Hard as it is for me to imagine, the ggobi windows stay open and functional
while R (in emacs) has crashed in the background after throwing the error
messages. As a perhaps naive Linux user, I thought that if a parent process
crashed any processes it spawned would crash too. I guess not in this cas
What were H & W? For png() they are (by default) in pixels, for pdf() in
inches.
You haven't told us your OS, but I guess Mac OS. Please update to R
2.7.0: that offers you two new png() devices for higher-quality plots, and
various other improvements. (If you want to use the cairo-based png
Duncan,
I know have two version of libcurl on my system, the Ubuntu installed 7.18.0
and my newly compiled from source 7.18.1 (which I installed after my
problems began with RCurl). I was afraid to uninstall 7.18.0 because
Synaptic wanted to uninstall half of my system if I did so via my package
m
On Tue, 6 May 2008, Kate wrote:
Thanks. I used the command you mention and it works fine in Linux. But I
need to get it to work for Windows XP as well (currently running
R-2.7.0). Any idea if it's possible?
Yes, same commmand works -- I've just tested it. You need the tools
installed, or to
>library(ggplot2)
>(p<- qplot(mpg, wt, data=mtcars))
What I am doing is to set color of the ticks to hide them.
>grid.gedit(gPath("xaxis", "ticks"), gp=gpar(col="white"))
It should be a better way to achieve the purpose. Thanks.
__
R-help@r-project.org
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi all
~ I'm glad this made it to R-help (or R-devel) so that I saw it
as this is the sort of problem that should be at least CC'ed to the
package maintainer.
~ Yes, there was a change to RCurl yesterday with one of the changes
being to synchron
Michael,
As a general rule of thumb (I believe this is in Jim Grace's book, if not
others) one should use 10-20 observations per variable. If you have 5
variables, and 18 observations, you should probably be a bit suspect of your
results. That said, if some of your paths are indeed non-signific
Martin,
Well, thanks for jumping in! We need all the help we can get ;)
I changed the execute bit as you suggested and recompiled, no luck, still
the same error message.
Below is the output you wanted me to look at, its a bit beyond me so I
include both a brief grep summary and then the whole en
Hi there,
Quick question about the output from the sem() function in the library
of the same name.
If I am getting probabilities >0.05 for some of my estimates of path
coefficients, I'm assuming the interpretation here is that the
coefficient is not significantly different from zero, correc
But see these posts:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/119079.html
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/119080.html
Simon.
On Tue, 2008-05-06 at 13:38 -0600, Greg Snow wrote:
> Fisher's exact test works with small cells. See ?fisher.test
>
> --
> Gregory (Greg) L. Sno
R News 4/1 has an article on this dates and the last page
contains a table with many date idioms.
On Tue, May 6, 2008 at 10:00 PM, Roslina Zakaria <[EMAIL PROTECTED]> wrote:
> Hi R-expert,
> If I have this daily rainfall data, how do call a particular set of data?
>
> Year,Month,Day,Amount
> 1981,
On May 6, 2008, at 10:43 PM, Alexy Khrabrov wrote:
I've used to have a script with a barplot command it in, preceded
by a png:
png(graph.file,height=H,width=W)
barplot(t,names.arg=breaks[2:(length(t)+1)],tck=gridlines)
-- worked before R 2.6.2. When I tried it in R 2.6.2, which I have
for
I've used to have a script with a barplot command it in, preceded by a
png:
png(graph.file,height=H,width=W)
barplot(t,names.arg=breaks[2:(length(t)+1)],tck=gridlines)
-- worked before R 2.6.2. When I tried it in R 2.6.2, which I have
for a while but didn't run with that script, it complain
R-users
E-mail: r-help@r-project.org
>My question is: Will this calculation be valid with the residual deviance
>returned by the glm() function using the quasibinomial family as
>reported in R?
Let me show you a simple example, assuming c=2.5:
function ()
{
xx <- c(1,2,3,4,5,6,7,8,9,10)
Hi everyone,
I want to fit the following mixed effect model
Y_ij = b_0i + b_1i * (t_ij*grp_ij == 1) + b_2i * (t_ij*grp_ij == 2) +
v_0i + v_1i*t_ij + e_ij
with a different covariance matrix of random effects for each group.
(Y is the response
t is time
grp is the group indic
raymond chiruka <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
> i am trying to carry out a categorical data analysis but my problem
> is that when in i use the chi squared test some of my expected
> values are less than 5. is there a test that can handle this
> situation. the data is no
Hi Esmail,
Try this:
# A simple example
set.seed(123)
X=matrix(rnorm(100*5),ncol=5)
colnames(X)=c("Y",paste("X",1:4,sep=""))
X[1:5,]
Y X1 X2 X3 X4
[1,] -0.56047565 -0.7104066 2.1988103 -0.7152422 -0.07355602
[2,] -0.23017749 0.2568837 1.3124130 -0.7526890 -1.16
Hi R-expert,
If I have this daily rainfall data, how do call a particular set of data?
Year,Month,Day,Amount
1981,1,1,0
1981,1,2,0
1981,1,3,78
1981,1,4,22
1981,1,5,2
1981,1,6,0
1981,1,7,0
1981,1,8,0
1981,1,9,0
1981,1,10,10
1981,1,11,0
1981,1,12,108
1981,1,13,328
1981,1,14,10
1981,1,15,0
1981,1,16,
Once again I need to tap into the collective knowledge here.
Let's say I have the following columns and data below
Y X1 X2 X3 X4
I would like to generate additional new columns and column names
(ie the data would be squared - and I'd like the column names to
reflect this) like:
Y X1 X2 X3 X4
>
> -Original Message-
> From: [EMAIL PROTECTED] on behalf of Yasir Kaheil
> Sent: Tue 5/6/2008 4:24 PM
> To: r-help@r-project.org
> Subject: Re: [R] Spatial join between two datasets using x and y co-ordinates
>
>
> vector dat1.select is the selected records from dat1 by dat2.
>
> > d
"Shubha Vishwanath Karanth" <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
> Thank you very much Mark! That worked Just a question, ?[ does
> give an error to me...how do I find it?
Try:
?"["
__
R-help@r-project.org mailing list
https://sta
Hi,
I am trying to use sqlSave to write a dataframe to an existing table in a
DB2 database. The database contains two schemas. My experience is the
following: (1) in the case that tablename is left empty in sqlSave, R
writes to the instance level schema (2) in the case that a tablename is
spec
Hi all,
I'm looking for a way to use loess model (?loess) with a different
percentile (instead of median).
I hope I made myself clear enough. Here's and example:
a<-data.frame(x=seq(1:200),y=200*rnorm(200))
lss<-loess(y~x,a)
predict(lss, 120) will give a local median of that dataset.
How about a
Hello R-list. I am a "long time listener - first time caller" who has
been using R in research and graduate teaching for over 5 years. I
hope that my question is simple but not too foolish. I've looked
through the FAQ and searched the R site mail list with some close hits
but no direct a
Hi Andrew,
Try also:
x1<-c(1824615,1823650,1821910)
y1<-c(5980732,5983220,5990931)
descript<-c("cat", "dog", "horse")
dat1<-data.frame(x1,y1,descript)
x2<-c(1824615,1823650)
y2<-c(5980732,5983220)
dat2<-data.frame(x2,y2)
colnames(dat2)=c('x1','y1')
merge(dat1,dat2,by=c('x1','y1'))
HTH,
Jorge
vector dat1.select is the selected records from dat1 by dat2.
> dat1.select<- dat1$x1 %in% dat2$x2 & dat1$y1 %in% dat2$y2
> dat1[dat1.select,]
x1 y1 descript
1 1824615 5980732 cat
2 1823650 5983220 dog
Andrew McFadden wrote:
>
> Hi R users
>
> I am trying to create a sp
In Dr. Wood's book on GAM, he suggests in section 4.1.6 that it might be
useful to shrink a single smooth by adding S=S+epsilon*I to the penalty
matrix S. The context was the need to be able to shrink the term to zero if
appropriate. I'd like to do this in order to shrink the coefficients towards
Andrew
?merge
HTH ..
Peter Alspach
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Andrew McFadden
> Sent: Wednesday, 7 May 2008 9:23 a.m.
> To: r-help@r-project.org
> Subject: [R] Spatial join between two datasets using x and y
> co-ordin
Thanks. I used the command you mention and it works fine in Linux. But I
need to get it to work for Windows XP as well (currently running R-2.7.0).
Any idea if it's possible?
On Tue, May 6, 2008 at 5:18 PM, Prof Brian Ripley <[EMAIL PROTECTED]>
wrote:
> How did you install it?
>
> You need to get
>
> I believe this function matches the description in OOO:
>
> mround <- function(number, multiple) multiple * round(number/multiple)
I've implemented a slightly more general form in the reshape package:
round_any <- function (x, accuracy, f = round) {
f(x/accuracy) * accuracy
}
Hadley
-
Hi R users
I am trying to create a spatial join between two datasets.
The first data set is large and contains descriptive data including x
and y co-ordinates.
The second dataset is small and has been selected spatially. The only
data contained within the second dataset is the x and y coordina
Dear Bill,
I expect that the problem is in the contrasts that your student used for A
and B, though I haven't thought specifically about the context of a mixed
model. If he or she used the default contr.treatment(), then the contrasts
for different factors (and the interaction) are not orthogonal
I am looking for a way to simulate genotypes of cases and control at a
disease locus in R. I am supposed to set the allele frequency as
control/cases. for each of the column below simulate 200 snp dataset.
I am looking at treesim function from popgen to stimulate the genotypes in
R.
Here is what
Thanks to all of you that helped me with the issues of bootstrapping
and downloading packages to a local disk.
As an starter I'm in the lower side of the learning curve, but this R
software is awesome. What I like most is this kind of forums when
people share their problems and we can find solution
0^(-0.2) = Inf, so you started with an infinite prediction for your first
point and hence an infinite sum of squares.
On Tue, 6 May 2008, Rick DeShon wrote:
Hi All.
I've run into a problem with the plinear algorithm in nls that is confusing
me.
Assume the following reaction time data over 15
Alex
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Alex Reynolds
> Sent: Wednesday, 7 May 2008 12:26 a.m.
> To: r-help@r-project.org
> Subject: [R] Dendrogram label size
>
> Is it possible to resize the labels in a dendrogram without
> applying
recall that 0 ^{-.2} = 1/0^{.2}, and that dividing by 0 gives Inf.
so when 0 is in trl, part of your model for RT is Inf:
> trl <- 0:14
> p <- -.2
> cbind(1,trl, trl^p)
trl
[1,] 1 0 Inf
[2,] 1 1 1.000
[3,] 1 2 0.8705506
[4,] 1 3 0.8027416
[5,] 1 4 0.7578583
[6,] 1
I suspect you have more than one version of R installed and are mixing
them up. Those symbols have been in package stats for quite a while.
Try starting R with --vanilla, and if that works, clean out your startup
files (see ?Startup). If not, remove all your R installations and
reinstall R 2
How did you install it?
You need to get A2R_0.0-4.tar.gz and do R CMD INSTALL A2R_0.0-4.tar.gz,
after reading the 'R Installation and Administration manual, especially
the sections for your OS.
The package in A2R/lastVersion/ was built for Unix on R 2.2.1. An expert
might be able to get it w
Hi All.
I've run into a problem with the plinear algorithm in nls that is confusing
me.
Assume the following reaction time data over 15 trials for a single unit.
Trials are coded from 0-14 so that the intercept represents reaction time in
the first trial.
trl RT
01132.0
1 630.5
2
On Tue, May 6, 2008 at 10:32 AM, Mark Kimpel <[EMAIL PROTECTED]> wrote:
> I am running 64-bit Ubuntu 8.04 and when I invoke rggobi the interactive
> graph displays but R crashes. See my sessionInfo() and a short example
> below. Ggobi and rggobi installed without complaints. Mark
>
> > sessionInfo
Hi Xavier,
Changing the grid settings has worked!
Thanks a lot!
Mikhail
PS Perhaps it'd still be really useful to be able to change text direction
in labels.
Hadley, what do you think?
--
View this message in context:
http://www.nabble.com/ggplo2%3A-x_discrete-labels-size-direction-tp1707747
Hi Birgit,
I'm not sure that I understand your question. I'll try to answer
anyways. Regression trees and therefore also RandomForests are invariant
to monotonic transformations in the independent variables. There are no
distributional assumptions for the independent variables. The dependent
Look at the addtable2plot function in the plotrix package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Anh Tran
> Sent: Tuesday,
Hi Anh,
Take a look at http://finzi.psych.upenn.edu/R/Rhelp02a/archive/128041.html
HTH,
Jorge
On Tue, May 6, 2008 at 3:46 PM, Anh Tran <[EMAIL PROTECTED]> wrote:
> Hi,
> Is there away to print a short table out along side with a plot?
> I'm thinking about doing a
>
> par(mfrow = c(1,2))
>
> T
Hi,
Is there away to print a short table out along side with a plot?
I'm thinking about doing a
par(mfrow = c(1,2))
Then, the plot is on one side, summary result on the other.
Is there any quick way to print out a data.frame in table format? Thanks
--
Regards,
Anh Tran
[[alternative HT
Jorge Ivan Velez wrote:
Hi Esmail,
Try this:
vars=c('X.1', 'X.2', 'X.3', 'X.4', 'X.5')
bits=c(1, 0, 1, 1, 0)
paste(vars[which(bits==1)],collapse="+")
HTH,
Jorge
Wow .. that is beautiful :-) .. and exactly what I was looking
for (and suspected existed).
I ended up doing this:
eqn=(paste
Thanks to all of you that helped me with the issues of bootstrapping
and downloading packages to a local disk.
As an starter I'm in the lower side of the learning curve, but this R
software is awesome. What I like most is this kind of forums when
people share their problems and we can find solution
Le mar. 06 mai à 14:23, Alberto Monteiro a écrit :
3) Bill Venables offered this about a week ago in this list:
--
This is probably as good a way as any way for this kind of problem.
First define a binary operator:
"%^%" <- function(x, n)
with(eigen(x), vectors %*% (values^
Fisher's exact test works with small cells. See ?fisher.test
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of raymond chiruka
> Sent:
Marc Schwartz wrote:
Esmail Bonakdarian wrote:
Hello,
Still a newbie with R, though I have learned a lot from reading
this list. I'm hoping someone can help with this question:
I have two vectors, one for variables, and one for bits.
I want to build a string (really a formula) based on the va
Hi there,
I've tried to install the A2R package using the files from
http://addictedtor.free.fr/packages/A2R/lastVersion/
This is the error I get when trying to load the library:
library(A2R)
Error in library(A2R) :
'A2R' is not a valid package -- installed < 2.0.0?
Can anyone please help? Th
I would actually go with this:
bits=c(1, 0, 1, 1, 0)
paste("X", which(bits==1), sep=".",collapse="+")
No need for the vars variable. Though admittedly it breaks down if
bits is identically 0.
Haris Skiadas
Department of Mathematics and Computer Science
Hanover College
On May 6, 2008, at 3:1
Esmail Bonakdarian wrote:
Hello,
Still a newbie with R, though I have learned a lot from reading
this list. I'm hoping someone can help with this question:
I have two vectors, one for variables, and one for bits.
I want to build a string (really a formula) based on the values in my
vector of 1
Hi Esmail,
Try this:
vars=c('X.1', 'X.2', 'X.3', 'X.4', 'X.5')
bits=c(1, 0, 1, 1, 0)
paste(vars[which(bits==1)],collapse="+")
HTH,
Jorge
On Tue, May 6, 2008 at 3:06 PM, Esmail Bonakdarian <[EMAIL PROTECTED]>
wrote:
> Hello,
>
> Still a newbie with R, though I have learned a lot from reading
Hello,
Still a newbie with R, though I have learned a lot from reading
this list. I'm hoping someone can help with this question:
I have two vectors, one for variables, and one for bits.
I want to build a string (really a formula) based on the values in my
vector of 1s and 0s in bits. If I have
probably you want:
stg <- c('a', 'b', 'c', 'a', 'c')
paste(unique(stg), collapse = ", ")
I hope it helps.
Best,
Dimitris
--
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16
Thanks Mark,
Your suggestion led me to this:
> !is.null(lapply(ls(pattern='bn'), function(y) cat(y, dim(get(y)),
"\t", names(get(y)), "\n")))
bn1993 2885 11 oplt rplt rsiz tree bd ht oaz odst raz rdst spr
bn1994 3158 7oplt tree bd ht spr stat dam
bn1995 734 7 oplt tree bd ht spr stat
Hi Pascal,
I think the function could be better but try this:
# Function: M is your matrix and n MUST be an integer>0
mat.pow<-function(M,n) {
result<-M
if(n>1){
for ( iter in 2:n) result<-M%*%result
result
}
else {result}
result
}
# The matrix
m <- rbind(c
Hi Anh,
Try this,
x=c('a','b','c','a','c')
paste(unique(x),collapse=", ")
[1] "a, b, c"
HTH,
Jorge
On Tue, May 6, 2008 at 2:22 PM, Anh Tran <[EMAIL PROTECTED]> wrote:
> I'm trying to use combine c('a','b','c','a','c') into 'a, b, c', the order
> does not matter.
>
> paste(c('a','b','c','a','c
> 3) Bill Venables offered this about a week ago in this list:
> --
> This is probably as good a way as any way for this kind of problem.
> First define a binary operator:
>
> > "%^%" <- function(x, n)
> with(eigen(x), vectors %*% (values^n * t(vectors)))
>
This example only w
Hello,
I have a set of one-liners (many thanks to previous responses from this
list) that I use to look at newly imported data sets with functions like
dim(), names(), str(), etc. within lapply(). Generally, these commands
work for me but, I am apparently still missing some aspect of list
manipula
I'm trying to use combine c('a','b','c','a','c') into 'a, b, c', the order
does not matter.
paste(c('a','b','c','a','c'), collapse=', ') yields 'a, b, c, a, c'.
Any idea?
--
Regards,
Anh Tran
[[alternative HTML version deleted]]
__
R-help@r-
Hello, I have come across a result that I cannot explain, and am hoping that
someone else can provide an answer. A student fitted a mixed model using
the lme function: out<- lme(fixed=Y~A+B+A:B, random=~1|Site). Y is a
continuous variable while A and B are factors. The data set is balanced
with
the data frames work as follows: columns= variables and rows= records. when
you plot with a data frame it's logical to plot one variable against another
not one record against another. with casting into a matrix and rotating then
casting back into a dataframe, you made your records variables.. thi
?gsub()
Match the start (_) followed by anything (.*) and replace by ""
gsub("_.*","",Name)
Weidong Gu,
Department of Medicine
University of Alabama, Birmingham
1900 University Blvd., Birmingham, Alabama 35294
PH: (205)-975-9053
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMA
Thanks for the help- It worked just fine:
This will cast my ignorance across the table, but here it goes. Why do I
need to make them a matrix? because they are in a row? if
d <- as.matrix(f)
f.t <- t(d)
f.m <- as.data.frame(f.t)
now I could use just the following
plot(f.m[c(rownumber), column],
#Install library rgl
#here is the function which you need to run first:
rgl.plot3d<-function(z, x, y, cols="red",axes=T,new=T)
{xr<-range(x)
x01<-(x-xr[1])/(xr[2]-xr[1])
yr<-range(y)
y01<-(y-yr[1])/(yr[2]-yr[1])
zr<-range(z)
z01<-(z-zr[1])/(zr[2]-zr[1])
if(new) rgl.clear()
if(axes)
I have a data frame such as
Name
DISC1_5916
DABC2_789
ABCD_1234
I want to substring column Name so that
DISC1
DABC2
ABCD
...
Thanks.
-
[[elided Yahoo spam]]
[[alternative HTML version deleted]]
__
Hi,
I'm having trouble plotting populations as separate colors and points in the
3d scatterplot package. I have a column with 4 different population names
and 3 columns with my data. I want to plot each population with a different
color and pch. In addition, I want to use the type="h" in my plotti
ComRades,
On Oct 16, 2006, I posted a question regarding how to find the parameter
values that made the likelihood function take a value equal to 1/8th of
the maximum likelihood value. There was no reply but I found a solution
and would like to know if there is a better solution using the funct
I am running 64-bit Ubuntu 8.04 and when I invoke rggobi the interactive
graph displays but R crashes. See my sessionInfo() and a short example
below. Ggobi and rggobi installed without complaints. Mark
> sessionInfo()
R version 2.7.0 Patched (2008-05-04 r45620)
x86_64-unknown-linux-gnu
locale:
L
Try also:
x1=as.numeric(f[2,4:26])
y1=as.numeric(f[1,4:26])
plot(x1,y1)
HTH,
Jorge
On Tue, May 6, 2008 at 12:43 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> f <- (structure(list(X = structure(96:97, .Label = c("119DAmm", "119DN",
> "119DNN", "119DO", "119DOC", "119Flow", "119Nit", "119ON",
On 5/6/2008 12:43 PM, stephen sefick wrote:
f <- (structure(list(X = structure(96:97, .Label = c("119DAmm", "119DN",
"119DNN", "119DO", "119DOC", "119Flow", "119Nit", "119ON", "119OPhos",
"119OrgP", "119Phos", "119TKN", "119TOC", "148DAmm", "148DN",
"148DNN", "148DO", "148DOC", "148Flow", "148Nit
cast the two vectors as.matrix-- see here:
plot(as.matrix(f[2,4:26]), as.matrix(f[1,4:26]))
y
stephen sefick wrote:
>
> f <- (structure(list(X = structure(96:97, .Label = c("119DAmm", "119DN",
> "119DNN", "119DO", "119DOC", "119Flow", "119Nit", "119ON", "119OPhos",
> "119OrgP", "119Phos", "11
Subject pretty much says it all. I am running 64-bit Ubuntu 8.04, i.e. Hardy
Heron, have openmpi installed, and get the following error message with
attempted install of Rmpi. sessionInfo() follows.
Mark
checking for ANSI C header files... yes
checking for sys/types.h... yes
checking for sys/stat
Dear all,
With R.2.7.0 (on windows XP) I have encountered what seems to be a "negative
memory size". If I use gc() afterwards the R goes down. I use R_alloc for
*most* memory allocations in my C-routines:
> memory.size()
[1] 11.08132
> dyn.load("rconipm.dll")
> dyn.unload("rconipm.dll")
Hi Ottorino,
You could just use the modulus operator "%%" as follows:
> x<-c(1803.02, 193.51, 3.47);
> x-x%%c(50,5,1) #just using the modulus operator
[1] 1800 1903
thanks
Dr. Ottorino-Luca Pantani wrote:
>
> Dear R-users,
> I have the following problem
>
> In a lab experiment I have to
Try:
x <- unlist(f[2, 4:26])
y <- unlist(f[1, 4:26])
plot(x, y)
On Tue, May 6, 2008 at 12:43 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> f <- (structure(list(X = structure(96:97, .Label = c("119DAmm", "119DN",
> "119DNN", "119DO", "119DOC", "119Flow", "119Nit", "119ON", "119OPhos",
> "119OrgP
f <- (structure(list(X = structure(96:97, .Label = c("119DAmm", "119DN",
"119DNN", "119DO", "119DOC", "119Flow", "119Nit", "119ON", "119OPhos",
"119OrgP", "119Phos", "119TKN", "119TOC", "148DAmm", "148DN",
"148DNN", "148DO", "148DOC", "148Flow", "148Nit", "148ON", "148OPhos",
"148OrgP", "148Phos",
On 5/6/2008 12:07 PM, Dr. Ottorino-Luca Pantani wrote:
Dear R-users,
I have the following problem
In a lab experiment I have to mix three solutions to get different
concentrations of various molecules in a cuvette
I've used R to calculate the necessary µliters for each of the level of
the ex
hie all
i am trying to carry out a categorical data analysis but my problem is that
when in i use the chi squared test some of my expected values are less than
5. is there a test that can handle this situation. the data is not a 2*2
table. its more from the social sciences where you have f
My summary of Bates' comments cited below is as follows:
1. ANOVA is an excellent tool but requires nested models.
You can do this fairly easily, but it is not so easily automated.
2. The standard definition of R^2 loses its meaning with
nonlinear models. Adjusted
Hi all,
I have issues using some basic functions in R such as these ones :
> pp.test(R) (where is a vector of returns)
Error in .C("R_approx", as.double(x), as.double(y), as.integer(nx), xout =
as.double(xout), :
C symbol name "R_approx" not in DLL for package "base"
>boxcox(reg,plotit=T)
Dear R-users,
I have the following problem
In a lab experiment I have to mix three solutions to get different
concentrations of various molecules in a cuvette
I've used R to calculate the necessary µliters for each of the level of
the experiment and I must confess that it is more useful and e
Indeed both options are salable, though I agree the latter may be more
convenient.
Merci!
Patrick Burns wrote:
>
> As the answers you've received suggest, you
> can use a list. Or you could have two
> vectors: one with the data, the other with the
> group identity. The latter format is likely
On 5/6/2008 11:46 AM, A Ezhil wrote:
Hi,
I have 3 vectors,
x=rnorm(10); y=rnorm(20); z=rnorm(30).
I would like to plot 3 vectors side by side (like a
bar plot) with scatter plot something similar to the
following:
.. *** ;;;
.. ** ;;;
.. *** ;;;
.. *** ;;;
... *** ;;
x
Try this:
subset(dat, format(date, "%Y-%m") == "1999-11")
On Tue, May 6, 2008 at 12:25 PM, <[EMAIL PROTECTED]> wrote:
> Dear list:
>
> I ask for your help in a simple problem in which I'm not figuring out
> the solution
>
> My data looks like:
> dat<- data.frame(date=c("12/12/1980", "03/11/199
Dear useRs,
This is to announce that the maintainers of the various distributions
have decided to provide "experimental" up-to-date versions of the
following R related packages on Debian stable and Ubuntu (i386 and
amd64 architectures):
littler
rkward
python-rpy
Hi,
I have 3 vectors,
x=rnorm(10); y=rnorm(20); z=rnorm(30).
I would like to plot 3 vectors side by side (like a
bar plot) with scatter plot something similar to the
following:
.. *** ;;;
.. ** ;;;
.. *** ;;;
.. *** ;;;
... *** ;;
x y z
How can I do this with Plot()?
T
Hi Eleni --
Although samr is not a Bioconductor package, you might have more luck
asking on the Bioconductor mailing list, http://bioconductor.org. The
obvious place to start, and probably you have already done this, is to
ensure that the class of the objects passed to the function agree with
the
Dear list:
I ask for your help in a simple problem in which I'm not figuring out
the solution
My data looks like:
dat<- data.frame(date=c("12/12/1980", "03/11/1994", "15/11/1999",
"31/10/2000", "20/03/2007", "05/01/2001"),
var1=c("A", "A", "B", "D", "C", "A"), var2=runif(6))
I was wondering if
Thank you very much Mark! That worked Just a question, ?[ does give an
error to me...how do I find it?
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com
-Original Message-
Hi,
Suppose
a=matrix(1:9,3,3)
> a
[,1] [,2] [,3]
[1,]147
[2,]258
[3,]369
Now,
> class(a[1:2,])
[1] "matrix"
> class(a[1:3,])
[1] "matrix"
> class(a[,1:2])
[1] "matrix"
> class(a[,1:3])
[1] "matrix"
But,
> class(a[1,])
[1] "i
On 5/6/2008 10:31 AM, Megh Dal wrote:
Hi all,
I have following problem :
a = b = seq(1, 5, by=500)
v = matrix(0, nrow=length(a), ncol=length(a))
for (i in 1:length(a))
{
for (j in 1:length(a))
{
d = c(17989*a[i], -18109*b[j])
v[i,j] = t(d) %*% matrix
Wonderful...This works...
lapply(list(x1,x2,x3,x4),plot,type="l")
Thanks a lot!
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of David Winsemius
Sent: Tuesday, May 06, 2008 7:55 PM
To: [EMAIL PROTECTED]
Subject: Re: [R] single plot statement, multiple plot
Hi all,
I have following problem :
a = b = seq(1, 5, by=500)
v = matrix(0, nrow=length(a), ncol=length(a))
for (i in 1:length(a))
{
for (j in 1:length(a))
{
d = c(17989*a[i], -18109*b[j])
v[i,j] = t(d) %*% matrix(c(0.0001741, 0.0001280, 0.0001280, 0.000
"Shubha Vishwanath Karanth" <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
> Hi R,
>
> par(mfrow=c(2,2))
>
> x1=(1:5)^1; x2=(1:5)^2; x3=(1:5)^3; x4=(1:5)^4
>
> I need to write a single plot statement, which creates 4 plots (for
> x1, x2, x3 and x4) in the graphics window, without using '
Thank you very much Gabor...Zoo is very powerful...
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent:
Try plot.zoo in which case you don't need the par:
library(zoo)
plot(zoo(cbind(x1, x2, x3, x4)), nc = 2)
or
plot(zoo(outer(1:5, 1:4, "^")), nc = 2)
See ?plot.zoo, ?xyplot.zoo and the three vignettes in
the zoo package.
On Tue, May 6, 2008 at 9:47 AM, Shubha Vishwanath Karanth
<[EMAIL PROTECTED
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