f Of Pedro Gavronski.
Sent: Friday, February 23, 2024 5:00 AM
To: Rui Barradas
Cc: r-help@r-project.org; r-help-requ...@r-project.org
Subject: Re: [R] help - Package: stats - function ar.ols
[External Email]
Hello,
Thanks for the reply Rui and for pointing out that I forgot to attach my code.
P
Às 16:34 de 22/02/2024, Pedro Gavronski. escreveu:
Hello,
My name is Pedro and it is nice to meet you all. I am having trouble
understanding a message that I receive when use function ar.ols from
package stats, it says that "Warning message:
In ar.ols(x = dtb[2:6966, ], demean = FALSE,
Thank you Ege, I appreciate your response.
I have move this to r-sig-geo.
WHP
Proprietary
-Original Message-
From: Ege Rubak
Sent: Monday, May 18, 2020 6:41 AM
To: Poling, William ; r-help@r-project.org
Subject: [EXTERNAL] Re: [R] Help with spTransform() function and final plot
You are more likely to get help with specific problems related to
spTransform() on the dedicated list r-sig-geo.
You should provide a minimal reproducible example. Your code refers to
e.g. the object `tmp1b`, which we don't have. I think the spTransform()
part will work with this correction:
xy
Hello again.
I also found this discussion on non-finite transformation error, however, I am
not sure what to look for in the output after I apply my data?
https://stackoverflow.com/questions/14880294/non-finite-transformation-detected-in-sptransform-in-rgdal-r-package
str(sample)
GPS.Points
Hello, I have found an additional problem.
I should be getting 3 columns back in the xy@data at some point.
$ ID : int 1 2 3 4 5 6 7 8 9 10 ...
# $ Clust : int 1 1 1 1 1 1 1 1 1 1 ... This is always - 1
# $ Clust.1 : int 1 1 1 1 1 1 1 1
#RStudio Version Version 1.2.1335
sessionInfo()
# R version 4.0.0 Patched (2020-05-03 r78349)
#Platform: x86_64-w64-mingw32/x64 (64-bit)
#Running under: Windows 10 x64 (build 17763)
Hello. I am running my data through a routine I found that finds clusters of
data points based on distance rule.
Good morning Steve. Terrific, so kind of you to follow-up.
I will add that to my ever growing R bag of tips and tricks.
Cheers.
WHP
William H. Poling, Ph.D., MPH | Manager, Revenue Development
Data Intelligence & Analytics
Zelis Healthcare
-Original Message-
From: S Ellison
Sent:
> tb2a$TID2 <- gsub(tb2a$TID, pattern="-[0-0]{0,7}", replacement = "")
Just to add something on why this didn't work ...
It looks like you were trying to match a hyphen followed by a number up to
seven digits. by mistake(?) you gave the digit range as [0-0] so it would
repmatch a hyphen
Yep, thank you Jeff, consequence of the first url I landed on asking how to do
it and rushing off.
All set now.
Appreciate your help.
WHP
From: Jeff Newmiller
Sent: Friday, March 15, 2019 4:00 PM
To: r-help@r-project.org; Bill Poling ; r-help
(r-help@r-project.org)
Subject: Re: [R] Help
)
Subject: Re: [R] Help with gsub function
If you want to remove just the hyphen, why not do
sub("-", "", tb2a$TID)
sub("-", "", "73-017323")
[1] "73017323"
Am I missing something?
Peter
On Fri, Mar 15, 2019 at 12:46 PM Bill Poli
Your pattern seems ... way overboard? Why not
gsub("-", "", tb2a$TID)
On March 15, 2019 12:45:27 PM PDT, Bill Poling wrote:
>Good afternoon.
>
>sessionInfo()
>#R version 3.5.3 (2019-03-11)
>#Platform: x86_64-w64-mingw32/x64 (64-bit)
>#Running under: Windows >= 8 x64 (build 9200)
>
>I am using
If you want to remove just the hyphen, why not do
sub("-", "", tb2a$TID)
sub("-", "", "73-017323")
[1] "73017323"
Am I missing something?
Peter
On Fri, Mar 15, 2019 at 12:46 PM Bill Poling wrote:
>
> Good afternoon.
>
> sessionInfo()
> #R version 3.5.3 (2019-03-11)
> #Platform:
On Fri, 15 Mar 2019 19:45:27 +
Bill Poling wrote:
Hello Bill,
> tb2a$TID2 <- gsub(tb2a$TID, pattern="-[0-0]{0,7}", replacement = "")
Is the pattern supposed to mean something besides the "-" you want to
remove? For the problem you describe, pattern="-" should be enough. It
should locate
Good afternoon.
sessionInfo()
#R version 3.5.3 (2019-03-11)
#Platform: x86_64-w64-mingw32/x64 (64-bit)
#Running under: Windows >= 8 x64 (build 9200)
I am using gsub function to remove a hyphen in a 9 character column of values
in order to convert it to integer.
Works fine except where the
Hi,
Locally I am using windows:
R version 3.5.1 (2018-07-02)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1
Locally on my laptop using RStudio I normally set my working directory and read
in my csv files.
setwd("C:/WHP/Appeals")
.@gmail.com]
> Sent: Thursday, December 14, 2017 10:29 AM
> To: DIGHE, NILESH [AG/2362] <nilesh.di...@monsanto.com>
> Cc: r-help <r-help@r-project.org>
> Subject: Re: [R] help with recursive function
>
> If you are trying to understand why the "stopifnot" condit
.
Nilesh
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: Thursday, December 14, 2017 11:26 AM
To: DIGHE, NILESH [AG/2362] <nilesh.di...@monsanto.com>
Cc: Eric Berger <ericjber...@gmail.com>; r-help <r-help@r-project.org>
Subject: Re: [R] help with recursive function
Sent: Thursday, December 14, 2017 10:29 AM
> To: DIGHE, NILESH [AG/2362] <nilesh.di...@monsanto.com>
> Cc: r-help <r-help@r-project.org>
> Subject: Re: [R] help with recursive function
>
> If you are trying to understand why the "stopifnot" condition is met yo
] <nilesh.di...@monsanto.com>
Cc: r-help <r-help@r-project.org>
Subject: Re: [R] help with recursive function
If you are trying to understand why the "stopifnot" condition is met you can
replace it by something like:
if ( any(dat2$norm_sd >= 1) )
browser()
This will put y
ot;, "treatment"),
>>
>> "\\.") %>% spread(treatment, value) %>% mutate(outlier = NA)
>>
>> stopifnot(!(any(data1$norm_sd >= 1)))
>>
>> if (!(any(data1$norm_sd >= 1))) {
>>
>> df1 <
gt; return(df1)
>
> }
>
>else {
>
> df2 <- recursive_funlp()
>
> return(df2)
>
> }
>
> }
>
> df3 <- recursive_funlp(dataset = dat1, func = funlp2)
>
> df3
>
> }
>
&
recursive_funlp(dataset = dat1, func = funlp2)
df3
}
From: DIGHE, NILESH [AG/2362]
Sent: Thursday, December 14, 2017 9:01 AM
To: 'Eric Berger' <ericjber...@gmail.com>
Cc: r-help <r-help@r-project.org>
Subject: RE: [R] help with recursive function
Eric: Thanks for taking
ic Berger [mailto:ericjber...@gmail.com]
Sent: Thursday, December 14, 2017 8:17 AM
To: DIGHE, NILESH [AG/2362] <nilesh.di...@monsanto.com>
Cc: r-help <r-help@r-project.org>
Subject: Re: [R] help with recursive function
My own typo ... whoops ...
!( any(dat2$norm_sd >= 1 ))
On Thu,
My own typo ... whoops ...
!( any(dat2$norm_sd >= 1 ))
On Thu, Dec 14, 2017 at 3:43 PM, Eric Berger wrote:
> You seem to have a typo at this expression (and some others like it)
>
> Namely, you write
>
> any(!dat2$norm_sd) >= 1
>
> when you possibly meant to write
>
>
You seem to have a typo at this expression (and some others like it)
Namely, you write
any(!dat2$norm_sd) >= 1
when you possibly meant to write
!( any(dat2$norm_sd) >= 1 )
i.e. I think your ! seems to be in the wrong place.
HTH,
Eric
On Thu, Dec 14, 2017 at 3:26 PM, DIGHE, NILESH [AG/2362]
Hi, I need some help with running a recursive function. I like to run funlp2
recursively.
When I try to run recursive function in another function named "calclp" I get
this "Error: any(!dat2$norm_sd) >= 1 is not TRUE".
I have never built a recursive function before so having trouble executing
I haven't tried your code, but I've seen far too many attempts like this where
there is no simple call to the function with starting parameters to see if a
sensible answer is returned. If you get NaN or Inf or ... that isn't sensible,
then you know it is not a good idea to try further.
Beyond
Hi all,
Many thank in advance for helping me. I tried to fit Expectation Maximization
algorithm for mixture data. I must used one of numerical method to maximize my
function.
I built my code but I do not know how to make the optim function run over a
different value of the parameters.
Please note that Kirsten is cross-posting to stats.stackexchange.com
creating extra work for everyone.
--
Frank E Harrell Jr Professor and Chairman School of Medicine
Department of *Biostatistics*
Please bear with me, I am very new to R.
My question is regarding the use of the improveProb function in the Hmisc
package. I have two logistic models, the only difference being that the
second model contains my novel marker of interest. I am trying to calculate
NRI and IDI to compare models.
I
On Oct 1, 2015, at 2:26 AM, kirsada wrote:
> Please bear with me, I am very new to R.
>
> My question is regarding the use of the improveProb function in the Hmisc
> package. I have two logistic models, the only difference being that the
> second model contains my novel marker of interest. I am
Hi,
Just do the following:
tran-c(7.2)
tgrid-c(7.1,7.4,7.3,7.1,7.3)
tgrid-tgrid-tran
tgrid
[1] -0.1 0.2 0.1 -0.1 0.1
abs(tgrid[tgrid0.1])
[1] 0.2
Andrés
El 12/06/2015, a las 11:01, Jeff Newmiller jdnew...@dcn.davis.ca.us
escribió:
FAQ 7.31
FAQ 7.31
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#..
Please read R FAQ 7.31.
On Fri, Jun 12, 2015 at 11:39 AM, Nelson, Gary (MISC)
gary.nel...@state.ma.us wrote:
I have come across some odd behavior (to me) using the abs() function that I
wonder if anyone can explain.
Using R version 3.2.0, I created a vector of absolute values using the
I have come across some odd behavior (to me) using the abs() function that I
wonder if anyone can explain.
Using R version 3.2.0, I created a vector of absolute values using the
following code:
tran-c(7.2)
tgrid-c(7.1,7.4,7.3,7.1,7.3)
dgrid-abs(tgrid-tran)
dgrid
[1] 0.1 0.2 0.1 0.1 0.1
Your problem is that PatientID, FatherID, MotherID are factors. The authors of kinship2
(myself and Jason) simply never thought of someone doing this. Yes, that is an oversight.
We will correct it by adding some more checks and balances. For now, turn your id
variables into character or
Jean: Thanks a lot!! The changes you made to the code gave me what I needed.
I truly appreciate your time in correcting the code.
Nilesh
From: Adams, Jean [mailto:jvad...@usgs.gov]
Sent: Tuesday, May 12, 2015 2:14 PM
To: DIGHE, NILESH [AG/2362]
Cc: r-help@r-project.org
Subject: Re: [R] help
Hi,
I have an anonymous function called function(x) that will run anova, run
HSD.test on the model, and then sort the results. I am passing this anonymous
function to the by function to get results by Isopair factor which is my
index variable. Since I want to run the anova on multiple
Nilesh,
I found a couple errors in your code. First, in your by() statement you
have a function to operate on the selected subset of data, which you refer
to as
x
but then, in your aov statement you refer to
data_set
not
x
Second, your funC() statement is a function of
Dear R-help,
I am interested in plotting some pedigrees and came across the kinship2
package. What follows is an example of the pedigrees I am working with.
Now, when running
## check package availability
if(!require(kinship2)) install.packages('kinship2')
require(kinship2)
## data to plot
d
Hi,
I have used this function before successfully. I could help you if you could
provide your code.
Thanks Regards,Arnab
From: hms Dreams cute_loo...@hotmail.com
To: r-help@r-project.org r-help@r-project.org
Sent: Saturday, February 14, 2015 6:27 AM
Subject: [R] help please
Kingston ON Canada
-Original Message-
From: cute_loo...@hotmail.com
Sent: Sat, 14 Feb 2015 15:27:58 +0300
To: r-help@r-project.org
Subject: [R] help please metro_hastings function
Hi :)anybody can help me please I'm trying to use Metro_Hastings (
MHadaptive package)the proplem
Hi :)anybody can help me please I'm trying to use Metro_Hastings (
MHadaptive package)the proplem is: How can I know the covariance matrix(
prop_sigma ) to enter it in Metro_Hastings:
mcmc_r=Metro_Hastings(li_func=baysianlog, pars=c(1,1,1), prop_sigma
that there is a
function named vcf somewhere in the parent environment and you
can't subset a function.
Jim
__
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
Hi all,
I need help with a function. I'm trying to write a function to apply
to varying number of columns in a lot of files - hence the function...
but I'm getting stuck. Here it is:
gt- function(x) {
alleles - sapply(x, function(.) strsplit(as.character(.), /))
gt - apply(x,
Just an update to this:
gtal - function(d) {
alleles - sapply(d, function(.) strsplit(as.character(.), /))
gt - unlist(lapply(alleles, function(x)
ifelse(identical(x[[1]], vcf[,3]) identical(x[[2]], vcf[,3]), 'RR',
ifelse(identical(x[[1]], vcf[,4])
Hi,
This is the input data frame:
###
df.1 = read.table(header=T,text=
id gender WMC_alcohol WMC_caffeine WMC_no.drug RT_alcohol RT_caffeine
RT_no.drug
1 1 female 3.7 3.7 3.9 488 236 371
2 2 female 6.4 7.3 7.9 607 376 349
3 3 female 4.6 7.4 7.3 643 226
I see your desired output has rather fewer data than the input data frame.
Instead of making us pore over a bunch of numbers, can you explain exactly
what filtering you wish to do to get the specific subset
of {male/female} {alcohol/caffeine} you're trying to get?
BHM wrote
Hi,
This is
On Nov 29, 2013, at 9:42 AM, Burhan ul haq wrote:
Hi,
This is the input data frame:
###
df.1 = read.table(header=T,text=
id gender WMC_alcohol WMC_caffeine WMC_no.drug RT_alcohol RT_caffeine
RT_no.drug
1 1 female 3.7 3.7 3.9 488 236 371
2 2
Hi,
First, a big thanks to all those who replied.
I am including all the replies in one email for easier reference later:
# Input from David
#
reshape(df.1, idvar=1:2, sep=_, direction=long,
varying=names(df.1)[3:8])
#
# Input from Dennis
#
dfr1 - reshape(df.1, idvar = c(id, gender),
j...@bitwrit.com.au
Cc : anoumou teko_maur...@yahoo.fr; r-help@r-project.org
Envoyé le : Mercredi 5 décembre 2012 16h26
Objet : Re: [R] Help for a function
Hello,
Also, t1 and min(xt) do not vary inside the loop so if it enters the
loop it never exits.
And
res1[j] -(a*h)
res2 -sum
Hi Everyone,
I am working with the R function dataEllipse. I plot the 95% confidence
ellipses for several different samples in the same plot and I color-code
the ellipse of each sample, but I do not know how to specify a different
line pattern for each ellipse. I can only modify the pattern for
Dear Jana,
The lty argument to dataEllipse() (in the car package) isn't vectorized. It
could be, and I'll add that as a feature request. Actually, lty isn't an
explicit argument to dataEllipse(); it's simply passed through to the lines()
function, which draws the ellipses.
You should be able
On 4/25/2013 8:00 PM, Jana Makedonska wrote:
Hi Everyone,
I am working with the R function dataEllipse. I plot the 95% confidence
ellipses for several different samples in the same plot and I color-code
the ellipse of each sample, but I do not know how to specify a different
line pattern for
Dear all,
I'm trying to merge 2 dataframes, but I'm not being entirely successful and
I can't understand why.
Dataframe x1
State_prov Shape_name bob2009 bob 2010 bob2011
Nova ScotiaAnnapolis 0 0 1
Nova ScotiaAntigonish0
Hello,
The following seems to do the trick.
x1 -
structure(list(State_prov = c(Nova Scotia, Nova Scotia, Nova Scotia
), Shape_name = c(Annapolis, Antigonish, Gly), bob2009 = c(0L,
0L, NA), bob2010 = c(0L, 0L, NA), bob2011 = c(1L, 0L, NA)), .Names =
c(State_prov,
Shape_name, bob2009, bob2010,
Hello, Thank you for your help. However the dataframes I gave you were only
examples, the actual dataframes are very big. Does this mean I have to
write every range of data for each variable??
On Fri, Apr 26, 2013 at 2:25 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
The following
-project.org
Cc:
Sent: Friday, April 26, 2013 1:10 PM
Subject: [R] Help with merge function
Dear all,
I'm trying to merge 2 dataframes, but I'm not being entirely successful and
I can't understand why.
Dataframe x1
State_prov Shape_name bob2009 bob 2010 bob2011
Nova Scotia Annapolis
Hello,
I don't understand the question, what range? I've just changed the 'all'
argument to 'all.y', without doing anything special to the variables.
Can you explain what you mean?
Rui Barradas
Em 26-04-2013 19:30, Catarina Ferreira escreveu:
Hello, Thank you for your help. However the
#3 Nova Scotia Gly 0 0 NA 2 1
A.K.
From: Catarina Ferreira catferre...@gmail.com
To: arun smartpink...@yahoo.com
Sent: Friday, April 26, 2013 2:23 PM
Subject: Re: [R] Help with merge function
Hello,
I didn't realize
...@yahoo.com
Sent: Friday, April 26, 2013 4:20 PM
Subject: Re: [R] Help with merge function
here they are. As you see the NS_update is data for only 1 province and I want
it to add this data to the bigger file (data), merging the common columns and
adding the new columns. But what it is doing
Hi,
I am fairly new to R and have encountered an issue with the lmRob function that
I have been unable to resolve. I am trying to run a robust regression using the
lmRob function which runs successfully, but the results are rather strange. I'm
not sure it's important, but my model has 3
hello all
I am writing a quite simple script to study dental wear patterns in humans
and I wrote this function
sqrt(var(Y1)+var(Y2))^2-4(var(Y1)*(var(Y2)-cov(Y1,Y2)^2)) but appear this
error message
Error: attempt to apply non-function
alternatively I wrote this
On 17-03-2013, at 16:47, Miguel Eduardo Delgado Burbano
mdelgadoburb...@gmail.com wrote:
hello all
I am writing a quite simple script to study dental wear patterns in humans
and I wrote this function
sqrt(var(Y1)+var(Y2))^2-4(var(Y1)*(var(Y2)-cov(Y1,Y2)^2)) but appear this
error
Hi,
Y1- 1:4
Y2- 5:8
sqrt(var(Y1)+var(Y2)^2)-4*((var(Y1)*(var(Y2)-cov(Y1,Y2)^2)))
#[1] 9.515593
A.K.
- Original Message -
From: Miguel Eduardo Delgado Burbano mdelgadoburb...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Sunday, March 17, 2013 11:47 AM
Subject: [R] help with simple
Hello,
You are missing a '*' in your first try.
(As for the second, in R parenthesis are round, not [])
sqrt(var(Y1) + var(Y2))^2 - 4*(var(Y1)*(var(Y2) - cov(Y1, Y2)^2))
This written as a function becomes
fun - function(Y1, Y2)
sqrt(var(Y1) + var(Y2))^2 - 4*(var(Y1)*(var(Y2) - cov(Y1,
Thanks for that. Eliano
2013/3/7 David Winsemius [via R] ml-node+s789695n4660565...@n4.nabble.com
On Mar 6, 2013, at 3:44 PM, Eliano wrote:
Thanks. Btw are you able to help with my issue? Thanks, Eliano
I'm sorry, I was too busy answering the question from 'Eliano' over
on
Hi, can I understand why this message was rejected ?
Thanks,
Eliano
Sent from my iPhone
On 6 Mar 2013, at 19:18, Eliano eliano.m.marq...@gmail.com wrote:
Hi everyone,
I am writing some code to generate a function. I am passing that code to a
dataset which i'm importing in R, e.g.
On Mar 6, 2013, at 11:25 AM, Eliano Marques wrote:
Hi, can I understand why this message was rejected ?
Thanks,
Eliano
First hit on a Markmail search:
http://markmail.org/message/5xog3ayx4amprsdx?q=list:org%2Er-project%2Er-help+nabble+rejected
--
David.
Sent from my iPhone
On 6 Mar
Thanks. Btw are you able to help with my issue? Thanks, Eliano
Sent from my iPhone
On 6 Mar 2013, at 23:41, David Winsemius [via R]
ml-node+s789695n4660547...@n4.nabble.com wrote:
On Mar 6, 2013, at 11:25 AM, Eliano Marques wrote:
Hi, can I understand why this message was rejected ?
On Mar 6, 2013, at 3:44 PM, Eliano wrote:
Thanks. Btw are you able to help with my issue? Thanks, Eliano
I'm sorry, I was too busy answering the question from 'Eliano' over on
StackOverflow. I didn't have time to address this one.
(Please do note that cross-posting questions to Rhelp is
Hello,
Also, t1 and min(xt) do not vary inside the loop so if it enters the
loop it never exits.
And
res1[j] -(a*h)
res2 -sum( res1[j])
is equivalent to
res2 - a*h
so the inner-most loop is not needed at all.
Hope this helps,
Rui Barradas
Em 05-12-2012 04:20, Jim Lemon
De : Rui Barradas ruipbarra...@sapo.pt
À : Jim Lemon j...@bitwrit.com.au
Envoyé le : Mercredi 5 décembre 2012 16h26
Objet : Re: [R] Help for a function
Hello,
Also, t1 and min(xt) do not vary inside the loop so if it enters the loop it
never exits.
And
res1[j] -(a*h
-project.org
Envoyé le : Mercredi 5 décembre 2012 16h45
Objet : Re: [R] Help for a function
Thanks you all
Maurice TEKO
Biostatisticien,Doctorant.
Université de Liège
Département des Sciences et Gestion de l'environnement
185 avenue de Longwy
6700 Arlon (Belgique)
:anoumou.tekoahate
Hello all,
I need a help.
I am modeling a disease and a create a R function like that:
Lambda-function (x,date1,r,h,a){
ndate1 - as.Date(date1, %d/%m/%Y)
t1 - as.numeric(ndate1)
x[order(x$i),]
t -x[,t]
i -x[,i]
CONTAGIEUX -x[,CONTAGIEUX]
while ( t1 min(t) ){
for (i in 1:length(i)
What are you expecting?
What do you get?
What is the problem?
J
On 4 December 2012 06:01, anoumou teko_maur...@yahoo.fr wrote:
Hello all,
I need a help.
I am modeling a disease and a create a R function like that:
Lambda-function (x,date1,r,h,a){
ndate1 - as.Date(date1, %d/%m/%Y)
t1
On 12/05/2012 01:01 AM, anoumou wrote:
Hello all,
I need a help.
I am modeling a disease and a create a R function like that:
...
But i do not get the results,i try by all means but i d'ont understant the
problem.
Hi anoumou,
Your function provides almost no indication of what two of its five
Hello,
Why mail a question just to me? Post to the list and the odds of getting
more answers (and better) are bigger.
As for your question, the problem is in the call to glm, you don't need
the prefix 'train$' in the formula, the argument 'data' solves that and
when predicting R will look for
That did it!
Thanks so much as always.
I emailed the question to you because I think you are an R expert based
on all the suggestions, feedback and codes I have received from you in
the past.
Yes, I do look for answers in the open
forum but when it comes to a question for which the
Hi,
I carried out an experiment, using a repeated measures design, in which I
broadcast 6 playbacks to 15 nests. I am trying to run a posthoc pairwise
comparison on a binomial GLMM with missing data points to determine which
playback treatments differ.
The response variable (Response) is
tmuman mumantariq at gmail.com writes:
Hi, am very new to R and I've written an optim function, but can't
get it to work
least.squares.fitter-function(start.params,gr,
low.constraints,high.constraints,model.one.stepper,data,scale,ploton=F)
{
result-optim(par=start.params,
Hi, am very new to R and I've written an optim function, but can't get it to
work
least.squares.fitter-function(start.params,gr,low.constraints,high.constraints,model.one.stepper,data,scale,ploton=F)
{
I encountered this problem when I was learning nonlinear regression with R :
there is this function plotfit in package nlstools,which displays
a superimposed plot of the dependent variable versus one the
independent variables together with the fitted model.,but when I
execute the example given in
hello,
I have a question regarding the function WebCorpus..When I am using this
function its showing error: couldnt find the function...I have downloaded
the tm package...Can you help me in this regard? Also I want to know how we
can data/information about a product or topic from google,facebook
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 Steve E.
Sent: Friday, June 08, 2012 2:33 PM
To: r-help@r-project.org
Subject: [R] help with rle function on paired data
On 09.06.2012 03:57, Ashley Stasko wrote:
Hello,
I'm using R code that includes a residual permutation that was written as a
supplement to the paper:
Turner et al. 2010. A general hypothesis-testing framework for stable isotopes
ratios in ecological studies. Ecology 91:2227-2233.
The
Dear Ashley,
I am one of the co-authors for these scripts. I just downloaded the example
files from the ESA archives and ran in naively on some data and it worked
fine. My guess is that the problem lies in a portion of the script that you
did not provide. If I had to be more certain with my
Dear R Community - I hope you might be able to provide some guidance
regarding the use of the rle function. I have a set of time-series data
where a measured value is recorded every 30 seconds after the start of an
experiment. Many of the measured values repeat and I am interested only in
the
On 2012-06-08 14:33, Steve E. wrote:
Dear R Community - I hope you might be able to provide some guidance
regarding the use of the rle function. I have a set of time-series data
where a measured value is recorded every 30 seconds after the start of an
experiment. Many of the measured values
Hello,
I'm using R code that includes a residual permutation that was written as a
supplement to the paper:
Turner et al. 2010. A general hypothesis-testing framework for stable isotopes
ratios in ecological studies. Ecology 91:2227-2233.
The supplemental code is available at:
Hi,
I am stuck on something for a couple days, I am almost about to give up.
This looks simple, but I can't figure out. I hope I can get some help here.
I am trying to do some symbolic and numerical derivations. Let me explain
the problem. Let's say, I have a matrix as follows:
load -
I missed a couple line of codes in the previous e-mail. Here is whole code
again:
load - matrix(c(3,0,1,4,1,3),nrow=3,ncol=2,byrow=TRUE)
l - matrix(nrow=nrow(load),ncol=ncol(load))
for(i in 1:nrow(load)) {
for(j in 1:ncol(load)) { l[i,j]=paste(l,i,j,sep=)}}
for(i in 1:nrow(load)){
for(j in
-boun...@r-project.org] On
Behalf
Of Cengiz Zopluoglu
Sent: Friday, May 18, 2012 11:33 AM
To: r-help@r-project.org
Subject: Re: [R] Help for numericDeriv function
I missed a couple line of codes in the previous e-mail. Here is whole code
again:
load - matrix(c(3,0,1,4,1,3),nrow=3,ncol=2
Something weird must be going on in your s641_social object. Can you
just simply check that the vertex names look OK with
'V(s641_social)$name'?
If they look good, then can you send me the s641_social object in
private? (Or part of it, assuming a part is enough to reproduce the
problem.)
Best,
Thank for your response. It is oddly working now.
Thanks again,
Brenda
On 5/14/12 10:36 AM, Gábor Csárdi-2 [via R]
ml-node+s789695n4629973...@n4.nabble.com wrote:
Something weird must be going on in your s641_social object. Can you just
simply check that the vertex names look OK with
I am using the code below to output some network measures:
central_social - data.frame(V(s641_social)$name, indegree_social,
outdegree_social, incloseness_social, outcloseness_social,
betweenness_social, eigen_social)
and I get the following error:
Error in Re(z) : non-numeric argument to
You don't provide a reproducible example, or even str(), but I'd guess you need
to match ^15 instead of just 15.
Sarah
On Apr 5, 2012, at 10:38 PM, ieatnapalm era...@tulane.edu wrote:
Hey, sorry if this has been addressed before, but I'm really new to R and
having trouble with the gsub
Hey, sorry if this has been addressed before, but I'm really new to R and
having trouble with the gsub function. I need a way to make this function
exclude certain values from being substituted:
ie my data looks something like (15:.0234,10:.0157) and I'm trying to
replace the leading 15 with
On Apr 5, 2012, at 10:38 PM, ieatnapalm wrote:
Hey, sorry if this has been addressed before, but I'm really new to
R and
having trouble with the gsub function. I need a way to make this
function
exclude certain values from being substituted:
ie my data looks something like
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