Many thanks. Still, if anyone knows of an automated way of achieving this
result, I would appreciate hearing about it.
Tolga
Boks, M.P.M. [EMAIL PROTECTED]
08/07/2008 20:45
To
[EMAIL PROTECTED], r-help@r-project.org
cc
Subject
RE: [R] Automatic placement of Legends
You may want to
Hi Pavel,
First, annonations should have the same cex-size on each axis. That said,
the way that this is implemented is not too cexy (ouch!). You need to plot
without axes, e.g. plot(obj, axes=F), then you add your axes afterwards
using your own specifications.
?axes
Also see ?par (sub ann)
Hi Pavel,
And perhaps read the entry for cex.axis a little more carefully. And bear in
mind that labels, main, and sub are distinct, having their own cex.-
settings.
HTH, Mark.
Mark Difford wrote:
Hi Pavel,
First, annonations should have the same cex-size on each axis. That said,
the
Dear R users.
Recently I wanted to update my R distribution to the current one (R-2.7.1). I
am running a Fedora core 8 distirbution. The installation went fine, but when I
tried to add some additional packages the majority made an exit with an error.
Only a few least demanding (e.g.
Hi R users,
I would known if any attempt of building psychological experimental paradigm
with R as Brainard/Pelli Psych Toolbox on Matlab or e-prime was done.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
guillaume chaumet guillaumechaumet at gmail.com writes:
Hi R users,
I would known if any attempt of building psychological experimental paradigm
with R as Brainard/Pelli Psych Toolbox on Matlab or e-prime was done.
As far as I know, the answer is no. I think that someone would
have to write
Dear all,
Probably a very basic question but I need some help.
I have a data frame (made by read.table from a text file) of microarray data,
of which the first column is a factor and the rest of the columns are numeric.
The factor column contains chromosome names, so values 1 through 22 plus
Hello,
I am creating a plot and I would like to know how to put this expression
to the y axis
µmol/10^6 cells
I've tried some combinations using the expression() function, but none
of them worked.
Any idea?
Best,
Dani
--
Daniel Valverde Saubí
x=c(1:25)
x[23]=X
x
x.new=ifelse(x==X,23,x)
x.new=as.numeric(x.new)
Best,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im
Auftrag von Booman, M
Gesendet: Wednesday,
x=rnorm(100,0,10)
y=rnorm(100,0,10)
plot(y~x,xlab=µmol/10^6)
Is that it?
Best,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im
Auftrag von Dani Valverde
Gesendet:
E.g.
plot(1:10,1:10,xlab=NA)
title(xlab=expression(mu*mol/10^6* cells))
Gabor
On Wed, Jul 09, 2008 at 11:21:46AM +0200, Dani Valverde wrote:
Hello,
I am creating a plot and I would like to know how to put this expression
to the y axis
Dear R users,
I have a big text file formatted like this:
x x_string
y y_string
id1id1_string
id2id2_string
z z_string
w w_string
stuff stuff stuff
stuff stuff stuff
stuff stuff stuff
//
x x_string1
y y_string1
z z_string1
w w_string1
stuff
Hi,
R seems to use the 1/n-1-factor calculating the standard-deviation sd().
If i wat to get the non-sample standard-deviation i use
sqrt(sd(x)^2*((n-1)/n))
Is there a parameter to get the sd()-function using the 1/n factor
directly?
Or is there any other function to do so?
Thank you in
On Tue, 2008-07-08 at 17:38 +1000, Fiona Johnson wrote:
Hi
I have just upgraded from R2.6.0 to R2.7.1 (running on Windows) and a part
of my code that previously ran ok now gives an error. The following is a
simple example to demonstrate my problem.
a -
Hi,
if given the value of, say, 15000, I would like to be able to divide
that value recursively by, say, 5, and to get a vector of a determined
length, say 9, the last value being (set to) zero- i.e. like this:
15000 3000 600 120 24 4.8 0.96 0.192 0
These are in fact concentration values from
I don't think so, but you can easily write a function for that:
vec=c(1,2)
pop.sd=function(x)(sqrt(var(x)*(length(x)-1)/length(x)))
pop.sd(vec) ##as compared to
sd(vec)
Best,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche
Sorry, you want it in the y-axis. So put ylab instead of xlab. Generally,
type ?plot in the R-prompt and hit enter. Click the little par link in the
middle of the page. There you see all the options (or at least very many)
that you can pass to your plot command.
Best,
Da.
Daniel Malter
R 2.7.2
PPC Mac OS X 10.4.11
library mice 1.13.1
I try to use mice for multivariate data imputation.
My variables are numeric, factors, count data, ordered factors.
First I created a vector for the methods to use with each variable
ImpMethMice-c(rep(logreg, 62), rep(polyreg,1),
your.number=15000
your.denominator=5
your.favorite.n=9
your.vector=NULL
for(i in 0:your.favorite.n){
your.vector[i+1]=your.number/your.denominator^i
your.vector[your.favorite.n+1]=0
}
your.vector ##check
Best,
Daniel
-
cuncta stricte discussurus
On Tue, 2008-07-08 at 19:31 +0100, [EMAIL PROTECTED] wrote:
Dear R-Users,
I am looking for a way to get legends placed automagically in an empty
spot on a graph. Additional complication comes through my useage of
multiple graphs on the same plot through mfrow.
Is there a way to achieve
a.start - 15000
a.step - 5
a.length - 9
c(a.start*a.step^-(0:(a.length-2)),0)
Anne-Marie Ternes [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Hi,
if given the value of, say, 15000, I would like to be able to divide
that value recursively by, say, 5, and to get a vector of a
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 version and enable the
repository again.
I have the
On Wed, 2008-07-09 at 11:40 +0200, Anne-Marie Ternes wrote:
Hi,
if given the value of, say, 15000, I would like to be able to divide
that value recursively by, say, 5, and to get a vector of a determined
length, say 9, the last value being (set to) zero- i.e. like this:
15000 3000 600 120
Maybe something like this:
f1 with loop, f2 without.
f1 - function(start,len, div) {
x - rep(start,len)
for (i in 2 : (len-1)) {
x[i] - x[i-1]/div
}
x[len] - 0
return(x)
}
f2 - function(start,len, div) {
x - rep(start,len)
y - div^(0:(len-1))
x - x/y
x[length(x)] - 0
return(x)
}
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
Many thanks in advance,
Tobias
### minimal example code ###
library(grid)
pushViewport(plotViewport(c(5,4,4,2)))
Jim Lemon wrote:
On Tue, 2008-07-08 at 19:31 +0100, [EMAIL PROTECTED] wrote:
Dear R-Users,
I am looking for a way to get legends placed automagically in an empty
spot on a graph. Additional complication comes through my useage of
multiple graphs on the same plot through mfrow.
Is there a
Keith,
I am simply baffled! Didn't think a second about doing it this way, tsss -
Great!
Thanks also for Daniel, Jim's and Bart's proposals!
R is cool, I realise it every day again :-)
Thanks!!
On Wed, Jul 9, 2008 at 12:33 PM, Jim Lemon [EMAIL PROTECTED] wrote:
On Wed, 2008-07-09 at 11:40
See ?Reduce
Reduce(/, rep(5, 9), 15000, acc = TRUE)
On Wed, Jul 9, 2008 at 5:40 AM, Anne-Marie Ternes [EMAIL PROTECTED] wrote:
Hi,
if given the value of, say, 15000, I would like to be able to divide
that value recursively by, say, 5, and to get a vector of a determined
length, say 9, the
G'day all,
On Wed, 09 Jul 2008 20:33:39 +1000
Jim Lemon [EMAIL PROTECTED] wrote:
On Wed, 2008-07-09 at 11:40 +0200, Anne-Marie Ternes wrote:
Hi,
if given the value of, say, 15000, I would like to be able to divide
that value recursively by, say, 5, and to get a vector of a
determined
Many thanks all,
Tolga
Frank E Harrell Jr [EMAIL PROTECTED]
09/07/2008 12:03
To
Jim Lemon [EMAIL PROTECTED]
cc
[EMAIL PROTECTED], r-help@r-project.org
Subject
Re: [R] Automatic placement of Legends
Jim Lemon wrote:
On Tue, 2008-07-08 at 19:31 +0100, [EMAIL PROTECTED] wrote:
Dear
Try this; what you want to do is to change the 'levels' of the factor.
x - factor(c(1:10,'x','y','xy'))
str(x)
Factor w/ 13 levels 1,10,2,3,..: 1 3 4 5 6 7 8 9 10 2 ...
x
[1] 1 2 3 4 5 6 7 8 9 10 x y xy
Levels: 1 10 2 3 4 5 6 7 8 9 x xy y
# your error
x[x == 'x'] - 23
Warning
This should do what you want: (it uses loops; you can work at
replacing those with 'lapply' and such -- it all depends on if it is
going to take you more time to rewrite the code than to process a set
of data; you never did say how large the data was). This also grows
a data.frame, but you have
Greetings R users!
I am working with the ENSEMBLE climate data (10 min resolution daily
temperatures images for all of Europe 1950-2006). The data comes
packaged in a single netCDF file. I would like to read the data in and
export a subset (2002-2006) as geotiffs (one image per day). So far, I
Dear R users.
Recently I wanted to update my R distribution to the current one (R-2.7.1). I
am running a Fedora core 8 distirbution. The installation went fine, but when I
tried to add some additional packages the majority made an exit with an error.
Only a few least demanding (e.g.
Hi Anne-Marie,
maybe its not particularly elegant, but this function does the trick:
dilute-function(val,div,len){
+ res-rep(val,len)
+ res-res/div^c(0:(len-1))
+ res[len]-0
+ res
+ }
dilute(15000,5,9)
Cheers, René
-Ursprüngliche Nachricht-
Von: Anne-Marie Ternes [EMAIL
n -
c(f,m,a,m,j,j,a,s,o,n,d,j,f,m,a,m,j,j,a,s,o,n,d,j)
plot(x.zoo[, 95], xaxt = n, ylim=c(1,2))
rng - range(time(x.zoo))
axis(1, at = seq(rng[1], rng[2], 1/12), labels = n, tcl = -0.3)
#why do I have to put in the y-lim explicitly? If you try the below code I
get a warning- Error in
Dear R users,
I have a question about the plot with the package gam.
I need to plot different main effect functions, related to different
gam models, in the same graphics (i.e. the same covariate about
different models).
I used the plot.gam e preplot.gam documentations. Using preplot.gam I
can
Dear all,
I have basically two questions were some help would be very useful.
The first question is whether there exists a package that performs Furness
iterations.
In the field of transportation, growth factor models are estimated using
this iterative procedure.
It encompasses and iterative
Dear all,
I've come across a problem using strptime, can anyone explain what's
going on? I'm using version 2.7.0 on Windows XP.
Thank you
Caroline
First read in a data file using read.table
alldata = read.table(file, header=F, skip=4, colClasses =
c(character,numeric))
dim(alldata)
[1]
jholtman wrote:
Working code would help. I would probably use 'lapply' since it
appears that you want to return a variable number of items for each
condition.
On Tue, Jul 8, 2008 at 2:23 PM, hesicaia [EMAIL PROTECTED] wrote:
Hello,
The quick version of my question is how can I
Hi there,
I'm trying to connect in an Oracle database. I am able to do it with
Java, but when trying this code in R:
library(DBI)
library(RJDBC)
drv = JDBC(oracle.jdbc.driver.OracleDriver,
C:\\Documents and
Settings\\rodri\\workspace\\AtlasQueryingTest\\lib\\ojdbc14.jar,
two alternative solutions with a more functional-style taste:
dilute = function(init, div, count)
sapply(1:count, function(i) init/div^(i-1))
dilute = function(init, div, count)
if (count == 1) init
else c(init, dilute(init/div, div, count-1))
vQ
René Capell wrote:
Hi
BB - structure(list(Yearmonth = structure(c(12L, 24L, 1L, 13L, 14L,
3L, 15L, 4L, 16L, 5L, 17L, 6L, 18L, 7L, 19L, 8L, 20L, 9L, 21L,
10L, 22L, 11L, 23L), .Label = c(2006-02, 2006-03, 2006-04,
2006-05, 2006-06, 2006-07, 2006-08, 2006-09, 2006-10,
2006-11, 2006-12, 2007-01, 2007-02, 2007-03, 2007-04,
This was fixed in the zoo devel version (to be zoo 1.5-4) just last week. See:
http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php/*checkout*/pkg/NEWS?rev=487root=zoo
You can either wait for that, use the workaround you found or source
the fixed version
of plot.zoo:
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 correspondence. One solution is to reprint the entire key, but
this is
ctu at bigred.unl.edu writes:
I have a silly question. I don't know how to express the loglikelihood
function of
1/(x!) where x=x1,x2,xn in R.
Not clear what you mean. If you just want the log of the
factorial, just use lfactorial(x) ...
Ben Bolker
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 looking for something that is very similar to the descriptive
statistics tool in excel; i.e. Mean, Std. Error, Std.
hi
Can i use a for loop with in the lapply..if so could u plz let me know the
syntax.
Ramya
--
View this message in context:
http://www.nabble.com/lapply-tp18363288p18363288.html
Sent from the R help mailing list archive at Nabble.com.
__
There are describe functions in prettyR and Hmisc packages and
doSummary in doBy.
On Wed, Jul 9, 2008 at 11:15 AM, nmarti [EMAIL PROTECTED] 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
Dear nmarti,
See ?basicStats in fBasics.
HTH,
Jorge
On Wed, Jul 9, 2008 at 11:15 AM, nmarti [EMAIL PROTECTED] 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
Yes. It is the same syntax. Can you provide an example of what you want to do.
lapply(1:5, function(num){
+ .ret - numeric(num)
+ for (i in 1:num) .ret[i] - i*i
+ .ret
+ })
[[1]]
[1] 1
[[2]]
[1] 1 4
[[3]]
[1] 1 4 9
[[4]]
[1] 1 4 9 16
[[5]]
[1] 1 4 9 16 25
On Wed, Jul
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(x),
Kurtosis=kurtosis(x))
}
Gabor
On Wed, Jul 09, 2008 at 08:15:00AM -0700, nmarti wrote:
I think the function describe() in the package psych will give you
want you want. There are other similar functions in the library Simple
as well.
Harold
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
This is hopefully a simple question. I am trying to escape single
quotes like so:
abc'sabc\'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 the output?
Thanks,
Sean
At 11:26 AM -0400 7/9/08, Gabor Grothendieck wrote:
There are describe functions in prettyR and Hmisc packages and
doSummary in doBy.
As well as describe and describe.by in the psych package.
On Wed, Jul 9, 2008 at 11:15 AM, nmarti [EMAIL PROTECTED] wrote:
I'm looking for a function
Dear List
I tried using rJava respectively JRI to run R code from within Java but
got an error message reproducible. I have broken down the problem to
this simple piece of code:
import org.rosuda.JRI.*;
public class rtest {
public static void main(String[] args) {
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 much
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 correct.
On Wed,
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 what you were thinking it
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|*|
+
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'sabc\'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 the output?
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
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). The
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-squares
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))
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
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$U)
not an upper triangular matrix?
u3 -
matrix(c(1,1,1,1,1,1,-1,1,0,0,0,0,0,-1,1,0,0,0,-1,0,1,0,0,0,0,0,-1,1,0,0),5,6,byrow=T)
elu -
It is not an R package, but rather a collection of Fortran functions
that R uses from netlib:
http://www.netlib.org/port/
On Wed, 9 Jul 2008, Jos Kaefer wrote:
Hi
When I type:
?nls
I come across this section:
algorithm: character string specifying the algorithm to use. The
A little more googling reveals:
The Port 3 Library is now available via netlib and licensing
arrangements are specified here:
http://www.netlib.org/port/readme
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox:
Hi Alan
How your lm model looks like?
Are your data stored on a data.frame ? Case yes, send us a str(df).
Send us a short sample of the code and a short subset of the data.
Cheers,
miltinho astronauta
brazil
On Tue, Jul 8, 2008 at 5:36 AM, Alan Kelly [EMAIL PROTECTED] wrote:
Hi, while using
# I would like to outline the squares in the legend with a black line. Does
anyone know how to do this?
x.t - structure(c(5987.387, 4354.516, 3685.789, 6478.592, 5924.315,
NA, 8386, 5559.468, NA, 4651.273, 3967.5, NA, 4339.167, 5053.56,
NA, 4631.978, 4808.694, NA, 5217.306, 4017.632, NA,
On Wed, Jul 9, 2008 at 12:01 PM, Ulrike Grömping [EMAIL PROTECTED] wrote:
Dear R-helpers,
I have a question regarding LU-decomposition with function lu in package
Matrix. The following simple example confuses me: Why is as.matrix(elu$U)
not an upper triangular matrix?
u3 -
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!
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
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 looking for
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 frame
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
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 say
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
1997-03-30
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:
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!
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 columns
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 along
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
__
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?
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
graphicsFun, fig=TRUE, echo=FALSE, height=10, width=5, eval=TRUE=
this works fine. The file
figures/figure-graphicsFun.pdf
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
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.
ACroske Audy3272 at 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.
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)
thnaks
y
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)
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
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,
On Wednesday 09 July 2008, Ben Bolker wrote:
ACroske Audy3272 at 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
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-
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 ENSEMBLE
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Dylan Beaudette wrote:
| On Wednesday 09 July 2008, Ben Bolker wrote:
| ACroske Audy3272 at 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
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 original plot. This works
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
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
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