To my knowledge, fixed and random effect models may be estimated for the
logit model and only the random effect model for the probit model
(because of the incidental parameter problem).
I think clogit in the survival package fits the model that is called the
fixed effect logit model in the
Apologies.
Let me clarify. I have included my code below :
data.table.b represents the medical nomenclature, whereas data.table.a is a
patient derived database.
data.table.b$CATEGORY categorizes features (e.g. 'cardiac', 'respiratory'),
whereas data.table.b$FEATURE is a corresponding disease
On 11 May 2010 00:52, Achim Zeileis achim.zeil...@uibk.ac.at wrote:
On Mon, 10 May 2010, RATIARISON Eric wrote:
I'm using maxlik with functions specified (L, his gradient hessian).
Now I would like determine some robust standard errors of my estimators.
So I 'm try to use vcovHC, or hccm or
Le 11/05/10 04:16, Elizabeth Lawson a écrit :
Hi,
I just bought new macbook pro 10.6.3. I am trying to use some old c code
that used to work. I tried to recompile the code. In the directory with
code I used R CMD SHLIB hello.c and get the error
make: Nothing to be done for `all'.
I have
Greg Orm wrote:
Dear r-help list members,
I am quite new to R, and hope to seek advice from you about a problem I have
been cracking my head over. Apologies if this seems like a simple problem.
I would not call it exactly simple... but mostly because your data
representation is so obscure.
R Friends,
I have data from which I would like to learn a more general
(smoothened) trend by applying data smoothing methods. Data points
follow a positive stepwise function.
|x
x
|
| xx
Hello,
adding the line:
space = 'inside'
to my xyplot() call provides a partial solution to the problem I
have.. However, this command puts the key in the left upper corner
when I want in the left upper corner. Looking at the help provides no
additional options for the inside, but surely this
Hi R-experts,
I try to find a way to transfer a matrix to a data.frame that is used as input
of aov.
can you give me advice for that?
mdat - matrix(c(1,2,3, 11,12,13), nrow = 2, ncol=3, byrow=TRUE, dimnames =
list(c(row1, row2), c(Col1, Col2, Col3)))
mdat
Col1 Col2 Col3
row1 1 2
Set up a function for the fisher.test on a 2x2 table and then include
this in the apply function for columns as in the example below. The
result is a list with names A to Z
# set up a dummy data set with 100 rows
Cat-LETTERS[sample(1:6,100, replace=T)]
GL-sample(1:6, 100, replace=T)
Elisabeth,
You should listen to Ted (Harding). He answered your question with:
the vertical axis is scaled logarithmically with the
numerical annotations corresponding to the *raw* values of Y,
not to their log-transformed values. Therefore it does not matter
what base of logarithms is
Hi:
Instead of space = , use corner = c(x, y) instead. For upper right, c(1, 1);
for lower left, c(0, 0), and anywhere in between...
Here's an illustration, slightly modified from an earlier post this evening:
df - data.frame(x = rnorm(30), y = rnorm(30))
library(reshape)
df2 - melt(df)
I thought that function var() and first element of function acf(object,
type=c('covariance')) should give me the same results. But they differ. Can
someone share an explanation.
--
View this message in context:
http://r.789695.n4.nabble.com/Variance-vs-covariance-tp2173501p2173501.html
Sent
Hi:
mdf3 - as.data.frame.table(as.table(mdat))
mdf3
Var1 Var2 Freq
1 row1 Col11
2 row2 Col1 11
3 row1 Col22
4 row2 Col2 12
5 row1 Col33
6 row2 Col3 13
HTH,
Dennis
On Mon, May 10, 2010 at 10:38 PM, Yuan Jian jayuan2...@yahoo.com wrote:
Hi R-experts,
I try to find a way
On Tue, 11 May 2010, Arne Henningsen wrote:
On 11 May 2010 00:52, Achim Zeileis achim.zeil...@uibk.ac.at wrote:
On Mon, 10 May 2010, RATIARISON Eric wrote:
I'm using maxlik with functions specified (L, his gradient hessian).
Now I would like determine some robust standard errors of my
Dear group,
Here is my df :
df3 -
structure(list(DESCRIPTION = c(COPPER May/10, COTTON NO.2 Jul/10,
CRUDE OIL miNY May/10, GOLD Jun/10, ROBUSTA COFFEE (10) Jul/10,
SOYBEANS Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 May/10,
WHEAT Jul/10, SPCL HIGH GRADE ZINC USD, STANDARD LEAD USD,
CORN May/10,
Hi:
To condense the output a bit, here's a slight modification that returns, for
each variable
tested, the p-value, confidence limits and odds ratio estimate:
ff-function(x,y){
m - fisher.test(table(x,y))
c(m$p.value, m$conf.int, m$estimate)
}
res - apply(data1[,LETTERS[1:26]], 2, ff,
On 05/11/2010 08:17 AM, Chaudhari, Bimal wrote:
Are there functions/packages which support plots (bar and/or line) where
I provide the point estimate and some error measure rather than the raw
data?
I often have to summarize/present data from multiple sources where the
original data is
Hi all,
Just like Fortran creates EXE files for a code, can R also do the same? Thanks
for the info.
Thanks and Regards,
Shubha Karanth | Amba Research
Ph +91 80 3980 8283 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com
On 11.05.2010 10:50, Shubha Vishwanath Karanth wrote:
Hi all,
Just like Fortran creates EXE files for a code, can R also do the same? Thanks
for the info.
No, R is an interpreter, not a compiler.
Uwe Ligges
Thanks and Regards,
Shubha Karanth | Amba Research
Ph +91 80 3980 8283 |
Am 11.05.2010 10:37, schrieb arnaud Gaboury:
Dear group,
Here is my df :
df3 -
structure(list(DESCRIPTION = c(COPPER May/10, COTTON NO.2 Jul/10,
CRUDE OIL miNY May/10, GOLD Jun/10, ROBUSTA COFFEE (10) Jul/10,
SOYBEANS Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 May/10,
WHEAT Jul/10, SPCL
df3$DESCRIPTION =
sub(' [a-z]{3}/[0-9]{2}','',df3$DESCRIPTION,ignore.case=TRUE)
- Phil Spector
Statistical Computing Facility
Department of Statistics
On 10-May-10 20:11:39, Jue \(Andy\) Wang wrote:
I am writing to ask if R has a build- in function to calculate this
polylogarithm Li_n(z) function , also known as the Jonquière's function
defined as
Li_n(z)=sum_(k=1)^infty(z^k)/(k^n)
Thanks
Andy
There is an R package 'gsl' which is a
TY for the two answers. Both work.
-Original Message-
From: Phil Spector [mailto:spec...@stat.berkeley.edu]
Sent: Tuesday, May 11, 2010 11:21 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] Regex and gsub
df3$DESCRIPTION =
sub('
Hi Ralf,
I can't offer you many resources, but the few I came across are:
1) loess (or the older version: lowess)
2) smooth
3) rollapply (from the zoo pacakge)
I used a combination of 1 and 3 when creating an R implementaion for a
(simplistic) quantile loess, you might find the code useful:
On 05/11/2010 02:05 PM, Greg Orm wrote:
Dear r-help list members,
I am quite new to R, and hope to seek advice from you about a problem I have
been cracking my head over. Apologies if this seems like a simple problem.
I have essentially two tables. The first (Table A) is a standard patient
Hello,
I have two tab-delimited files which I would like to combine.
In the first one I have gene IDs (Unique) on column 1 and than various
experimental results from microarray analysis (see attached files list1 )
the second arrays have the same genes IDs (more and in a different order,
some are
Elisabeth and I have been corresponding off-list about this, and
came to a potential solution which is on the lines also outlined
by Mark Difford.
Where Elisabeth (rather, her tutor) may have become confused may
lie in the fact that, with a simple plot(...,log=y), R will
(by default) make its own
This removes runs of length 1 and 2. It replaces the values in any
such run with NA and then uses na.locf from the zoo package to fill
those NA's by carrying forward the last occurrence of a non-NA. In
this example the run consisting of a single 2, the run consisting of
two 3's and the run
Reduce the non-unique case to the unique case in the first line and
then form the ids ensuring that ids has names. Finally sapply over
the ids summing across rows using drop = FALSE so that the two cases
in your code are both handled at once. Adding 0 converts to numeric.
ag -
?merge
Here is what you would do:
x1 - read.table('/tmp/list1.txt', header=TRUE, sep='\t', as.is=TRUE)
x2 - read.table('/tmp/list2.txt', header=TRUE, sep='\t', as.is=TRUE)
combined - merge(x1, x2, by.x=Probe.Id, by.y=Probe.ID)
Unfortunately your data did not have any IDs in common between the
?paste
On Mon, May 10, 2010 at 6:21 PM, Jonathan Greenberg
greenb...@ucdavis.eduwrote:
Rhelpers:
I'd like to modify this RSQLite statement:
rs_stations-dbSendQuery(con_stations, select * from stations)
so that stations is actually an R variable, e.g.:
stations=c(stationA,stationB)
How
Dear group,
I have 3 data frames I would like to merge.
Here they are:
pose16 -
structure(list(DESCRIPTION = structure(c(1L, 2L, 3L, 4L, 5L,
6L, 7L, 8L, 9L, 12L), .Label = c(COPPER May/10, COTTON NO.2 Jul/10,
CRUDE OIL miNY May/10, GOLD Jun/10, ROBUSTA COFFEE (10) Jul/10,
SOYBEANS Jul/10,
I'm surprised no one except Ralf mentioned tree-based methods. Basic
regression trees are fitting exactly the type of functions (piecewise
constant) that Ralf is asking about. So, either tree() or rpart() or
whatever is in party should fit the bill.
Another possibility is wavelets with the Haar
If you just want to combine them together (rbind), try this:
str(avprix16)
'data.frame': 16 obs. of 4 variables:
$ DESCRIPTION: chr COPPER May/10 CORN May/10 COTTON NO.2 Jul/10
CRUDE OIL miNY May/10 ...
$ prix : num -716.9 -1082 -80 -84.8 2295.5 ...
$ POSITION : num -2 -3 -1 -1
On 5/11/10, chen jia chen_1...@fisher.osu.edu wrote:
Are there any functions that specifically deal with fixed-effects?
Other than plm and its vignette, you may want to check this document [1].
Liviu
[1] http://cran.r-project.org/doc/contrib/Farnsworth-EconometricsInR.pdf
Dear Daniel
On 5/11/10, Daniel Malter dan...@umd.edu wrote:
R-squared of interest is typically the within R-squared, not the overall or
Could you point to an example on how to compute the within R-squared
in R, either via lm() or plm()?
Thank you
Liviu
Hi -
I am using pls package for some pcr computations. There is a data set called
gasoline.
Would someone be able to tell me what command(s) could be used to produce
this graph in R? I am not sure where the log(1/R) - Y-axis - are coming
from
Thanks much
Ravi
Hello,
I saw statistical method of stepwise multivariate analysis of variance in
several literatures.
I now want to how to do it in R software. Thanks in advance!
Guoqiang Li
http://agri520.cn
__
R-help@r-project.org mailing list
Hi all,
I am running a set of logistic regressions, where we want to use some
weights, and I am not sure whether what I am doing is reasonable or
not.
The dependent variable is turnout in an election - i.e. survey
respondents were asked whether or not they voted. The percentage of
those who say
Thanks for the answers, I finally managed!
Still need to learn a lot but with such fast help will be easy :)
http://n4.nabble.com/file/n2173565/Perfect_one.png
Thanks again
A
--
View this message in context:
Thank you all of you,;
I submit you my case:
( Pseudo likehood for exponential family with offset )
loglik - function(param) {
b-param
m=as.vector(of+z%*%b)
ll - sum(v*m-exp(m)) }
gradlik - function(param) {
b-param
m=as.vector(of+z%*%b)
gg-(v-exp(m))*z }
In fact, that is one solution. I can then do a
ddply(x,DESCRIPTION,summarise, pl=sum(prix),quantity=sum(POSITION)). It
does work. I was just wandering if it exists another solution.
From: jim holtman [mailto:jholt...@gmail.com]
Sent: Tuesday, May 11, 2010 2:14 PM
To: arnaud Gaboury
Cc:
THX Jim,
that works just fine. I had to play with it a little, so taht it fitted my
expectations but the idee of using merge was very good.
Assa
On Tue, May 11, 2010 at 13:55, jim holtman jholt...@gmail.com wrote:
?merge
Here is what you would do:
x1 - read.table('/tmp/list1.txt',
When I try
$ rm hello.o hello.so
I get the error
-bash: $: command not found
What does that mean?
--
View this message in context:
http://r.789695.n4.nabble.com/make-Nothing-to-be-done-for-all-tp2173220p2173679.html
Sent from the R help mailing list archive at Nabble.com.
dear R wizards---
I am looking for a reference that explains how to work with formula
objects. for example, say, I have a formula which I want to use in an
NLS. I want to test what this formula to see if my function was
defined correctly. Is there a way to apply a formula to data? For
Le 11/05/10 13:40, Elizabeth Lawson a écrit :
When I try
$ rm hello.o hello.so
I get the error
-bash: $: command not found
What does that mean?
Did you actually type the '$' ? You should not have.
--
Romain Francois
Professional R Enthusiast
+33(0) 6 28 91 30 30
thanks for your help. maybe I have poor statistics level, I can not well
understand your means.
wishes
kevinbrbrå¨2010-05-11ï¼Bert Gunter gunter.ber...@gene.com åéï¼
(Near) non-identifiability (especially in nonlinear models, which include
linear mixed effects models, Bayesian
Dear useRs and expeRts,
I would like to ask your views on an issue for which there possibly
exists an established policy of the R-help list, but I am not aware
of it.
In 2008, I have spotted some errors in a package, one which is likely
to have many users (I am not one myself). The more serious
You could send a reminder to the author. Its happened to me that
nothing occurred and 6 months later I sent another email and that
seemed enough to get it fixed.
You could also ask the maintainer if the package has been abandoned
and offer to take over as maintainer.
If you just want to record
The Mantel-Byar test is a simple modification to the logrank test. Can
R perform this test or any other similar test?
The logrank test = score test from coxph
The Mantel-Byar is the score test from a Cox model with a time dependent
covariate.
Terry Therneau
On Tue, 11 May 2010, RATIARISON Eric wrote:
Hi,
I think my meatfunction is wrong.
Yes.
But the estfun(resMaxlik) doesn't work;
Because you haven't supplied one.
It's as I wrote in my previous mails: You need to write a bread() and an
estfun() method as specified in my previous mail.
Does anyone know if R has a function for splines under tension. I know there
are numerous packages for spline interpolation within R i just can't find
one that lets you determine the tension factor.
Any help would be much appreciated!
Sam
--
View this message in context:
Dear group,
I have these 2 following data frame:
allcon -
structure(list(DESCRIPTION = structure(1:17, .Label = c(COFFEE C Jul/10,
COPPER May/10, CORN Jul/10, CORN May/10, COTTON NO.2 Jul/10,
CRUDE OIL miNY May/10, GOLD Jun/10, HENRY HUB NATURAL GAS May/10,
ROBUSTA COFFEE (10) Jul/10, SILVER
Hello,
I am trying to make a barplot of means where the difference between
the barheights is a different color. My goal is to convey the same
information as a dodged barplot, but with only four bars. Below is
the data I am working with, and the two plots I have tried. The
stacked plot would be
first, apologies for so many posts yesterday and today. I am
wrestling with nls() and nls2(). I have tried to whittle it down to a
simple example that still has my problem, yet can be cut-and-pasted
into R. here it is:
library(nls2)
options(digits=12);
y=
Got it!
First, here is the correct line :
y=merge(value,allcon,all.y=T)
Then, I got NA because of some white space (blancs) in names of elements of
my data frame VALUE.It came from the .csv file from where I obtained the
VALUE df. TY Excel again!
In fact, as a general advice, I would
Dear Joshua,
You can accomplish that by setting the position argument in geom_bar to
position_dodge(width = 0).
ggplot(plot.means, aes(x=index, y=means, fill=factor(hilow))) +
geom_bar(position = position_dodge(width = 0.1), stat=identity)
But such a plot can be very misleading. Therefore I
Dear Thierry,
Thank you, that worked excellently, and good point about being
potentially misleading. That had not occurred to me.
Thanks again,
Josh
On Tue, May 11, 2010 at 8:05 AM, ONKELINX, Thierry
thierry.onkel...@inbo.be wrote:
Dear Joshua,
You can accomplish that by setting the
Is it the tick labels that you want to change?
-Original Message-
From: Elisabeth Bjerke Rastad ebr...@post.uit.no
To: r-help@r-project.org r-help@r-project.org
Sent: 5/10/10 11:20 AM
Subject: [R] [Fwd: Re: Plotting log-axis with the exponential base to a plot
with the default logarithm
Is there a way to recover the model matrix that is being used by the plm
package? For example, if you run a panel regression with fixed effects, the
model matrix would contain the generated dummy variables for the groups.
Thanks.
Abiel Reinhart
This communication is for informational
Hi,
I think my meatfunction is wrong.
But the estfun(resMaxlik) doesn't work;
Error message :
Erreur dans UseMethod(estfun) :
pas de méthode pour 'estfun' applicable pour un objet de classe c('maxLik',
'maxim', 'list')
-Message d'origine-
De : RATIARISON Eric
Envoyé : mardi 11 mai
Hello,
I am VERY new to R, just picking it up infact. I have got my head around the
basics of ANOVA with post hoc tests but I am struggling with regression,
especially with ANCOVAs.
I have two sets of data, one of type A, one of type B. Both have been placed
in a wind tunnel and sampled every
I am using R to read from an excel(csv) file. Within the excel file is a column
with the date set that looks likes this:
53:40.2
and in the Insert function box it looks likes this:
9/21/2006 4:53:40 PM
I tired separating the time and date using the function below and then plotting
again
I think the problem with nls() is the model specification. Look:
y - c(0.4334,0.3200,0.5848,0.6214,0.3890,0.5233,0.4753,
+0.2104,0.3240,0.2827,0.3847,0.5571,0.5432,0.1326,0.3481)
x - c(0.3521,0.4334,0.3200,0.5848,0.6214,0.3890,0.5233,
+
Is the problem that you want to use the formula as if it were a
function? In that case:
1. nls2 in the nls2 package is like nls but its extensions include a
brute force algorithm which simply computes the formula at the
starting values:
my.formula - y ~ a+b*x^2
y - x - 1:10
library(nls2)
I am trying to do an exhaustive model search and am having some sort of input
error.
leaps ( x= At[,3:8], y = At [,2],method=Cp)
Error in `colnames-`(`*tmp*`, value = 1) :
attempt to set colnames on object with less than two dimensions
- where At is my data matrix
Any suggestions on what
As an end-user, I wonder about Revolution R. Is the relationship
between Revolution R and the R community at-large a positive one? Do
the former contribute to the development efforts of the latter? Is
there a competitive aspect? is their forum competitive with r-help?
any other thoughts?
On Tue, 11 May 2010, Jos Elkink wrote:
Hi all,
I am running a set of logistic regressions, where we want to use some
weights, and I am not sure whether what I am doing is reasonable or
not.
The dependent variable is turnout in an election - i.e. survey
respondents were asked whether or not
On 05/11/2010 08:20 AM, Gabor Grothendieck wrote:
You could send a reminder to the author. Its happened to me that
nothing occurred and 6 months later I sent another email and that
seemed enough to get it fixed.
You could also ask the maintainer if the package has been abandoned
and offer to
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of dlu...@yeah.net
Sent: Tuesday, May 11, 2010 3:23 AM
To: Bert Gunter
Cc: r-help@r-project.org
Subject: Re: [R] How to estimate whether overfitting?
thanks for your help. maybe I have
R-users,
I have the following piece of code which I am trying to run on a dataframe
(aga2) with about a half million records. While the code works, it is
extremely slow. I've read some of the help archives indicating that I should
allocate space to the p1 and ags1 vectors, which I have done,
On Tue, May 11, 2010 at 8:08 AM, ivo welch ivo...@gmail.com wrote:
As an end-user, I wonder about Revolution R. Is the relationship
between Revolution R and the R community at-large a positive one? Do
the former contribute to the development efforts of the latter? Is
there a competitive
On May 11, 2010, at 9:00 AM, Terry Bassett wrote:
I am using R to read from an excel(csv) file. Within the excel file
is a column with the date set that looks likes this:
53:40.2
and in the Insert function box it looks likes this:
9/21/2006 4:53:40 PM
This is really an Excel question,
Hello,
I need to select a sample from a small population using simple random
sampling without replacement. I've found two possibilities in R, the
function sample() and the function S.SI() in the package
TeachingSampling, but I don't know which one is better.
Can someone help me?
Thank you
Instead of looping on each row, try the following
p1 - as.character(aga$AP)
# skew by one on the paste
p1 - ifelse(aga2$first.exon, p1, paste(c(, tail(ags, -1)), aga2$AP,
sep=','))
ags - as.character(aga$AS)
ags - ifelse(aga2$first.exon, ags, paste(c(, tail(ags, -1)), aga2$AS,
sep=',')
On Tue,
Hi,
At each iteration in my program,I need to generate tree vectors,X1,X2,X3,
from exponential distribution with parameters a1,a2,a3. Can you help me
please how can I do it such that it take a little time?
thank you
khazaei
__
R-help@r-project.org
On 2010.05.11 18:30:57, Lourdes Molera wrote:
Hello,
I need to select a sample from a small population using simple random
sampling without replacement. I've found two possibilities in R, the
function sample() and the function S.SI() in the package
TeachingSampling, but I don't know which
It was supposed to be 'head(p1, -1)' instead of 'tail(p1, -1)'
On Tue, May 11, 2010 at 12:17 PM, Mark Lamias mlam...@yahoo.com wrote:
R-users,
I have the following piece of code which I am trying to run on a dataframe
(aga2) with about a half million records. While the code works, it is
sprintf(%d,4)
[1] 4
for(i in 1:4) sprintf(%d,4)
for(i in 1:4) print(4)
[1] 4
[1] 4
[1] 4
[1] 4
Why doesn't sprintf like the for loop here
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
# density function
desp - function(x,lambda) {
return(lambda * exp(-lambda*x)*(x=0))
}
# cum. probability function
pesp - function(q,lambda) {
return((1-exp(-lambda*q)))
}
# quantile function
qesp - function(p,lambda) {
return(-lambda^{-1}*log(1-p))
}
# random sample
resp -
Hi,
On Tue, May 11, 2010 at 12:36 PM, khaz...@ceremade.dauphine.fr wrote:
Hi,
At each iteration in my program,I need to generate tree vectors,X1,X2,X3,
from exponential distribution with parameters a1,a2,a3. Can you help me
please how can I do it such that it take a little time?
So, at
On 11-May-10 16:53:56, Matt Young wrote:
sprintf(%d,4)
[1] 4
for(i in 1:4) sprintf(%d,4)
for(i in 1:4) print(4)
[1] 4
[1] 4
[1] 4
[1] 4
Why doesn't sprintf like the for loop here
Because, as a general rule, output generated inside loops,
and within execution of functions, etc., does
Not always the result of a function is printed on screen.
Use:
for(i in 1:4) print(sprintf(%d,4))
ciao!
mario
Matt Young wrote:
sprintf(%d,4)
[1] 4
for(i in 1:4) sprintf(%d,4)
for(i in 1:4) print(4)
[1] 4
[1] 4
[1] 4
[1] 4
Why doesn't sprintf like the for loop here
Dear all
I would like to plot a Gantt chart type plot, but with more than one
colour on a line. The gantt.chart function will allow me to add more
than one bar on a line, but it is in the same colour as the first bar
on that line. For example (using the code helpful provided in the R
graphics
I'm having difficulty installing RWinEdt in WinEdt6. I receive the following
message:
Error : .onAttach failed in attachNamespace() for 'RWinEdt', details:
call: getWinEdt()
error:
WinEdt is not installed properly.
I have tried following the manual instructions in the readme.txt without
I nominate Duncan's last statement:
If you don't want informative help files, it's really not much work to
make uninformative ones.
Duncan Murdoch
For the fortunes package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
I will clarify my problem as others has asked for more detail:
I have a dataframe, aga2, that looks like this:
Row.ID AgilentProbe GeneSymbol GeneID Exons AgilentStart first.geneid
first.exon last.geneid last.exon
8 1348 A_23_P116898 A2M 2 34 9112685 TRUE
On May 11, 2010, at 11:52 AM, D Holl wrote:
Dear all
I would like to plot a Gantt chart type plot, but with more than one
colour on a line. The gantt.chart function will allow me to add more
than one bar on a line, but it is in the same colour as the first bar
on that line. For example (using
I didn't see the cumulative paste before,
so my 'ifelse' suggestion doesn't make much
sense, sorry.
But if what you show is representative in that
there are very few 'FALSE' values in 'first.exon',
then you could use 'ifelse' and then go back and
fix the wrong ones with a loop over the rows that
Hello:
I have 5 years of weekly passage data and want to predict fish passage
for the following year. I don't have a model to use to predict data for
the sixth year. Can I somehow still predict based on these five years?
I just want to see on the graph what the predicted year would look like
and
Ted,
Regarding the addition of a 'line' to a plot with log-y axis,
there is a better way: curve() with 'add=TRUE' will respect
the current plot's log setting:
plot((1:10), log=y, yaxt=n)
axis(side=2, at=c(1,2,5,10))
f - function(x, a=0, b=1) {a + b*x}
curve(f, add = TRUE)
I recommend the ancova function in the HH package.
install.packages(HH)
library(HH)
example(ancova)
?ancova
For your example,
tmp - textConnection(
day A B
0 10.010.0
7 9.0 9.1
14 8.0 8.2
21 7.0 7.3
28 6.0 6.4
35 5.0 5.5
42
On 11/05/2010 12:23 PM, David Smith wrote:
On Tue, May 11, 2010 at 8:08 AM, ivo welch ivo...@gmail.com wrote:
As an end-user, I wonder about Revolution R. Is the relationship
between Revolution R and the R community at-large a positive one? Do
the former contribute to the development
You are spending most of the time in the loop accessing the dataframe. Put
the data you want out to a vector and then process that:
x - read.table(textConnection( Row.ID AgilentProbe GeneSymbol
GeneID Exons AgilentStart first.geneid first.exon last.geneid last.exon
+ 8 1348
Felipe Carrillo mazatlanmexico at yahoo.com writes:
## snip
In the absence of any other information, I would say your
best bet would just be to take the weekly average across the
previous years. There are lots of ways to do this (tapply,
aggregate, etc.), but cast() works:
fallavg -
Dear list,
since I have upgraded openOffice to version 3.2 I have some
trouble to open very simple ODT files generated by odfweave:
the file is apparently corrupted (but recovery is fine).
I have observed this under windows and mac OS 10.4.11 with
R 2.10.0, odfWeave_0.7.11, XML_2.6-0,
I forgot to reply to all: This is what I sent earlier to him:
I would look at the data. You don't really have any information other than
what appears to be the weekly passage. I would look to see if there is any
relationship between the current value of passage and previous values of
passage.
On Tue, May 11, 2010 at 11:24 AM, Walmes Zeviani
walmeszevi...@hotmail.com wrote:
I think the problem with nls() is the model specification. Look:
y - c(0.4334,0.3200,0.5848,0.6214,0.3890,0.5233,0.4753,
+ 0.2104,0.3240,0.2827,0.3847,0.5571,0.5432,0.1326,0.3481)
x -
In the spirit of open software, and in the spirit of R which encourages review
of software, if you find an error I think you should let the R community know
of the error. Of course you should give the package maintainer the first crack
at fixing the error, but if no fix is forthcoming, please
On 11.05.2010 18:01, Gregoire wrote:
I'm having difficulty installing RWinEdt in WinEdt6. I receive the following
message:
Error : .onAttach failed in attachNamespace() for 'RWinEdt', details:
call: getWinEdt()
error:
WinEdt is not installed properly.
I have tried following the manual
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