On Wed, Sep 14, 2011 at 11:56 PM, Daniel Malter wrote:
> Not satisfactory in which sense?
>
> The survreg(Surv(Value,Censoring)~indvars+strata(id)) should/may work. For a
or cluster(year) if you have a new sample of people in each year
> discussion of Tobit fixed effects, see also Greene's websi
Dear all,
I am trying to create a map for central asian countries(Kazakhstan,
Uzbekstan, Kyrgyzstan, Turkmenstan, and Tajikstan). I tried google for
shapefiles and Rdata files of central asia, but I can't seem to find them.
There is only a world map and individual maps for each countries. If I use
I have created an R-package with datasets which I want my students to install
(the package is not on CRAN).
1) I've put the package on the web in a directory called 'data' and I thought I
could do:
> install.packages("http://gbi.agrsci.dk/statistics/courses/2011-ISMLS-course/data/LiSciData_0.0
On 15.09.2011 10:34, Søren Højsgaard wrote:
I have created an R-package with datasets which I want my students to install
(the package is not on CRAN).
1) I've put the package on the web in a directory called 'data' and I thought I
could do:
install.packages("http://gbi.agrsci.dk/statistic
On Thu, Sep 15, 2011 at 8:41 AM, Salaam Batur wrote:
> Dear all,
>
> I am trying to create a map for central asian countries(Kazakhstan,
> Uzbekstan, Kyrgyzstan, Turkmenstan, and Tajikstan). I tried google for
> shapefiles and Rdata files of central asia, but I can't seem to find them.
> There is
Salaam,
If you would like to display the chart online than the googleVis package might
be helpful.
The package provides you with an interface to the Google Visualisation API.
Here is an example with a map of the population on the central Asian countries:
library(googleVis)
## googleVis comes wit
On Thu, 15 Sep 2011 09:47:35 +1200, Rolf Turner
wrote:
On 15/09/11 07:21, Torbjørn Ergon wrote:
Dear list,
I'm looking for a function to generate (simulate) a random Weibull
point process. Can anyone help?
Cheers,
Torbjørn Ergon, University of Oslo
Do you mean a renewal process with the
Many thanks to all of you! AV plots are what I am trying to plot. Perhaps to
reduce confusion I can give you an example of what I am doing:
I am looking at behaviour of re-sighted individuals over two time points. I
use lm() on these data and obtain the residuals.
Then I am interested to know wh
Thanks Ken!
This was not exactly what I was after - see my reply to Rolf Turner.
Sorry for not explaining this well enough.
Cheers,
Torbjørn
On Wed, 14 Sep 2011 16:37:48 -0400, Ken Hutchison
wrote:
I don't know if this settles the matter, but if you are modeling
Weibull as the Log-Inten
Dear R helpers
I would like to move the x-axis labels, which plot automatically at the base
of a dot plot to the top of the plot. Is there a way to do this?
Code snippet below
with(Cal_dat,
dotplot(reorder(paste(Mine,Company), Resc_Gt) ~ Resc_Gt,
fill_var = Commodity,
Any insight maybe (sorry for the bump, need this info), JT.
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Hello,
I have a question concerning courses in R in *Vienna, Austria*.
Does anyone know where courses in R are offered, that you can do after
having attended a course for the basics?
Thanks in advance for your help.
Marion
[[alternative HTML version deleted]]
_
Let me explain my problem: I have a data set of municipalities from 2002 to
2010. For each municipality I have the amount of money per capita invested
in infra-structure for each year. This variable is truncated at zero, as
there is no negative values for investment and there are many cases in whic
Dear R helpers
I wish to move the main title, which appears on a dotplot to be right
aligned with the left axis. Is there are parameter associated with dotplot
'main' that allows the title to be placed where I want it?
Code snippet relating to dotplot is below.
with(Cal_dat,
dotplot(reorder
Thank you so much!
foreL<-8
b0f<-matrix(nrow=9, ncol=foreL)
ct<-1 ### use this as the index of b0f
for(ar.ord in 1:3){
for(ma.ord in 1:3){
b0f[ct,]<-c(predict(arima(para_qtr[1:(n-8),1],order=c(ar.ord,1,ma.ord)),
n.ahead=foreL)$pred)
ct<-ct+1 ### increment the counter
}
}
this one works! Best rega
hi R Community
Is it possible to plot a model using the ggplot2 package.
Something like:
x<-rnorm(100)
y<-rnorm(100)
fit<-lm(y~x)
plot(fit)
but with ggplot2
qplot(fit)?
thanks
michael
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Dear R helpers
I would like to be able to colour code the y-axis labels on a complex dot
plot by a variable known as company (of which there are only two). The code
is below and data attached.
Thanks
MarkM
library("lattice")
library(latticeExtra) # for mergedTrellisLegendGrob()
# set size of
problem solved
?fortify -> examples
thanks
Michael
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Hello
I have just installed Tinn R 2.3.7.1. on my new computer. But the R
toolbar has turned grey, which never happened with my previous computer.
I don't know why it turned grey. Any help?
Thank you.
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On 15/09/11 19:24, Torbjørn Ergon wrote:
On Thu, 15 Sep 2011 09:47:35 +1200, Rolf Turner
wrote:
On 15/09/11 07:21, Torbjørn Ergon wrote:
Dear list,
I'm looking for a function to generate (simulate) a random Weibull
point process. Can anyone help?
Cheers,
Torbjørn Ergon, University of Oslo
Hi,
i am a student of tecnical mathematics in austria, and my english is not
sooo good, German would be easier, but an answer in english is perfect too.
I would need any help to understand the exact output of the function
"garchFit". Maybe it would help me just to know what the coefficients are
On 09/14/2011 10:37 PM, Diviya Smith wrote:
> Hi there,
>
> I have a complex math equation which does not have a closed form solution.
> It is -
>
> y <- (p*exp(-a*d)*(1-exp((d-p)*(a-x[1]/((p-d)*(1-exp(-p*(a-x[1]
>
> For this equation, I have all the values except for x[1]. So I need to so
Dear Felipe
On 15 September 2011 09:58, Felipe Nunes wrote:
> My technical problems were:
>
> (1) tob3 <- censReg(transfers.cap ~ factor(mayor) + vot.weight + vot.per +
> transfers.cap.lag + pib.cap + cluster(factor(year)), left=0, data = pool)
>
> Using 'censReg' my results never converge. Std.
Hi,
I have a dataset where the residual variance decreases with on one of
the predictors (population size).
Currently, the full model looks like this:
prior<-list(R=list(V=1e-16, nu=-2),G1=list(V=diag(2), nu=2))
m<-MCMCglmm(response~poly(population size,2)*poly(other
predictor,2)+time, random=~
Please see the R Machine Learning Task View
(http://cran.r-project.org/web/views/MachineLearning.html) for a
starting point on decision trees.
On 9/14/2011 7:11 PM, Lorenzo Isella wrote:
> Dear All,
> I am recycling a previous email of mine where I asked some questions
> about clustering mixed
The 'name' attribute is not used for vertex labels by default. If you
want to use it, you need to set it as the 'label' attribute as well,
or supply them in the function call:
tkplot(graph, vertex.label=V(graph)$name)
Best,
Gabor
On Wed, Sep 14, 2011 at 7:11 PM, karena wrote:
> Hi,
>
> I am usi
Many thanks!
Adam
2011/9/14 Achim Zeileis
> On Wed, 14 Sep 2011, Adam Copella wrote:
>
> Dear All,
>>
>> Is there any function in R for fitting linear probability model, as my
>> response variable is a uniformly distributed.
>>
>
> If it is really uniformly distributed, I think there is not muc
Hello useRs!
Recently, I migrated from Debian i386 to Fedora 15 64-bit, after a long
and happy experiences with Debian distributions (first Ubuntu, later
Debian itself). I installed R with yum install R, and then necessary
packages. Everything seems to work fine, for now, except that exporting
plot
On Sep 15, 2011, at 7:06 AM, Petar Milin wrote:
> Hello useRs!
> Recently, I migrated from Debian i386 to Fedora 15 64-bit, after a long
> and happy experiences with Debian distributions (first Ubuntu, later
> Debian itself). I installed R with yum install R, and then necessary
> packages. Everyth
What is the correct way to combine multiple calls to odfCat,
odfItemize, odfTable etc. inside a function?
As an example lets say I have a function that needs to write two
paragraphs of text and a list to the resulting odf-document (the real
function has much more complex logic, but I don'
optim()
optimx package
minpackLM package
and several others
JN
On 09/15/2011 06:00 AM, r-help-requ...@r-project.org wrote:
> Message: 77
> Date: Wed, 14 Sep 2011 20:44:16 +0100
> From: Liam Brown
> To: r-help@r-project.org
> Subject: [R] Nonlinear Regression
> Message-ID:
>
> Content-Typ
Atle Torvik Kristiansen gmail.com> writes:
> I have a dataset where the residual variance decreases with on one of
> the predictors (population size).
>
> Currently, the full model looks like this:
>
> prior<-list(R=list(V=1e-16, nu=-2),G1=list(V=diag(2), nu=2))
>
> m<-MCMCglmm(response~poly(p
Dear all,
I have a matrix that provides, for a series of data points, the probability
that each of these points belongs to a certain group.
Take the following example, which represents 20 data points and their group
membership probability to five groups (A-E):
set.seed(1)
probs <- matrix(runif(10
Hi Mauricio
On 9/14/2011 2:10 PM, Mauricio Cornejo wrote:
Thanks Z.
Let me clarify my problem a bit further ... as I don't think I could use cut()
or spine() to solve it.
I have a data frame with three columns (A, B, Value). 'A' and 'B' are
categorical and 'Value' is continuous and non-nega
>From: Madeleine Seeland in.tum.de>
>Subject: help with hclust
>Date: 2011-09-14 08:14:48 GMT
>Hello,
>
>I have two questions regarding hclust:
>
>1) First I would like to cut a hclust tree at a specific height and
receive the cluster (di) similarity at this
>height.
With an example:
How can I apply difftime to a vector of sorted dates? I can do this just
fine with diff, but difftime doesn't seem to take in a vector.
> diff(r$BOOKING_DATE)
Works. Great!
> difftime(r$MY_DATE, units="days")
Error in as.POSIXct(time2) : argument "time2" is missing, with no default
Thanks,
Bra
Hello,
If I understand good, I can't have p-value for the nls model.
I have 2 vectors. And I'am doing
model <- nls(crf ~a*(1-exp(-x/b)) + c, data= d,
start = list(a=1, b=3, c=0))
and I want to know If my result is significat, if I can't have p-value, how can
I know it?
Thank y
Hello,
I want to understand how to tell if a model is significant.
For example I have vectX1 and vectY1.
I seek first what model is best suited for my vectors and
then I want to know if my result is significant.
I'am doing like this:
model1 <- lm(vectY1 ~ vectX1, data= d),
model2 <- nls(vectY1
This sounds like you need a mixed effects model (e.g. lme or lmer
instead of lm) instead of the possibly spurious adhocery that you
describe. I suggest you post your message on the SIG-mixed-effects
list and/or get help from a local statistician
Cheers,
Bert
On Thu, Sep 15, 2011 at 1:49 AM, RCull
Tatiana:
It sounds like you are in way over your head statistically. This is
not a statistics tutorial site (though sometimes good folks do help
witjh this). I suggest you try http://stats.stackexchange.com/ .
Better yet, work with your local statistician.
Cheers,
Bert
On Thu, Sep 15, 2011 at 6
Hello All,
I posted a similar question before, but the direction was driven to whether
my case is suitable for a wilcoxon test. After research about the
appropriateness, I am pretty sure that a wilcoxon test is the right tool for
my case. But how to compute the power of the test is still an unansw
Hi Joseph (and Martin),
Don't mean to beat a dead horse, but I wanted to add one last comment
to this thread in case someone stumbles upon this via google/gmane (or
you) and gives it a shot.
I neglected to mention a very important step that you'd have to do to
in order to avoid shooting yourself
You need to supply two vectors:
> x <- seq(Sys.time(), by = '10 min', length = 10)
> x
[1] "2011-09-15 09:58:30 EDT" "2011-09-15 10:08:30 EDT" "2011-09-15
10:18:30 EDT"
[4] "2011-09-15 10:28:30 EDT" "2011-09-15 10:38:30 EDT" "2011-09-15
10:48:30 EDT"
[7] "2011-09-15 10:58:30 EDT" "2011-09-15 11
On Wed, 2011-09-14 at 18:06 -0500, Drew Tyre wrote:
> I am trying to reproduce plots in Chapter 3 of Zuur et al Mixed
> Effects models and extensions in Ecology. For pedagogical reasons,
> they make a series of plots with gam(...) in package gam. I encounter
> errors that trace back to the predict.
There are examples in the package directory that explain this.
On Thu, Sep 15, 2011 at 8:16 AM, Jan van der Laan wrote:
>
> What is the correct way to combine multiple calls to odfCat, odfItemize,
> odfTable etc. inside a function?
>
> As an example lets say I have a function that needs to write
Dear all,
I am running a simulation study to test variable imputation methods for Cox
models using R 2.9.0 and Windows XP. The code I have written (which is rather
long) works (if I set nsim = 9) with the following starting values.
>bootrs(nsim=9,lendevdat=1500,lenvaldat=855,ac1=-0.19122,bc1=-
On Sep 15, 2011, at 15:47 , tn85 wrote:
> Hello All,
>
> I posted a similar question before, but the direction was driven to whether
> my case is suitable for a wilcoxon test. After research about the
> appropriateness, I am pretty sure that a wilcoxon test is the right tool for
> my case. But h
Thank you, Peter. They are empirical quantities, that's why I felt confused as
being requested to give the power of my test. Could you please clarify the
reason why power calculations are not fit for empirical quantities? So that I
can defend my test for such request.
BTW, the correlation of th
Hi Laura,
On Thu, Sep 15, 2011 at 10:53 AM, Bonnett, Laura
wrote:
> Dear all,
>
> I am running a simulation study to test variable imputation methods for Cox
> models using R 2.9.0 and Windows XP. The code I have written (which is
> rather long) works (if I set nsim = 9) with the following sta
One additional thing to note is that %>% will have different precedence than >
(something that was pointed out to me based on %<% that is in the TeachingDemos
package).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Or
Hi,
You can use scales() and parametrize it with a list where you can define
colors, fonts, etc.
I could not test it on your code because variable "Commodity" is not present
in the dataset you provided.
Regards,
Carlos Ortega
www.qualityexcellence.es
On Thu, Sep 15, 2011 at 11:03 AM, markm0705
Hello useRs,
I am trying to put my funcs in a package to avoid clutter in my workspace as
the funcs are now in .Rprofile.
All plain R code no compilations. Using win XP with a full cygwin install and
R2.12
I first did
package.skeleton("mypack",list="getdfv", namespace=T) # just a single func
w
Hi Steve,
Thanks for your response. The slight issue is that I need to use a different
starting seed for each simulation. If I use 'lapply' then I end up using the
same seed each time. (By contrast, I need to be able to specify which starting
seed I am using).
Thanks,
Laura
-Original M
On 15.09.2011 17:32, Bond, Stephen wrote:
Hello useRs,
I am trying to put my funcs in a package to avoid clutter in my workspace as
the funcs are now in .Rprofile.
All plain R code no compilations. Using win XP with a full cygwin install and
R2.12
I first did
package.skeleton("mypack",list=
Uwe,
That gave me the same error like CMD install
> install.packages("C:/Temp/mypack_1.0.tar.gz", repos=NULL, type="source")
install.packages("C:/Temp/mypack_1.0.tar.gz", repos=NULL, type="source")
Loading required package: stats
Loading required package: utils
Loading required package: graphics
Hi all,
I was recently writing a script to identify the value and id of the maximum
observation in a sliding window when I ran into some unexpected behavior. I
have included an example.
> test <- c()
> test$elev <- c(1:200)
> test$i <- 1
> test$window <- 10
The following works for me:
> check.
Hi everyone,
I'm trying to create a scatterplot (with ggplot2) with different groups and
a faceting (see code below... much clearer I think)
But when I want to change the color label to something else using the
scale_colour_manual function (based on
http://had.co.nz/ggplot2/scale_manual.html), it
If I enclose x$i+x$window in parentheses it works:
check.max<- function(x){obs.max<- which.max(x$elev[x$i:(x$i+x$window)]);
obs.max}
x$elev[x$i:x$i+x$window] means:
x$elev[(x$i:x$i)+x$window]
Simple.
mario
check.max(test)
[1] 11
What am I missing?
Here is sessionInfo (I
rjswift gmail.com> writes:
>
> I'm trying to select a model under PCA using independent contrasts. Since
> PICs need to be forced through the origin I've been using lmorigin for the
> original regression, but it doesn't appear that stepAIC recognizes it. I
> keep receiving an error message - "Er
On 15.09.2011 17:47, Bond, Stephen wrote:
Uwe,
That gave me the same error like CMD install
install.packages("C:/Temp/mypack_1.0.tar.gz", repos=NULL, type="source")
install.packages("C:/Temp/mypack_1.0.tar.gz", repos=NULL, type="source")
Loading required package: stats
Loading required pack
Thanks for all the input. You were right. It was not csv file in the
correct format. As it turned out, it's not a R problem, more a MAC
problem. For some reason, my MAC doesn't correctly download csv file
from internet, which caused all the subsequent problems. I will have
to take it to a m
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Bond, Stephen
> Sent: Thursday, September 15, 2011 8:48 AM
> To: Uwe Ligges
> Cc: r-help@r-project.org
> Subject: Re: [R] how to install a locally built package
>
> Uwe,
>
> Tha
I have two variables, both numerical. I would like to find the unique
values of the pairs, in other words, unique coordinates if I were to
plot them.
I also need to know how many pairs there are, but I guess I can use
length() if I can somehow isolate the unique pairs first?
Thanks a lot!
Hi Bonnie,
?unique probably would have done the trick for you. But here's an
example:
> z <- data.frame(x = rep(1:5, each=2), y = rep(1:2, each=5))
> z
x y
1 1 1
2 1 1
3 2 1
4 2 1
5 3 1
6 3 2
7 4 2
8 4 2
9 5 2
10 5 2
> unique(z)
x y
1 1 1
3 2 1
5 3 1
6 3 2
7 4 2
9 5 2
> nrow(unique(z
Bonnie:
1. As usual, see FAQ 7.31. unique() may not mean what you think it
means for finite precision numbers:
> unique(c(1/3, sqrt(2)*1/3/sqrt(2)))
[1] 0.333 0.333
But modulo that: unique(paste(v1,v2))
will give you what you're looking for I think.
-- Bert
On Thu, Sep 15, 2011 at 9:27
On Sep 15, 2011, at 12:27 PM, bby2...@columbia.edu wrote:
I have two variables, both numerical. I would like to find the
unique values of the pairs, in other words, unique coordinates if I
were to plot them.
I also need to know how many pairs there are, but I guess I can use
length() if
I should try reading the Help page! Please ignore my prior "advice"
-- except for the party about FAQ 7.31, which may still apply, of
course.
-- Bert
On Thu, Sep 15, 2011 at 9:34 AM, Sarah Goslee wrote:
> Hi Bonnie,
>
> ?unique probably would have done the trick for you. But here's an
> exampl
Hello,
There are some specific examples included in the help of barplot() that
answer your question.
Regards,
Carlos Ortega
www.qualityexcellence.es
On Wed, Sep 14, 2011 at 10:14 PM, Allie818 wrote:
> I've made a barplot that has several bars. I'd like the bars to be colored
> according to the
David and Sarah,
Both suggestions work beautifully. Thanks!
Bonnie
> CC: r-help@r-project.org
> From: dwinsem...@comcast.net
> To: bby2...@columbia.edu
> Subject: Re: [R] how to find unique pairs of variables?
> Date: Thu, 15 Sep 2011 12:41:03 -0400
>
>
> On Sep 15, 2011, at 12:27 PM, bby2...@c
What null hypothesis are you trying to test? There is a standard null for
linear models that makes sense in a large number of cases, but what the null is
for non-linear regression is not obvious (and the coefficient = 0 may not even
be possibly, let alone interesting). If you can state what yo
On 15/09/2011 10:40 AM, Jason Curole wrote:
Hi all,
I was recently writing a script to identify the value and id of the maximum
observation in a sliding window when I ran into some unexpected behavior. I
have included an example.
> test<- c()
> test$elev<- c(1:200)
> test$i<- 1
> test$wind
Thank you so much, Gabor.
It works well!
--
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h
Dear List Members,
I have created a function to run a simulation based on a given set of
values within a vector, such that I run the function like this:
new.data<-sapply(vector, function)
In which I run 'function' on every value within a vector that I created.
The result is a matrix of 1000
I have Q
how can I do integrate to BS
where BS=1/m sum{(s(t|x)^2/g(t))+(1-s(t|x)^2)/g(t)}
lower=1, upper = 5
Thanks
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PLEASE
What does any of this mean?
You might try integrate()
Michael Weylandt
On Thu, Sep 15, 2011 at 1:46 PM, tarq khhh wrote:
> I have Q
>
> how can I do integrate to BS
> where BS=1/m sum{(s(t|x)^2/g(t))+(1-s(t|x)^2)/g(t)}
>
> lower=1, upper = 5
> Thanks
>[[alternative HTML version deleted
It would seem you want something like this:
new.data <- sapply(seq(10, 100, 10), function(i) sapply(vector, yourFunc, i))
On Thu, Sep 15, 2011 at 1:50 PM, Mike Treglia wrote:
> Dear List Members,
>
> I have created a function to run a simulation based on a given set of values
> within a vector,
I got it working by typing a string in getdfv.Rd after \title{
\title{ getdf
%% ~~function to do ... ~~
}
Strange why the skeleton would not do that given it did
\name{getdfv}
\alias{getdfv}
Anyway, I'm happy now.
Stephen Bond
-Original Message-
From: Uwe Ligges [mailto:lig...@stati
On Sep 15, 2011, at 1:50 PM, Mike Treglia wrote:
Dear List Members,
I have created a function to run a simulation based on a given set
of values within a vector, such that I run the function like this:
new.data<-sapply(vector, function)
In which I run 'function' on every value within a ve
On 15 September 2011 20:09, Felipe Nunes wrote:
> I did both ways (pooled and pdata.frame), but in none I got a result. The
> coefficients are estimated, but not the std. errors. I'm using BFGS method,
> but I didn't increase the number of iterations yet. Let me try!
... and I highly recommend to
Max,
Thank you for your answer. I have had another look at the examples (I
already had before mailing the list), but could find the example you
mention. Could you perhaps tell me which example I should have a look at?
Regards,
Jan
On 09/15/2011 04:47 PM, Max Kuhn wrote:
There are examples
I'm plotting x and y and there are a couple of outliers. I want to
show the value of z for these couple of outliers. Is there an easy way
to do this?
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PLEASE do read th
?identify
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of bby2...@columbia.edu
> Sent: Thursday, September 15
On Sep 15, 2011, at 17:15 , tn85 wrote:
> Thank you, Peter. They are empirical quantities, that's why I felt confused
> as being requested to give the power of my test. Could you please clarify the
> reason why power calculations are not fit for empirical quantities? So that I
> can defend my
Thank you all- with your suggestions I was able to get this simulation
to run- sorry, next time I'll post some reproduceable code- it had been
a while since I posted, so I forgot about that.
Best,
Mike
On 9/15/2011 1:40 PM, David Winsemius wrote:
On Sep 15, 2011, at 1:50 PM, Mike Treglia wrot
I am trying to use RgoogleMaps but when I try to run the script example from
"RgoogleMaps-intro.pdf -page 3" (see below) I get the following error
"Error en plot.xy(xy.coords(x, y), type = type, ...) "
I have tried to solve it but I am going crazy with this, because sometime in
the past I made i
Hi Arne,
I did both ways (pooled and pdata.frame), but in none I got a result. The
coefficients are estimated, but not the std. errors. I'm using BFGS method,
but I didn't increase the number of iterations yet. Let me try!
Best,
*Felipe Nunes*
CAPES/Fulbright Fellow
PhD Student Political Science
Hi folks,
Please let me know what I am doing wrong. I want to have a legend with
symbols that are filled with same color as the drawn line, but I failed
to do that:
plot(1:100, 1:100, pch=21, bg="red")
legend("bottomright", "test", bty='n', pch=21, bg="red", col="red")
It looks to me that th
What type of singularity exactly, if you're working with counts is it a special
case? If using a Monte Carlo generation scheme, there are various workarounds
such as while(sum(vec)!=0) {sample} for example. More info on the error
circumstances would help.
Good luck!
Ken Hutchison
On Sep
On 9/15/11 4:40 PM, Duke wrote:
Hi folks,
Please let me know what I am doing wrong. I want to have a legend with
symbols that are filled with same color as the drawn line, but I
failed to do that:
plot(1:100, 1:100, pch=21, bg="red")
legend("bottomright", "test", bty='n', pch=21, bg="red", c
Dear folks:
Let’s suppose I want a function to print return the name of the object
passed to it.
> myname <- function(object) {out<-deparse(substitute(object)); out}
This works fine on a single object:
> O1 <-c(1:4)
> myname(O1)
[1] "O1"
However it does not work if you use lapply to pass it th
My guess is that anyone willing to help will want more information.
What version of R do you have(?), for example.
BiodiversityR Depends on R version ≥ 2.13.1, and vegan ≥ 1.17-12
SG wrote:
>
> I have installed R on windows 7 machine.
> In the install packages I can't find BiodiversityR pac
I have a problem with lattice log scales that I could use some help with.
I'm trying to print log y-axis scales without exponents in the labels.
A similar thread with Deepayan' recommendation is here:
http://tolstoy.newcastle.edu.au/R/e11/help/10/09/9865.html. For
example, this code using xyplot
Hi group,
I am trying to right a code to do the following
This is how the test file looks like:
Chr start end sample1 sample2
chr2 9896633 9896683 0 0
chr2 9896639 9896690 0 0
chr2 14314039 14314098 0 -0.35
chr2 14404467 14404502 0 -0.35
chr2 14421718 14421777 -0.43 -0.35
chr2 16031710 16031769 -0.
Dear R gurus,
I am looking for a way to fit a predictive model for a contingency table which
has counts. I found that glm( family=poisson) is very good for figuring out
which of several alternative models I should select. But once I select a model
it is hard to present and interpret it, especia
Hi,
I'm having some problems with the aggregate() function in the {stats}
package, and the documentation doesn't address them.
1) Why would the first line work, but the second not? According to the
help file, it accepts a "data=" argument.
with(tsrc, aggregate(x=DistRatio, by=list(Conditi
I have installed R on windows 7 machine.
In the install packages I can't find BiodiversityR package.
I was wondering if I can get some help with the installation process.
Thanks.
--
View this message in context:
http://r.789695.n4.nabble.com/installation-of-BiodiversityR-package-tp3816807p381680
You'll never figure it out if you don't "play around" with your data.
Assuming you have been able to import the data, a good place to start is to
look at what tools you have available.Check out this:
http://cran.r-project.org/doc/contrib/Short-refcard.pdf
check out things like
?which
?max
?
On Sep 15, 2011, at 4:07 PM, Jon Zadra wrote:
> Hi,
>
> I'm having some problems with the aggregate() function in the {stats}
> package, and the documentation doesn't address them.
>
> 1) Why would the first line work, but the second not? According to the help
> file, it accepts a "data=" argu
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of andrewH
> Sent: Thursday, September 15, 2011 2:11 PM
> To: r-help@r-project.org
> Subject: [R] Returning the name of an object passed directly or from a list
> by lapply
>
> Dea
Hi Dear all,
I have some gene expression data samples from different tissue types
---
- 120 samples from blood (B)
- 20 samples from Liver (L)
- 15 samples from Kidney (K)
- 6 samples from heart (H)
---
All the
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