Hi:
The input to jarque.bera.test() is a numeric vector or time series. Try
running the function str() on your input object to see if it is of the
correct type. If you have a vector that is not numeric or a time series
object, you need to convert it to one with something like as.numeric(myvec).
Th
Hi Nick,
I've used MCMC to fit change point regressions to a variety of
ecological data and prefer this approach to strucchange and similar
because I feel I have more control over the model, ie. I find it
easier to tailor the form of the model to biological / demographic
processes. I also find the
I figured it out myself.
Again, Michael and Petr, many thanks to both of you!!! :)
--
View this message in context:
http://r.789695.n4.nabble.com/Sampling-problem-tp3043804p3046161.html
Sent from the R help mailing list archive at Nabble.com.
__
R-h
Here is a better approach that will keep the axis ticks as well on the two axes.
## define a lattice "axis function"
axis.L <-
function(side, ..., line.col)
{
if (side %in% c("bottom", "left")) {
col <- trellis.par.get("axis.text")$col
axis.default(side, ..., line.col = col
As long as the names are unique, there is not a problem to shorten them.
El mié, 17-11-2010 a las 01:02 +0100, José Fernando Zea Castro escribió:
> Hello.
>
> First, I'm thankful about your wonderful project.
>
> However, I have serious worries about the reliability of R. I found
> the next bug
Hi:
Try this:
# Function to generate one sample from the data frame
sampler <- function(df) {
s1 <- sample(nrow(df), 1, replace = FALSE)
s2 <- sample(setdiff(1:nrow(df), s1), 2, replace = FALSE)
list(sample1 = df[s1, grep('^C', names(df))],
sample2 = df[s2, grep('^W', nam
On Nov 16, 2010, at 7:02 PM, José Fernando Zea Castro wrote:
Hello.
First, I'm thankful about your wonderful project.
However, I have serious worries about the reliability of R. I found
the next bug which I consider important because in my job everytime We
work with datanames like next. Pleas
2010/11/17 Kiana Basiri
> Hello,
> I'm so confused why I can't run Jarque-Bera test on my data. I have 9968
> observation and I want to run Jarque-Bera test on them, but no matter how
> hard I am trying I can't get it work. please let me know what should I do.
>
> Best,
> Kiana
>
>
Did you check
> Date: Tue, 16 Nov 2010 17:39:57 -0800
> From: peter.langfel...@gmail.com
> To: jbass...@cs.gmu.edu
> CC: r-help@r-project.org
> Subject: Re: [R] Non-positive definite cross-covariance matrices
>
> > Peter,
> >
> > I see your point. As it turns out
Hello.
First, I'm thankful about your wonderful project.
However, I have serious worries about the reliability of R. I found
the next bug which I consider important because in my job everytime We
work with datanames like next. Please see below:
b=data.frame(matrix(1:9,ncol=3))
names(b)=c("q99
Okay here is a solution that works in less than 60 minutes but i feel likes
its messy, if anyone has an alternative solution i would very much
appreciate your insights.
#Create test data
TNode<-c(1:20,21)
FNode<-c(rev(1:20),22)
Volume<-c(rep(100,20),200)
ClassCode=c(rep("Local",20),rep("Freeway
Hello,
I'm so confused why I can't run Jarque-Bera test on my data. I have 9968
observation and I want to run Jarque-Bera test on them, but no matter how
hard I am trying I can't get it work. please let me know what should I do.
Best,
Kiana
[[alternative HTML version deleted]]
__
Ben,
Thank you for your assistance.
Going back to basics and using the data set as you suggested has resulted in a
win.
Set A works!
using +offset(log(variable)) or ,offest=(log(Eff)) is the same as using
exposure(variable) program stata.
I went back and isloated a problem with code be
Hi,
I hope this is a simple question. I am having trouble changing the scale of a
secondary y-axis on a barplot. When I run the code below the limits set for the
first axis are always applied to the second axis as well. I am using the latest
R version 2.12.0.
For example, if I have 3 vectors
R-helpers,
I have had difficulty installing the "pcvsuite" package on R version 2.12.0
(2010-10-15). The pcvsuite package is not available on CRAN, but is located
for download at the following website at the University of Washington:
Windows version
http://labs.fhcrc.org/pepe/dabs/pcvsuite_1.
Fitting curves to an ECDF will result in a fit that has the same precision as
the ECDF if variances are calculated correctly. So why not stop with the
ECDF as your estimator?
Frank
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context:
http://r
I think the problem is with the (2,2) element in your hessian, unless
that is a typo.
Cheers
David Cross
d.cr...@tcu.edu
www.davidcross.us
On Nov 16, 2010, at 8:33 PM, Jimmy Martina wrote:
Hi, R-folks:
I have been tryin many combination of parameter to make Matern
variogram to work,
Hello there,
My name is Alireza. I am interested in utilizing R-program for measurement of
earnings management (accounting related issue). However, there are two source
programs which should be used simultaneously with the R-program to be able to
measure earnings management. Previously (i.e. in
Hi, R-folks:
I have been tryin many combination of parameter to make Matern variogram to
work, but I can't find the available one. I'm near to be crazy.
I tiped:
Año2003Selg.lf<-likfit(Año2003Selg,cov.model="matern",ini.cov.pars=c(1.5,14),kappa=2.5,fix.kappa=FALSE,nugget=0.08,lambda=0.008,fix.l
> Peter,
>
> I see your point. As it turns out though, what I'm trying to
> calculate is heritability using a slightly modified version of an
> equation from multivariate quantitative genetics. Theoretically I
> suppose a heritability matrix could be non-positive definite, but in
> practice it al
On Tue, Nov 16, 2010 at 1:49 PM, Peter Langfelder
wrote:
>
> It is easy to come up with examples where Cov(A, B) + Cov(B, A) is not
> positive definite. As an extreme example, consider a matrix A (say 10
> columns, 100 rows) such that the off-diagonal covariances are all zero
> and the columns are
Hi Diana,
Yes, this seems to be a little bug in the setparts function. The
following is a modified version which should work for any x > 0.
You'll see I've just changed a couple of lines...
setparts2 <- function (x)
{
if (length(x) == 1) {
if (x < 1)
stop("if single value, x
G'day John,
On Tue, 16 Nov 2010 14:02:57 -0500
"Prof. John C Nash" wrote:
> Are the xxxPR routines now deprecated (particularly for 64 bit
> systems) or still OK to use?
They are still OK to use, and I use them occasionally.
> If OK, can anyone point to documentation and examples?
Section
Another approach would be
> Y <- list(sqrt, sin, function(u) u/2)
> Ybar <- function(u) rowMeans(sapply(Y, function(fun) fun(u)))
>
> integrate(Ybar, 0, 1)
0.4587882 with absolute error < 5.6e-05
>
i.e. make the function vectorized directly.
Note, however, that if you had
Y[[4]] <- function(
Thank you very much! Works like a charm!
On Tue, Nov 16, 2010 at 10:24 PM, Phil Spector wrote:
> Eduardo -
> Thanks for the reproducible example!
>
>> Y<-list()
>> Y[[1]]<-function(u) sqrt(u)
>> Y[[2]]<-function(u) sin(u)
>> Y[[3]]<-function(u) 1/2*u
>> Ybar = function(u)mean(sapply(Y,function(
Eduardo -
Thanks for the reproducible example!
Y<-list()
Y[[1]]<-function(u) sqrt(u)
Y[[2]]<-function(u) sin(u)
Y[[3]]<-function(u) 1/2*u
Ybar = function(u)mean(sapply(Y,function(fun)fun(u)))
Since integrate requires a function which accepts a vector
and returns a vector, we'd need to use Ve
HI, Dear R community,
I have used the following codes this morning, but this afternoon, I got the
following errors:
> x <- seq(0,10, by=1)
> y <- c(0.952, 0.947, 0.943, 0.941, 0.933, 0.932, 0.939, 0.932, 0.924,
0.918, 0.920) # missense
> z <- c(0.068, 0.082, 0.080, 0.099, 0.108, 0.107, 0.101, 0.1
Thanks, guys... but it seems these suggestions won't work.
Let me try to be more specific with a simple example:
Y<-list()
Y[[1]]<-function(u) sqrt(u)
Y[[2]]<-function(u) sin(u)
Y[[3]]<-function(u) 1/2*u
I wanted something equivalent to
Ybar<-function(u){
1/3*(Y[[1]](u) + Y[[2]](u) + Y[[3]](
Hi all,
#
test=data.frame(x=1:26,y=-23.5+0.45*(1:26)+rnorm(26))
rownames(test)=LETTERS[1:26]
attach(test)
#test
test.lm=lm(y~x)
plot(test.lm,2)
identify(test.lm$res,,row.names(test))
# not working
plot(x,y)
identify(x,y,row.names(test))
# works fine
ident
Elijah DePalma ucr.edu> writes:
>
> Greetings,
>
> May you please suggest a package or function to use for fitting a GLMM
> (generalized linear mixed model) with spatially correlated random effects?
>
> Thank you,
> Elijah DePalma
Not easy, and I hope you have a lot of data.
Your choices
On Nov 16, 2010, at 6:18 PM, Phil Spector wrote:
Eduardo -
I'd guess that
Ybar = function(u)mean(sapply(Y,function(fun)fun(u)))
I had an example to which this was offered and all it did was make
Ybar a function;
# Polymorphic Fn-object Version 1
Fs <- list(mode="language")
> Fs
$mode
[
The documentation for agrep says it uses the Levenshtein edit distance,
but it seems to get this wrong in certain cases when there is a
combination of deletions and substitutions. For example:
> agrep("abcd", "abcxyz", max.distance=1)
[1] 1
That should've been a no-match. The edit distance bet
Thank you both.
casper
--
View this message in context:
http://r.789695.n4.nabble.com/plot-linear-model-problem-tp3045763p3045932.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/
Thank you so much
--
View this message in context:
http://r.789695.n4.nabble.com/Counting-tp3045756p3045917.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
Thank you so much
--
View this message in context:
http://r.789695.n4.nabble.com/Counting-tp3045756p3045918.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
It's implemented in the metafor package.
Using the example from the pdf that Marc pointed out:
library(metafor)
ai <- c(53, 121, 95, 103, 64, 7, 0)
bi <- c(2, 3, 14, 27, 51, 29, 13)
ci <- c(61, 152, 114, 66, 81, 28, 0)
di <- c(1, 5, 7, 12
Eduardo -
I'd guess that
Ybar = function(u)mean(sapply(Y,function(fun)fun(u)))
will do what you want, but without a reproducible example,
it's hard to tell.
- Phil Spector
Statistical Computing Facility
Fencl, Martin eawag.ch> writes:
>
> Helllo,
> I am having trouble with running the library Playwith in the R-2.12.0. running
under 32bit Windows XP.
> After calling the library the error message "The procedure entry point
gdk_cairo_reset_clip could not
> be located in the dynamic library libgdk-
Try this:
u <- 1:10
mean(sapply(Y, function(f)match.fun(f)(u)))
On Tue, Nov 16, 2010 at 9:00 PM, Eduardo de Oliveira Horta <
eduardo.oliveiraho...@gmail.com> wrote:
> Hello there
>
> I have a list, Y, and each component of that list is a real-valued function
> (that is, Y[[i]](u) returns a numbe
Hello there
I have a list, Y, and each component of that list is a real-valued function
(that is, Y[[i]](u) returns a number).
I was wishing to build the mean function and the first thing I thought of
was
Ybar<-function(u){
mean(Y[[1:n]](u))
}
but obviously this doesn't work, since Y[[1:n]]
Your outlier has row.names "1". If this is selected in the bootstrap sample
once, it will also have row.names "1". If it is selected more than once the
row.names of the successive entries will begin with "1."
Here is a possibility you may wish to consider.
> txt <- textConnection("
+
Thanks very much for your help. This was a case of some weird network card IP
reset hardware error that did something to permissions or something during the
install. Not exactly sure if this was part of the problem. But getting the
hardware corrected and reinstalling R 2.12.0 solved the prob
First, I would rename your function "samples" to "mksample"
to avoid confusion with the R function "sample".
Next, I would modify the function so that you are returning
a list of samples, instead of a list containing a list of
samples:
mksamples = function(data,num)
lapply(1:num,f
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 11/16/2010 03:08 PM, Columbine Caroline Waring wrote:
> Officially I tried:
**A**
>> glm(count~md+ms+rf+sg+offset(log(Eff)), family=poisson,data=DepthHabGen)
>> glm(count~md+ms+rf+sg, offset=(log(Eff)), family=poisson,data=DepthHabGen)
> (which of
thank you very much for your idea,
if i write code as;
my data name is data.
samples<-function(data,num){
resamples<-lapply(1:num,function(i) sample(data,n,replace=TRUE))
list(resamples=resamples)}
>n=10
data<-rnorm(n=10,mean=5,sd=2)
data[1]=100
obj<-samples(data,1000)
i generate 1000 sample, i
I am trying to understand my population abundance data and am looking into
analyses of change point to try and determine, at approximately what point
do populations begin to change (either decline or increasing).
Can anyone offer suggestions on ways to go about this?
I have looked into bcp and st
I have a growth curve, which is essentially an ECDF: Statistically,
it's F(x)...
> GrowthCurve
[1] 0.06919932 0.24154761 0.42206402 0.61412408 0.72228295 0.79727292
0.86605315 0.91271120 0.98258397 1.
I'd like to fit a Weibull Curve (then a LogLogistic) to this ECDF, and
have no clue ho
Dennis and all,
Thank you for the help as I try to get this method for importing and batch
processing files organized. I currently have this set-up to import data from
two files in my working directory. "Var1" specifies data from file 1 and
file 2.
filenames=list.files()
library(plyr)
import
Hi All -
Doug, thanks for your reply. The context I'm looking at is a Poisson
GLMM with random (intercept,slope) for each subject. The
variance-covariance matrix is 2x2. By unstructured, I meant a 3
parameter matrix (sig1^2,sig2^2,sig12), as compared to a (reduced)
alternative diagonal structur
Hi,
I did this exact thing for my masters, with intertidal fish, I just used a PCA?
have you tried that?
Sent from my iPhone
On 16 Nov 2010, at 17:01, Mike Gibson wrote:
>
> My objective is to look at differences in two species of fish from
> morphometric measurements. My morphometric me
There may be an easier way to do this, but you could always just do it
the long way.
Ex.
plot(residuals(test.lm)~fitted.values(test.lm))
Andrew Miles
On Nov 16, 2010, at 5:01 PM, casperyc wrote:
Hi all,
Say I fit a linear model, and saved it as 'test.lm'
Then if I use plot(test.lm)
i
On Nov 16, 2010, at 5:01 PM, casperyc wrote:
Hi all,
Say I fit a linear model, and saved it as 'test.lm'
Then if I use plot(test.lm)
it gives me 4 graphs
How do I ask for a 'subset' of it??
?plot.lm # The answer is in the first sentence.
say just want the 1st graph,
the residual vs
You could try something like this:
Loop through your bootstrapped samples and store which ones have the
outlier you are looking for using code like:
count = c(count, outlier.value %in% boot.sample$outlier.variable)
Then subtract the count variable from the total number of samples to
get th
Hi all,
Say I fit a linear model, and saved it as 'test.lm'
Then if I use plot(test.lm)
it gives me 4 graphs
How do I ask for a 'subset' of it??
say just want the 1st graph,
the residual vs fitted values,
or the 1,3,4th graph?
I think I can use plot(test.lm[c(1,3,4)]) before,
but now, it's
Hi dear all,
i have a data (data.frame) which contain y and x coloumn(i.e.
y x
1 0.58545723 0.15113102
2 0.02769361 -0.02172165
3 1.00927527 -1.80072610
4 0.56504053 -1.12236685
5 0.58332337 -1.24263981
6 -1.70257274 0.46238255
7 -0.88501561 0.89484429
8
I found the problem.
For some reason, when I converted the list object with the data in it to
numeric, the values changed. This resulted in different clustering
results. Once that was fixed, the clustering was the same.
Thanks for the responses!
On Mon, Nov 15, 2010 at 2:37 PM, Peter Langfeld
Hi Lei,
Here is one option relying on is.finite()
## Messy data for means
dat <- data.frame(values = c(rnorm(7), 1:7, c(1, 2, 3, NA, 4, 5, 6),
c(1, 2, Inf, 4, 100, -Inf, NaN)), group = rep(letters[1:4], 7))
## use is.finite() to select for only finite numbers
tapply(dat$values, dat$group, fun
Dear r-help,
I want to use tapply to calculate means for a variable. But there were several
infinite values in the observations.
How can I calculate means not considering these infinite values?
Thanks in advance.
Regards,
Lei
[[alternative HTML version deleted]]
__
I am cc:ing the r-sig-mixed-mod...@r-project.org mailing list on this
reply as such questions are often answered more quickly on that list.
On Tue, Nov 16, 2010 at 2:00 PM, Daniel Jeske wrote:
> Dear R Help,
> I believe the glmer() function in lme4 automatically fits an
> unstrucruted covariance
Can you try it with version 7.16 on R-Forge? Use
install.packages("odfWeave", repos="http://R-Forge.R-project.org";)
to get it.
Thanks,
Max
On Tue, Nov 16, 2010 at 8:26 AM, Søren Højsgaard
wrote:
> Dear Mike,
>
> Good point - thanks. The lines that caused the error mentioned above are
> sim
Thanks
On 11/16/2010 12:41 PM, Seth Falcon wrote:
> Hi Abhijit,
>
> [I've cc'd R-help to keep the discussion on the list]
>
> On Tue, Nov 16, 2010 at 8:06 AM, Abhijit Dasgupta
> wrote:
>
>> Seth,
>>
>> I was looking for something like this too. I've a question. If
>> you're reading the data f
Have you tried filter()?
filter(a, rep(1,7)/7)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ray Brownrigg
> Sent: Tuesday, November 16, 2010 12:05 PM
> To: r-help
Definitely out of sequence - it should be
[,1] [,2]
[1,]4 21
[2,]5 22
[3,]6 23
[4,]7 24
[5,]8 25
[6,]9 26
[7,] 10 27
[8,] 11 28
[9,] 12 29
[10,] 13 30
[11,] 14 31
On 16 November 2010 20:12, David Winsemius wrote:
>
> On No
Dear R Help,
I believe the glmer() function in lme4 automatically fits an
unstrucruted covariance matirx for the random effects.
Is that true?If so, do I have an option to somehow ask for a
diagonal structured covariance matrix?
Thank you,
Daniel Jeske
Department of Statistics
University of
On Nov 16, 2010, at 2:33 PM, Paolo Rossi wrote:
Hi,
Can anyone suggest a clever way to compute a rolling weekly average
of the
columns in a matrix? The column bit is straightforward use apply
given a
function which does what you want on a column. With regard to a
particular
column, th
On Nov 16, 2010, at 2:32 PM, Jannis wrote:
Dear List,
this may be a Newbi question and may have been asked several times,
but i am too stupid to find the posts.
I have a plot of values against POSIXct time steps. If I want to add
a second x axis to the top margin of the plot, only numb
Ben,
Thank you, it WAS a typo of sorts.
Officially I tried:
> glm(count~md+ms+rf+sg+offset(log(Eff)),family=poisson,data=DepthHabGen)
> glm(count~md+ms+rf+sg, offset=(log(Eff)),family=poisson,data=DepthHabGen)
(which of course are the same as eachother)
> glm(count~md+ms+rf+sg, offset=(
On Wed, 17 Nov 2010, Paolo Rossi wrote:
> Hi,
> Can anyone suggest a clever way to compute a rolling weekly average of the
> columns in a matrix? The column bit is straightforward use apply given a
> function which does what you want on a column. With regard to a particular
> column, the obviou
A coefficient of -0.4 means that survival times are multiplied by
exp(-0.4), that is, people survival only 67% as long.
-thomas
On Wed, Nov 17, 2010 at 4:32 AM, Vincent Vinh-Hung wrote:
> Thanks for sharing the questions and responses!
>
> Is it possible to appreciate how much the coefficie
Greetings,
May you please suggest a package or function to use for fitting a GLMM
(generalized linear mixed model) with spatially correlated random effects?
Thank you,
Elijah DePalma
[[alternative HTML version deleted]]
__
R-help@r-project.org
I am using \Sexpr to include a variable in a title of a Sweave document:
\documentclass[a4paper]{article}
<>=
#mytitlevar <- "Stuff" # case 1, everything is find
mytitlevar <- "Stuff_first" # case 2, f is turned into sub-text
@
\title{MyTitle: \\ \Sexpr{mytitlevar} }
\begin{document}
\maketitle
\e
Hi Josh,
I thought of giving up and started writing code in excel using
VBA but then I saw ur message and gave a try in R. I got it, the location
from which I am calling the package "plyr" is not working out so called from
different CRANmirror location and then I was able load the pac
Tena koe Nasrin
Try points() instead of plot() in your second and subsequent calls to plot().
points() and lines() adds to the current plot by default. Of course you may
have difficulties with setting the x and y limits by that's another matter.
HTH
Peter Alspach
> -Original Messag
assuming your data takes the form of
locationlatitudelongitude
string num num
string2 num num
try:
sub <- dat[sample.int(length(dat$location), 1000),]
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kea
Hi Josh,
Thanks for your reply.
While I am loading the package, it says package plyr is
required and I tried to install plyr package but I am unable to do so
because it is giving an error message as shown below.
Please help me.
library(reshape)
Loading required pa
Try this:
DF[sample(seq(nrow(DF)), 1000),]
Where DF is your data
On Tue, Nov 16, 2010 at 3:13 PM, Mariana wrote:
>
> Hi there:
> I am a total beginner in R, and I have a simple question:
> I have a table with thousands of lines that represent locations, and two
> columns: latitude and longitud
Hi,
Can anyone suggest a clever way to compute a rolling weekly average of the
columns in a matrix? The column bit is straightforward use apply given a
function which does what you want on a column. With regard to a particular
column, the obvious way is to run a for loop indexing the last 7 day
On second glance, while it works for 'stack', it doesn't appear to work for
'overlay':
> overlay(sapply(obs,get))
Error in function (classes, fdef, mtable) :
unable to find an inherited method for function "overlay", for signature
"list", "missing"
At this point, this may be more of a 'rast
Using David's examples:
do.call(stack, lapply(obs, get))
On Tue, Nov 16, 2010 at 4:04 PM, Kevin Ummel wrote:
> Sorry, I shouldn't have used 'sum' as an example; I am looking for a
> solution in the case of functions that do not result in simple vectors or
> matrices.
>
> The "real-world" examp
Dear List,
this may be a Newbi question and may have been asked several times, but i am
too stupid to find the posts.
I have a plot of values against POSIXct time steps. If I want to add a second x
axis to the top margin of the plot, only numbers are at the tickmarks. Is there
a straightforw
Thanks, David. That does, indeed, work. It didn't occur to me that a list would
do the job as an argument.
Thanks for the fix!
kevin
On Nov 16, 2010, at 6:58 PM, David Winsemius wrote:
>
> On Nov 16, 2010, at 1:04 PM, Kevin Ummel wrote:
>
>> Sorry, I shouldn't have used 'sum' as an example;
On Nov 16, 2010, at 12:13 PM, Mariana wrote:
Hi there:
I am a total beginner in R,
Time to read the Posting Guide. (Especially since Nabble appears to
deficient in providing an introductory document, at least judging from
the email behavior of some of its users.)
and I have a simple q
Helllo,
I am having trouble with running the library Playwith in the R-2.12.0. running
under 32bit Windows XP. After calling the library the error message "The
procedure entry point gdk_cairo_reset_clip could not be located in the dynamic
library libgdk-win32-2.0-0.dll." occurs and the R asks fo
Fabulicious! It worked!!!
One more question, in the following data frame as posted above:
SubIDCSE1 CSE2 CSE3 CSE4 WSE1 WSE2 WSE3 WSE4
1 6 5 6 2 6 22 4
2 6 4 7 2 6 6
My objective is to look at differences in two species of fish from morphometric
measurements. My morphometric measurements are head length, eye diameter,
snout length, and measurements from tail to each fin. I want to use
discrimanant function analyis to determine if there are differences betw
Hi there:
I am a total beginner in R, and I have a simple question:
I have a table with thousands of lines that represent locations, and two
columns: latitude and longitude. I need to randomly sample 1000 lines. How
do I do it? I know the command "sample", but it samples elements
independently, no
Sorry, I shouldn't have used 'sum' as an example; I am looking for a solution
in the case of functions that do not result in simple vectors or matrices.
The "real-world" example is something like the following using the 'raster'
package, though (I think) any command producing an S4 object could
My problem is that I have a data set for every day of measurement in a
seperate file and I want to plot one parameter of the data for all the days
in one graph. I tried to use for loop but only the last data remains in the
program memory, I don`t know how to plot each day`s data continusly after
th
Tianchan -
Your X is not a matrix -- it's a dataframe. Probably the
simplest solution is to use
lm(y~as.matrix(X))
but you should also learn the difference between a data frame
and a matrix.
- Phil Spector
Try this:
y <- rnorm(100)
X <- matrix(runif(100 * 10), ncol = 10)
lm(y ~ ., data = cbind.data.frame(y, X))
On Tue, Nov 16, 2010 at 5:07 PM, Tianchan Niu wrote:
> Dear All,I would like to do multiple regression in R. I used: lm(y~X),
> where y is a n by 1 vector, and X is a n by m matrix. I kep
Ravi Varadhan and I have been looking at UCMINF to try to identify why it gives
occasional
(but not reproducible) errors, seemingly on Windows only. There is some
suspicion that its
use of DBLEPR for finessing the Fortran WRITE() statements may be to blame.
While I can
find DBLEPR in Venables an
Dear All,I would like to do multiple regression in R. I used: lm(y~X),
where y is a n by 1 vector, and X is a n by m matrix. I kept getting the error
message:Error in model.frame.default(formula = y ~ X, : invalid type (list)
for variable 'X'. However, when I used: lm(y~X[,1]+X[,2]+X[,3]+…+X[,m]
On Nov 16, 2010, at 1:04 PM, Kevin Ummel wrote:
Sorry, I shouldn't have used 'sum' as an example; I am looking for a
solution in the case of functions that do not result in simple
vectors or matrices.
The "real-world" example is something like the following using the
'raster' package, th
Try this:
aggregate(TEMP ~ Year + Month + Day, DF, mean)
On Tue, Nov 16, 2010 at 4:48 PM, facehappywy wrote:
>
> Thanks every one first.
>
> I have a dataset like this
>
> YearMonth Day HourDIR SPD (m/s) SKYCOVER
> TEMP (C)RH(%)
> 20091 1 0
On Tue, Nov 16, 2010 at 9:40 AM, Jeff Bassett wrote:
> Giovanni,
>
> Both matrices describing the points (A and B in my example) are the
> same size, so the resulting matrix will always be square. Also, the
> equation I'm using is essentially the following identity:
>
> Var(A + B) = Var(A) + Var(
Thanks every one first.
I have a dataset like this
YearMonth Day HourDIR SPD (m/s) SKYCOVERTEMP
(C)RH(%)
20091 1 0 310 13.858243 -5 23
20091 1 1 330 14.305283 -7.
Hi,
Did you also load the package (you'll need to load it every session
you want to use it)?
## load reshape package
library(reshape)
## now try
merge_all(Br, Ki, Lu, Pr, by="Genes")
If this does not resolve your problem, please run: sessionInfo()
at your console and report the output (it wil
Just increase the margins on the left side and add the rownames
x <- cor(matrix(rnorm(600), 60, 100))
rownames(x)<-paste("row", 1:100)
op<-par(mar=c(1,5,1,1), xpd=TRUE)
image(t(x[nrow(x):1,]), axes=FALSE)
text(-0.01, seq(0,1,length=nrow(x) ), rownames(x), pos = 2, offset = 0,
cex = .7)
Another
On Tue, Nov 16, 2010 at 10:58 AM, Aleksey Naumov wrote:
> Hi R experts,
>
> I am looking for a simple error handling approach, whereby I could stop
> function execution with a customized error message. For example:
>
> for (i in 1:10) {
> if (i == 5)
> # I'd like to be able to stop right he
It is often more effective to send questions about lmer or glmer to
the r-sig-mixed-mod...@r-project.org mailing list, which I am cc:ing
on this response.
On Tue, Nov 16, 2010 at 3:25 AM, Annika wrote:
>
> Dear list,
>
> I am new to this list and I am new to the world of R. Additionally I am not
1 - 100 of 164 matches
Mail list logo
