summary: spatial data to be input to a regional-scale environmental
model must (1) be converted to netCDF and then (2) "regridded" (cropped,
projected, increased resolution). In a public git repository
https://github.com/TomRoche/GEIA_to_NetCDF
I have R code (with bash drivers) that does the con
Please note:
1) your example is not working in the way you provided it (see
http://www.minimalbeispiel.de/mini-en.html)
2) you receive a warning, not an error
3) I'd try and debug qua.regressCOP2 to see why the warning appears
4) in case 3) does not help, contact the maintainer of copBasic (Willia
On 10/21/2012 09:03 AM, YAddo wrote:
Thanks Jim and Rui.
My apologies, i did not give enough info on my plot.
I am using : plot(x,y) for a line plot. I want to center the labels on the
x-axis for each tick.
Hi YAddo,
This is another guess, that you have more ticks than labels because they
Hello,
The function in my previous post gives neighbours in north, south, east
and west but also the corners, for a total of 8, not 4, neighbours.
Corrected:
is.border <- function(idx, DF){
i1 <- DF$ix %in% DF$ix[idx] + c(-1, 1) & DF$iy == DF$iy[idx]
i2 <- DF$iy %in% DF$iy[idx] + c(-1
Dear Berend,
Thank you very much for proposing the limSolve-package!
The lsei function of the package is exactly what I was looking for.
Best regards
Thomas
--
View this message in context:
http://r.789695.n4.nabble.com/Which-package-function-for-solving-weighted-linear-least-squares-with-ine
Dear Richard,
It is funny. I have to perform the approach of sediment fingerprinting for
my master thesis. Mr. Hasselman gave me the advice to take a closer look
into the limSolve package a few days ago.
http://cran.r-project.org/web/packages/limSolve/index.html
I guess, the lsei-function of thi
On 21-10-2012, at 13:37, Thomas Schu wrote:
> Dear Richard,
>
> It is funny. I have to perform the approach of sediment fingerprinting for
> my master thesis. Mr. Hasselman gave me the advice to take a closer look
> into the limSolve package a few days ago.
> http://cran.r-project.org/web/packa
I didn't get any responses to this question so just trying one more time ...
-Frank
Frank Harrell wrote
> The bild package appears to be an excellent package for serial binary
> responses. But it is for discrete time. I would like to specify a smooth
> function of time for the odds ratio connec
Dear All,
I am using a specific approach for my master thesis. In essence, a
supervised reclassification is used as an intermediate step to find chemical
parameters which are able to reclassify defined groups. These variables will
be used in a next step where location and scale estimators of the g
HI,
I am not sure whether this is what you want.
Mydata<-read.table(text="
idxy ix iy country col5
1 1 1 c1 x1
2 1 2 c1 x2
3 1 3 c1 x3
4 2 4 c1 x4
5 2 4 c2
Hi Marius,
I have tried debugging the qua.regressCOP2 function.
The error I'am getting is:
"Error in cop(u, v + delv, ...) : unused argument(s) (v + delv)".
Unable to decipher it.
And have mailed to william.asquith at ttu.edu>.
Thanks
indu
--
View this message in context:
http://r.789695.n4
Dear Berend,
Many thanks for taking your time to assist with this optimization problem.
I'll work on data this week and let you know how I get on.
Again, many thanks
Richard
--
View this message in context:
http://r.789695.n4.nabble.com/Optimization-in-R-similar-to-MS-Excel-Solver-tp4646759
yes, can you use the (plm) package for panel data estimation in general
but, i not know a package for dynamic panel data models
--
View this message in context:
http://r.789695.n4.nabble.com/R-Comparing-posterior-and-likelihood-estimates-for-proportions-off-topic-tp814022p4646910.html
Sent fro
Hello，I am a Chinese student, and I got the same problem.
I have learnd R for several weeks, I want to draw a PCA picture.
I have data in excel format like this:
ID d1 d2 d3 d4
A 1234
B 1234
C 1234
D 1234
E 1234
How to load these dat
You are right. I ran the code again and got the same error again. But ran it
for the second time, it didn't return an error and I got some values. It's
my first time doing this kind of coding and I'm still learning. I agree that
my code may look unorganized and wrong. I'll work on your comments a
Hello,
I'm doing a research on the impulse responses in VAR models and I'm having
troubles in interpretation of R results.
My question is what is the shock of impulse variable that is produced to
obtain the response? Is it one-standard-deviation positive shock? If it is
so how can I obtain the re
Apparently there is one or more concepts that I do not fully understand
from the descriptions of a function and the apply material. I have
been reading the mail from this forum and have learned much but, in this
case, what I have been reading here and from the manual isn't enough.
The followin
Hello,
Thanks for the dataset, Arun, I could test my function and it was still
wrong (apologies to the op).
Now I think I've got it.
is.border <- function(idx, DF){
i1 <- DF$ix %in% (DF$ix[idx] + c(-1, 1)) & DF$iy == DF$iy[idx]
i2 <- DF$iy %in% (DF$iy[idx] + c(-1, 1)) & DF$ix == DF$ix[
Greetings
I am trying to make a multiple panel plot from a tool (genoPlotR) that
is layered on grid. My code and session info are below. When I execute
the code, I get a single plot in the center of the graphics page. I
have tried various pop/push/down/seekViewport commands at the end of
I would like to use GLM to analysis a longitudinal data
any one know which library in R do so
thanks alot
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
Hello,
I am working on a simple non-parametric (Theil) regression function and and
am following Hollander and Wolfe 1999 text. I would like some help making
my function faster. I have compared with pre-packaged version from "MBLM",
which isnt very fast either, but it appears mine is faster with
Obviously sort() is not needed in the following line:
X <- median( sort( do.call(c, num) / do.call(c, dom) ) )
Brad Schneid wrote
> Hello,
>
> I am working on a simple non-parametric (Theil) regression function and
> and am following Hollander and Wolfe 1999 text. I would like some help
Thank you!!!
--
View this message in context:
http://r.789695.n4.nabble.com/MLE-of-negative-binomial-distribution-parameters-tp4646703p4646927.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://
On 21-10-2012, at 20:06, Brad Schneid wrote:
> Hello,
>
> I am working on a simple non-parametric (Theil) regression function and and
> am following Hollander and Wolfe 1999 text. I would like some help making
> my function faster. I have compared with pre-packaged version from "MBLM",
> which
Wow. Thank you greatly, that is amazing.
Thiel statistic ==> (Pedantic comment: it is Theil (swap the i and e)
Yes sir; I do that every time.
Dyslexia perhaps?
Thanks again.
Berend Hasselman wrote
> On 21-10-2012, at 20:06, Brad Schneid wrote:
>
>> Hello,
>>
>> I am working on a simple
My comments have nothing to do with speed of your code,
but with the correctness.
> > np.lm <-function(dat, X, Y, ...){
> > # Ch 9.2: Slope est. (X) for Thiel statistic
> > ...
> > num[[i]] <- dat[j.s[i],Y] - dat[i.s[i],Y]
> > dom[[i]] <- dat[j.s[i],X]
I understand, thank you. I think I wanted the output to look similar to
that from mblm for quick comparison; that is obviously a problem.
William Dunlap wrote
> My comments have nothing to do with speed of your code,
> but with the correctness.
>
>> > np.lm <-function(dat, X, Y, ...){
>> >
I plan to estimate the multi-factor model for Kalman Filter Mean Reverting,
Random Coefficient.
For example:
R(it)= Alpha(it)+ Beta(it)R(mt)+Gamma(it)(R(mt)^2)+delta(it)(R(mt)^3)+ V(it)
Note: (alphabar= Mean Alpha, Betabar= Mean Beta, Gamma= Mean Gamma,
Deltabar= Delta Mean)
KF Mean R
User has to type (input) x. After inputcode has to check X with
statement if
There is a statement IF.
If X=0 then Y=5/2 else =7;
How to code it please. I tried but my code does not work ;( I wanna see how
it looks like( thank u in advance(
--
View this message in context:
http://r.7896
Read the posting guide. We have no way to know what you are doing unless you
show the list readers with sample data and code.
If you are wondering about Vector Autoregressive models, I don't claim any
specific experience with them, but [1] says they are linear, and I do know that
linear models
Hello.
I am running 9 poisson regressions with 5 predictors each, using glm with
family=gaussian.
Gaussian distribution fits better than linear regression on fit indices,
and also for theoretical reasons (e.g. the dependent variables are counts,
and the distribution is highly positively skewed).
Hello,
You're right an _explicit_ for loop is not necessary in your case, but
note that the *apply functions are just for loops in disguise. They are
also the prefered R idiom.
In this case I've used ?replicate, a function of the *apply family. To
be more readable it uses a new function, calle
Hello,
First of all, you should _really_ try to read An Introduction to R, file
R-intro.pdf that comes with every installation of R. The answer to your
question is in chapter 9 Grouping, loops and conditional execution.
It's the very first example:
"9.2.1 Conditional execution: if statements
Hi,
With sapply():
fun3<-function(n){
paste(t(sapply(random.string(n),randomizeString)),collapse="")
}
DNA3<-replicate(20,fun(21))
DNA3
#[1] "cttgcaatgtttaatcgttggagcagt"
#[2] "tgcgttatcgcgcagagggccgtgagggat"
#[3] "ctggccgcatatg
Others have said what system time is, time spent in the kernel of the operating
system doing things on behalf of your program. Also, switching between user
and system mode can be time consuming, so the number of system calls can
be important, even if each one takes little time. I don't know how
HI,
Try this:
fun1<-function(X){
Y<-ifelse(X==0,5/2,7)
return(Y)}
fun1(5)
#[1] 7
fun1(0)
#[1] 2.5
fun1(2)
#[1] 7
A.K.
- Original Message -
From: Rlotus
To: r-help@r-project.org
Cc:
Sent: Sunday, October 21, 2012 3:03 PM
Subject: [R] Help me please to code
User has to type (
Hallo,
I would like to run a model (differential equations)
with a whole bunch of parameter combinations.
I think the best way of doing this is to put the model
function into a loop and set a matrix with the parameters:
Each row will represent the parameters for one loop.
I know how to compile
Hi,
I am trying to assign values to records based conditionally on other records
within a dataframe. For example, in the following dataframe the "NA" value
in dat$age would be replaced if the age status for that individual and
specific year can be found in another record. Ideally this would then a
I am running 9 negative binomial regressions with count data.
The nine models use 9 different dependent variables - items of a clinical
screening instrument - and use the same set of 5 predictors. Goal is to
find out whether these predictors have differential effects on the items.
Due to various
Hi,
Try this:
new1<-as.data.frame(expand.grid(a=a,b=b,c=c))
mat1<-as.matrix(new1[order(new1$a),])
row.names(mat1)<-1:nrow(mat1)
mat1
# a b c
#1 1 5 1
#2 1 5 6
#3 2 5 1
#4 2 5 6
A.K.
- Original Message -
From: Simeon Lisovski
To: r-help@r-project.org
Cc:
Sent: Sunday, October 21, 2
Stephanie,
I'm working on an option for the plot method for Predict that will allow you
to do this. Note that this approach will not result in very readable
predictor category labels when they are long character strings. If you are
using linux I can get you a new version of rms with this option
Hi all,
Can some one tell me the difference between the following two formulas?
1. epiG.rf <-randomForest(gamma~.,data=data, na.action = na.fail,ntree =
300,xtest = NULL, ytest = NULL,replace = T, proximity =F)
2.epiG.rf <-randomForest(gamma~.,data=data, na.action = na.fail,ntree =
300,xtest = NU
Sorry, the previous was not right post.
I want to know the difference between following to methods of random forest.
1. epiG.rf <-randomForest(gamma~.,data=data, na.action = na.fail,ntree =
300,xtest = NULL, ytest = NULL,replace = T, proximity =F)
2. epiG.rf <-randomForest(x = data,,y = data$gam
https://stat.ethz.ch/pipermail/r-help/2012-October/326837.html
> summary: spatial data to be input to a regional-scale environmental
> model must (1) be converted to netCDF and then (2) "regridded"
> (cropped, projected, increased resolution). In a public git
> repository
> https://github.com/Tom
Hello,
First of all, two notes on the way you create your data.frame
1. Use age <- c("Adult", NA, ...etc...) _without_ quotes around NA. If
you use quotes it will seen as a character, not as the value NA.
2. Do not cbind and then coerce to data.frame, use data.frame only.
So the data and an
Hello all,
I'm trying to group several consecutives rows (and assigning them the same
value) while leaving some of the rows empty (when a certain condition is not
fulfilled).
My data are locations (xy coordinates), the date/time at which they were
measured, and the time span between measures. Some
I'm studying alone the R language for data preparation. I found a course at
MIT for data preparation that uses python but I'm using R to learning. The
first exercise is the preparation of data from a database that shows the
contributions made to candidates for U.S. president. The database is
descri
Hello,
I am looking at a two-way ANOVA dataset, and would like to count the rows in
the dataframe with the same level of the first factor ("Gender") and the
second factor ("Dosage"). In other words, I am interested in the number of
observations per each "cell" in a (not necessarily balanced) two-w
I have this code:
x=seq(80-3*15,80+3*15,length=200);
y=dnorm(x,80,15)
plot(x,y,type="l",lwd=2,col="red")
x=seq(80-3*15,70,length=100)
y=dnorm(x,80,15)
polygon(c(0,x,70),c(0,y,0),col="gray")
text(70,0.005,"70")
I'd like to draw the number "70" in red underneath a vertical arrow pointing
to the pos
On Sun, 21 Oct 2012, Eiko Fried wrote:
I am running 9 negative binomial regressions with count data.
The nine models use 9 different dependent variables - items of a clinical
screening instrument - and use the same set of 5 predictors. Goal is to
find out whether these predictors have different
50 matches
Mail list logo
