Dear R users
I'm looking for algorithms that assist in spreading out crowded labels, e.g.
labels of points in a scatter plot, in order to obtain a nicer visual
appearance and better legibility.
I'm probably just stuck because I didn't find the right key words for a
successful search on the R
Hi all,
I installed the package inline (windows-version) but can not compile any
code, I alway get an error message
"ERROR(s) during compilation : source code errors or compiler
configuration errors!"
Unfornutanely there is no description where the package finds a
c-compiler nor where so set
I am struggling with a simple repeated-measure model:
fit<-lme(trait~year * A, random = ~1|subj/year)
A being a factor with three levels. I got have the following results
for anova(fit):
numDF denDF F-value p-value
(Intercept) 1 126 2471.4720 <.0001
year 2060
Hi R users:
Is there a function to obtain the correlation within groups
from this very simple lme model?
> modeloMx1
Linear mixed-effects model fit by REML
Data: barrag
Log-restricted-likelihood: -70.92739
Fixed: fza_tension ~ 1
(Intercept)
90.86667
Random effects:
Formula: ~1 | mo
Dear R users,
I have been using a dynamic data extraction from raw files strategy at
the moment, but it takes a long long time.
In order to save time, I am planning to generate a data set of size
1500 x 2 with each data point a 9-digit decimal number, in order
to save my time.
I know R is limi
If DF is your data frame:
DF2 <- edit(DF)
and then make the changes manually in the spreadsheet that
pops up.
On Tue, May 19, 2009 at 3:50 PM, tsunhin wong wrote:
> Dear R users,
>
> I have 1 data.frame of 1500x80 - data1. I found out that there are a
> few cells of data that I have misplace, a
Exactly what are you trying to do? Are you trying to just change a subset
of the values? 'subset' does not have an 'assignment' operator. Maybe you
want something like this (but it is not clear from your description. Also
it is not clear if you have exactly the same set of matching values in th
?file.copy
On Tue, May 19, 2009 at 9:51 PM, XinMeng wrote:
> There's 10 files in c:\\
> I wanna copy 3 of them to d:\\
>
> How to do it via R?
>
>
> Thanks!
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLE
There's 10 files in c:\\
I wanna copy 3 of them to d:\\
How to do it via R?
Thanks!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and p
Thank you for the advice. For the density08 variable, that is
population density in year 2008. I also have population densities for
year 2000, so I could put them both in, and interpolate between them
for the times that are covered by the panel (2000-2008), and then just
have a "density" column tha
Dear useRs,
A new version of actuar is available since last Friday. This is mainly
a bugfix release. From the NEWS file:
Version 1.0-2
=
USER-VISIBLE CHANGES
o m() and lev() now return Inf instead of NaN for infinite
moments. (Thanks to David Humke for the idea.)
BUG FIXES
o N
Try this (note dot after ~):
lm(response[, 3] ~., as.data.frame(spectra.spec[, 2:20]))
On Tue, May 19, 2009 at 6:21 PM, MikSmith wrote:
>
> Hi
>
> I'm fairly new to R and am trying to analyse some large spectral datasets
> using stepwise regression (fairly standard in this area). I have a field
On Tue, 19 May 2009, Stephen J. Barr wrote:
Ah, thank you for the help, and for the explanation of what is going
on. I suppose I will have to reload my data with plm.data set such
that RATE is not a factor.
plmWithDensity$RATE <- as.numeric(as.character(plmWithDensity$RATE))
should suffice.
On Mon, May 18, 2009 at 11:47 AM, Dimitri Liakhovitski wrote:
> Hello!
> I have a question about my lattice barchart that I am trying to build
> in Section 3 below. I can't figure out a couple of things:
> 1. When I look at the dataframe "test" that I am trying to plot, it
> looks right to me (the
Ah, thank you for the help, and for the explanation of what is going
on. I suppose I will have to reload my data with plm.data set such
that RATE is not a factor. For my time index, will
2000,2000.25,2000.5, etc. work? Meaning 2000 quarter 1, 2000 quarter
2, etc? Or is there some special way that I
See ?toString
x <- 0:10
toString(x)
See ?sQuote for cases where the vector is a character and needs to be quoted.
Jason Law
Statistician
City of Portland
Bureau of Environmental Services
Water Pollution Control Laboratory
6543 N Burlington Avenue
Portland, OR 97203-5452
jason@bes.ci.portland
On Tue, 19 May 2009, Stephen J. Barr wrote:
Hello,
I am working on a data set (already as a plm.data object) located
here: http://econsteve.com/arch/plmWithDensity.Robj
With the following R session:
> library(plm)
...
>load("plmWithDensity.Robj")
>model <- plm(RATE ~ density08, data=plmWithD
Hello,
I am working on a data set (already as a plm.data object) located
here: http://econsteve.com/arch/plmWithDensity.Robj
With the following R session:
> library(plm)
...
>load("plmWithDensity.Robj")
>model <- plm(RATE ~ density08, data=plmWithDensity)
Error: subscript out of bounds
I am not
Hi
I'm fairly new to R and am trying to analyse some large spectral datasets
using stepwise regression (fairly standard in this area). I have a field
sampled dataset, of which a proportion has been held back for validation. I
gather than step() needs to be fed a regression model and lm() can prod
My favorite tool for finding things like this is
"RSiteSearch.function" in the "RSiteSearch" package. For the objects
you mention, I get the following:
library(RSiteSearch)
hits(a.s <- RSiteSearch.function("auto.stats")) # 0
hits(sx <- RSiteSearch.function("saving.x")) # 0
hits(rn <- R
There are 17 different help pages in 5 different packages citing
"Agresti and Coull". This is quickly displayed using the "RSiteSearch"
package as follows:
library(RSiteSearch)
HTML(RSiteSearch.function("Agresti and Coull"))
I have not checked all these 17, but they doubtless hel
On Tue, May 19, 2009 at 2:01 PM, Michael Hannon wrote:
>
> Greetings. I'm trying to learn to program in R. (I'm definitely NOT new to
> programming, just to R.) A colleague suggested that I have a look at the
> book:
>
> An Introduction to S and S-Plus
> by:
> Phil Spector
>
> I've glance
On 19-May-09 19:52:20, deanj2k wrote:
> dlogl <-
> -(n/theta)-sum((y/(theta)^2)*((1-exp(y/theta))/(1+exp(y/theta)))
>
> d2logl <- (n/theta^2) -
> sum((-2y/theta^3)*(1-exp(y/theta))/(1+exp(y/theta)))
> - sum(((2*y/theta^4)*exp(y/theta))/((1+exp(y/theta))^2))
>
> returns the error message:
> Error:
You're missing a ")" off end of the first line. You should consider
using an editor (e.g. ESS/Emacs) that does parentheses matching. I
found this in less than 5 sec (less time than I'm taking to write you
a note) by cut and pasting in Emacs.
--sundar
On Tue, May 19, 2009 at 12:52 PM, deanj2k wro
On Mon, May 18, 2009 at 9:22 AM, Thomas Lumley wrote:
> On Mon, 18 May 2009, Debbie Zhang wrote:
>> Based on a set of binomial sample data, how would you utilize the "nlm"
>> function in R to estimate the true proportion of the population?
> I can't see why anyone would want to use nlm() for thi
dlogl <- -(n/theta)-sum((y/(theta)^2)*((1-exp(y/theta))/(1+exp(y/theta)))
d2logl <- (n/theta^2) - sum((-2y/theta^3)*(1-exp(y/theta))/(1+exp(y/theta)))
- sum(((2*y/theta^4)*exp(y/theta))/((1+exp(y/theta))^2))
returns the error message:
Error: unexpected symbol in:
"dlogl <- -(n/theta)-sum((y/(the
while (theta1!=theta) {...}
gives the error message:
Error in while (theta1 != theta) { :
missing value where TRUE/FALSE needed
but when i extract theta1!=theta and paste it into the console it comes up
with the output TRUE which contradicts the error message- im not sure what I
am doing wron
Dear R users,
I have 1 data.frame of 1500x80 - data1. I found out that there are a
few cells of data that I have misplace, and I need to fix the ordering
of them.
In an attempt trying to swap column 22 & 23 of the Subject with
misplaced data, I did the following:
> data2 <- data1
> subset(data1,(S
Stavros Macrakis wrote:
>
> You might think that you can check names(xxx) to see if the slot has been
> explicitly set, but it depends on *how* you have explicitly set the slot to
> NULL:
>
>> xxx$hello <- 3
>> xxx$hello <- NULL
>> names(xxx)
>character(0) # no name
On Tue, May 19, 2009 at 12:07 PM, routík wrote:
> > SmoothData <- list(exists=TRUE, span=0.001)
> > exists("SmoothData$span")
> FALSE
>
As others have said, this just checks for the existence of a variable with
the (strange) name "SmoothData$span".
In some sense, in R semantics, xxx$yyy *alway
Maybe you should just bypass that book for one of these?
http://www.springer.com/series/6991
-Ro
On Tue, May 19, 2009 at 12:01 PM, Michael Hannon wrote:
>
> Greetings. I'm trying to learn to program in R. (I'm definitely NOT new to
> programming, just to R.) A colleague suggested that I have
Greetings. I'm trying to learn to program in R. (I'm definitely NOT new to
programming, just to R.) A colleague suggested that I have a look at the
book:
An Introduction to S and S-Plus
by:
Phil Spector
I've glanced at the book, and it does indeed seem to be the kind of thing I
wanted
Hi,
R>=2.9.0 ships grepl, which lets you do that:
any( grepl( TEXT, OBJECT) )
You can also:
install.packages( "operators" )
require( operators )
OBJECT %~+% TEXT
Romain
Dennis Fisher wrote:
Colleagues
R2.8.1 in OSX
I often combine two commands as follows:
length(grep(TEXT, OBJECT)) >
Colleagues
R2.8.1 in OSX
I often combine two commands as follows:
length(grep(TEXT, OBJECT)) > 0
to see if a particular snippet of text exists within an object.
Is there a single command that would accomplish this?
Dennis
Dennis Fisher MD
P < (The "P Less Than" Company)
Phone: 1-866-P
pompon wrote:
Hi,
Thank you very much for the answer.
However, I have still some misunderstandings.
from the output, can we say that plant and leaf age are significant but not
their interaction?
And the last question I promise, what would you advise me to write in the
paper to explain the diffe
see ?paste
e.g.
x <- seq(0,10,1)
paste(x, collapse=", ")
2009/5/19 Katharina May :
> Hi,
>
> how do I create a string of the comma-separated content of a vector?
>
> I've got the vector i with several numeric values as content:
>>str(i)
> num 99
>
> and want to create a SQL statement to look li
# Here are two options:
p <- ggplot(mtcars, aes(wt, mpg)) + geom_point() + geom_text(aes(x = 5, y =
30, label = "A Label"))
#or
response <- c(2,4)
xvar <- c(1,2)
label <- response;
myData <- data.frame(response,xvar,label)
p <- ggplot(myData, aes(y=response, x=xvar))
p + geom_bar(position="dodge
Hi,
Thank you very much for the answer.
However, I have still some misunderstandings.
from the output, can we say that plant and leaf age are significant but not
their interaction?
And the last question I promise, what would you advise me to write in the
paper to explain the different method and
Thanks Peter,
You are correct! After I sent the previous message, I realized that I was
comparing the sign test against the Wilcoxon test. I would have replied
sooner, but I realized that while I was out walking my dogs.
CHV
 
Charles H Van deZande
---Original Message---
From: P
Linlin Yan wrote:
> SmoothData$span is not an object which can be checked by exists(), but
> part of an object which can be checked by is.null().
>
>
is.null is unhelpful here, in that lists can contain NULL as a named
element, and retrieving a non-existent element returns NULL:
foo = list
Dear Dieter,
I tried that. But it rescales one of the axis. The resulting graph is
still square. But now 1 cm Y-axis equal 2.5 cm on the X-axis. This seems
not te be the documented behaviour.
ggplot(ds, aes(x = x, y = y)) + geom_point() + coord_equal(ratio = 2/5)
>From sessionInfo()
R 2.9.0 o
Žroutík wrote:
>
>> SmoothData <- list(exists=TRUE, span=0.001)
>> SmoothData
>>
> $exists
> [1] TRUE
>
> $span
> [1] 0.001
>
>
>> exists("SmoothData")
>>
> TRUE
>
>
>> exists("SmoothData$span")
>>
> FALSE
>
>
'SmoothData$span' = 'foo'
exists("SmoothData$span")
Žroutík wrote:
Dear R-users,
in a minimal example exists() gives FALSE on an object which obviously does
exist. How can I check on that list object anyway else, please?
SmoothData <- list(exists=TRUE, span=0.001)
SmoothData
$exists
[1] TRUE
$span
[1] 0.001
exists("SmoothData")
On 5/19/2009 12:07 PM, Žroutík wrote:
Dear R-users,
in a minimal example exists() gives FALSE on an object which obviously does
exist. How can I check on that list object anyway else, please?
SmoothData <- list(exists=TRUE, span=0.001)
SmoothData
$exists
[1] TRUE
$span
[1] 0.001
exists("Sm
SmoothData$span is not an object which can be checked by exists(), but
part of an object which can be checked by is.null().
On Wed, May 20, 2009 at 12:07 AM, Žroutík wrote:
> Dear R-users,
>
> in a minimal example exists() gives FALSE on an object which obviously does
> exist. How can I check on
Hi,
how do I create a string of the comma-separated content of a vector?
I've got the vector i with several numeric values as content:
>str(i)
num 99
and want to create a SQL statement to look like the following where
the part '(2, 4, 6, 7)' should be
the content of the vector i:
select * from
Dear R-users,
in a minimal example exists() gives FALSE on an object which obviously does
exist. How can I check on that list object anyway else, please?
> SmoothData <- list(exists=TRUE, span=0.001)
> SmoothData
$exists
[1] TRUE
$span
[1] 0.001
> exists("SmoothData")
TRUE
> exists("SmoothData
Eric McKibben wrote:
Hi.
I am very new to R and have been diligently working my way through the manual
and various tutorials. I am now trying to work with some of my own data and
have encountered a problem that I need to fix. I have a dataframe with 8
columns and approximately 850 rows. I h
Eric McKibben wrote:
>
> Within column 6 (Question) the numbers 1:33 repeat down the entire column.
> Occasionally, however, another value (-32767) appears. I need to locate
> this value everytime it appears and in its place insert 33 rows that are
> numbered 1:33 in column Question.
> Addi
Hi.
I am very new to R and have been diligently working my way through the manual
and various tutorials. I am now trying to work with some of my own data and
have encountered a problem that I need to fix. I have a dataframe with 8
columns and approximately 850 rows. I have provided an excerpt
Dear R users,
I try to use a very large file (~3 Gib) with the filehash package. The
length of the dataset is around 4,000,000 obs. I get this message from R
while trying to "load" the dataset (named "cc084.csv"):
> dumpDF(read.csv("cc084.csv", header=T), dbName="db01")
Erreur : impossible d'all
I read that Spearman rho can be used to detect the presence of trend in a time
series.
However, I cannot figure out how to use such a test to thsi purpose. First of
all which one
of the available functions and how to pass my mono-channel time series which
contains both
positive and negative v
Dear all:
I'm trying to fit the optimal Box-Cox
transformation related to nls (see the code
below) for the demand of money data in Green (3th
Edition) but in the last step R gives the next
error message.
Error en
`[.data.frame`(eval(object$data), ,
as.character(formula(object)[[
Hi R-users!
I am trying to learn how to use the glmpath package. I have a dataframe like
this
> dim(data)
[1] 605 109
and selected the following
> response <- data[,1]
> features<-as.matrix(data[,3:109])
> mymodel <- glmpath(features,response, family = binomial)
Error in if (lambda <= min.lam
I don't get the error you mention:
> site1_data<-1
> site2_data<-2
> site3_data<-3
> for (i in 1:3) paste("site",i,"_data",sep="")
>
In my example, another way is: rm(list=paste("site",1:3,"_data",sep=""))
Or you can use rm(list=ls(pattern=you pattern)), in my example, it is:
rm(list=ls(pattern=
thanks to all your solutions, works out perfectly!
2009/5/19 Henrique Dallazuanna :
> Try this:
>
> rm(list=ls(patt="site[0-9]$"))
>
> On Tue, May 19, 2009 at 7:47 AM, Katharina May
> wrote:
>>
>> Hi,
>>
>> how can I use rm() on objects named like:
>> paste("site",i,"_data",sep="") while looping
That is very unusual, but the C stack can be increased under Windows
by recompiling R (see src/gnuwin32/front-ends/Makefile). On most other
OSes it is much easier, just adjust the setting via ulimit or limit in
your shell.
But I suspect the problem is that your model is too complex.
Incidenta
Try this:
rm(list=ls(patt="site[0-9]$"))
On Tue, May 19, 2009 at 7:47 AM, Katharina May wrote:
> Hi,
>
> how can I use rm() on objects named like:
> paste("site",i,"_data",sep="") while looping
> through i?
> I tried rm(paste("site",i,"_data",sep="")) but I get the error that
> rm() must contai
Charles Van deZande wrote:
Thanks Peter,
There are 8 measurements less than 8.5, so calculating the probability
(binomial) of 8, or fewer, happening by chance with n = 20 and p = 0.50
gives P = 0.25-- the book answer. I've tried several problems in other
textbooks and in each case I get vast
Easy enough. What if some of the matrix elements contained missing values?
Then how could you still calculate the means? Example code below:
mat1 <- matrix(c(1,2,3,4,5,NA,7,8,9),3,3)
mat2 <- matrix(c(NA,6,1,9,0,5,8,2,7),3,3)
mat3 <- matrix(c(5,9,1,8,NA,3,7,2,4),3,3)
Gabor Grothendieck wrote:
Thanks Peter,
There are 8 measurements less than 8.5, so calculating the probability
(binomial) of 8, or fewer, happening by chance with n = 20 and p = 0.50
gives P = 0.25-- the book answer. I've tried several problems in other
textbooks and in each case I get vastly different P-values than I get
ONKELINX, Thierry inbo.be> writes:
>
> I'm plotting some points on a graph where both axes need to have the
> same scale. See the example below. Coord_equal does that trick but in
> this case it wastes a lot of space on the y-axis. Setting the limits of
> the y-axis myself was no avail.
>
> Any
Dear gabor,
Many thanks for your answer. I indeed didn't check the read me text.
Bests
> From: ggrothendi...@gmail.com
> Date: Tue, 19 May 2009 08:44:13 -0400
> Subject: Re: [R] problem with installing a local zip file : GFCURE
> To: marc_bern...@hotmail.co.uk
> CC: r-help@r-project
> "RT" == Rolf Turner
> on Tue, 19 May 2009 11:02:08 +1200 writes:
RT> On 19/05/2009, at 10:20 AM, Steve Lianoglou wrote:
>> Hi all,
>>
>> I've (tried) to look through the bug tracker, and gmane-search the
>> R list to
>> see if this has been mentioned befo
Katharina May wrote:
Hi,
how can I use rm() on objects named like:
paste("site",i,"_data",sep="") while looping
through i?
I tried rm(paste("site",i,"_data",sep="")) but I get the error that
rm() must contain names or
text strings which is confusing me as I thought paste() would create
somethi
Have you try principal component analysis to reduce space variables?
2009/5/19 Nora Pérez
>
>
>
>
>
>
>
>
> Dear colleagues,
>
> I am trying a glm.nb for the distribution of a plant species with 93
> environmental variables. I execute the instruction and I get the following
> message: "Error: C
Dear colleagues,
I am trying a glm.nb for the distribution of a plant species with 93
environmental variables. I execute the instruction and I get the following
message: "Error: C stack usage is too close to the limit".
How can I increase the memory of R?
Your sincerely,
Nora.
_
On Tue, May 19, 2009 at 6:17 AM, marc bernard
wrote:
>
> Dear all,
>
>
>
> I am trying to install a package called "GFCURE" from a local zip file. This
> package fits a cure survival model and has been downloaded from:
>
You are assuming its in the form of an R *package* but its not.
Unzip it a
Try this also:
rm(list=names(which(unlist(eapply(globalenv(), function(a)all(is.na(a) ||
is.null(a)))
On Tue, May 19, 2009 at 9:07 AM, Katharina May wrote:
> Thanks Jim, the removal of objects which are NA works perfectly!
>
> For my second problem it didn't express myself correctly:
> I
Note that this could be done like this for ordinary
vectors:
> x <- seq(1:4)^2
> apply(embed(x, 2), 1, function(x, f) f(rev(x)), f = diff)
[1] 3 5 7
> apply(embed(x, 2), 1, function(x, f) f(rev(x)), f = sum)
[1] 5 13 25
or a method to rollapply in zoo could be added for ordinary vectors.
Here i
On Tue, 19 May 2009, marc bernard wrote:
Dear all,
I am trying to install a package called "GFCURE" from a local zip
file. This package fits a cure survival model and has been
downloaded from:
http://post.queensu.ca/~pengp/software.html
However, it is not an R package. Read the Readme.t
Look to the glm function then pass the output to the step function
Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034
steve_fried...@nps.gov
Office (305) 224 - 4282
Fax (305) 224 - 4147
Dear R Users,
A would like to fit a loglinear analysis to a three dimensional contingency
table. But I Don't want to run a full saturated modell. Is there any package
in R that could handle somekind of stepwise search to choose out the best
soultion? And how can I fit a non fully saturated modell,
Thanks Jim, the removal of objects which are NA works perfectly!
For my second problem it didn't express myself correctly:
I actually meant objects with rows (attributes?) but no data in it
but I solved this
adjusting your approach:
for(object in objects()) if(is.null(dim(get(object))[1]) ||
dim
Katharina May wrote:
Hi,
how can I remove all empty objects (which are NA or have zero rows)
from my workspace?
Hi Katharina,
To remove objects that are all NA:
for(object in objects()) if(all(is.na(get(object rm(list=object)
If by "zero rows" you mean objects that do not have a dimen
If you use coord_equal on data where the range on the x-axis is larger
than the range on the y-axis, then of course you'll observe extra
space on the y-axis. What did you expect?
Also, this post may be better suited to the ggplot2 mailing list:
http://had.co.nz/ggplot2/
On Tue, May 19, 2009 at 7:
Hi,
how can I use rm() on objects named like:
paste("site",i,"_data",sep="") while looping
through i?
I tried rm(paste("site",i,"_data",sep="")) but I get the error that
rm() must contain names or
text strings which is confusing me as I thought paste() would create
something like that...?
Thanks,
Hi,
how can I remove all empty objects (which are NA or have zero rows)
from my workspace?
Thanks,
Katharina
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.o
Dear all,
I'm plotting some points on a graph where both axes need to have the
same scale. See the example below. Coord_equal does that trick but in
this case it wastes a lot of space on the y-axis. Setting the limits of
the y-axis myself was no avail.
Any suggestions to solve this problem?
l
Dear all,
I am trying to install a package called "GFCURE" from a local zip file. This
package fits a cure survival model and has been downloaded from:
http://post.queensu.ca/~pengp/software.html
The problem is that when I try to install this package from a local zip file
using R,
Dear Christine,
(Month|Block) and (1|Block) + (1|Month) are completely different random
effects. The first assumes that each Block exhibits a different linear trend
along Month. The latter assumes that each block has a random effect, each month
has a random effect and that the random effects of
I just tried it in Minitab and got
--
Test of median = 8.500 versus median not = 8.500
N for Wilcoxon Estimated
N Test Statistic P Median
C1 20 19 80.5 0.573 8.460
-
One tailed gave me closer to the textbook, but still not
cvandy wrote:
When I use wilcox.test, I get vastly different p-values than the problems
from Statistics textbooks.
For example:
The following problem comes from "Applied Statistics and Probability for
Engineers", 2nd Edition, by D. C. Montgomery. Page736, problem 14.7. The
problem is to compare
Wacek Kusnierczyk wrote:
>
> btw., the error message here is confusing:
>
> lag = 1:2
> diff(1:10, lag=lag)
> # Error in diff.default(1:10, lag = lag) :
> # 'lag' and 'differences' must be integers >= 1
>
> is.integer(lag)
> # TRUE
> all(lag >= 1)
> # TRUE
>
> wh
Dear list,
Has anyone used the 'copula' or 'fCopulae' package with empirical
distributions. I have two distributions (10.000 samples each) which I need
to combine using archimedean copulas (probably Clayton and/or Frank).
Is this possible? Is there an existing empirical distribution function
def
Thanks. I did try using quasipoisson and a negative binomial error but am
unsure of the degree of overdispersion and whether it is simply due to
missing values. I am investigating to see if I can replace these missing
values so that I can have a balanced orthogonal design and use lme or aov
instea
Wacek Kusnierczyk wrote:
> Stavros Macrakis wrote:
>
>>
[...]
>> I am not talking about creating a new class with an analogue to the
>> subtraction function. I am talking about a function which applies another
>> function to a sequence and its lagged version.
>>
>> Functional arguments are use
On Tue, 19 May 2009 14:04:19 +1000 Kon Knafelman
wrote:
KK> i have the sample variances for 1000 samples, and i want to fit it
KK> to a chi-squared distribution.
KK> can someone please help me fit this to a chi-squared distribution
KK> with n degrees of freedom. Thanks a lot
Dear Kon,
1. ple
Stavros Macrakis wrote:
> On Mon, May 18, 2009 at 6:00 PM, Gabor Grothendieck
>> wrote:
>>
>
>
>> I understood what you were asking but R is an oo language so
>> that's the model to use to do this sort of thing.
>>
>>
>
> I am not talking about creating a new class with an analogue
is there a command like "mat.or.vec" for an array that I have to create with a
cicle for?
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Dear Christine,
The poisson family does not allow for overdispersion (nor
underdispersion). Try using the quasipoisson family instead.
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Res
On Mon, 2009-05-18 at 11:48 -0700, William Paterson wrote:
> Hi,
>
> I am using GAMMs to show a relationship of temperature differential over
> time with a model that looks like this:-
>
> gamm(Diff~s(DaysPT)+AirToC,method="REML")
>
> where DaysPT is time in days since injury and Diff is repeat
Your problem is statistical and has nothing particularly to do with R.
It looks like homework to me.
You may care to look at it this way:
###
> fm <- glm(outcome ~ age, binomial, surgery)
> summary(fm)
Call:
glm(formula = outcome ~ age, family = binomial, data = surgery)
Dev
Johannes Hüsing wrote:
>
>
> Am 19.05.2009 um 05:39 schrieb phen_ys:
>
>>
>>> surgery <- data.frame(outcome = c(0, 0, 0, 0, 0, 0, 0, 0, 0,
>> + 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0,
>> + 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0), age = c(50, 50, 51,
>> + 51, 53, 54, 54, 54, 55, 5
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