Hi
r-help-boun...@r-project.org napsal dne 05.08.2010 10:47:03:
>
> I have some code that is working.
>
> The code calculates the error from the real vale when it does a run.
>
> The error in metres in called
>
> errorxy
>
>
> I want to do 10 runs of the code and everytime it does a run I w
In an upcoming release of the rms package, all fit objects can be
printed using LaTeX if putting LaTeX code directly to the console
(this is optimized for Sweave). You will be able to say print(fit,
latex=TRUE).
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Dear UseRs,
I have fitted a logistic regression using glm and want a 95% confidence
interval on a response probability. Can I use
predict(model, newdata, se.fit=T)
Will fit +/- 1.96se give me a 95% of the logit? And then
exp(fit +/- 1.96se) / (exp(fit +/- 1.96se) +1) to get the probabilities?
I made some anonymous function which performs some process in matlab code..
But I want to perform it in R program...
Is there any method???
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Hi all,
I would like to flag each record in the data according to certain conditions
as specified below.
For example,
If "Close_date" in *dat* is between ("Open" & "Close") or ("Open1" &
"Close1") or ("Open2" & "Close2") in *oc, *flag the records as "Valid",
otherwise "Invalid"
I would like to
I have only achieved a half improvement.
x <- array(1:2400*1, dim = c(200,300,400))
y <- 1:400*1
ptm <- proc.time()
z <- x*as.vector(t(replicate(dim(x)[1]*dim(x)[2], y[1:dim(x)[3]])))
"replicate:"
proc.time() - ptm
x <- array(1:2400*1, dim = c(200,300,400))
y <- 1:400*1
ptm <- proc.time
thanks, it works just great.
David Biau.
De : "Abhijit Dasgupta, PhD"
Cc : r help list
Envoyé le : Jeu 5 août 2010, 22h 15min 37s
Objet : Re: [R] How to extract se(coef) from cph?
if the cph model fit is m1, you can try
sqrt(diag(m1$var))
This is coded i
On Aug 5, 2010, at 7:40 PM, noclue_ wrote:
I have a 64-bit windows box -
Intel Xeon CPU E7340 @ 2.4GHz 31.9GB of RAM
I have R 2.11.1 (64bit) running on it.
Dear noclue_;
What does this return?:
.Machine$sizeof.pointer
My csv data is 3.6 GB (with about 15 million obs, 120 variables.)
Hi all~
I am wondering if it is possible to alter the stopping precision for parameters
estimated using the 'optim'?
If it helps, I am minimizing the log-likelihood of a function using constraints
(i.e. L-BFG-S).
-Jeremy
__
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Dear R-users,
I would like to model data where the response variable consists of many minus
ones and many different positive values that seem to follow an apparently
separate distribution (ie. -1, -1, 0.5, -1, 3, 3.5, 1.2, -1, -1, 0.4, etc); no
values of the response can be less than minus on
I have a 64-bit windows box -
Intel Xeon CPU E7340 @ 2.4GHz 31.9GB of RAM
I have R 2.11.1 (64bit) running on it.
My csv data is 3.6 GB (with about 15 million obs, 120 variables.)
I have successfully imported the data above into R. No problem.
> Hi R-users,
>
> I'm using R 2.11.1, mgcv 1.6-2 to fit a generalized additive mixed model.
> I'm new to this package...and just got more and more problems...
>
> 1. Can I include correlation and/or random effect into gam( ) also? or
> only
> gamm( ) could be used?
Only in gamm()
> 2. I want to e
I sometimes have to work with vectors/matrices with > 2^31 - 1
elements. I have found the bigmemory package to be of great help. My
lab is also going to learn sqldf package for getting bits of big data
into/out of R. Learning both of those packages should help you work
with large datasets in R.
Th
Assigning the value directly to result[i] would give a null to result. Like
that:
result[[i]]<-taxondiveO(abd,taxa)
result[[i]]<-as.data.frame(result[[i]])[1,]
In fact your code's result should have null elements in it. You lost your
nulls through do.call process.
I can't find a way to assign t
Works perfectly. Thank you!
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https:/
On Aug 5, 2010, at 6:22 PM, Jun Shen wrote:
I believe this has been discussed many times in the archives. Here is
one way to do it.
Yes, you are so right.
Use function "sink" to direct the output to a file you specify
sink( file='your file.txt')
after you execute this command, all the ou
I believe this has been discussed many times in the archives. Here is
one way to do it.
Use function "sink" to direct the output to a file you specify
>sink(file='your file.txt')
after you execute this command, all the output will be saved to your
file and you won't see it on the screen. Say
>sum
Hi Colm,
the replaced code below works (its probably not the most elegant way,
but rather the result of hours trial and error with eval, paste, bquote
and Co).
But if you want to keep all of your model results you should read the
help for 'assign' and insert something like
assign(paste("lme.",mean
hi pete,
try
download.file("http://ir.eia.gov/wpsr/psw09.xls","C:\\Temp\\psw09.xls",mode="wb";)
since xls are binary files.
cheers
Am 05.08.2010 23:25, schrieb Pete B:
>
> I am trying to download an Excel file from the the internet with the
> following R command
>
> download.file("http://ir.eia
I am trying to download an Excel file from the the internet with the
following R command
download.file("http://ir.eia.gov/wpsr/psw09.xls","C:\\Temp\\psw09.xls";)
The file is downloaded but when I try and open it I get a message from Excel
telling me that the file is corrupt and it cannot be open
Thanks very much! Works fine now. I'll read up on R's evaluation model.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented,
Hi everybody,
I'm having trouble getting an eval-parse combination to work with lme in a for
loop.
Consider the code below. The call to lme outside the loop works. The call to
aov inside the loop works, but the call to to lme inside the loop does not.
The latter fails with
Error in model.fram
OK:
> comp <- function(x) Reduce(Funcall, list(plus2,plus3,plus4), init=x,
> right=TRUE)
> comp(3)
[1] 12
but of course, Gabor's answer is more insightful and cleaner.
-- David
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Mog
Jennifer Hou wrote:
Thank you very much for your kind reply, I have found that the error
was in the stdCoeff function and not in the linear model. summary()
works pretty well on my model, I will simply use another function to
compute my coefficients. Best regards, Jennifer
See ?coef , summar
Thank you very much for your kind reply, I have found that the error was in the
stdCoeff function and not in the linear model. summary() works pretty well on
my model, I will simply use another function to compute my coefficients.
Best regards,
Jennifer
> Date: Wed, 4 Aug 2010 15:27:40 -0500
>
On Thu, Aug 5, 2010 at 4:31 PM, Mog wrote:
> Hi All,
>
> I'd like to be able to specify the ordered composition of several
> functions. So I defined compose:
>
> compose <- function(f,g){
> function(x){f(g(x))}
> }
>
> and some simple test functions:
>
> plus2 <- function(x){x+2}
> plus3 <- funct
I realize that the Funcall example with an init parameter will work,
but I'd like to be able to get a one-argument function/closure out of
that reduction, if at all possible. Sorry, I under-specified in my
question.
Thanks!
__
R-help@r-project.org maili
> Funcall <- function(f, ...) f(...)
> Reduce(Funcall,list(plus2,plus3,plus4),init=3,right=TRUE)
[1] 12
Basically from the examples from ?Reduce.
HTH,
David L. Reiner
XR Trading LLC
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of M
On Aug 5, 2010, at 3:59 PM, karena wrote:
library(GenABEL)
gs.b <- gs
> library(GenABEL)
Loading required package: MASS
GenABEL v. 1.6-0 (June 21, 2010) loaded
> gs.b <- gs
Error: object 'gs' not found
You are not making this easy. I asked what str(gs) returned. Instead
you gave us code b
Hi All,
I'd like to be able to specify the ordered composition of several
functions. So I defined compose:
compose <- function(f,g){
function(x){f(g(x))}
}
and some simple test functions:
plus2 <- function(x){x+2}
plus3 <- function(x){x+3}
plus4 <- function(x){x+4}
> (compose(plus2,compose(
On Thu, 2010-08-05 at 10:59 +0100, Liviu Andronic wrote:
> On Thu, 05 Aug 2010 11:13:00 +0200
> Olga Lyashevska wrote:
> > I do not have any desktop environment (running R on a server), can I
> > still try xfce?
> >
> I am not very sure. If you are running gnome-terminal, I assume that
> you hav
Excellent!
Yes, FH has a function to get LateX tables, but I not malleable enough.
Thanks,
David Biau.
De : David Winsemius
Cc : r help list
Envoyé le : Jeu 5 août 2010, 22h 11min 20s
Objet : Re: [R] How to extract se(coef) from cph?
On Aug 5, 2010, at
if the cph model fit is m1, you can try
sqrt(diag(m1$var))
This is coded in print.cph.fit (library(rms))
On 08/05/2010 04:03 PM, Biau David wrote:
Hello,
I am modeling some survival data wih cph (Design). I have modeled a predictor
which showed non linear effect with restricted cubic splines.
library(GenABEL)
gs.b <- gs
ok <- complete.cases(g...@phdata[,c('sex','age','b.dbp','b.bmi')])
gs.b <- gs.b[ok]
g...@phdata <- g...@phdata[,c('id','sex','age','b.dbp','b.bmi')]
index=1:g...@gtdata@nsnps
ran.snp=sample(index,261,replace=F)
gs.b.gkin=ibs(gs.b[,ran.snp], weight="freq")
attach(g...@p
On Aug 5, 2010, at 4:03 PM, Biau David wrote:
Hello,
I am modeling some survival data wih cph (Design). I have modeled a
predictor
which showed non linear effect with restricted cubic splines. I
would like to
retrieve the se(coef) for other, linear, predictors.
The cph object has a "var
Dear all,
I run a loop wrapped in try(), and for each of the rows where
"try-error" is true I want to fill that row with NA (at the moment it is
omitted). So I would expect to get a dataframe with 1000 rows some of
which would be empty, but instead I get a dataframe with 995 rows. In
this case mi
Hello,
I am modeling some survival data wih cph (Design). I have modeled a predictor
which showed non linear effect with restricted cubic splines. I would like to
retrieve the se(coef) for other, linear, predictors. This is just to make nice
LateX tables automatically. I have the coefficients w
On Aug 5, 2010, at 1:22 PM, yankeetilidie wrote:
>
> Hello,
>
> I am attempting to create a bar plot that contains a range of possible
> response values on the x-axis of 1 to 5 and contains barplots for the number
> of responses even in the event that there are 0 responses. For example, I
> hav
Suppose
x <- array(c(1,1,1,1,2,2,2,2), dim = c(2,2,2))
y <- c(5, 10)
Now I would like to multiply x[, , 1] with y[1] and x[, , 2] with y[2].
Possible solution is a for-loop:
for (i in 1:2) {
x[, , i] * y[i]
}
Another possible solution is this construction:
as.vector(t(replicate(nrow(
On Aug 5, 2010, at 3:44 PM, karena wrote:
thank you, but even after I tried 'complete.cases' function,
How did you try?
I still get the
same error messages.
What does str(gs) tell you?
help.
Plaintive cries for help are considerably less useful than details
about the
thank you, but even after I tried 'complete.cases' function, I still get the
same error messages.
help.
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Thanks for the help so far-- one interesting thing about this particular page
is that the data displayed on the website actually differs from the data you
can access with the "download" link. The XML package command works, but the
table it produces in R has the following column names:
> x1 =
>
On Aug 5, 2010, at 3:05 PM, David Winsemius wrote:
On Aug 5, 2010, at 2:48 PM, Archana Dayalu wrote:
Hello,
I am using read.table to read files directly from a public ftp
site. I have
a general list of files that may or may not exist in the ftp
directory, but
my hope was that R would rea
Hi,
Thank you, you're right. I should have read the website more carefully:
"Download and install R (installers for Linux, Mac OS X universal, Windows,
...; use at least R 2.10.1)."
http://www.sciviews.org/SciViews-K/index.html
In Windows, I use that version but I didn't think about it when tryi
I'm trying to export the results of my summary data for the object
horton.nlme, but failing miserably. Running summary(horton.nlme) works
fine, but both write.table and write.csv return the error "cannot coerce
class 'c("summary.lme", "nlme", "lme")' into a data.frame".
I know I can copy and past
On Aug 5, 2010, at 2:48 PM, Archana Dayalu wrote:
Hello,
I am using read.table to read files directly from a public ftp site.
I have
a general list of files that may or may not exist in the ftp
directory, but
my hope was that R would read the file if it existed and ignored it
if it
didn't
Hello,
I am using read.table to read files directly from a public ftp site. I have
a general list of files that may or may not exist in the ftp directory, but
my hope was that R would read the file if it existed and ignored it if it
didn't exist and move on to the next one. However, when R arrives
Hello,
I am attempting to create a bar plot that contains a range of possible
response values on the x-axis of 1 to 5 and contains barplots for the number
of responses even in the event that there are 0 responses. For example, I
have a data set that contains values of 2, 3, 4, and 5 but I would
Hi David,
I would like to apologize for what I wrote earlier. It was late and I
was frustrated. Please give me time to adapt to the formal structures
of the forum.
Best,
Ralf
On Thu, Aug 5, 2010 at 7:32 AM, David Winsemius wrote:
>
> On Aug 5, 2010, at 4:10 AM, Ralf B wrote:
>
>> This is unbeli
Try this:
library(XML)
readHTMLTable('
http://www.invescopowershares.com/products/holdings.aspx?ticker=PGX', which
= 13, header = TRUE)
On Wed, Aug 4, 2010 at 6:07 PM, AndrewPage wrote:
>
> Hi all,
>
> I am using RCurl to try and download data from a website, but I'm having
> trouble finding ou
On 05.08.2010 19:48, rod84 wrote:
Thank you for your prompt response. I am not used to the process of building a
package. And I am using Windows as an OS.Do you have some steps I should follow
to be able to build the package.Thank you for your time and recommendations.
See the manual R
Thank you for your prompt response. I am not used to the process of building a
package. And I am using Windows as an OS.Do you have some steps I should follow
to be able to build the package.Thank you for your time and recommendations.
Date: Thu, 5 Aug 2010 09:19:18 -0700
From: ml-node+2315087
Thank you for your prompt response,I am using Windows.
> Date: Thu, 5 Aug 2010 12:32:19 -0500
> Subject: Re: [R] Help installation lme4a, Error Message: lme4a is not a valid
> installed library
> From: ba...@stat.wisc.edu
> To: ngueye...@hotmail.com
> CC: r-help@r-project.org
>
> On Thu
Hi,
httpdPort arrived with R 2.10.0, apparently Sciviews-K relies on this,
so you need to upgrade R to a newer version.
Romain
Le 05/08/10 19:16, Albert-Jan Roskam a écrit :
Hi,
I'm trying to install Sciviews-K on Linux Ubuntu 9.10 (karmic) but I'm not able
to establish the connection bet
On Thu, Aug 5, 2010 at 9:31 AM, rod84 wrote:
> Dear R users,
> I recently downloaded the library lme4a by
> svn checkout svn://svn.r-forge.r-project.org/svnroot/lme4.
> I tried to install the library lme4a by copying the downloaded document in
> the location where all the R libraries are saved.
On 8/4/2010 2:07 PM, AndrewPage wrote:
Hi all,
I am using RCurl to try and download data from a website, but I'm having
trouble finding out what URL to use. Here is the site:
http://www.invescopowershares.com/products/holdings.aspx?ticker=PGX
See how in the upper right, above the displayed s
Correct.
--
Wolfgang Viechtbauerhttp://www.wvbauer.com/
Department of Methodology and StatisticsTel: +31 (0)43 388-2277
School for Public Health and Primary Care Office Location:
Maastricht University, P.O. Box 616 Room B2.01 (second floor)
6200 MD Maastricht,
Hi,
I'm trying to install Sciviews-K on Linux Ubuntu 9.10 (karmic) but I'm not able
to establish the connection between Komodo and R. Here;s the error I get, plus
some diagnostic info:
R version 2.9.2 (2009-08-24)
R is SciViews ready!
Error in get(name, envir = asNamespace(pkg), inherits = FALS
Dear Angelo,
rma(yi=o, sei=se, mods=~s+t-1, method="REML")
is *a* way to run the arm-based pairwise meta-analysis. Whether it is the
*correct* way is a question I cannot answer.
lme(o~s+t-1, random=~t-1 | s, weights=(~ se^2))
is a different model. First of all, it adds a random effect only to
It is not clear what question you are trying to answer. Perhaps if you can
give us an explanation of your overall goal then we can be more helpful.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
>
The warning (with an error you would not see any results) means that there are
ties in your data, the theory behind the ks test says that the probability of
seeing ties is 0, so your data and the theory do not match, therefore the
p-value is suspect (though an ok approximation for some uses).
T
Hi folks,
I originally tried R-SIG-Mixed-Models for this one
(https://stat.ethz.ch/pipermail/r-sig-mixed-models/2010q3/004170.html),
but I think that the final steps to a solution aren't mixed-model
specific, so I thought I'd ask my final questions here.
I used gamm4 to fit a generalized additive
On Aug 5, 2010, at 12:23 PM, karena wrote:
Hi,
I am using a function 'polygenic' in the package called 'GenABEL'.
h2.gs <- polygenic(b.dbp~age+age2+age3+sex+b.bmi, kin=gs.gkin,
data=gs)
Have you considered us
I am not new to R, but I am new to .grib files. I am downloading some
climate data and I would like to analyze it in R. R has a nice netcdf
package, but I don’t see any package available to deal specifically
with grib files. I see a few posts from other people using grib files
in R. However, I
Dear R users,
I recently downloaded the library lme4a by
svn checkout svn://svn.r-forge.r-project.org/svnroot/lme4.
I tried to install the library lme4a by copying the downloaded document in
the location where all the R libraries are saved.
When I try to load the library, I obtain the message
Alison,
Check out the options for the function bxp(), they include control over
the colors of all parts of the boxplot, e.g., whiskcol for whisker color.
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East
Hi,
I am having a hard time getting what the summary.lm-output for an ANOVA /
ANCOVA means.
Examples I find always seem to deal with simpler cases than what I meet in my
data.
My main problem is understanding the output when getting significant
INTERACTION TERMS (what never occurs in examples
Hi,
I am using a function 'polygenic' in the package called 'GenABEL'.
> h2.gs <- polygenic(b.dbp~age+age2+age3+sex+b.bmi, kin=gs.gkin, data=gs)
--
rod84 wrote:
Dear R users,
I recently downloaded the library lme4a by
svn checkout svn://svn.r-forge.r-project.org/svnroot/lme4.
I tried to install the library lme4a by copying the downloaded document in
the location where all the R libraries are saved.
When I try to load the library, I obtain
A related hint, Google doesn't let you search for %nin%, because it
ignores % symbols (and most other punctuation), but cuil does allow
you to search:
http://cuil.com/search?q=%25nin%25+R
On 5 August 2010 08:53, David Winsemius wrote:
> The examples in the help page for "%in%" (shared by "match")
Dear R users,
I recently downloaded the library lme4a by
svn checkout svn://svn.r-forge.r-project.org/svnroot/lme4.
I tried to install the library lme4a by copying the downloaded document in
the location where all the R libraries are saved.
When I try to load the library, I obtain the message
Yeah, and %w/o% seems to have reinvented setdiff(). =)
-Ken
On 8/5/10 10:53 AM, "David Winsemius" wrote:
> The examples in the help page for "%in%" (shared by "match") has the
> definition of a "%w/o%" binary operator.
>
> "%w/o%" <- function(x,y) x[!x %in% y] #-- x without y
> since:
> "%
The examples in the help page for "%in%" (shared by "match") has the
definition of a "%w/o%" binary operator.
"%w/o%" <- function(x,y) x[!x %in% y] #-- x without y
since:
"%in%" <- function(x, table) match(x, table, nomatch = 0) > 0
It appears that you have just re-invented the without-wheel.
For curiosity's sake, and perhaps closer in keystrokes to R home, here's
another version,
`%ni%` <- Negate(`%in%`)
baptiste
On Aug 5, 2010, at 5:30 PM, Ken Williams wrote:
> Ha! Thanks. I should have a closer look at Hmisc in general.
>
> -Ken
>
>
> On 8/5/10 10:25 AM, "David Huffer" wr
Dear all
It seems that it is not possible [1] to generate several graphs in a
loop within an Sweave document. For example,
<>
for (i in 1:4) plot(rnorm(100)+i)
@
will not work. Since this limitation dates from old times (at least
2005), I was curious whether workarounds other than
"\includegr
Ha! Thanks. I should have a closer look at Hmisc in general.
-Ken
On 8/5/10 10:25 AM, "David Huffer" wrote:
> See Harrell's Hmisc package
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Ken Williams
> Sent: Thursday,
Just FYI, the Hmisc package has had an implementation of %nin% for some
time now.
Ken Williams wrote:
Sometimes I write code like this:
qf.a <- subset(qf, pubid %in% c(104, 106, 107, 108))
qf.b <- subset(qf, !pubid %in% c(104, 106, 107, 108))
and I get a little worried that maybe I've remem
See Harrell's Hmisc package
--
David Huffer, Ph.D.
Deputy Director
CSOSA/ORE
Washington, DC
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Ken Williams
Sent: Thursday, August 05, 2010 11:20 AM
To: r-help@r-project.org
Subject:
Sometimes I write code like this:
> qf.a <- subset(qf, pubid %in% c(104, 106, 107, 108))
> qf.b <- subset(qf, !pubid %in% c(104, 106, 107, 108))
and I get a little worried that maybe I've remembered the precedence rules
wrong, so I change it to
> qf.a <- subset(qf, pubid %in% c(104, 106, 107, 10
On 03/08/10 21:50, GL wrote:
If I have a column with 2 levels, but one level has no remaining
observations. Can I remove the level?
Like this?
d <- data.frame(a = factor(rep("A", 3), levels = c("A", "B")))
levels(d$a)
# [1] "A" "B"
d$a <- d$a[,drop=TRUE]
levels(d$a)
# [1] "A"
Hope this h
Hello,
I'm trying to illustrate the relationships between various trait and
environment data gathered from a number of sites. I've created a GAM to do
this: gam1=gam(trait~s(env1)+s(env2)+te(env1,env2)) and I know how to create
a 3D plot using vis.gam. I want to be able to show points on the 3D p
Am 05.08.2010 12:57, schrieb spigo:
> I don't know how to do it. I need to use reshape? how?
> thanks for your reply.
example(reshape)
Stefan
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PLEASE do read the posting
I believe Wu Gong has given you a solution. As a note you were probably
reading in the first two columns as either factors or characters and the last
one as numeric. You might want to try it again and then do a str() on the
resulting data.frame to see what was happening.
It can be confusing
Dear all,
As I couldn't find any thread on the internet I hope the help-list might help
me out.
I've tried to update Rtools from R210 used in combination with R2.9.1 to R211
in combination with R2.11.1. However, I do not succeed.
I have R2.11.1 running, as well as Inno Setup 5, HTML help and M
Try this:
unstack(DF, Avariato ~ TIPO)
On Thu, Aug 5, 2010 at 7:57 AM, spigo wrote:
>
> Dear all,
> I'm new here, I'm starting to learn R (oh my God!!). So, now I'm in panic
> because I have this situation:
>
> TIPO Avariato
> A 0.05
> B 0.09
> A 9
> B 8
> A 9
>
Hello,
For that simple case you could just do...
newdf <- data.frame(A=df[df$TIPO == "A",2], B=df[df$TIPO == "B",2])
where df is the name of your existing data.frame
Michael
On 5 August 2010 20:57, spigo wrote:
>
> Dear all,
> I'm new here, I'm starting to learn R (oh my God!!). So, now I'm
On Aug 5, 2010, at 6:50 AM, Giuseppe Amatulli wrote:
Hi,
can somebody tell me why R is not able to calculate a linear model
written in this way?
lm (seq(1:100)~seq(1:100))
Call:
lm(formula = seq(1:100) ~ seq(1:100))
Coefficients:
(Intercept)
50.5
Warning messages:
1: In model.matrix
On 08/05/2010 05:50 AM, Giuseppe Amatulli wrote:
Hi,
can somebody tell me why R is not able to calculate a linear model
written in this way?
lm (seq(1:100)~seq(1:100))
Call:
lm(formula = seq(1:100) ~ seq(1:100))
Coefficients:
(Intercept)
50.5
Warning messages:
1: In model.matrix.def
It would really help if you could reduce the information below to a small
reproducible example, that is, some small bit of code that a reader can paste
into R and poke around at.
For sending a small sample dataset have a look at ?dput
I believe that this problem has been discussed on the R-he
Hello,
Please read the posting guide found at the bottom of every post to
this list. We need to be able to see a small, reproducible example
of code that illustrates your question.
It sounds like you might be looking for ?replicate.
On 08/05/2010 03:47 AM, Turn & Fall wrote:
I have some code
On 木, 2010-08-05 at 07:32 -0400, David Winsemius wrote:
> I do no
> see where that is "posting cheeky R code". I saw it as trying to be
> constructive. Using it would only be part of the recommended actions
> to take before posting
Ralf B., Dr. Wisemius is one of the most prolific, unselfis
Dear all,
I'm new here, I'm starting to learn R (oh my God!!). So, now I'm in panic
because I have this situation:
TIPO Avariato
A 0.05
B 0.09
A 9
B 8
A 9
B 3
and this is what I would like to do:
AvariatoA B
Avariato0.050.09
Avari
I have some code that is working.
The code calculates the error from the real vale when it does a run.
The error in metres in called
errorxy
I want to do 10 runs of the code and everytime it does a run I want to an
output of the errorxy, so that it can form an array.
I am guessing I should u
Hi,
can somebody tell me why R is not able to calculate a linear model
written in this way?
> lm (seq(1:100)~seq(1:100))
Call:
lm(formula = seq(1:100) ~ seq(1:100))
Coefficients:
(Intercept)
50.5
Warning messages:
1: In model.matrix.default(mt, mf, contrasts) :
the response appeared on
Hi all,
I have 6 datasets(dataframes Assem_ContigsLen7 through all_ContigsLen12)
containing 3 columns (contig_id, contig_length, read_count).
Each dataset is composed of 3 types of contigs (assemblies of genomic
fragments), 1- all Bacterial fragments, 2 - all Viral fragments, 3 -
mixed fragments.
Try this:
n <- 3
w <- as.numeric(format(Sys.Date(), '%w'))
fut <- c(Sys.Date() - 0:(n + ifelse(w - n > 6, (w - n) - 6, 0)),
Sys.Date() + 1:(n + 1 + ifelse(w + n > 6, (w + n) - 6, 0)))
sort(fut[!format(fut, '%w') %in% c(6, 0)])
On Thu, Aug 5, 2010 at 9:09 AM, Sergey Goriatchev wrote:
Thank you David, will check it out, as I use chron and zoo!
Regards,
Sergey
On Thu, Aug 5, 2010 at 14:29, David Winsemius wrote:
>
> On Aug 5, 2010, at 8:09 AM, Sergey Goriatchev wrote:
>
>> Hi, Henrique
>>
>> Thank you for trying, but that is not what I want. You get WEEKdays, I
>> need WORKday
On Aug 5, 2010, at 8:09 AM, Sergey Goriatchev wrote:
Hi, Henrique
Thank you for trying, but that is not what I want. You get WEEKdays, I
need WORKdays, and preferably sorted in order from future to the past.
Perhaps you would be pleased to know that there is an is.holiday
function in chron
Hi, Henrique
Thank you for trying, but that is not what I want. You get WEEKdays, I
need WORKdays, and preferably sorted in order from future to the past.
Best,
Sergey
On Thu, Aug 5, 2010 at 14:06, Henrique Dallazuanna wrote:
> Try this:
>
> c(Sys.Date() + 0:3, Sys.Date() - 0:3)
>
> or
>
> 0:3
Dear list members,
I have a question concerning multiple comparisons after using glm.
My response variable is days until emergence of an insect species. The
explanatory variables are sex (two levels), parasitoids added (two levels) and
populations (34 levels). I would like to know now which po
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