Hi
[EMAIL PROTECTED] napsal dne 21.02.2008 22:57:31:
>
> useR's,
>
> I want to apply this function to the columns of a data frame:
>
> u[u >= range(v)[1] & u <= range(v)[2]]
Do you really mean this function?
range(v) gives you an overall range in data frame v e.g. (1,5)
and u[your ranges]
Hi there,
I am working on clustering genomic data. I am using HOPACH method for
clustering.
I have a query here. How can i get SS (split silhouette) value for
individual clusters. What I am getting as MSS value, is Mean of all SS value
of individual cluster. Hence I am getting single value for a
> # 2. create one new row for each case in level.1 and level.2
>
> # the new reshaped data.frame would should look like this:
>
> # indiv factorcovar case.id
> # A level.1 4.6141051
> # A level.1 4.6141052
> # A level.2 31.0644051
> # A level.2 31.064405
On 21/02/2008 3:13 PM, Kenneth Lo wrote:
> Thank you for your reply. Hope that it will be
> available soon.
It won't happen unless someone does it. What you should be asking is
how to learn to do it, or how to encourage someone else to do it.
Duncan Murdoch
>
> Kenneth
>
>
> --- Duncan Mur
Jim's solution showed me that mine should be simplified to:
mapply(data.frame,value=g,nm=names(g),SIMPLIFY=FALSE)
This has the slight advantage of automatically naming the list.
Cheers,
Bert Gunter
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of jim hol
Here is one way of doing it:
> lapply(names(g), function(z)cbind(x=g[[z]], var1=z))
[[1]]
x var1
1 1a
2 2a
3 3a
[[2]]
x var1
1 4b
2 5b
3 6b
[[3]]
x var1
1 7c
2 8c
3 9c
On Thu, Feb 21, 2008 at 1:22 PM, Lauri Nikkinen <[EMAIL PROTECTED]> wrote:
> R users
mapply(data.frame,value=g,nm=lapply(names(g),rep,length(g)),SIMPLIFY=FALSE)
Cheers,
Bert Gunter
Genentech
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of John Kane
Sent: Thursday, February 21, 2008 2:05 PM
To: Lauri Nikkinen; [EMAIL PROTECTED]
Subject: R
On Thu, Feb 21, 2008 at 4:51 PM, MassimoH <[EMAIL PROTECTED]> wrote:
>
> How do I plot data like this:
>
> Date,Series,Value
> 1/5/2007,Blue,300
> 1/20/2007,Blue,400
> 1/9/2007,Red,200
> 1/15/2007,Red,500
>
> Two things I'm having trouble figuring out from online help & tutorials:
>
> 1) How does R
> So what could be the difference between the last example and it running
> in a loop, that the former generates a proper jpeg file and the lattter
> does not?
>
> Does something more special than dev.off() need to be done with qplot
> output when in a loop?
As well as Henrik's advice, you co
Hi,
In the boot package, the original statistic is simply the statistic
function evaluated on the original data (called t0).
However, in Harrell et al 1996 "Multivariable prognostic models..."
Stats Med vol 15, pp. 361--387, it is different (p. 372):
The statistic function evaluated on the origin
useR's,
I want to apply this function to the columns of a data frame:
u[u >= range(v)[1] & u <= range(v)[2]]
where u is the n column data frame under consideration and v is a data frame
of values with the same number of columns as u. For example,
v1 <- c(1,2,3)
v2 <- c(3,4,5)
v3 <- c(2,3,4)
v
I agree with what others have said, the R core team is a great and
unique group.
There are a couple of ideas that I would like to add that may have
played a part in the level of growth that R has had.
I think timing has played a part. The field of statistics has matured
along with the computer.
Many of the doctors that I work with also have teaching appointments at
the local university and use the bookstore to buy academic versions of
SAS, SPSS, Minitab, and others. When they have asked me to use those
(sometimes offered to buy me a student copy) I have refered them to the
licence agreem
I just realised I made a bad mistaqke (see below)
On 21-Feb-08 21:39:56, Ted Harding wrote:
> On 21-Feb-08 20:58:25, Evgenia wrote:
>>
>> Dear R users,
>> I would like to sample from a mixture distribution
>> p1*f1+p2*f2+p3*f3 with f1,f2,f3 three different forms
>> of distributions. I know that
Hi,
I'm new to R and am hoping someone might be able to help with the following lme
problem.
I am trying to fit an ellipse equation to some spatial human factors data,
varying the major and minor axes randomly and specifying an exp~ variogram for
errors. Using normally-distributed random effects
I don't see any difference
--- Lauri Nikkinen <[EMAIL PROTECTED]> wrote:
> R users,
>
> I have a simple lapply question.
>
> g <- list(a=1:3, b=4:6, c=7:9)
> g <- lapply(g, function(x) as.data.frame(x))
> lapply(g, function(x) cbind(x, var1 = rep(names(g),
> each=nrow(x))[1:nrow(x)]))
>
> I ge
Here's a similar variant on what has been proposed, but is simpler. It
relies on the fact that plot() doesn't need a ~.
a<-1:100
b<-seq(1,length(a),5)
plot(1:20, a[1:b])
Alternately, if you were using a data frame, as long as you knew the column
names, you could do something like
plot(my.dat
How do I plot data like this:
Date,Series,Value
1/5/2007,Blue,300
1/20/2007,Blue,400
1/9/2007,Red,200
1/15/2007,Red,500
Two things I'm having trouble figuring out from online help & tutorials:
1) How does R handle dates? It seems there is no built-in support for dates,
but some standard add-ons
Hello list,
I'm new to this list, so please forgive my ignorance. I have searched
R-help for some hints into what might be my problem, but I truly have no
idea where to go from here.
I have an object of approximately 15,000 rows and 2 columns. There are
many duplicates in the first column
On 21-Feb-08 20:58:25, Evgenia wrote:
>
> Dear R users,
> I would like to sample from a mixture distribution
> p1*f1+p2*f2+p3*f3 with f1,f2,f3 three different forms
> of distributions. I know that in the case of two
> distributions I have to sample the mixture compoment
> membership.
>
> How can
On Thu, 21 Feb 2008, Luís Paulo F. Garcia wrote:
In nnet.formula(classe ~ ., data = treinamento_data, size = 10, : group
Class is empty
What does this mean?
Well, you are the best placed to tell us as we don't have a reproducible
example (see the footer) and you do.
But at a guess, 'class
On Thu, 21 Feb 2008, Elizabeth Purdom wrote:
> Hello,
>
> I am running some commands in batch on a server that I SSH into; some of
> the commands call jpeg(). If I continue to stay signed on (with my
> xwindows working on my machine) then the jpeg() command works (with a
> single caveat, below). I
Hi.
I think the devices provided in 'Cairo' package solves your problem.
[ Alternatively you can use the bitmap() device which utilizes
Ghostscript to generate bitmap images. In the 'R.utils' package there
are wrappers png2() and jpeg2() calling bitmap() but that imitates the
png() and jpeg() ar
Hello,
I am running some commands in batch on a server that I SSH into; some of
the commands call jpeg(). If I continue to stay signed on (with my
xwindows working on my machine) then the jpeg() command works (with a
single caveat, below). If I leave it and sign off (with a nohup command
in fr
Dear R users,
I would like to sample from a mixture distribution p1*f1+p2*f2+p3*f3 with
f1,f2,f3 three different forms of distributions.
I know that in the case of two distributions I have to sample the mixture
compoment membership.
How can I weight my three distributions with their respective
Try this:
plot(x[,1], ylim=range(x[c(1,3)]), xaxt="n", xlab="", pch=16, col="blue")
points(x[,3], pch=16, col="red")
axis(1, at=1:nrow(x), labels=rownames(x), las=2)
segments(1:nrow(x), x[,1]-x[,2], 1:nrow(x), x[,1]+x[,2], col="blue")
segments(1:nrow(x), x[,3]-x[,4], 1:nrow(x), x[,3]+x[,4], col="r
First,
please provide us with enough unambiguous information so we don't have
to second guess that you are using qplot() in the 'ggplot' package.
Report sessionInfo() is recommended.
It has nothing to do with jpeg() or any other device driver. You get
the same problem with:
library(ggplot)
x <-
Chris Rhoads wrote:
>
>
> I wish to find the root of a function of two variables that is defined by
> an integral which must be
> evaluated numerically.
>
> So the problem I want to solve is of the form: Find k such that f(k)=0,
> where f(y) = int_a^b
> g(x,y) dx. Again, the integral
> invo
Hi Georg,
Georg Ehret wrote:
> Dear R community,
>I would wish to color the background of my histogram differently to the
> left and right of an abline... Can you please help?
maybe not very elegant or beautiful, but I hope it will get you started:
### Start of Code Example
n
Thank you for your reply. Hope that it will be
available soon.
Kenneth
--- Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 20/02/2008 4:54 PM, Kenneth Lo wrote:
> > Hi all,
> >
> > I encountered difficulties when I tried to install
> the
> > rgl package. I'm using R devel (2.7.0) on Mac OS
>
Henrique Dallazuanna schrieb:
Perhaps:
plot(x[c(1,3)], pch=16)
segments(x[,1],x[,3]-x[,2], x[,1], x[,3]+x[,4], col="red")
On 21/02/2008, Paul Hammer <[EMAIL PROTECTED]> wrote:
hi members,
i try to plot two columns of a matrix as points with standard deviation.
a legend should also be th
To answer my own question, the notes that I was drawing from made it seem as
through the mean I was using for my sum of squares was unweighted. In
reality, I needed to weight the contribution of each contributing mean by
the sample size. Once I did that, I got a value that agreed with SAS.
So, the
In nnet.formula(classe ~ ., data = treinamento_data, size = 10, : group
Class is empty
What does this mean?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read t
Hi All,
In 2005, I noted that when I used read.xport function in the foreign
package to import a SAS transport file (.xpt) with multiple files
imbedded in it, data frames from some of the files were not created.
Since then, I discovered that the information from the missing files is
being read in
Dear R community,
I would wish to color the background of my histogram differently to the
left and right of an abline... Can you please help?
Thankin you,
Georg.
Georg Ehret
JHU
Baltimore - USA
[[alternative HTML version deleted]]
Perhaps:
plot(x[c(1,3)], pch=16)
segments(x[,1],x[,3]-x[,2], x[,1], x[,3]+x[,4], col="red")
On 21/02/2008, Paul Hammer <[EMAIL PROTECTED]> wrote:
> hi members,
>
> i try to plot two columns of a matrix as points with standard deviation.
> a legend should also be there. the one points should be
Hi,
I understand for 1 d classifiers, you can use ROCR package.
Is there a package you can plot ROC curve for 2d classifiers? One of
my colleagues asked me about this. I have been quite puzzled,
conceptually, how you can do the ROC curve for 2d classifiers. Can
someone share his/her knowledge
That's perfect! The subset way is very easy to use and
I have to play around a bit more with label() which
seems quite complex.
Thanks a million,
Werner
--- Gabor Grothendieck <[EMAIL PROTECTED]>
schrieb:
> On Thu, Feb 21, 2008 at 6:09 AM, Werner Wernersen
> <[EMAIL PROTECTED]> wrote:
> > Hi,
hi members,
i try to plot two columns of a matrix as points with standard deviation.
a legend should also be there. the one points should be green and the
others red...
column 1 and 3 are the values to plot and column 2 and 4 are the
attendant standard deviation (sd)... as xlab should show the
Hello all,
I'm stuck with a strange issue with writing jpegs of plots to a folder
in a loop.
This works:
for (step in 1:length(steps)) {
jpeg(filename=paste("frame_",sprintf("%05d",step),".jpg",sep=""))
plot(steps[[step]])
dev.off()
}
But if I use qplot to generate the plot (which is my a
On Thursday 21 February 2008 (19:22:40), Lauri Nikkinen wrote:
> R users,
>
> I have a simple lapply question.
>
> g <- list(a=1:3, b=4:6, c=7:9)
> g <- lapply(g, function(x) as.data.frame(x))
> And I would like to have
>
> $a
> x var1
> 1 1 a
> 2 2 a
> 3 3 a
>
> $b
> x var1
> 1 4
Try this:
g <- list(a=1:3, b=4:6, c=7:9)
with(stack(g), split(stack(g), ind))
On 21/02/2008, Lauri Nikkinen <[EMAIL PROTECTED]> wrote:
> R users,
>
> I have a simple lapply question.
>
> g <- list(a=1:3, b=4:6, c=7:9)
> g <- lapply(g, function(x) as.data.frame(x))
> lapply(g, function(x) cbin
Hello!
I have posted this message before, but no answer. Now, after a few
upgrades, I still have the same warning messages. I am a bit worried,
since I do not know does this affect or could affect R. I use R under
Linux (Ubuntu 6.06).
After start, and after loading few libraries, I got following
I sent a message a couple days ago about doing calculations for power of the
ANOVA. Several people got back to me very quickly which I really
appreciated.
I'm working now on a similar problem, but instead of a balanced ANOVA, I
have an unbalanced one. The first part of the question was:
You assum
R users,
I have a simple lapply question.
g <- list(a=1:3, b=4:6, c=7:9)
g <- lapply(g, function(x) as.data.frame(x))
lapply(g, function(x) cbind(x, var1 = rep(names(g), each=nrow(x))[1:nrow(x)]))
I get
$a
x var1
1 1a
2 2a
3 3a
$b
x var1
1 4a
2 5a
3 6a
$c
x var1
Em Qui 21 Fev 2008, David Winsemius escreveu:
> Ronaldo Reis Junior <[EMAIL PROTECTED]> wrote in
>
> news:[EMAIL PROTECTED]:
> > I try this update:
> >
> > mmaa <- update(mma,biomass~qvartemp)
>
> That does not look like it has proper arguments to update.formula. As
> the update() help page suggest
Try the triax.plot in plotrix package
http://rss.acs.unt.edu/Rdoc/library/plotrix/html/triax.plot.html
Zhaoming
-Original Message-
From: WCD [mailto:[EMAIL PROTECTED]
Sent: Thursday, February 21, 2008 12:21 PM
To: r-help@r-project.org
Subject: [R] triangle.plot - change the axes orientat
Great! Now it works. Many thanks.
Dani
Daniel Valverde Saubí
Grup de Biologia Molecular de Llevats
Facultat de Veterinària de la Universitat Autònoma de Barcelona
Edifici V, Campus UAB
08193 Cerdanyola del Vallès- SPAIN
Centro de Investigación Biomédica en Red
en Bioingeniería, Biomateriales y
Hello, I need to chenge axes orirentation in triangle plot. (function
triangle.plot in ade4 package)
I want to plot elasticities of some species in demographic triangle, where
axes values commnly increace "clockwise".
If some better imangination is needed, see
http://www.open.ac.uk/science/biosci/
> spect2 <- insert(spect1, ats=pos, values=as.list(intensities))
> str(spect2)
num [1:13112] -0.457 -0.457 0.300 1.781 -0.381 ...
/Henrik
On Thu, Feb 21, 2008 at 9:26 AM, Henrik Bengtsson <[EMAIL PROTECTED]> wrote:
> On Thu, Feb 21, 2008 at 4:30 AM, Dani Valverde <[EMAIL PROTECTED]> wrote:
>
On Thu, Feb 21, 2008 at 4:30 AM, Dani Valverde <[EMAIL PROTECTED]> wrote:
> Hello,
> I still having problems with insert() from R.utils package. I provide a
> code with no error:
>
> x <- seq(1:10909)
> x1 <- c(13112-10909)
> spect1 <- rnorm(10909)
>
> interpol <- approx(x,spect1,xout=c(seq(fro
Hi,
the package 'survival' provides the possibility to estimate Cox PH
models with a frailty term.
library(survival)
?frailty
If this is not what you are looking for, there is also a package called
'frailtypack'. But I have no experience with that.
I hope this helps.
Best,
Roland
P.S. Boccon
Look at using a list to store the data, something like this:
> results <- list()
> for (year in 2002:2008){
+ results[[as.character(year)]] <- matrix(year,10,10)
+ }
> results
$`2002`
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 2002 2002 2002 2002 2002 2002 2002 2002 2002 2
Here are two ways of doing it with POSIXct and one with chron.
Read RNews 4/1, ?read.zoo and the three zoo vignettes:
Lines <- '"Data","MAX","MIN","Note"
"07-01-2008 08:00:00", 135, 90, "Eccessi feste, inizio dieta"
"07-01-2008 18:00:00", 135, 85, ""
"08-01-2008 08:00:00", 125, 75, ""
'
library(zo
Also consider using a 'list' to store the results.
On 2/21/08, Paul Hammer <[EMAIL PROTECTED]> wrote:
> Paul Hammer schrieb:
> > jim holtman schrieb:
> >> Exactly what do you mean by additional text? Have you tried "paste"?
> >>
> >> On 2/21/08, Paul Hammer <[EMAIL PROTECTED]> wrote:
> >>
> >>> d
You need to convert to POSIXct since POSIXlt is a vector of size 9.
So do the following:
miedate <- as.POSIXct(strptime(as.character(pressione[,1]),
format="%d-%m-%Y %H:%M:%S"))
There is a newsletter (I forget the issue) that you might want to
refer to on using 'dates'.
On 2/21/08, vittorio <
?assign
On 2/21/08, Paul Hammer <[EMAIL PROTECTED]> wrote:
> Paul Hammer schrieb:
> > jim holtman schrieb:
> >> Exactly what do you mean by additional text? Have you tried "paste"?
> >>
> >> On 2/21/08, Paul Hammer <[EMAIL PROTECTED]> wrote:
> >>
> >>> dear members,
> >>>
> >>> i would like to wr
Paul Hammer schrieb:
> jim holtman schrieb:
>> Exactly what do you mean by additional text? Have you tried "paste"?
>>
>> On 2/21/08, Paul Hammer <[EMAIL PROTECTED]> wrote:
>>
>>> dear members,
>>>
>>> i would like to write a variable in a plot title (main="") but i don't
>>> know the right syn
jim holtman schrieb:
> Exactly what do you mean by additional text? Have you tried "paste"?
>
> On 2/21/08, Paul Hammer <[EMAIL PROTECTED]> wrote:
>
>> dear members,
>>
>> i would like to write a variable in a plot title (main="") but i don't
>> know the right syntax:(...i tried a lot of differ
On Thu, 2008-02-21 at 11:06 -0500, b.r. clark wrote:
> > hist.with.normal<-function(x,xlab=deparse(substitute(x)),...)+ (+
> h<-hist(x, plot=F, ...)+ s<-sd(x)Error: unexpected symbol
> in:"h<-hist(x, plot=F, ...)s">
>
> I am using ISwR to teach myself R. Above is what happens when I try to
> dup
In the text file pressione2008.csv I have the following
"Data","MAX","MIN","Note"
"07-01-2008 08:00:00", 135, 90, "Eccessi feste, inizio dieta"
"07-01-2008 18:00:00", 135, 85, ""
"08-01-2008 08:00:00", 125, 75, ""
which is a collection of blood pressure data at different time of the day.
I woul
Dear R-help,
I am trying to estimate a Cox model with nested effects, or better
h(t,v,w)=v*w*h0(t)*exp(B'x)
where h(t,v,w) is the individual hazard function
w and v are both frailty terms (gamma or normal distributed)
I have 12 clusters and for each one of them I would like to associate a
> hist.with.normal<-function(x,xlab=deparse(substitute(x)),...)+ (+ h<-hist(x,
> plot=F, ...)+ s<-sd(x)Error: unexpected symbol in:"h<-hist(x, plot=F, ...)s">
I am using ISwR to teach myself R. Above is what happens when I try to
duplicate the example on page 32. What went wrong? Note: I am u
Exactly what do you mean by additional text? Have you tried "paste"?
On 2/21/08, Paul Hammer <[EMAIL PROTECTED]> wrote:
> dear members,
>
> i would like to write a variable in a plot title (main="") but i don't
> know the right syntax:(...i tried a lot of different ways without success.
>
> here
It seems that is not posible to send R file in the messages, well, then I
resend the message with the script included.
Hello, I'm creating a loop to work with vegan, to get a species abundance
curve. Here I send the script I've created and also an excel file to prove
what it can do.
Well, I ha
Hello, I'm creating a loop to work with vegan, to get a species abundance
curve. Here I send the script I've created and also an excel file to prove what
it can do.
Well, I have a database with 20 years, and each year we have sampled 19
stratum, and in each estratum we have carry out some sumpli
dear members,
i would like to write a variable in a plot title (main="") but i don't
know the right syntax:(...i tried a lot of different ways without success.
here my example:
y=30
z=33
for (i in 10:length(tissue)) {
png(filename = tissues[i], width = 1024, height = 768, pointsize = 12,
bg =
"Angelo Passalacqua" <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
> I entered the following:
>
> formula<-nst~age+soc+inc+reg+imp
>
> pnstlm<-lm(formula,nst)
>
> summary(pnstlm)
>
> imp and soc are ordered categorical variables but the summary does
> not give an output of the overall
Here is a very clumsy way to do it but I think it
works
fact1 <- rep("level.1", length(mydat[,1]))
fact2 <- rep("level.2", length(mydat[,1]))
lels <- c(fact1,fact2)
nams <- c("indiv", "case.id", "covar")
set1 <- mydat[, c(1,2,3)] ; names(set1) <- nams
set2 <- mydat[,c(1, 4,5)] ; names(set2) <-
On Thu, 2008-02-21 at 14:25 +, Angelo Passalacqua wrote:
> I entered the following:
>
> formula<-nst~age+soc+inc+reg+imp
>
> pnstlm<-lm(formula,nst)
>
> summary(pnstlm)
>
> imp and soc are ordered categorical variables but the summary does not give
> an output of the overall p-values, just
Perhaps:
txt <- list(A=bquote(a == .(a)~'%'), B=bquote(b==.(b)~km^2))
plot(c(0,100), c(0,100), xaxs="i", yaxs="i", type="n")
text(95-cumsum(strheight(txt)), 95-cumsum(strheight(txt)), parse(text=txt))
On 21/02/2008, Bálint Czúcz <[EMAIL PROTECTED]> wrote:
> Dear HelpeRs,
>
> I would like to do
Ronaldo Reis Junior <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
> I try this update:
>
> mmaa <- update(mma,biomass~qvartemp)
That does not look like it has proper arguments to update.formula. As
the update() help page suggests:
?update.formula
You should have included the call tha
I entered the following:
formula<-nst~age+soc+inc+reg+imp
pnstlm<-lm(formula,nst)
summary(pnstlm)
imp and soc are ordered categorical variables but the summary does not give
an output of the overall p-values, just individual comparisons. I can't
find help for this in the manual. Is there a co
Julien Martin usherbrooke.ca> writes:
>
> Hi
>
> I want to do use AIC for model selection on mixed model. However, before
> going deeper in the model selection, I want to assess is there is
> overdispersion with the full model in order to decide if I should use QAIC
> instead of AIC.
>
> Is t
On Tue, Feb 19, 2008 at 11:07 PM, Chris Rhoads
<[EMAIL PROTECTED]> wrote:
> To start, let me confess to not being an experienced programmer, although I
> have used R fairly
> extensively in my work as a
> graduate student in statistics.
>
> I wish to find the root of a function of two variable
Josep Maria Campanera Alsina ub.edu> writes:
>
> Dear all,
> I wonder which R algorithm could perform a search of local maxima in
> an spatial grid, in other words, having the coordinates of a map
> (x,y,z... up to 6 coordinates) and then the altitude/height at each
> point (h) (in total 7 numer
On Thu, Feb 21, 2008 at 6:09 AM, Werner Wernersen
<[EMAIL PROTECTED]> wrote:
> Hi,
>
> I have two loosely related questions which could make
> my live again a bit easier:
>
> 1) Is there a simple way to select a range of columns
> in a data frame using column names?
> I am thinking of something lik
Dear HelpeRs,
I would like to do sg similar to:
plot(c(0,100), c(0,100), xaxs="i", yaxs="i", type="n")
txt <- paste("a =", a, "%\n b =", b, " km2", sep="")
text(95, 95, txt, adj=c(1,1))
just with the "km2" formatted with 2 in a superscript.
I thought
txt <- substitute("a = "*a*" %\n b ="*b*" km
Hello,
I still having problems with insert() from R.utils package. I provide a
code with no error:
x <- seq(1:10909)
x1 <- c(13112-10909)
spect1 <- rnorm(10909)
interpol <- approx(x,spect1,xout=c(seq(from=1, by=((10909 - 1)/(x1 -1)),
length.out=x1)))
pos <- round(interpol$x,0)
intensities <- inte
Try this:
mydf <- data.frame(s8v2=rnorm(10), s12v3=runif(10), s1a5=rnorm(10),
s6v1=rnorm(10))
pos <- match(c('s8v2', 's1a5'), names(mydf))
mydf[,seq(pos[1], pos[2])]
On 21/02/2008, Werner Wernersen <[EMAIL PROTECTED]> wrote:
> Thank you very much, Henrique! The comment function is
> exactly what
Thank you very much, Henrique! The comment function is
exactly what I was looking for.
Regarding 1) why example was bad: The column names
don't follow an easy pattern but are more like "s8v2",
"s12v3", "s6v1" etc.
Kind regards,
Werner
--- Henrique Dallazuanna <[EMAIL PROTECTED]> schrieb:
>
Edna Bell wrote:
>
> ...
> I'm using the find command, (surprise), but I only want to see the
> output where it exists, not all of the other stuff.
>
> Is there an option that I could select, please?
>
Hi Edna,
When you say you want to see the output, I assume you mean output files
created by R.
Maybe also:
$ whereis R
In Debian based distributions works
On 20/02/2008, Edna Bell <[EMAIL PROTECTED]> wrote:
> Dear R Gurus:
>
> I'm trying to find R on another Linux system.
>
> I'm using the find command, (surprise), but I only want to see the
> output where it exists, not all of the oth
Try this:
1)
mydf[1,paste("col", 4:8, sep="")]
2) You can use comment function:
comment(mydf) <- c("Column number1", "Column number2")
or
comment(mydf$col1) <- "Column number1 etc"
then
comment(mydf) and comment(mydf$col1) to see the labels
On 21/02/2008, Werner Wernersen <[EMAIL PROTECTED
Hi,
I have two loosely related questions which could make
my live again a bit easier:
1) Is there a simple way to select a range of columns
in a data frame using column names?
I am thinking of something like mydf[1,"col4":"col8"]
2) I have a data frame with many columns and they all
have short
Dear all,
I wonder which R algorithm could perform a search of local maxima in
an spatial grid, in other words, having the coordinates of a map
(x,y,z... up to 6 coordinates) and then the altitude/height at each
point (h) (in total 7 numerical variables) I would like to localise
the peaks (local ma
> Am I right in thinking that the correct way of working out how many
possible
> permutations there can be in this instance is: N! / (n1!) (n2!) (n3!)...
> (ni!). Where N is the number of taxa (14) and (for character 2) n1 could
be
> 'number of state zeros' (i.e., 2), n2 could be 'number of state
bartjoosen wrote:
>
>
> Greg Snow-2 wrote:
>>
>>> write.table(my.data, 'clipboard', sep="\t")
>> Then in Excel just do a paste and the data is there, this saves a couple
>> of steps from saving as a .csv file and importing that into excel. This
>> would probably be fine for a few tables.
>
Hi,
I have a situation here.
I try this update:
mmaa <- update(mma,biomass~qvartemp)
but I have this message:
Error in eval(expr, envir, enclos) : object "qvartemp" not found
but this object exist:
[1] "cont" "i" "levelsord" "mma" "qvar" "qvarmma"
[7] "qvartemp" "t
Em Qua 20 Fev 2008, Henrique Dallazuanna escreveu:
> Try this:
> factor(pH,levels=names(sort(tapply(Riqueza,pH,mean
>
Thanks
it work
Ronaldo
--
> Prof. Ronaldo Reis Júnior
| .''`. UNIMONTES/Depto. Biologia Geral/Lab. de Biologia Computacional
| : :' : Campus Universitário Prof. Darcy Ribei
Thanks Bill,
How about situations where time steps are not hourly?
Best regards,
Lauri
2008/2/21, [EMAIL PROTECTED] <[EMAIL PROTECTED]>:
> Since both sequences are in hourly steps, there is a fairly easy way to
> do this:
>
> > before <- colSums(outer(seq1, seq2, "<"))
> > sort(c(seq1[before], s
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