On Tue, Jul 14, 2009 at 8:20 AM, Michael Knudsen wrote:
> What do you mean? It looks a like a very general solution to me.
Just got an email suggesting using the functions col and row. For example
temp = matrix(c(1:36),nrow=6)
which(col(temp)>row(temp))
This gives the indices (in the matrix vie
On Tue, Jul 14, 2009 at 8:03 AM, Moshe Olshansky wrote:
> Make it
> for (i in 1:9)
Thanks. That's also how I solved the problem myself. I just somehow
think it makes my code look rather clumsy and opaque. Maybe I just
have to get used to this kind of nasty tricks.
> This is not the general solut
Make it
for (i in 1:9)
This is not the general solution, but in your case when i=10 you do not want to
do anything.
--- On Tue, 14/7/09, Michael Knudsen wrote:
> From: Michael Knudsen
> Subject: [R] Nested for loops
> To: r-help@r-project.org
> Received: Tuesday, 14 July, 2009, 3:38 PM
> Hi,
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Michael Knudsen
> Sent: Monday, July 13, 2009 10:39 PM
> To: r-help@r-project.org
> Subject: [R] Nested for loops
>
> Hi,
>
> I have spent some time locating a quite subtle (at
Hi All,
I am using raw data.
Following is an example of my data frame
P1 P2 P3
H1 1 1
M2 4 2
L 2 2 2
H,M,L relate to the three cases, and P1 to P3 relate to the
raters/participants.
The problem I am having is if all the values are 1 in the above data frame
(
Hi,
I have spent some time locating a quite subtle (at least in my
opinion) bug in my code. I want two nested for loops traversing the
above-diagonal part of a square matrix. In pseudo code it would
something like
for i = 1 to 10
{
for j = i+1 to 10
{
// do something
}
}
However,
There are many more things available than just 'nls'. An overview
is available at "www.r-project.org" -> CRAN (select your favorite
mirror) -> "Task Views" (on the left, third line under CRAN) ->
"Optimization Optimization and Mathematical Programming".
For a possibly more di
On Jul 13, 2009, at 9:31 PM, kfort...@email.unc.edu wrote:
Hello All, Thank you for taking my question. I am looking for
information on how R handles interaction terms in a multiple
regression using the “lm” command. I originally noticed something
was unusual when my R output did not mat
One (awkward) way to do this is:
x <- matrix(c(c(test),c(test2)),ncol=2)
y <- rowMeans(x,na.rm=TRUE)
testave <- matrix(y,nrow=nrow(test))
--- On Tue, 14/7/09, Tish Robertson wrote:
> From: Tish Robertson
> Subject: [R] averaging two matrices whilst ignoring missing values
> To: r-help@r-proje
On 14/07/2009, at 1:46 PM, Tish Robertson wrote:
Hi folks,
I'm trying to do something that seems like it should easy, but it
apparently isn't. I have two large matrices, both containing a
number of missing values at different cells. I would like to
average these matrices, but the NAs ar
Hello All, Thank you for taking my question. I am looking for
information on how R handles interaction terms in a multiple regression
using the lm command. I originally noticed something was unusual
when my R output did not match the output from JMP for an identical
test run previously. Bot
Hi folks,
I'm trying to do something that seems like it should easy, but it apparently
isn't. I have two large matrices, both containing a number of missing values
at different cells. I would like to average these matrices, but the NAs are
preventing me. I get a "non-numeric argument to bi
any ideas? has anybody tried to recover RData files before?
Duncan Murdoch-2 wrote:
>
> On 7/13/2009 3:38 PM, sten...@go.com wrote:
>>Is there a way to identify the beginning and end of an .RData file?
>>I am trying to restore lost files on opensolaris where the table of
>> contents
>
Or if you need it to be fast, try data.table. X[Y] is a join when X and Y
are both data.tables. X[Y] is a left join, Y[X] is a right join. 'nomatch'
controls the inner/outer join i.e. what happens for unmatched rows. This
is much faster than merge().
"Gabor Grothendieck" wrote in message
Thanks for your help. I found another cmd as follows:
R CMD BATCH infile outfile
what's the difference between this expression and commandArgs()? I try to
refer to the instruction of commandArgs but it seems not very helpful.
Henrique Dallazuanna wrote:
>
> See commandArgs() function.
>
Thanks for your help. I found another cmd as follows:
R CMD BATCH infile outfile
what's the difference between this expression and your suggestion?
cls59 wrote:
>
>
> edisonying wrote:
>>
>> I am a beginner in R and know only a little about it yet. I have a script
>> written in R langua
Dear R experts,
I am observing undesired behavior of predict(fit, newdata), in case when fit
object is produced by lm() involving a poly(). Here is how to reproduce:
x <- c(1:10)
y <- sin(c(1:10))
fit <- lm(formula=y~poly(x, 5, raw=TRUE))
predict(fit, newdata=data.frame(x=c(1:10))) ## this works
Hello to everybody,
I need to know the Smoothing Parameter to obtain Home Range of an animal
through the Area Kernel. I have 200 locations with x and y. How can I obtain
the Smoothing Parameter with R for LSCV, CV and Href method???
Thank you.
--
View this message in context:
http://www.nab
Hello group,
recently I read about the SweaveListingUtils package and now I would like to
try it out. However I can not make it run...
Below is a minimal example. The problem seems the following line, generated
by the package:
\ifthenelse{\boolean{swe...@gin}}{\setkeys{gin}{width=0.6\textwidth}}{
edisonying wrote:
>
> I am a beginner in R and know only a little about it yet. I have a script
> written in R language, named as "a.txt" for example. I am using a Linux
> machine, at present I only know that I can type "R" in the terminal and
> then copy-paste the content in "a.txt" to the R's
Hi - since Rgraphviz was officially moved over to Bioconductor, this might
be a misguided post, but it's worth a shot:
I'm plotting a graph using Rgraphviz and in an attempt to force the edges to
be behind the nodes (on a fairly complex graph), I set the "outputorder"
graph attribute to "edgesfirst
On Mon, Jul 13, 2009 at 4:26 PM, saurav pathak wrote:
> I am using R 2.9.1,
That's good!
> I am not sure about the version of sampleSelection and maxLik
This is important! Please check the version numbers, e.g. with
R> help(package="maxLik")
R> help(package="sampleSelection")
BTW: Did you insta
Hi,
I was wondering whether there is any Fortran function or associated library
for evaluating the quantiles of a set of values (something which the
R-function quantile() does). Any help will be much appreciated.
Thanks and regards,
Dhiman Bhadra
[[alternative HTML version deleted]]
_
Hi Henrique & other R-helpers,
Thank you for helping me last week. I used Henrique's suggestion to develop
some code (below) to combine two rows in my dataframe into a third row, and
then delete the original two rows. It works well.
My solution is not very elegant however; if there's a function (o
Hi Uyen,
You do not have enough information to estimate 4 parameters in your
nonlinear model. Even though you have 12 data points, only 6 are
contributing independent information (you essentially have 6 replicate
points). If you plot the fittted function you will see that it fits your
data real
On Mon, 13 Jul 2009, Charles C. Berry wrote:
On Mon, 13 Jul 2009, Heinz Tuechler wrote:
At 20:18 13.07.2009, Charles C. Berry wrote:
> On Mon, 13 Jul 2009, Heinz Tuechler wrote:
>
> > Dear All,
> >
> > since years I am struggling with Surv objects in data.frames. The
> > following seem
On Mon, 13 Jul 2009, Heinz Tuechler wrote:
At 20:18 13.07.2009, Charles C. Berry wrote:
On Mon, 13 Jul 2009, Heinz Tuechler wrote:
> Dear All,
>
> since years I am struggling with Surv objects in data.frames. The
> following seems to have to do with it.
> See below the modified example from
AgusSusanto wrote:
>
> Dear all,
> I would like to fit a linear regression with replication (on each year,
> observation is replicated, e.g 4 times). The independent variable ranges
> for instance 1-5 year, so I expect to have a linear fit of 5 points.
> For that purpose I do these (with dummy
Your basic problem is that you are trying to change the value of a
global object (TotalCover.df) from within a function. A better
approach is to pass in the object that you are changing and then
return it as the value of the function. You can then assign it back
to the original object if you want
At 20:18 13.07.2009, Charles C. Berry wrote:
On Mon, 13 Jul 2009, Heinz Tuechler wrote:
Dear All,
since years I am struggling with Surv objects in data.frames. The
following seems to have to do with it.
See below the modified example from the help page of survSplit. The
original works, as ex
Thomas Petzoldt schrieb:
Hi,
using a variable named .trPaths is a feature and a frequent nuisance
of recent Tinn-R and also explained in the Tinn-R docs.
Solution 1:
In Tinn-R select the menu:
R --> Configure --> and then "Temporarily" or "Permanent"
Solution 2:
Manually edit file /etc/
On 7/13/2009 3:27 PM, Rebecca Sela wrote:
Thank you! (That was easy to fix.)
How does one deal with quoting (in \reference)? The following line causes
problems:
\references{Sela, Rebecca J., and Simonoff, Jeffrey S., \dQuote{RE-EM Trees: A
New Data Mining Approach for Longitudinal Data}.}
Th
On 7/13/2009 3:38 PM, sten...@go.com wrote:
Is there a way to identify the beginning and end of an .RData file?
I am trying to restore lost files on opensolaris where the table of contents
is overwritten and will be reading the raw disk byte by byte.
I looked at several files in hex a
On 7/13/2009 3:21 PM, Mark Knecht wrote:
On Mon, Jul 13, 2009 at 12:08 PM, David Winsemius wrote:
In R a function only returns the last evaluation, so you need to wrap up all
of the local results into a list at the end of the function.
David Winsemius, MD
Heritage Laboratories
West Hartfo
Hello,
recently I read about the SweaveListingUtils package and now I want to try
it out. However, I can not make it work...
Below a minimal example. The problem seems to be the following line
(generated by SweaveListingPreparations()?):
\ifthenelse{\boolean{swe...@gin}}{\setkeys{gin}{width=0.6\t
Hi,
First up, thanks again for all the help I'm getting on this list.
I'm making great headway in analyzing my experimental data on an
experiment by experiment basis. No way I could have done this in the
time I've done it without your help.
This email is partially a question about R but is a
Hello R users,
I have developed code in R that generates populations of non-overlapping
random polygons ("polygonal maps") for testing certain stereological
estimation methods. I have tried to look for articles or papers online on
the generation of random polygonal maps and have found very littl
Hi, I am turning to you with a (hopefully simple?) stats question. I would
like to test equality of two correlation coefficients in a setting with
three variables X,Y,Z, i.e. equality of r(X,Y) and r(Z,Y). I have found a
formula to transform the "2 dependent correlations difference" to
t-distribut
How important it is to wrap the list in a return statement, ala
return(list(ShrubCover.df, TreeCover.df, TotalCover.df))
or
answer <- list(ShrubCover.df, TreeCover.df, TotalCover.df)
return(answer)
---
Completely Un.
Consult the R Docs, especially the R Language Definition manual, for answers
Dear all,
I would like to fit a linear regression with replication (on each year,
observation is replicated, e.g 4 times). The independent variable ranges
for instance 1-5 year, so I expect to have a linear fit of 5 points.
For that purpose I do these (with dummy variables x and y):
x<-rep(seq(1:5
Hi Brandy,
The `vmmin' refers to a variable metric algorithm, which is a quasi-Newton
method for optimization. This algorithm is used when `method="BFGS"' in the
optim() call. The quasi-Newton methods iteratively build an approopriate
Hessian matrix, which is of dimension p x p, where p is the p
Is there a way to identify the beginning and end of an .RData file?
I am trying to restore lost files on opensolaris where the table of contents
is overwritten and will be reading the raw disk byte by byte.
I looked at several files in hex and the end seems different, while most of
Thank you! (That was easy to fix.)
How does one deal with quoting (in \reference)? The following line causes
problems:
\references{Sela, Rebecca J., and Simonoff, Jeffrey S., \dQuote{RE-EM Trees: A
New Data Mining Approach for Longitudinal Data}.}
The error given is:
Warning in parse_Rd("./man
See commandArgs() function.
On Mon, Jul 13, 2009 at 4:15 PM, edisonying wrote:
>
> I am a beginner in R and know only a little about it yet. I have a script
> written in R language, named as "a.txt" for example. I am using a Linux
> machine, at present I only know that I can type "R" in the termi
On Mon, Jul 13, 2009 at 12:08 PM, David Winsemius wrote:
> In R a function only returns the last evaluation, so you need to wrap up all
> of the local results into a list at the end of the function.
>
>
>
> David Winsemius, MD
> Heritage Laboratories
> West Hartford, CT
>
How important it is to
Hi,
using a variable named .trPaths is a feature and a frequent nuisance of
recent Tinn-R and also explained in the Tinn-R docs.
Solution 1:
In Tinn-R select the menu:
R --> Configure --> and then "Temporarily" or "Permanent"
Solution 2:
Manually edit file /etc/Rconfig.site
The minimal l
I am a beginner in R and know only a little about it yet. I have a script
written in R language, named as "a.txt" for example. I am using a Linux
machine, at present I only know that I can type "R" in the terminal and then
copy-paste the content in "a.txt" to the R's interface to execute the
progr
In R a function only returns the last evaluation, so you need to wrap
up all of the local results into a list at the end of the function.
On Jul 13, 2009, at 1:23 PM, Chip Maney wrote:
I have a function (see below). This function has one object, ID.
If I run the loops by itself using a
On Mon, 13 Jul 2009, Heinz Tuechler wrote:
Dear All,
since years I am struggling with Surv objects in data.frames. The following
seems to have to do with it.
See below the modified example from the help page of survSplit. The original
works, as expected. If, however, a Surv object is added to
I have a function (see below). This function has one object, ID. If I run
the loops by themselves using a character value (ie,"VFFF1-7") instead of
the function object, then the loops work fine. However, when I try to
insert the character value via the function call, it doesn't work. I don't
ge
On Mon, Jul 13, 2009 at 5:31 PM, serbring wrote:
> excuse me for my english, i am using R on windows and i have to do several
> graphs with axis labels and the axis text thicks has a specified font type,
> (Arial) and a specified font size. How can i do these? Thank you in advance
Interesting que
Hi people,
I have a text file like this one posted:
snp_id genechromosome distance_from_gene_center
positionpop1pop2pop3pop4pop5pop6pop7
rs2129081 RAPT2 3 -129993 "upstream" 0.439009
1.169210NA 0.2330200.093042
I have a function (see below). This function has one object, ID. If I run the
loops by itself using a character value (ie,"VFFF1-7"), then the loops work
fine. However, when I try to insert the character value via the function call,
it doesn't work. I don't get an error, but the TotalCover.d
Dear Lars,
One way would be
anova(model1, model2)
See ?anova for more information.
HTH,
Jorge
On Mon, Jul 13, 2009 at 7:04 AM, Lars Bergemann
wrote:
>
> Hi.
>
>
>
> I would like to use a likelihood-ratio test to compare a linear model
> (lm()) to another linear model containing the first on
On Mon, 13 Jul 2009, Jeff Xu wrote:
Hi Lars,
Using the definition of Likelihood-ratio test is straightforward,
To which I would add, following the instructions in the _posting guide_ is
equally straightforward. viz
help.search("loglikelihood")
takes you to
?stats::logL
Le lundi 13 juillet 2009 à 13:04 +0200, Lars Bergemann a écrit :
> Hi.
>
>
>
> I would like to use a likelihood-ratio test to compare a linear model (lm())
> to another linear model containing the first one to see if the extra factors
> are needed - but I was unable to find any help on how to
Thanks Ron.
I apologize for not properly specifying which class/package I was using. I
am using the timeSeries class from the timeSeries Package, version 2100.83
from rmetrics.org
The code you supplied works for me! -- thanks again.
Warm regards,
Andrew
--
Ron Burns-2 wrote:
>
> Here is w
Hi,
excuse me for my english, i am using R on windows and i have to do several
graphs with axis labels and the axis text thicks has a specified font type,
(Arial) and a specified font size. How can i do these? Thank you in advance
--
View this message in context:
http://www.nabble.com/graph%3
Hi Lars,
Using the definition of Likelihood-ratio test is straightforward,
Jeff
Lars Bergemann wrote:
>
>
> Hi.
>
>
>
> I would like to use a likelihood-ratio test to compare a linear model
> (lm()) to another linear model containing the first one to see if the
> extra factors are needed
Simon,That produced exactly what I was looking for. Thanks so much for the
humble help.
KC
On Mon, Jul 13, 2009 at 9:10 AM, Simon Wood wrote:
> You can get some idea by doing something like the following, which compares
> the r^2 for models b and b2, i.e. with and without s(x2). It keeps the
Dear All,
since years I am struggling with Surv objects in data.frames. The
following seems to have to do with it.
See below the modified example from the help page of survSplit. The
original works, as expected. If, however, a Surv object is added to
the data.frame, each record gets doubled.
I tried to specify a model in dse1 but something isn't right. Anybody
have any tips?
model<-SS(F=f,G=g,H=h,Q=q,z0=z,P0=p)
Error in locateSS(model$R, constants$R, "R", p, p, plist) :
The dimension of something in the SS model structure is bad.
> dim(f)
[1] 5 5
> dim(g)
[1] 5 1
> dim(h)
[1] 1 5
I am using R 2.8.1 and lattice to produce xyplots conditioned on
two factors. What I would like is to have the scales be free between values
of one factor, but some within. Thus, in this example,
xyplot(mpg ~ disp | factor(gear) + factor(cyl), mtcars,
scales=list(x=list(relation="free")))
Dear Rafael,
Try the following which seems to be close to what you asked for:
plot(1, ylab = expression( Delta ~ hat(y) [t] ) )
HTH,
Jorge
On Mon, Jul 13, 2009 at 10:56 AM, Rafael Moral
wrote:
> Dear useRs,
>
> I want to put a math expression in the ylab in my plot which should be a
> Delta
Uwe Ligges schrieb:
I have to admit that I have no idea what we are talking about here
(yes, I tend to forget many things these days) - and you have not
cited the original message, unfortunately (nor have you specifies R
versions, Tinn-R versions and both OS versions, but just one) ...
Be
It appears you are conflating beta coefficients (individual covariate
effect measures) with overall model fit measures. Beta coefficients
are not directly comparable to R-squared measures in ordinary least
squares analyses, so why would they be so in gam models?
I cannot tell whether you ac
You can get some idea by doing something like the following, which compares
the r^2 for models b and b2, i.e. with and without s(x2). It keeps the
smoothing parameters fixed for the comparison. (s(x,fx=TRUE) removes
penalization altogether btw, which is not what was wanted).
dat <- gamSim(1,n
Dear useRs,
I want to put a math expression in the ylab in my plot which should be a Delta
and a 'y' with a trace and a hat above it and a 't' as subscription.
How could I manage to do it?
Thanks in advance,
Regards,
Rafael.
__
On 7/13/2009 10:42 AM, Jason Rupert wrote:
Yes. You are right that is the trick:)
Here is what I'm seeing:
input_path<-c("C:/WINDOWS/system32")
file.info(list.files(input_path))
size isdir mode mtime ctime atime exe
$winnt$.inf NANA
1025 NANA
...
xsi.j
On 7/13/2009 10:21 AM, Mark Knecht wrote:
Thanks Jim.
How does one search the help system for info on simple logic like this?
Look in the manuals (the pdf ones), rather than the man pages. For
example, expressions like the ones below are in the Introduction to R,
section 2.4, Logical Vector
I have what I *think* should be a simple problem in R, and hope
someone might be able to help me.
I'm working with cancer survival data, and would like to calculate
adjusted survival figures based on the age of the patient and the
tumour classification. A friendly statistician told me I shou
MathZero wrote:
>
> Hi, I am trying to use the nls() function to closely approximate a vector
> of values, colC and I'm running into trouble. I am not sure how if I am
> asking the program to do what I think its doing, because the same
> minimization in Excel's Solver does not run into problem
Yes. You are right that is the trick:)
Here is what I'm seeing:
input_path<-c("C:/WINDOWS/system32")
file.info(list.files(input_path))
size isdir mode mtime ctime atime exe
$winnt$.inf NANA
1025 NANA
...
xsi.jar NA NA
xsi.zip NA NA
zipfldr
Here is what the train() function in the caret package does by default
(you can change this behavior; see below).
Using the entire data set, estimate the RBF parameter using the
sigest() function in the kernlab package (which, if I recall correctly
involves the median of a sample of kernel matrix
You can find the operators with:
?"|"
This will show you logical operators. You can do:
?"+"
to get the others.
On Mon, Jul 13, 2009 at 10:21 AM, Mark Knecht wrote:
> Thanks Jim.
>
> How does one search the help system for info on simple logic like this?
>
> On Mon, Jul 13, 2009 at 6:59 AM, j
The dyn and dynlm packages can handle time series in lm and glm.
(dyn can also handle many additional regression functions as well)
In the case of dyn just write dyn$lm instead of lm like this:
> library(dyn)
> a <- ts(c(1, 2, 4))
> dyn$lm(a ~ diff(a))
Call:
lm(formula = dyn(a ~ diff(a)))
Coeffi
Thanks Jim.
How does one search the help system for info on simple logic like this?
On Mon, Jul 13, 2009 at 6:59 AM, jim holtman wrote:
> 1) X <- subset(A, (t < 1000) | (t > 1200))
> 2) X <- subset(A, (t > 1000) & (t < 1200))
>
> On Mon, Jul 13, 2009 at 9:47 AM, Mark Knecht wrote:
>> How woul
Dear all,
I am struggling with the weighted Efron approximation. I really need some
help.
I am confused by the explanation regarding applying weights to Efron
approximation on the paper of “A Package for Survival Analysis in S” (Terry
M. Therneau, Feb 1999). In the section 3.3.7 Weighted Cox mo
Many thanks for the advice David. I would really like to figure out, though,
how to get the contribution of each factor to the Rsq - something like a
Beta coefficient for GAM. Ideas?
KC
On Sun, Jul 12, 2009 at 5:41 PM, David Winsemius wrote:
>
> On Jul 12, 2009, at 5:06 PM, Kayce Anderson wrote
Duh! Thanks and good advice. I was using 2.7.2 because it was, until
recently, the latest version working with RPy (http://rpy.sourceforge.net/).
Also didn't realize plm was still actively developed.
Interesting that since plm now correctly handles diff and lag operations, it
actually breaks with
On Mon, Jul 13, 2009 at 11:18 AM, Pathak,
Saurav wrote:
> Dear Arne
> I have gone through the paper and I have tried it at my end, I would really
> appreciate if you could address the following:
>
> 1. Based upon your suggestion I used the following:
>
> regmod2 <- selection(s ~ age + gender + gem
dear anna,
if you are not interested in point estimate and SE of the parameter of
the aforementioned categorical variable, I believe the conventional
glm(..,family=binomial) is correct. In particular, the returned deviance
is reliable and also it is the relevant likelihood ratio test..
hope t
At 6:48 PM +1200 7/13/09, mehdi ebrahim wrote:
Hi All,
I am using fleiss kappa for inter rater agreement. Are there any know
issues with Fleiss kappa calculation in R? Even when I supply mock data
with total agreement among the raters I do not get a kappa value of 1.
instead I am getting negat
Belated answer:
A few remarks regarding your questions:
Your running max problem could be solved in the following way:
(which is a soution based o Duncan Murdoch's suggestion,
but a little bit more general.
foldOrbit<-function(x,fun){
res<-numeric(length(x))
res[1]<-x[1]
rSymPy and Ryacas can do algebraic manipulation.
See
http://rsympy.googlecode.com
http://ryacas.googlecode.com
> library(rSymPy)
Loading required package: rJava
> sympy("var('a b m r y x')")
[1] "(a, b, m, r, y, x)"
> sympy("solve(a*b*m*y*(1-x)-r*x, x)")
[1] "[-a*b*m*y/(-r - a*b*m*y)]"
> library
1) X <- subset(A, (t < 1000) | (t > 1200))
2) X <- subset(A, (t > 1000) & (t < 1200))
On Mon, Jul 13, 2009 at 9:47 AM, Mark Knecht wrote:
> How would I write the two selections each in a single subset command?
>
> 1) Two non-overlapping time ranges I want to collect together - before
> 10AM an
Hi
I am trying to perform a bootstrap estimate of classification accuracy of a
logistic regression using the 'Daim' package in r using the code at the bottom
of this post, this all works great and I get the .632+ misclassification
accuracy, specificity, sensitivity, AUC etc etc but what I woul
Hi all
Maybe someone knows a way to solve this anomaly in sample():
I like to compute a sample (n=100) with replications from a population of
2500 units but if I draw repeated samples from it I dont get what seems to
be a representative sample if I look at other partitions of the population.
Encl
How would I write the two selections each in a single subset command?
1) Two non-overlapping time ranges I want to collect together - before
10AM and after noon. Should be an OR function:
X = subset(A, t<1000) + subset(A, t>1200)
2) One range between two defined times like after 10AM and before
Hi,
can anyone tell me if R can be used to rearrange very complicated equations
in terms of one of the variables?
I have:
dx/dt = a*b*m*y*(1-x)-r*x
and, having set:
dy/dx = 0,
need to rearrange in terms of x. The problem I have is that I don't know how
to rearrange equations when the variabl
Hello Everyone
I am calculating Fleiss Kappa, I have 28 raters, 5 Subjects and 5 ratings
(i.e. A dataset with 28 columns and 5 rows). The problem is that there are 2
missing values in the data. (Right now I have manually deleted the values in
csv file and then imported it in R
1) Would it better
If you apply the function to a simple dataframe or show your code, you might
be able to get more accurate help. I've used the IRR package in the past
and haven't noticed any problems (maybe I overlooked them ?)
david freedman
mehdi ebrahim wrote:
>
> Hi All,
>
> I am using fleiss kappa for in
Seems to me if splitting once for all the bias will be big and if splitting
once for each choice of parameters the variance ill be big.
In LibSVM, for each choice of (c, gamma), the searching script grid.py calls
svm_cross_validation() which has a random split of the dataset. So seems to me
Why do we get Partial correlation values greater than 1?
I have used the default function pcor.mat :--
I have manipulated the default pcor.mat function a bit so ignore tha
variables corr_type,element1_in_no,element2_in_no,P.Please ignore the
“pairwise” section and have a look at athe “listwise ”
Try this:
# First read in the data using chron "times" class
Lines.Bary <- "XY Date Time Depth
-2.620748 48.75121 01/00/00 12:07:16.000 -25.6
-2.620714 48.75121 01/00/00 12:07:17.000 -25.8
-2.620698 48.75121 01/00/00 12:07:17.000 -26.0
-2.620682 48.
On Jul 13, 2009, at 5:37 AM, anna.bucharova wrote:
Dear all.
I am sort of beginner with R. I do logistic regression with binomial
response variable and several continuous and categorical variables.
In one
categorical variable, zero cell occures (2x2 table looks like
7 - 0
23 - 25
This leads
Dear all,
I want make correction depth of a bathymetric data set.
To do so, I have the depth data set sample every second (a depth at each
second) in one hand, and in the other hand, I have a tide variation
level data set sample every 250 ms. The time register in each data sets
(tide and bathy
Dear all.
I am sort of beginner with R. I do logistic regression with binomial
response variable and several continuous and categorical variables. In one
categorical variable, zero cell occures (2x2 table looks like
7 - 0
23 - 25
This leads to overestimating of odds ratio and inflated confidence
Hi there,
Can anyone please help me because I am going to get crazy with the pbc data
set. I just want to apply simple cox regression in the data set. I am a
beginner in R but I don't think I am doing anything wrong.
I have the book of Fleming and Harrington 1990. I perform cox regression by
Dear all,
I am using package.skeleton to build a small packages of misc function
for personal use. I have recently discovered that the option
force=TRUE doesn't seem to do what is meant to do. Here's what I'm
doing:
> setwd("/Users/danielk/Documents/R/packages/dk")
> files <- paste("codeb
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