HI,
Tried it on 1e5 row dataset:
l1<- letters[1:10]
s1<-sapply(seq_along(l1),function(i) paste(rep(l1[i],3),collapse=""))
set.seed(24)
x1<-data.frame(x=paste(paste0(sample(s1,1e5,replace=TRUE),sample(1:15,1e5,replace=TRUE)),paste0(sample(s1,1e5,replace=TRUE),sample(1:15,1e5,replace=TRUE)),paste0(s
Hello,
Below is a reproducible example to generate the output by using Dr. Young's R
code on the above subject . As commented below, the issue is that part of
the code (regtol.int and plottol) does not seem to work.
I would appreciate receiving your advice toward resolving the issue.
Thanks
I still argue for na.rm=FALSE, but that is cute, also substantially faster
f1 <- function(x1, x2, x3) do.call(paste0, list(x1, x2, x3))
f2 <- function(x1, x2, x3) pmax(3*x3, 2*x2, es, 0, na.rm=FALSE)
f3 <- function(x1, x2, x3) Reduce(`+`, list(x1, x2, x3))
f4 <- function(x1, x2, x3) rowSums(cbind(
HI,
May be this helps:
res<-data.frame(x=x,read.table(text=gsub("[A-Za-z]","",x[,1]),sep="_",header=FALSE),stringsAsFactors=FALSE)
res
# x V1 V2 V3
#1 aaa1_bbb1_ccc3 1 1 3
#2 aaa2_bbb3_ccc2 2 3 2
#3 aaa3_bbb2_ccc1 3 2 1
A.K.
- Original Message -
From: Dimitri Liakho
I would do this to get the highest non-missing level:
x <- pmax(3*cg, 2*hs, es, 0, na.rm=TRUE)
rock chalk...
-nfultz
On Fri, Jun 07, 2013 at 06:24:50PM -0700, Joshua Wiley wrote:
> Hi Paul,
>
> Unless you have truly offended the data generating oracle*, the
> pattern: NA, 1, NA, should be a da
Hello!
I have a column in my data frame that I have to split: I have to distill
the numbers from the text. Below is my example and my solution.
x<-data.frame(x=c("aaa1_bbb1_ccc3","aaa2_bbb3_ccc2","aaa3_bbb2_ccc1"))
x
library(stringr)
out<-as.data.frame(str_split_fixed(x$x,"aaa",2))
out2<-as.data.
Hi Paul,
Unless you have truly offended the data generating oracle*, the
pattern: NA, 1, NA, should be a data entry error --- graduating HS
implies graduating ES, no? I would argue fringe cases like that
should be corrected in the data, not through coding work arounds.
Then you can just do:
x <-
In our Summer Stats Institute, I was asked a question that amounts to
reversing the effect of the contrasts function (reconstruct an ordinal
predictor from a set of binary columns). The best I could think of was to
link together several ifelse functions, and I don't think I want to do this
if the e
> Very interesting. But what explicitly happens with ***source()***
> in Splus???
Here are some examples where I use the exprs argument instead of making a
tempory file and source the file. I don't know what you are looking for.
(The OP never showed any calls to the functions in defined in his s
Hello,
I'm optimizing a log-likelihood function using the built-in optim()
function with the default method. The model is essentially a Weibull
distribution where the rate parameter changes based on a single covariate,
coded 1 or 0 (that is, when the indicator is 1, the rate parameter changes
to a
See in line below.
On 08/06/13 00:00, Hertzog Gladys wrote:
Dear all,
I’m new in R and I’m trying to estimate the covariance matrix of a bivariate
normal distribution by maximizing the log-likelihood. The maximization really
has to be performed with the non-linear minimization routine (nlm).
Very interesting. But what explicitly happens with ***source()***
in Splus???
cheers,
Rolf
On 08/06/13 11:33, William Dunlap wrote:
foo <- function(x) {
sin(42)
x^2
}
foo(3)
[1] 9
The value of sin(42) is never seen.
This is true in b
> foo <- function(x) {
> sin(42)
> x^2
> }
>
> foo(3)
> [1] 9
>
> The value of sin(42) is never seen.
This is true in both R and S+. Only the return value of a function is available
for autoprinting and sin(42) is not the return value. Perhaps you are
rem
On Fri, 7 Jun 2013, Janh Anni wrote:
Try contacting Dr Dennis Helsel, the developer at
dhel...@practicalstats.com
I don't know what OS David's using, nor if R-3.0.0 is an upgrade or a new
installation, but I had no issues when upgrading R from the last 2.x
version. The existing NADA in the s
On 07/06/13 23:05, Duncan Murdoch wrote:
> On 13-06-06 6:22 PM, Rolf Turner wrote:
>> On 07/06/13 03:19, Scott Raynaud wrote:
>>> I actually had tried placing arguments in the call but it didn't
>>> work. However, I did
>>> not think about writing it to a variable and printing. That seems
>>>
Try contacting Dr Dennis Helsel, the developer at dhel...@practicalstats.com
On Fri, Jun 7, 2013 at 1:47 PM, David Doyle wrote:
> Hello folks,
>
> Im trying to install the NADA package in R 3.0.0
>
>
> It has been archived so I tried to downloading it and installing it
> locally. I get
>
> >
Dear all mailing listers,
Does Anyone have the same problem as mine when using the fast99
(extended-FAST method) to perform SA of model with norm distribution inputs?
See the simple example given following.
Any suggestion will be greatly appreciated.
Thank you!
Marino
# Simple example
# 1.
Hi,
If i replace `lst3` with:
lst1<-split(final3,list(final3$year,final3$industry))
lst2<-lst1[lapply(lst1,nrow)>0]
lst3<-lapply(lst2,function(x) lapply(x$dimension,function(y) {x1<-x[(y<
(x$dimension+x$dimension*0.1)) & (y> (x$dimension-x$dimension*0.1)),];
x1[abs(diff(x1$dimension))<200,]})) #
Maybe the final result is the same, but I want to impose it on final3, when we
are matching the firms.
Thanks,
Cecília
De: arun [smartpink...@yahoo.com]
Enviado: sexta-feira, 7 de Junho de 2013 23:04
Para: Cecilia Carmo
Cc: R help
Assunto: Re: [R] match
HI,
Do you want to impose this on 'res' on 'final3'?
If it is on `res`:
res1<-do.call(rbind,lapply(split(res,(1:nrow(res)-1)%/%2+1),function(x)
x[abs(diff(x$dimension))<200,]))
row.names(res1)<-1:nrow(res1)
nrow(subset(res1,dummy==0))
#[1] 14
nrow(subset(res1,dummy==1))
#[1] 14
head(res1)
# fi
On Fri, 2013-06-07 at 13:12 -0700, William Shadish wrote:
> Dear R-helpers,
>
> I'd like to understand how to test the statistical significance of a
> random effect in gamm. I am using gamm because I want to test a model
> with an AR(1) error structure, and it is my understanding neither gam
>
Thank you very much.
I apologize but I want to ask only one more thing:
How to do if I want in addition to impose that the diffence between
dimensions doesn't exced an absolute value, for example, 200.
Firm 1 continues matching with firm 5 in year 2000 because
2189-2120=69, and 69 is < 10%x212
Dear R-helpers,
I'd like to understand how to test the statistical significance of a
random effect in gamm. I am using gamm because I want to test a model
with an AR(1) error structure, and it is my understanding neither gam
nor gamm4 will do the latter.
The data set includes nine short interr
hey i need help with a case in my R project.
i have a vector Rt with returns from day 1 till day t now i want to form a
new vector r= which is rt = Xt − Xt−1
then i want to definde positive and negative aggregates
r(n)+ _t= 1/n (rt +...+rt−n) I {(rt+...+ rt−n) ≥0}
r(n)− _t= 1/n (rt +...+rt−n) I
myconn<-odbcConnect("testdata")
sql.select<-paste("select UNIT_ID from UNITS where (UNIT_TYPE='",unit,"'
and COMMUNITY='",property,"')",sep="")
unit_ids<-sqlQuery(myconn,sql.select,as.is=TRUE)
This should works if myconn and sql.select are defined properly
Andrija
On Jun 7, 2013 9:58 PM, "Sneha
tried as.is ,gives an error,
[1] "01000 10054 [Microsoft][ODBC SQL Server
Driver][DBNETLIB]ConnectionWrite (send())."
[2] "[RODBC] ERROR: Could not SQLExecDirect 'select UNIT_ID from UNITS
where (UNIT_TYPE='1X1' and COMMUNITY='SAN1193')'"
On Fri, Jun 7, 2013 at 3:21 PM, andrija djurovic wrote:
Thanks again, Brian.
Indeed, I had Java for 32 bits. Replaced it and rJava is working now.
On Thu, Jun 6, 2013 at 10:58 AM, Prof Brian Ripley wrote:
> On 06/06/2013 15:52, Dimitri Liakhovitski wrote:
>
>> Thank you very much, Brian.
>> It's clearly christal clear - but one needs a christal ball
On Jun 7, 2013, at 8:38 AM, Shane Carey wrote:
> Hi,
>
> I have a plot with horizontal lables on the x-axis. For example:
> Devonian volcanic rocks
> n=2
>
> with n=2 on a new line. How do I centre n=2 under the " Devonian volcanic
> rocks "
> label?
>
> This was my code: text(axis_text, par(
Hope to get some clarity from the creators of a (wonderful) package
ChoiceModelR. I am using the main function 'choicemodelr'.
Everything works.
?choicemodelr says "average of MCMC draws of unit-level model coefficients
are written to Xbetas.csv"
Question 1: Is this a typo? The real file created
?sqlQuery
as.is - argument
Andrija
On Jun 7, 2013 9:10 PM, "Sneha Bishnoi" wrote:
> Hey all!
>
> I am trying to select a bunch of id's (data type -character) from a table
> and store them in a variable in R
> But when i do this, it automatically truncates the leading zero's in id's
> even though
Thanks for all your help. I will try to fix this.
Very helpful.
Best,
Daofeng
On Fri, Jun 7, 2013 at 11:25 AM, Sarah Goslee wrote:
> As Marc already pointed out, take a close look at this part of your loop:
>
> R> i <- 6
> R>
> R> y <- as.numeric(data[i,-1])
> R> y
> [1] 3 3 3 3 4 4 4 4
> R> g
Thank you for the reference. Very helpful
Mikhail
On Friday, June 07, 2013 12:06:22 R Heberto Ghezzo, Dr wrote:
> You better see the original book
> Permutation methods - Mielke Jr Paul J, Berry Kenneth J. The programs in
> Fortran for the tests are in www.stat.colostate.edu/permute
>
> R.Hebert
Thanks Marc...still didn't get it...
Tried this...
> str(group)
num [1:8] 1 1 1 1 0 0 0 0
> fit=glm.nb(y~group)
Error in while ((it <- it + 1) < limit && abs(del) > eps) { :
missing value where TRUE/FALSE needed
> group <- c('1','1','1','1','0','0','0','0')
> str(group)
chr [1:8] "1" "1" "1" "
Thank you Sarah and Marc for your fast and nice response.
Apology for didn't include all information.
I have a input file like following:
gene1 18 15 13 13 16 9 20 24
gene2 15 8 8 7 0 12 18 4
gene3 10 9 8 12 9 11 12 12
gene4 4 0 4 3 0 5 0 0
gene5 0 1 0 0 1 5 1 0
gene6 3 3 3 3 4 4 4 4
gene7 0 4 0
Hey all!
I am trying to select a bunch of id's (data type -character) from a table
and store them in a variable in R
But when i do this, it automatically truncates the leading zero's in id's
even though they are of character type.
code is :-
>myconn<-odbcConnect("testdata")
>sql.select<-paste("
Sorry Sarah.
> dput(dat)
structure(list(gene = structure(c(1L, 3L, 4L, 5L, 6L, 7L, 8L,
9L, 10L, 2L), .Label = c("gene1", "gene10", "gene2", "gene3",
"gene4", "gene5", "gene6", "gene7", "gene8", "gene9"), class = "factor"),
b1 = c(18L, 15L, 10L, 4L, 0L, 3L, 0L, 4L, 11L, 6L), b2 = c(15L,
8L,
I shall use this from now on. Thank you.
On Friday, June 07, 2013 19:19:55 Pascal Oettli wrote:
Hello,
You also can use the package "sos"
>findFn('non-parametric multivariate
ANOVA')
Regards,Pascal
2013/6/7 Mikhail Umorin
Hello,
I am comparing treatments by comparing within group to be
From: peter dalgaard
Cc: "r-help@r-project.org"
Sent: Friday, June 7, 2013 11:12 AM
Subject: Re: [R] Clogit R and Stata
Here is the R output:
Call:
coxph(formula = Surv(rep(1, 1404L), sftpcons) ~ sftptv2a3 + sftptv2a4 +Â
  sftptv2a5 + sftptv2a2 + sf
Yes. vegan and TreMineR. Thank you!
On Friday, June 07, 2013 10:52:02 Jose Iparraguirre wrote:
> Hi Mikhail,
>
> After a cursory online search using simply " A new method for non-parametric
> multivariate analysis" + CRAN + R, I got the following three packages which
> include this paper in their
Hello folks,
Im trying to install the NADA package in R 3.0.0
It has been archived so I tried to downloading it and installing it
locally. I get
> utils:::menuInstallLocal()
package NADA successfully unpacked and MD5 sums checked
But then when I load it I get
> library(NADA)
Error in l
Hi there
does anybody know if the p-values in coxme are given one- or two-tailed by
default?
Thank you
David
-Messaggio originale-
Da: r-help-boun...@r-project.org per conto di r-help-requ...@r-project.org
Inviato: gio 06/06/2013 11.00
A: r-help@r-project.org
Oggetto: R-help Digest, Vo
Sorry, it included a case with both values for dummy were 1.
lst7<-lapply(lst6,function(x) {lst<-lapply(x,function(y)
y[sum(y$dummy)==1,]);lst[lapply(lst,nrow)>0]})
res<-do.call(rbind,lapply(lst7,function(x) do.call(rbind,x)))
row.names(res)<-1:nrow(res)
nrow(subset(res,dummy==0))
#[1] 22
nro
As Marc already pointed out, take a close look at this part of your loop:
R> i <- 6
R>
R> y <- as.numeric(data[i,-1])
R> y
[1] 3 3 3 3 4 4 4 4
R> group
[1] 1 1 1 1 0 0 0 0
R> fit <- glm.nb(y~group)
Error in while ((it <- it + 1) < limit && abs(del) > eps) { :
missing value where TRUE/FALSE neede
Sorry,
Something is not ok, because when I do
> nrow(subset(res,res$dummy==0))
[1] 22
> nrow(subset(res,res$dummy==1))
[1] 24
There the number o observatios in the two subsets is not equal.
Thanks again,
Cecília
De: arun [smartpink...@yahoo.com]
En
> that's exactly what I was looking for!
That code has some problems if your time series do not cross at the time
points. E.g., Duncan's code with some diagnostic plotting is:
f0 <- function (site1, site2, x = time(site1), check = FALSE)
{
stopifnot(length(x) == length(site1), length(
Hi,
On Fri, Jun 7, 2013 at 11:36 AM, Daofeng Li wrote:
> Thank you Sarah and Marc for your fast and nice response.
> Apology for didn't include all information.
>
> I have a input file like following:
>
> gene1 18 15 13 13 16 9 20 24
> gene2 15 8 8 7 0 12 18 4
> gene3 10 9 8 12 9 11 12 12
> gene4
On Jun 7, 2013, at 10:36 AM, Daofeng Li wrote:
> Thank you Sarah and Marc for your fast and nice response.
> Apology for didn't include all information.
>
> I have a input file like following:
>
> gene1 18 15 13 13 16 9 20 24
> gene2 15 8 8
That's it, the first one!
Thank you very much,
Cecília Carmo
De: arun [smartpink...@yahoo.com]
Enviado: sexta-feira, 7 de Junho de 2013 16:43
Para: Cecilia Carmo
Cc: R help
Assunto: Re: [R] matched samples, dataframe, panel data
Hi,
Not sure if this is
Hi,
Not sure if this is what you wanted.
res<-do.call(rbind,lapply(lst6,function(x) do.call(rbind,x)))
row.names(res)<-1:nrow(res)
# this combines the list of lists to a data.frame
res[1:4,]
# firm year industry dummy dimension
#1 1 2000 20 0 2120
#2 5 2000 20 1
Hi,
I have a plot with horizontal lables on the x-axis. For example:
Devonian volcanic rocks
n=2
with n=2 on a new line. How do I centre n=2 under the " Devonian volcanic
rocks "
label?
This was my code: text(axis_text, par("usr")[three], labels = paste(LABELS,
" \n ", " n =", t(t(name.count[,t
Hi,
May be this helps:
lst1<-split(final3,list(final3$year,final3$industry))
lst2<-lst1[lapply(lst1,nrow)>0]
lst3<-lapply(lst2,function(x) lapply(x$dimension,function(y) x[(y<
(x$dimension+x$dimension*0.1)) & (y> (x$dimension-x$dimension*0.1)),]))
lst4<-lapply(lst3,function(x) x[lapply(x,nrow)==2
On Jun 7, 2013, at 9:44 AM, Daofeng Li wrote:
> Dear R Community,
>
> I have encountered a problem while using the R function glm.nb.
> The code that produce the error was following two lines:
>
> group=c(1,1,1,1,0,0,0,0)
> fit=glm.nb(y~group)
>
> While the y contains 8 sets of number like:
>
Hi,
On Fri, Jun 7, 2013 at 10:44 AM, Daofeng Li wrote:
> Dear R Community,
>
> I have encountered a problem while using the R function glm.nb.
> The code that produce the error was following two lines:
>
> group=c(1,1,1,1,0,0,0,0)
> fit=glm.nb(y~group)
>
> While the y contains 8 sets of number li
On Jun 7, 2013, at 15:34 , Richard Beckett wrote:
> Sorry to once again write a message but I'm once again stumped and am having
> no luck finding a solution anywhere else.
>
>
> This question requires some finesse in both R and STATA so hopefully I will
> be able to get an answer here. I am
Dear R Community,
I have encountered a problem while using the R function glm.nb.
The code that produce the error was following two lines:
group=c(1,1,1,1,0,0,0,0)
fit=glm.nb(y~group)
While the y contains 8 sets of number like:
gene2750 1 0 0 1 5 1
Trying to plot an ROC curve to determine effectiveness of a neural network
model in R.
using the neuralnet package I created nn1a:
nn1a<-neuralnet(warstns ~ aigenlz+
p2l+anoclinv2+flinstab+mill+minconl+fllgdppclz+lpoplz+floil+flmtnestz+c_peaceyears,
data=na.omit(dfram1), hidden=4)
then usin
On 06/07/2013 07:16 AM, arun wrote:
Hi,
You would get better response if you post at Bioconductor mailing list.
http://www.bioconductor.org/help/mailing-list/
Agreed, though in this case I think
ReadAffyData <- function( filename ) {
...
Data <- ReadAffy( sampleNames=as.character( Cov$
Hi,
?which()
which(A==2)
#[1] 3 7 10
A.K.
Suppose there is a vector
A <- c(1,3,2,6,7,8,2,1,3,2).
Now, I want to get the subscript of elements of A which equal 2.
How can I do it ?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman
Great, thanks Sarah
On Fri, Jun 7, 2013 at 3:46 PM, Sarah Goslee wrote:
> Hi Shane,
>
> On Fri, Jun 7, 2013 at 10:36 AM, Shane Carey wrote:
> > Hi,
> >
> > Does anybody know how to increase boxplot outlier symbol size? And also
> the
> > line from min/max to outlier
>
> If you read the help fo
Hi Shane,
On Fri, Jun 7, 2013 at 10:36 AM, Shane Carey wrote:
> Hi,
>
> Does anybody know how to increase boxplot outlier symbol size? And also the
> line from min/max to outlier
If you read the help for boxplot, it will send you to the help for bxp
for more information on custom plotting. ?bxp
Hi,
Does anybody know how to increase boxplot outlier symbol size? And also the
line from min/max to outlier
Thanks
--
Shane
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-he
Hi,
On Fri, Jun 7, 2013 at 7:38 AM, R_Antony wrote:
> here i have a dataframe
>
> for eg:-
> DATETIMECOL_A COL_B COL_C COL_D
> 1/1/2007 0:01 0 3 0 0
> 1/1/2007 0:02 0 0 3 0
> 1/1/2007 0:03 0 3 0 0
> ...
Hi,
You would get better response if you post at Bioconductor mailing list.
http://www.bioconductor.org/help/mailing-list/
A.K.
- Original Message -
From: Payal Urs
To: r-help@r-project.org
Cc:
Sent: Friday, June 7, 2013 6:12 AM
Subject: [R] Bionconductor help
Hi
I am trying to do so
Thank you. That answers my question.
-- Bert
On Fri, Jun 7, 2013 at 7:09 AM, Prof Brian Ripley wrote:
> On 07/06/2013 14:49, Bert Gunter wrote:
>>
>> Folks:
>>
>> Feel free to provide me a link or reference instead of an answer.
>>
>> Preamble:
>>
>> f <- function() function(x)rnorm(x)
>> g <- f
here i have a dataframe
for eg:-
DATETIMECOL_A COL_B COL_C COL_D
1/1/2007 0:01 0 3 0 0
1/1/2007 0:02 0 0 3 0
1/1/2007 0:03 0 3 0 0
... . ... ...
On 07/06/2013 14:49, Bert Gunter wrote:
Folks:
Feel free to provide me a link or reference instead of an answer.
Preamble:
f <- function() function(x)rnorm(x)
g <- f()
g(3)
## [1] -0.4448492 -0.2379978 -0.4537394
## But
rnorm <- function()1 ## nasty nasty
g(3)
## Error in rnorm(x) : unused a
aah perfect, exactly what i was looking for!
Thank you
On Fri, Jun 7, 2013 at 3:24 PM, arun kirshna [via R]
wrote:
> HI,
> Try:
> ?split()
>
> source("http://www.openintro.org/stat/data/cdc.R";)
> str(cdc)
> #'data.frame':2 obs. of 9 variables:
> # $ genhlth : Factor w/ 5 levels "excel
Arun,
Perfect, this is what I was looking for.
Thanks,
Sudha Krishnan
-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: Friday, June 07, 2013 2:36 AM
To: Sudha Krishnan
Cc: R help
Subject: Re: [R] generating a bar chart with two axis for co-linear variable
HI,
Not s
Hello,
I try to install pbdMPI.
Compilation successful, but load fails with segfault.
Is anyone can help me?
R version 3.0.0
pbdMPI version 0.1-6
Intel compiler version 13.1.1
OpenMPI version 1.6.4-1
CPU Intel x86_64
# R CMD INSTALL pbdMPI_0.1-6.tar.gz
..
checking for gcc... icc -std=gnu99
Hi
I am trying to do some data analysis using R and Bioconductor. I have a
function to read my data called "ReadAffyData" and a function to plot the
data called "preqc". I want to know if there is any way I can extract
information (read: Data) from the ReadAffyData function into preqc in order
t
Hello again,
I am having problem with ODBC connection using the RODBC package.
I am basically trying to read the attached Excel-2003 file using RODBC
package. Here is my code:
> head(sqlFetch(odbcConnectExcel("d:/1.xls"), ""), 30); odbcCloseAll()
Criteria s d fd f
Sorry to once again write a message but I'm once again stumped and am having no
luck finding a solution anywhere else.
This question requires some finesse in both R and STATA so hopefully I will be
able to get an answer here. I am much more adept in R and am trying to
replicate the results of
At 12:39 07/06/2013, Bernd Weiß wrote:
Am 07.06.2013 13:17, schrieb Michael Dewey:
At 19:34 06/06/2013, Bernd Weiss wrote:
Dear all,
I am struggling to add a prediction interval to a forest plot that was
created with forest.meta(), package "meta".
I checked the source of forest.meta() and rea
Again my problem, better explained.
#I have a data panel of thousands of firms, by year and industry and
#one dummy variable that identifies one kind of firms (1 if the firm have an
auditor; 0 if not)
#and another variable the represents the firm dimension (total assets in
thousand of euros)
#
HI,
Please dput() the example data.
"""
Could anyone give a help ASAP ?"
You have been posting for long time and your many posts show this "ASAP". I
know that you got comments to that and also advised to dput() the data.
Formatting your data took some time. Would it be better to follow th
Folks:
Feel free to provide me a link or reference instead of an answer.
Preamble:
f <- function() function(x)rnorm(x)
g <- f()
g(3)
## [1] -0.4448492 -0.2379978 -0.4537394
## But
rnorm <- function()1 ## nasty nasty
g(3)
## Error in rnorm(x) : unused argument(s) (x)
## of course f <- function
HI,
Try:
?split()
source("http://www.openintro.org/stat/data/cdc.R";)
str(cdc)
#'data.frame': 2 obs. of 9 variables:
# $ genhlth : Factor w/ 5 levels "excellent","very good",..: 3 3 3 3 2 2 2 2 3
3 ...
# $ exerany : num 0 0 1 1 0 1 1 0 0 1 ...
# $ hlthplan: num 1 1 1 1 1 1 1 1 1 1 ...
On 07/06/2013 12:21, Aakanksha Dahiya01 wrote:
Hi
Could just anyone explain me the coefficients in the output of arima model
The person who wrote the help page already did, but that is hardly 'just
anyone'.
timeseriesarima <- arima(series, order=c(1,1,2))
timeseriesarima
Series: series
Am 07.06.2013 13:17, schrieb Michael Dewey:
At 19:34 06/06/2013, Bernd Weiss wrote:
Dear all,
I am struggling to add a prediction interval to a forest plot that was
created with forest.meta(), package "meta".
I checked the source of forest.meta() and realized that it is heavily
relying on grid
Hi
Could just anyone explain me the coefficients in the output of arima model
timeseriesarima <- arima(series, order=c(1,1,2))
> timeseriesarima
Series: series
ARIMA(1,1,2)
Coefficients:
ar1 ma1 ma2
0.9744 -1.7695 0.7873
s.e. 0.0310 0.0481 0.0426
sigma^2 estimated
At 19:34 06/06/2013, Bernd Weiss wrote:
Dear all,
I am struggling to add a prediction interval to a forest plot that was
created with forest.meta(), package "meta".
I checked the source of forest.meta() and realized that it is heavily
relying on grid. I am lacking any experience with grid graph
On 13-06-06 6:22 PM, Rolf Turner wrote:
On 07/06/13 03:19, Scott Raynaud wrote:
I actually had tried placing arguments in the call but it didn't work.
However, I did
not think about writing it to a variable and printing. That seems to have done
the
trick. Funny, I don't remember having to
Please, please do not use html formatted mail.
It is clearly requested by the mailing list
The code of your last mail is a mess and when replying it becomes even more of
a mess.
I told you to do
siml[g,] <- optm(perm=20)
See the comma after g.
and not what you are now doing with siml[g]<-op
On 07-06-2013, at 11:57, Laz wrote:
> Dear Berend,
>
> I have made some changes but I still get errors:
>
> For reproducibility,
> Rspatswap() is a function which originally returns a single value. For
> example Rspatswap(...) and you get 0.8
>
> So, run Rspatswap() 20 times and store all
Hello,
You also can use the package "sos"
>findFn('non-parametric multivariate
ANOVA')
Regards,
Pascal
2013/6/7 Mikhail Umorin
> Hello,
>
> I am comparing treatments by comparing within group to between group
> distances like
> described in
>
> MJ Anderson. 2001. A new method for non-param
Hi,
I am almost getting there, but still have errors. Thanks for your help.
I have tried improving but I get the following errors below:
>itn<-function(it){
+siml<-matrix(NA,ncol=5,nrow=it)
+for(g in 1:it){
+siml[g]<-optm(perm=20)[g]
+}
+siml
+}
>itn(3)
[,1] [,2] [,3] [,4] [,5]
[1,
Hi Mikhail,
After a cursory online search using simply " A new method for non-parametric
multivariate analysis" + CRAN + R, I got the following three packages which
include this paper in their references: vegan, mefa, and TraMineR. I don't know
whether they deal with the procedure you are after
Hi Duncan,
that's exactly what I was looking for! The series have common times, so that's
no problem. Thanks a lot!
Roland
Von: Duncan Murdoch [murdoch.dun...@gmail.com]
Gesendet: Donnerstag, 6. Juni 2013 18:45
An: Roland Pape
Cc: r-help@r-project.org
Bet
Dear Berend,
For reproducibility,
Rspatswap() is a function which originally returns a single value. For
example Rspatswap(...) and you get 0.8
So, run Rspatswap() 20 times and store all the values.
Then from these 20 values, calculate the calculate average,
sd,se,min,max to get something si
I R-helpers
#I have a data panel of thousands of firms, by year and industry and
#one dummy variable that separates the firms in two categories: 1 if the firm
have an auditor; 0 if not
#and another variable the represents the firm dimension (total assets in
thousand of euros)
#I need to creat
On 07-06-2013, at 10:59, Laz wrote:
> Dear R users,
>
> I am stuck here: My first function returns a vector of 5 values.
> In my second function, I want to repeat this, a number of times, say 10
> so that I have 10 rows and five columns but I keep on getting errors.
>
> See the code and resul
Hi there,
You need a function for your statistic:
> boot(x, function(x, index) mean(x[index]), R = 1000)
ORDINARY NONPARAMETRIC BOOTSTRAP
Call:
boot(data = x, statistic = function(x, index) mean(x[index]),
R = 1000)
Bootstrap Statistics :
original biasstd. error
t1* 0.40696
Dear R users,
I am stuck here: My first function returns a vector of 5 values.
In my second function, I want to repeat this, a number of times, say 10
so that I have 10 rows and five columns but I keep on getting errors.
See the code and results below:
optm <-function(perm, verbose = FALSE)
{
I am getting started with the boot package and boot command. As a first
step I tried the following. Something is wrong, but i can't see what. Any
advice would be much appreciated.
> x = runif(10)
> mean(x)
[1] 0.467626212374307
> boot(x, mean, R=100)
Error in mean.default(data, original, ...) :
Dear all,
I am estimating a mixed-model in Ubuntu Raring (13.04¸ amd64), with the
code:
fm0 <- lme(rt ~ run + group * stim * cond,
random=list(
subj=pdSymm(~ 1 + run),
subj=pdSymm(~ 0 + stim)),
data=mydat1)
When I check the
You can try this:
library(inflection)
#you have to instsall package inflection first
a<-findiplist(cbind(year),cbind(piproute),1)
a
The answer:
[,1] [,2] [,3]
[1,]5 35 1986.0
[2,]5 30 1983.5
shows that the total inflection point is between 1983 and 1986, if we treat
data as fi
Hello,
At an R prompt, type
?power.t.test
Hope this helps,
Rui Barradas
Em 06-06-2013 20:58, Rebecca Greenblatt escreveu:
Looking to determine sample sizes for both my experimental and control
groups (I want only a small portion of my participants in my experimental
condition) in order to co
There are 5 remaining places on the following course:
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When: 26 - 30 August 2013.
Where: Edmonton, Canada
For details, see: http://www.highstat.com/statscourse.htm
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