That is an intrinsic part of nonlinear optimization. Choose your starting point
wisely.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.
On Oct 20, 2012, at 8:01 PM, stats12 wrote:
Hi,
Thank you for your comment. I worked on the code again and was able to make
it work.
Does that mean you know what value is correct for certain cases? Is there an
overall strategy that is guiding this effort? I wrote earlier:
DW Your
On 21.10.2012 02:47 (UTC+2), J. Maxwell wrote:
R Compiliing or installation failure on FreeBSD
9.0-RELEASE FreeBSD ; I386 box.
configuring with the default settings, no 'options' =
./configure
Is there any reason that you are not using the FreeBSD ports system, in
this case
On 21-10-2012, at 05:01, stats12 wrote:
Hi,
Thank you for your comment. I worked on the code again and was able to make
it work. The only problem I am having right now is that nlm depends on the
initial value.
When the initial value is 1, I get the following estimates
0.1230414
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
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
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
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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
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.
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
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
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
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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
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yes, can you use the (plm) package for panel data estimation in general
but, i not know a package for dynamic panel data models
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http://r.789695.n4.nabble.com/R-Comparing-posterior-and-likelihood-estimates-for-proportions-off-topic-tp814022p4646910.html
Sent
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
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
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
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
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[idx]
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]]
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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 N
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!!!
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Sent from the R help mailing list archive at Nabble.com.
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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 isnt
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, ...){
# Ch
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
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(
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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,
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]
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 yerl...@hotmail.com
To: r-help@r-project.org
Cc:
Sent: Sunday, October 21, 2012 3:03 PM
Subject: [R] Help me please to code
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$age would be replaced if the age status for that individual and
specific year can be found in another record. Ideally this would then
I am running 9 negative binomial regressions with count data.
The nine models use 9 different dependent variables - items of a clinical
screening instrument - and use the same set of 5 predictors. Goal is to
find out whether these predictors have differential effects on the items.
Due to various
Hi,
Try this:
new1-as.data.frame(expand.grid(a=a,b=b,c=c))
mat1-as.matrix(new1[order(new1$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 simeon.lisov...@me.com
To: r-help@r-project.org
Cc:
Sent:
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 =
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 =
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
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
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