Dear R-helpers,
Please look at the following minimal code.
\documentclass[a4paper]{article}
\begin{document}
==
library(zoo)
a-zoo(1:4,order.by=Sys.time()+1:4)
str(a)
@
\end{document}
When I do R CMD Sweave,followed by pdflatex ,and view the final pdf, the
letter surrounding the phrase zoo,
a
Thank you. It works great.
Kristian
2011/9/16 R. Michael Weylandt michael.weyla...@gmail.com
I think changing nrow() to NROW() will get around that error. However,
this, as you can see, this goes a little astray at the end. Perhaps
something like this will work for you:
z.ext -
On 09/19/2011 12:49 PM, aurelien wrote:
Dear all,
I need to compute a joint histogram to compare 2 images (with image()
function). I wonder if there is any function available in R to do this
properly?
Thanks for your help,
Aurrélien
--
View this message in context:
Hi all,
I have a very simple problem, but i cant find a solution probably because i
search for the wrong keywords.
I have coordinates x and y. They look like this:
str(data)
'data.frame': 13979 obs. of 2 variables:
$ x : Factor w/ 3815 levels ,186011.4971280,..: 1528 1524
1524
this is an FAQ (forget which one), but the solution is:
as.numeric(as.character(data$x))
Sent from my iPad
On Sep 20, 2011, at 3:40, _Luc_ l.temarve...@nioo.knaw.nl wrote:
Hi all,
I have a very simple problem, but i cant find a solution probably because i
search for the wrong keywords.
I
I have a data set with many missing observations. When I run a
regression, R of course discards the observations (the whole row) that
have NA. I want to tabulate some baseline characteristics (column
means) but only for the observations that R used for the regression.
I tried to recreate this
Hi,
The str_locate function instringr package may do what you are looking for.
Hope this link will help...
http://en.wikibooks.org/wiki/R_Programming/Text_Processing
Taka
On Mon, Sep 19, 2011 at 7:15 PM, SNV Krishna kris...@primps.com.sg wrote:
Hi All,
I have a character vector by name
Hi
[R] Problem with converting factors to numbers
Hi all,
I have a very simple problem, but i cant find a solution probably
because i
search for the wrong keywords.
I have coordinates x and y. They look like this:
str(data)
'data.frame': 13979 obs. of 2 variables:
$ x
I am not sure who to send this to - so I am writing this here.
It seems that the index.html pages on CRAN are gone.
For example:
http://cran.r-project.org/web/packages/ggplot2/index.html
Is gone. But the directory is accessible:
http://cran.r-project.org/web/packages/ggplot2/
Who should this be
On Tue, 20 Sep 2011, Tal Galili wrote:
I am not sure who to send this to - so I am writing this here.
It seems that the index.html pages on CRAN are gone.
For example:
http://cran.r-project.org/web/packages/ggplot2/index.html
Is gone. But the directory is accessible:
On 09/19/2011 04:46 PM, Henri-Paul Indiogine wrote:
Greetings!
I am using the R library RQDA to assign certain codes to paragraphs of
documents in a collection. Several paragraphs are assigned more than
1 code. E.g. often the codes poverty and education will be
assigned to the same
hello,
thank you for your answer!
yes, now it is working!
marion
2011/9/19 Duncan Murdoch murdoch.dun...@gmail.com
On 11-09-19 7:30 AM, Marion Wenty wrote:
Hello,
could someone help me with this problem?:
I would like to create a latex-script inside of a character vector in
order
to
I have another question concerning the paste command:
now instead of a vector I would like to paste the elements of a matrix
together, which works in the same:
Mypastedmatrix - paste(Mymatrix,collapse=)
My problem now is that the program does this BY COLUMN, but I would like to
have the
Dear R- Splus experts,
In R, I have frequently used do.call with strsplit. and I have a hard time
with Splus.. any suggestions?
for example, the R code below:
do.call(rbind,strsplit(paste(letters[1:10],c(1:10)), ))
Thanks so much,
Santosh
On Fri, Dec 5, 2008 at 8:51 AM, William Dunlap
Hi marion,
just transpose the matrix:
mm-matrix(LETTERS[1:20],nrow=5)
paste(t(mm),collapse=)
or - if you want the result seperated by rows
apply(mm,1,paste,collapse=)
cheers
Am 20.09.2011 11:55, schrieb Marion Wenty:
I have another question concerning the paste command:
now instead of a
Hi,
I am trying to add a function in a linear quantile regresion to find a
breakpoint. The function I want to add is:
y=(k+ax)(xlt;B)+(k+(a-d)B+dx)(xgt;B)
How do I write it in the rq() function? Do I need to define the parameters
in any way and how do I do that? I'm a biologist new to R.
It isn't entirely clear to me what you want, but here are all the possibilities
I could think of. I hope one of them does what you want.
testmat - matrix(1:8, ncol=2)
testmat
[,1] [,2]
[1,]15
[2,]26
[3,]37
[4,]48
paste(testmat, collapse= )
[1] 1 2 3 4 5 6 7
On 09/20/2011 10:57 AM, Achim Zeileis wrote:
On Tue, 20 Sep 2011, Tal Galili wrote:
I am not sure who to send this to - so I am writing this here.
It seems that the index.html pages on CRAN are gone.
For example:
http://cran.r-project.org/web/packages/ggplot2/index.html
Is gone. But the
Szeptember 12-től 26-ig irodán kívül vagyok, és az emailjeimet nem érem el.
Sürgős esetben kérem forduljon Kárpáti Edithez (karpati.e...@gyemszi.hu).
Üdvözlettel,
Mihalicza Péter
I will be out of the office from 12 till 26 September with no access to my
emails.
In urgent cases please contact
On 11-09-20 3:08 AM, Ashim Kapoor wrote:
Dear R-helpers,
Please look at the following minimal code.
\documentclass[a4paper]{article}
\begin{document}
==
library(zoo)
a-zoo(1:4,order.by=Sys.time()+1:4)
str(a)
@
\end{document}
When I do R CMD Sweave,followed by pdflatex ,and view the final pdf,
ok great, as.numeric(as.character(data$x)) works perfectly.
I am sorry that i did not find this before making this post.
@ peter
what did not work was:
data$x2 - as.character(data$x)
data$x3 - as.numeric(data$x2)
--
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On 11-09-20 7:25 AM, _Luc_ wrote:
ok great, as.numeric(as.character(data$x)) works perfectly.
I am sorry that i did not find this before making this post.
@ peter
what did not work was:
data$x2- as.character(data$x)
data$x3- as.numeric(data$x2)
That should have worked. Can you put together
On Tue, Sep 20, 2011 at 4:31 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote:
On 11-09-20 3:08 AM, Ashim Kapoor wrote:
Dear R-helpers,
Please look at the following minimal code.
\documentclass[a4paper]{**article}
\begin{document}
==
library(zoo)
a-zoo(1:4,order.by=Sys.time()**+1:4)
Hi,
could anyone tell me how predict() predicts the meanError or
standardDerivation of a garchFit(1,1)-model,
knowing the coefficients mu, omega, alpha1, beta1 and of course all
datapoints?
Thanks and sorry for my poor english.
--
View this message in context:
On Sep 19, 2011, at 8:49 PM, justin jarvis wrote:
I have a data set with many missing observations. When I run a
regression, R of course discards the observations (the whole row) that
have NA. I want to tabulate some baseline characteristics (column
means) but only for the observations that
Jim Lemon wrote on 09/20/2011 04:15:46 AM:
On 09/19/2011 04:46 PM, Henri-Paul Indiogine wrote:
Greetings!
I am using the R library RQDA to assign certain codes to paragraphs of
documents in a collection. Several paragraphs are assigned more than
1 code. E.g. often the codes poverty
hi eik, mark and sarah,
thank you for your help!
i was looking for the t-command but in the end the apply-command did the
trick.
marion
2011/9/20 Sarah Goslee sarah.gos...@gmail.com
It isn't entirely clear to me what you want, but here are all the
possibilities
I could think of. I hope one
Good afternoon/morning readers. This is the first time I am trying to run
some Bayesian computation in R, and am experiencing a few problems.
I am working on a Poisson model for cancer rates which has a conjugate Gamma
prior.
1) The first question is precisely how I work out the parameters.
Hello all,
I am looking for an editor for R which has got functions beyond the normal R
editor that is included in the program.
I had a look at VIM but I think it's difficult if you are just starting
programming.
Could anyone recommend an editor that is suitable for beginners?
Thanks for your
RStudio
Hope this helps,
Michael Weylandt
On Tue, Sep 20, 2011 at 9:26 AM, Marion Wenty marion.we...@gmail.comwrote:
Hello all,
I am looking for an editor for R which has got functions beyond the normal
R
editor that is included in the program.
I had a look at VIM but I think it's
Hi everyone
I have got a quick question:
I the seqinr package:
*dist.alignment(x,identity)*
This is calculating the square root of pairwise distances. Does anyone know
whether/how gaps are counted in this function?
Thank you.
Best wishes,
Bettina
[[alternative HTML version
On Sep 20, 2011, at 4:50 AM, mael wrote:
Hi,
I am trying to add a function in a linear quantile regresion to find a
breakpoint. The function I want to add is:
y=(k+ax)(xlt;B)+(k+(a-d)B+dx)(xgt;B)
How do I write it in the rq() function? Do I need to define the
parameters
in any way and
On Tue, 20 Sep 2011, Marion Wenty wrote:
I am looking for an editor for R which has got functions beyond the normal
R editor that is included in the program.
I had a look at VIM but I think it's difficult if you are just starting
programming.
Marion,
VIM (Vi) is a line-oriented editor.
Marion:
You should always first search CRAN for such queries. Had you done so, you
would have been led to:
http://www.sciviews.org/_rgui/
There is an extensive list of editors/IDE's there.
-- Bert
On Tue, Sep 20, 2011 at 6:26 AM, Marion Wenty marion.we...@gmail.comwrote:
Hello all,
I am
Hi Jean and Jim!
Thanks for your suggestions.
Best,
Henri-Paul
--
Henri-Paul Indiogine
Curriculum Instruction
Texas AM University
TutorFind Learning Centre
Email: hindiog...@gmail.com
Skype: hindiogine
Website: http://people.cehd.tamu.edu/~sindiogine
Hi Samir,
I do vaguely understand your intention. However I'm not sure taking
the mean of acf's makes any sense.
I do not know what your final goal is (what do you want to do with the
data?) but here is a suggestion. Instead of trying to aggregate acfs
computed on different series, why don't you:
Hello list members,
I am working with simulated data for landscape pattern analysis. I have
1000 replicates of binary (2 colour) gridded landscapes at each combination
of 9 levels of class proportion and 11 levels of spatial autocorrelation.
The results are stored in an array as follows:
Hi everybody. I'm trying to fit a weibull survival model with a spline
basis for the predictor, using the survival library. I've noticed that it
doesn't seem to be possible to use the aic method to choose the degrees of
freedom for the spline basis in a parametric regression (although it's
fine
Marion Wenty wrote:
...Could anyone recommend an editor that is suitable for beginners?...
I had trouble with this for a long time. I tried several different programs,
but couldn't get any to work properly—they were either too complicated or
broken. I recently started using Notepad++ with
HI,
This code is part of a code I used to do a linear regression:
points(var1~var2,data=Regress,pch=21,bg=grey)
reg11-lm(var1~var2,data=Regress)
abline(lm(var1~var2,data=Regress),lty=2,lwd=2,col=grey)
legend(topleft,legend=
c(NDII from composite,
y= 0.0007x - 0.1156,expression(paste(r^2 ==
I have a list of data frames like the following:
set.seed(123)
a- data.frame(x=runif(10), y = runif(10), sample = seq(1,10))
b- data.frame(x=runif(10), y = runif(10), sample = seq(1,10))
L- list(a,b)
All data frames in the list have the same dimensions. I need to calculate
the sample means for
I am struggling to get GAM model predictions from the top models calculated
using model.avg in the package MuMIn.
My model looks something like the following:
gamp - gam(log10(y)~s(x1,bs=tp,k=3)+s(x2,bs=tp,k=3)+
s(x3,bs=tp,k=3)+s(x4,bs=tp,k=3)+s(x5,bs=tp,k=3)+
When sampling from a multi-variate truncated normal using rtmvnorm from
the tmvnorm-package, I experience extreme performance differences between
two of my computers. On my laptop computer, draws take ~5s, on my desktop
~30s. I need to run MCMCs with repeat calls to rtmvnorm on my desktop. The
Hello,
I am trying to estimate a multivariate regression of Y on X with
regression splines. Y is (nx1), and X is (nxd), with d1. I assume the
data is generated by some unknown regression function f(X), as in Y =
f(X) + u, where u is some well-behaved regression error. I want to
estimate f(X) via
Maybe by adding
col=c(grey,white,white), bg=white
but not sure ...
Have a nice end of day,
Ptit Bleu.
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Sent from the R help mailing list archive at Nabble.com.
Dear all,
Thanks a lot for the help. It worked very well in the end.
Best regards,
Marianne
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Maybe you can use Rcmdr
http://socserv.mcmaster.ca/jfox/Misc/Rcmdr/
It is a very easy program.
Da: r-help-boun...@r-project.org [r-help-boun...@r-project.org] per conto di
Jack Siegrist [jack...@eden.rutgers.edu]
Inviato: martedì 20 settembre 2011 16.07
You're passing a vector of length 3 to the legend argument of legend(), so R is
assuming that you have three things you want to display.
There are two alternatives:
1. Provide legend() a legend that's a single character vector.
2. Specify that you don't want symbols for the last two of the three
I just wanted to second Rstudio:
http://rstudio.org/
Honestly, try this first -- it's an easy install, it just works, and
there's no real learning curve to the editor itself. Very n00b
friendly.
-steve
On Tue, Sep 20, 2011 at 9:28 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
One possibility is
library(mgcv)
## isotropic thin plate spline smoother
b - gam(Y~s(X[,1],X[,2]))
predict(b,newdata=list(X=W))
## tensor product smoother
b - gam(Y~te(X[,1],X[,2]))
predict(b,newdata=list(X=W))
## variant tensor product smoother
b - gam(Y~t2(X[,1],X[,2]))
Hi
I have a list of data frames like the following:
set.seed(123)
a- data.frame(x=runif(10), y = runif(10), sample = seq(1,10))
b- data.frame(x=runif(10), y = runif(10), sample = seq(1,10))
L- list(a,b)
All data frames in the list have the same dimensions. I need to
calculate
the
You did ask for a loop, but if you are willing to consider sapply(), the
result you want can be obtained with a single (admittedly long) command:
slope-sapply(1:100, function(x) lm(c(rnorm(1, mean=.01, sd=.001),
rnorm(1, mean=.1, sd=.01))~c(10,400))$coefficients[2])
head(slope)
c(10, 400)
Rstudio and Rcmdr are very popular and for good reason, find a good book
while the latter is installing though.
IF you are using linux, rkward is fantastic to woRk with.
Ken Hutchison
On Tue, Sep 20, 2011 at 11:21 AM, Steve Lianoglou
mailinglist.honey...@gmail.com wrote:
I just
Thank you Sarah and Petit bleu for your help.
I solved my problem with the code of Sarah.
Komine
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First off, let me apologize for the elementary question. I'm obviously a
novice.
Here's a stripped version of my problem.
March
foreign id = 1234, my id = 1
foreign id = 1235, my id = 2
foreign id = 1236, my id = 3
So we are adding new people for April, and things don't necessarily come in
El Gorgonzola elgorgonzola at hotmail.com writes:
My question: how do you document methods of S4-classes in a package?
It works fine for the method show but it does not work for showall,
for which I had to define the generic function first. What can I do to
fix this, or is it better to just
A reproducible example would be useful, as yours isn't entirely clear.
But maybe something like this?
id.info - data.frame(foreign=sort(unique(foreign.id)),
local=1:length(unique(foreign.id)))
and then use merge() to combine that with your actual data.
Sarah
On Tue, Sep 20, 2011 at 11:50 AM,
something like this should work:
oldIDs - c(1234, 1235, 1236)
newIDs - c(5000, 1234, 7000, 1236)
# really new ones -- don't match the old ones
(reallyNew - setdiff(newIDs, oldIDs))
[1] 5000 7000
# assign these back to the oldIDs for the next month
(oldIDs - c(oldIDs, reallyNew))
[1] 1234
Tarmo Remmel wrote on 09/20/2011 09:51:45 AM:
Hello list members,
I am working with simulated data for landscape pattern analysis. I have
1000 replicates of binary (2 colour) gridded landscapes at each
combination
of 9 levels of class proportion and 11 levels of spatial
autocorrelation.
Dear all;
A very basic question. I have the following data:
A - 1/1000*c(347,328,129,122,18,57,105,188,57,257,53,108,336,163,
62,112,334,249,45,244,211,175,174,26,375,346,153,32,
Hello,
I am using the kernlab package to do regression.
I have a data frame called Data6 which looks like this:
head(Data6)
WA PO ZA ZB ZC ZD KL
1 2.955447 6.378324 14.10622 0.134343 0.247120 0.734810 4.05988
2 2.939718 6.344122 14.03528 0.127512
Hi
I am having a problem using varImpPlot in randomForest. I get the error
message Error in plot.window(xlim = xlim, ylim = ylim, log = ) : need
finite 'xlim' values
When print $importance, several variables have NaN under %IncMSE. There
are no NaNs in the original data. Can someone help
Yes, in over 3/4s of the data points A is B… which suggests the A measure is
reading higher than the B measuring system.
length(A[AB])/length(A)
On 20 Sep 2011, at 6:46 PM, Pedro Mardones wrote:
Dear all;
A very basic question. I have the following data:
On Sep 20, 2011, at 12:46 PM, Pedro Mardones wrote:
Dear all;
A very basic question. I have the following data:
A - 1/1000*c(347,328,129,122,18,57,105,188,57,257,53,108,336,163,
Hello there,
I am using NLS for fitting a complex model to some data to estimate a couple
of the missing parameters. The model is -
y ~ (C+((log(1-r))*exp(-A*d)*(-1+r+exp(d*(A-B)))/(r*(-A*d+d*B+log(1-r)
where A, B and C are unknown.
In order to test the model, I generate data by setting
Pedro Mardones wrote on 09/20/2011 12:46:54 PM:
Dear all;
A very basic question. I have the following data:
A - 1/1000*c(347,328,129,122,18,57,105,188,57,257,53,108,336,163,
As can be seen by plotting as follows:
plot(A,B,pch=+,col=blue) ## The raw data
plot(A,B-A,pch=+,col=blue) ## The differences versus A
lines(c(0,0.7),c(0,0))
Ted.
On 20-Sep-11 17:54:15, Timothy Bates wrote:
Yes, in over 3/4s of the data points A is B
which suggests the A
measure is
Hello R users,
I have a set of data frames for which I am tallying row numbers, as shown
below.
nrow(mC_Explant)
[1] 14480
nrow(mC_Callus)
[1] 23320
nrow(mC_RegenPlant)
[1] 8108
etc.
I want to create a new data frame which has the variable names as column
headings, and then a single row
Please keep the list in the loop for the archives:
The data provided looks like this:
V = structure(list(V1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L), .Label = 02/22/2011, class = factor), V2 = structure(c(1L,
1L, 2L,
Hi Kelly,
c() creates a vector. You need data.frame() instead.
dfIntron - c(mC_Explant=nrow(mC_Explant), mC_Callus=nrow(mC_Callus),
mC_RegenPlant=nrow(mC_RegenPlant)) # set colnames simultaneously.
Sarah
On Tue, Sep 20, 2011 at 2:23 PM, Vining, Kelly
kelly.vin...@oregonstate.edu wrote:
Hello
Diviya Smith wrote on 09/20/2011 01:03:22 PM:
Hello there,
I am using NLS for fitting a complex model to some data to estimate a
couple
of the missing parameters. The model is -
y ~
(C+((log(1-r))*exp(-A*d)*(-1+r+exp(d*(A-B)))/(r*(-A*d+d*B+log(1-r)
where A, B and C are unknown.
In S+ do.call's first argument must be a character string
that gives the name of the function, so replace
do.call(rbind, ...)
with
do.call(rbind, ...)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
From: Santosh [mailto:santosh2...@gmail.com]
Sent: Tuesday, September 20, 2011 2:55 AM
The following untested code might also be of some help to you:
x = data.frame(x = 1:5)
y = data.frame(y = 1:10)
z = data.frame(z = 1:3)
a = rnorm(50)
b = function(x) x^2
makeRowFrame - function(){
n - ls(envir = .GlobalEnv)
tempFunc - function(VARIABLE){
VARIABLEM - get(VARIABLE)
Thank you for the reply, it looks like the second option (te) will
work perfectly!
Max
On Tue, Sep 20, 2011 at 2:39 PM, Max Farrell ma...@umich.edu wrote:
One possibility is
library(mgcv)
## isotropic thin plate spline smoother
b - gam(Y~s(X[,1],X[,2]))
predict(b,newdata=list(X=W))
Hi,
I am doing an analysis to see if these is tissue specific effects on the
gene expression data .
Our data were collected from 6 different labs (batch effects). lab 1 has
tissue type 1 and tissue type 2, lab 2 has tissue 3, 4,5,6. The other labs
has one tissue type each. The 'sample' data is
Might be of interest to list subscribers:
EMBL Advanced Course
R Programming and Development
EMBL Heidelberg, Germany
Monday 28 November - Tuesday 29 November 2011
The course will focus on two aspects of R programming and development.
In the first part, we will introduce object-oriented
Hi there,
I have the following problem on my macbook air with mac os x lion on it.
when any program tries to load the tcltk library the R GUI or R command line
freezes. also happens if I just use library(tcltk)
changing versions of tcltk and/or R doesn't help. currently I have the
following:
R
Hi,
I'm having some difficulty installing R on our server which is running
RHEL (Red Hat Enterprise Linux) 5.2. I'm using RPM (Red Hat Package
Manager) and by default it installs R in /usr/bin. Unfortunately due
to a very small partition size, I have run out of space in this
directory. I was
We are contemplating the installation of R on a Windows server with Hyper-V but
we can't find any information on whether it can be done.
This transmission is intended solely for the person or organization to whom it
is addressed and it may contain
I dont think *r* is related to the problem. I am not trying to estimate *r* and
so basically I am giving the model the correct value of *r* and so log(1-r)
should not go to infinity.
For test data, I generate data from the same model and add noise (using *r
norm*), with the following parameters -
Greegings!
Any idea why this works from the command line, but not from a source
file? This is driving me (more) insane.
regexp - [Aa]ccountability
Thanks!
--
Henri-Paul Indiogine
Curriculum Instruction
Texas AM University
TutorFind Learning Centre
Email: hindiog...@gmail.com
Skype:
On Sep 20, 2011, at 4:02 PM, Darrel Barbato wrote:
Hi,
I'm having some difficulty installing R on our server which is running
RHEL (Red Hat Enterprise Linux) 5.2. I'm using RPM (Red Hat Package
Manager) and by default it installs R in /usr/bin. Unfortunately due
to a very small partition
Further to the plot suggested below, the plot
plot(log(A),log(B/A),pch=+,col=blue)
reveals an interesting structure to the data. Distinct curved
sequences are clearly visible. While their curved form is a
consequence of the fact that, for large A, A/B is close to 1
and so they tend to approach
What works? Assigning a string to a function, thus replacing the
function in your search
path?
Or is that what doesn't work?
What do you expect? What are you trying to do? Using what commands?
Sarah
On Tue, Sep 20, 2011 at 6:19 PM, Henri-Paul Indiogine
hindiog...@gmail.com wrote:
Greegings!
My apologies: not the name of a function, despite ?regexp having a return value.
But the rest of my questions are still relevant.
On Tue, Sep 20, 2011 at 6:32 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
What works? Assigning a string to a function, thus replacing the
function in your search
Hi Sarah!
2011/9/20 Sarah Goslee sarah.gos...@gmail.com:
What works? Assigning a string to a function, thus replacing the
function in your search
path?
Bad choice for character variable name on my part. How about this?
pattern - [Aa]ccountability
If I enter this at the CLI and then type
My objective is to define a pattern for a grep() statement.
Thanks,
Henri-Paul
--
Henri-Paul Indiogine
Curriculum Instruction
Texas AM University
TutorFind Learning Centre
Email: hindiog...@gmail.com
Skype: hindiogine
Website: http://people.cehd.tamu.edu/~sindiogine
Can you perhaps provide us a minimal working script in which this assignment
doesn't happen? If it really is just the line
pattern - [Aa]ccountability
in the main body of the script, I can't for the life of me guess why it
wouldn't work. General keywords that come to mind are things like
On Sep 20, 2011, at 6:49 PM, Henri-Paul Indiogine wrote:
My objective is to define a pattern for a grep() statement.
Thanks,
Henri-Paul
You are not illustrating the problem with enough detail to determine
the source of your errors. The errors do not appear when tested in
what appears to
I'm having difficulty finding the syntax to use to specify the beginning
and ending dates for the x-axis while plotting a zoo object. I thought that
I had seen a message on this list that used start=as.Date(...)
end=as.Date(...), but I cannot find that message. I've tried ?plot,
?plotxy,
Diviya Smith diviya.smith at gmail.com writes:
I dont think *r* is related to the problem. I am not trying to
estimate *r* and
so basically I am giving the model the correct value of *r* and so log(1-r)
should not go to infinity.
For test data, I generate data from the same model and
Greetings!
2011/9/20 David Winsemius dwinsem...@comcast.net:
There is probably an environment/scoping problem as you stated. I am
using ESS and org-mode.
I started a new R code block and the problem disappeared.
Thanks for your kind feedback.
Henri-Paul
--
Henri-Paul Indiogine
Curriculum
That still discards the other data columns. For example, in the data frame
V1 V2 V3 V4
1 1 1 NA 1
2 1 NA 1 1
3 1 NA 1 1
4 1 11 1
5 1 11 1
Suppose I was running a regression using V1 and V2. R will remove rows 2
and 3 due to the NA. I would like a way to look at only the
Pardon my ignorance, but why is the do.call necessary? why not just execute
the rbind function? What's the advantage in putting it in a do.call wrapper?
On Sep 20, 2011, at 2:44 PM, William Dunlap wdun...@tibco.com wrote:
In S+ do.call's first argument must be a character string
that
You could use the na.action function on the fitted
object to see which observations were omitted. E.g.,
let's make a data.frame that we can actually do some
regressions with and try na.action():
d - data.frame(V1=11:15, V2=log(c(1,NA,NA,4,5)), V3=sqrt((-1):3),
V4=sin(1:5))
Warning message:
I don't have access to your alon.txt file (see ?dput for future posts),
but...
I'm pretty sure info you want isn't in row.names(colon[1:2])
it should just be
text(x,y, label = colon[1:20])
??
HTH
baumeist wrote:
Hi,
I am new to R.
I have a matrix that I have assigned to the object
Look at the help file for do.call. It is
most useful when you don't know how many
arguments will be given to the function you
are calling. E.g., if you know that the list
x is always 3 long then you can do
rbind(x[[1]], x[[2]], x[[3]])
to make a matrix out of the components, just as
If different labs tested different tissue.types, I am not sure you can
effectively partition variance between labs(batches) and tissue.types.
Weidong Gu
On Tue, Sep 20, 2011 at 2:14 PM, karena dr.jz...@gmail.com wrote:
Hi,
I am doing an analysis to see if these is tissue specific effects on
On Sep 20, 2011, at 7:37 PM, justin jarvis wrote:
That still discards the other data columns. For example, in the
data frame
V1 V2 V3 V4
1 1 1 NA 1
2 1 NA 1 1
3 1 NA 1 1
4 1 11 1
5 1 11 1
Suppose I was running a regression using V1 and V2. R will remove
Hi
I recall running across a function a while back which would return
information about running processes (such as their cpu and memory
usage), but I cannot seem to locate it. Wondering if someone would be
kind enough to refresh my memory. I vaguely recall it was parsing the
output of the
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