, n = 1, size = 1), nrow = nr)
ACroske wrote:
I have a large matrix full of probabilities; I would like to convert each
probability to a 1 or a 0 using rbinom.
How can I do th
Dear all,
I have problem when reading a table into R. The total row of read in table
has is much less than the original saved table.
I built a 1,273,230 by 6 data set named "mydata2", it was saved in the
following command,
write.table(mydata2, "mydata2.txt", row.name=F,col.name=T,quote=F,sep="\t
yes use plotting chars 22, 21
replace your legend statement with:
legend(x="topright", legend=c("2006 mean", "2007 mean", "2008 mean",
"1964-2005 mean \n max and min flows "), pch=c(22, 22, 22, 21),
col=c(1,1,1,1), pt.bg=c(grey.colors(3, gamma=4),"black"), cex=1.5)
box(which="plot", lty="solid
ACroske yahoo.com> writes:
>
>
> I have a large matrix full of probabilities; I would like to convert each
> probability to a 1 or a 0 using rbinom.
> How can I do this on the entire matrix? The matrix was converted from a
> raster ArcMap dataset, so the matrix is essentially a map. Because of
Hi,
I'm trying to plot a field obtained from the atmospheric model WRF-NMM
which uses a "Rotated Lat-Lon¨ map projection.
The WRF documentation mentions that:
· Rotates the earth's lat/lon grid such that the intersection of the
equator and prime meridian is at the center of the model domain.
Hi, there may be a more elegant way of doing this, but at least it works.
You have to be careful about putting the same axis limits in both graphs and
to use axis-labels in only one of them.
##generate data
x1=c(1:100)
e1=rnorm(100,0,10)
e2=rnorm(100,0,30)
x1=x1+e1
x2=x1+e2
y1=100+x1+0.025*x1^2+e1
Hi,
I have a problem using figures in Sweave:
To save my figures, I use
\SweaveOpts{prefix.string=figures/figure}
I adjust the figure size for my pdf document using
<>=
this works fine. The file
figures/figure-graphicsFun.pdf
has the right size, and so has the figure in the final pdf
doc
Hello everyone -
I am currently modeling some data with ARIMA(p,d,q), and have successfully
used the "fracdiff" package to obtain estimates for d and the ARMA
parameters. However, I don't know how to get fracdiff to obtain
innovations for me. Can fracdiff even do this? Can any other package?
Plea
Hi,
I have a problem using garchFit, when I use :
x<-model$resid
fit = garchFit(~garch(1, 1), data = x, cond.dist="dst")
[EMAIL PROTECTED]
it gives me error : object "fit" not found
Why it doesn't recognize fit?
Thanks,
Shirin
__
R-help@r-pro
I am going to assume your data.frame is called x
#this transposes the matrix
x.t <- t(x)
rollmean(x.t)
On Wed, Jul 9, 2008 at 12:50 PM, Rheannon <[EMAIL PROTECTED]> wrote:
>
> Hello,
>
> I am trying to calculate a 31 day running mean in some temperature data
> along ROWS. Rollmean() works great a
See ?t
On Wed, Jul 9, 2008 at 12:50 PM, Rheannon <[EMAIL PROTECTED]> wrote:
>
> Hello,
>
> I am trying to calculate a 31 day running mean in some temperature data
> along ROWS. Rollmean() works great along columns, but how do I perform this
> same action on my rows?
>
> The data is a matrix of 365
I have a large matrix full of probabilities; I would like to convert each
probability to a 1 or a 0 using rbinom.
How can I do this on the entire matrix? The matrix was converted from a
raster ArcMap dataset, so the matrix is essentially a map. Because of this,
I have no column headings.
Thanks!
x=c(1:100)
your.number=5.43
which(abs(x-your.number)==min(abs(x-your.number)))
Best,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im
Auftrag von R_Learner
Gesendet: Wedne
Thank you, but why does this happen?
a =(1:223960)[is.na(datetimes)]
datetimes[a]
> [1] "1981-03-29 01:20:00" "1990-03-25 01:43:00" "1992-03-29 01:43:00"
> "1996-03-31 01:30:00" "1996-03-31 01:57:00" [6] "1997-03-30 01:02:00"
> "1997-03-30 01:14:00" "1997-03-30 01:27:00" "1997-03-30 01:44:00"
Try this:
which.min(abs(x - 5.43))
where x is your vector of numbers.
On Wed, Jul 9, 2008 at 12:28 PM, R_Learner <[EMAIL PROTECTED]> wrote:
>
> I have a long list of numbers [3.4,5.4,3.67,], and I basically want to
> find the index of the number closest to this number that I have, let's sa
Hello,
First, thanks for your help (and sorry for my english !)
I have a data frame in which each row represents a vote (in percent, only
20,40, 60,80,100) of one person on one content, with three columns : name
(the name of the voters), content_id, vote :
str(votesredac)
'data.frame': 1000 o
Seems that the following makes what I want :
attach(votesredac)
tapply(value, list(name, content_id), mean)
Only thing is, I don't need to make a mean - there is only one or no value.
VinceD wrote:
>
> Hello,
>
> First, thanks for your help (and sorry for my english !)
>
> I have a data fr
Thanks for your replies.
basicStats(x) in "fBasics" is exactly what I was looking for.
>
>
nmarti wrote:
>
> I'm looking for a function that lists a few summary stats for a column (or
> row) of data. I'm aware of summary(x), but that does not give me what I'm
> looking for.
> I'm actually looki
Hello,
I am trying to calculate a 31 day running mean in some temperature data
along ROWS. Rollmean() works great along columns, but how do I perform this
same action on my rows?
The data is a matrix of 365 columns (days of the year) by 5,000 rows
(lat/long coordinates).
I would like to perform
I have a long list of numbers [3.4,5.4,3.67,], and I basically want to
find the index of the number closest to this number that I have, let's say
5.43. How would I do this without writing a for loop (I have to do this many
times for several lists)? Is there a "lookup" function in R?
Thanks!
On Wed, Jul 9, 2008 at 12:01 PM, Ulrike Grömping <[EMAIL PROTECTED]> wrote:
> Dear R-helpers,
> I have a question regarding LU-decomposition with function lu in package
> Matrix. The following simple example confuses me: Why is as.matrix(elu$U)
> not an upper triangular matrix?
> u3 <-
> matrix(
# I would like to outline the squares in the legend with a black line. Does
anyone know how to do this?
x.t <- structure(c(5987.387, 4354.516, 3685.789, 6478.592, 5924.315,
NA, 8386, 5559.468, NA, 4651.273, 3967.5, NA, 4339.167, 5053.56,
NA, 4631.978, 4808.694, NA, 5217.306, 4017.632, NA, 5846.90
Hi Alan
How your lm model looks like?
Are your data stored on a data.frame ? Case yes, send us a str(df).
Send us a short sample of the code and a short subset of the data.
Cheers,
miltinho astronauta
brazil
On Tue, Jul 8, 2008 at 5:36 AM, Alan Kelly <[EMAIL PROTECTED]> wrote:
> Hi, while usi
On 7/9/08, David Afshartous <[EMAIL PROTECTED]> wrote:
>
>
> All,
>
> I'm plotting points and lines for various groups.
> I'd like subsequent plots done on subsets to maintain the color assignments
> from the original plot. This works fine, but the key for the subset doesn't
> maintain the co
A little more googling reveals:
The Port 3 Library is now available via netlib and licensing
arrangements are specified here:
http://www.netlib.org/port/readme
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox:
It is not an R package, but rather a collection of Fortran functions
that R uses from netlib:
http://www.netlib.org/port/
On Wed, 9 Jul 2008, Jos Kaefer wrote:
> Hi
>
> When I type:
> > ?nls
>
> I come across this section:
>
> algorithm: character string specifying the algorithm to use. The
>
Dear R-helpers,
I have a question regarding LU-decomposition with function lu in package
Matrix. The following simple example confuses me: Why is as.matrix(elu$U)
not an upper triangular matrix?
u3 <-
matrix(c(1,1,1,1,1,1,-1,1,0,0,0,0,0,-1,1,0,0,0,-1,0,1,0,0,0,0,0,-1,1,0,0),5,6,byrow=T)
elu <- e
A more accurate wording, I believe, would be "Port Library" see:
http://www.bell-labs.com/project/PORT/
Martin will correct me if there really is a package!!
Unfortunately, the licensing link is broken on the URL above and
it would be interesting to know what the status of licensing
Hi Caroline,
Because POSIXlt is a complicated structure: you are dealing with a list, not
with what you think you are. Maybe this will help you to see more clearly.
strptime(19800604062759, format="%Y%m%d%H%M%S")
[1] "1980-06-04 06:27:59"
str(strptime(19800604062759, format="%Y%m%d%H%M%S"))
P
Hi
When I type:
> ?nls
I come across this section:
algorithm: character string specifying the algorithm to use. The
default algorithm is a Gauss-Newton algorithm. Other
possible values are '"plinear"' for the Golub-Pereyra
algorithm for partially linear least-squar
It might be best to use Perl for this processing since it is better
equipped to work with text files of this nature.
On Wed, Jul 9, 2008 at 12:18 PM, Paolo Sonego <[EMAIL PROTECTED]> wrote:
> I apologize for giving wrong information again ... :-[
> The number of files is not a problem (30/40). T
I apologize for giving wrong information again ... :-[
The number of files is not a problem (30/40). The real deal is that some
of my files have ~10^6 lines (file size ~ 300/400M) :'(
Thanks again for your help and advices!
Regards,
Paolo
jim holtman ha scritto:
How much time is it takin
On 09-Jul-08 15:49:54, Sean Davis wrote:
> This is hopefully a simple question. I am trying to escape single
> quotes like so:
>
> abc's >> abc\'s
>
> However, I cannot find an easy way to do that with gsub:
>
> gsub("'","'","abc's")
># returns "abc\\'s"
>
> How can I get a single \ in t
You probably want POSIXct instead of POSIXlt:
x <-
read.table(textConnection("#TZUTC+0|*|SANR08002|*|SNAMENAUL|*|SWATERDELVIN|*|CNR98808|*|
+ #CNAMEQ|*|CTYPEn-min-ip|*|CMW1440|*|RTIMELVLhigh-resolution|*|
+ #CUNITm3/s|*|RINVAL-777|*|RNR-1|*|REXCHANGE98913|*|
+ #RTYPEinstantaneous values|*|
+ 198
On Wed, Jul 9, 2008 at 11:57 AM, jim holtman <[EMAIL PROTECTED]> wrote:
> It does have a single \; the printing just shows that it is escaped.
> If you 'cat' it to output, you will see:
>
>
>> gsub("'","'","abc's")
> [1] "abc\\'s"
>> cat(gsub("'","'","abc's"))
> abc\'s
>
> Which I think is
It does have a single \; the printing just shows that it is escaped.
If you 'cat' it to output, you will see:
> gsub("'","'","abc's")
[1] "abc\\'s"
> cat(gsub("'","'","abc's"))
abc\'s
Which I think is what you were thinking it would be. So when you
write it out to a file, it will be cor
How much time is it taking on the files and how many files do you have
to process? I tried it with your data duplicated so that I had 57K
lines and it took 27 seconds to process. How much faster to you want?
On Wed, Jul 9, 2008 at 10:57 AM, Paolo Sonego <[EMAIL PROTECTED]> wrote:
> Thanks so muc
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