On Sep 27, 2012, at 9:07 PM, Ben Harrison wrote:
> Hello,
>
> I have produced some segmented regressions with the segmented package by
> Viggo Mutteo. I have some example data and code below. I want to annotate
> the individual segments with the slope parameter (actually it would be
> nicer to a
On Sep 27, 2012, at 11:13 PM, Bhupendrasinh Thakre wrote:
>
> Hi Everyone,
>
> I am trying a very simple task to append the Timestamp with a variable name
> so something like
> a_2012_09_27_00_12_30 <- rnorm(1,2,1).
If you want to assign a value to a character-name you need to use ... `assig
Hello,
Try the following:
b <- unclass(Sys.time())
eval(parse(text=paste("c_",b," <- rnorm(1,2,1)",sep="")))
ls()
Regards,
Pascal
Le 28/09/2012 15:13, Bhupendrasinh Thakre a écrit :
Hi Everyone,
I am trying a very simple task to append the Timestamp with a variable name so
something like
Hi!
28.09.2012 09:13, Bhupendrasinh Thakre wrote:
Statement I tried :
b <- unclass(Sys.time())
b = 1348812597
c_b <- rnorm(1,2,1)
Do you mean this:
--- code ---
> df<-data.frame("x"=0,"y"=0)
> colnames(df)
[1] "x" "y"
> colnames(df)[2]<-paste("b",unclass(Sys.time()),sep="_")
> colnames(df)
Hi Everyone,
I am trying a very simple task to append the Timestamp with a variable name so
something like
a_2012_09_27_00_12_30 <- rnorm(1,2,1).
Tried some commands but it doesn't work out well. Hope someone has some answer
on it.
Session Info
R version 2.15.1 (2012-06-22)
Platform: i386-
Hi!
28.09.2012 08:41, Atte Tenkanen wrote:
Sorry. I should have mentioned that the order of the components is important.
So c(1,4,6) is accepted as a subvector of c(2,1,1,4,6,3), but not of
c(2,1,1,6,4,3).
How to test this?
How about this:
--- code ---
g1<- c(2,1,1,4,6,3)
g2<- c(2,1,1,6,4
HI,
May be this helps you:
set.seed(1)
mat1<-matrix(rnorm(60,5),nrow=5,ncol=12)
colnames(mat1)<-paste0("Var",1:12)
vec2<-format(c(1,cor(mat1[,1],mat1[,2:12])),digits=4)
vec3<-colnames(mat1)
arr2<-array(rbind(vec3,vec2),dim=c(2,3,4))
res<-data.frame(do.call(rbind,lapply(1:dim(arr2)[3],fun
Sorry. I should have mentioned that the order of the components is important.
So c(1,4,6) is accepted as a subvector of c(2,1,1,4,6,3), but not of
c(2,1,1,6,4,3).
How to test this?
Cc: R help
Aihe: Re: [R] How to test if there is a subvector in a longer
On Thu, Sep 27, 2012 at 8:21 AM, Rantony wrote:
> Hi,
>
> Can anyone please help to get "StartDay" and "End-day of a particular month"
> with time ?
>
> For eg:- Input wil be "2012-09-27"
>
> i need to get output as given below
>
> StartDt <- "2012-09-01 00:00:01"
> EndDt <- "2012-09-30 23:59:59"
Hello,
I have produced some segmented regressions with the segmented package by
Viggo Mutteo. I have some example data and code below. I want to annotate
the individual segments with the slope parameter (actually it would be
nicer to annotate with 1000*slope and add some small amount of text as
we
Hi,
Try this:
Not sure whether this is the fastest:
set.seed(932)
vec1<-sample(1:10,6,replace=TRUE)
vec2<-sample(1:7,3,replace=TRUE)
vec2[vec2%in%vec1]
#[1] 5
library(rbenchmark)
benchmark(isTRUE(all(vec2%in%vec1)),replications=1e4)
# test replications elapsed relative us
HI Joshua,
Thanks for providing much easier solution.
But, I am getting the output from dat1 as:
# inputDate startDate endDate
#1 2012-09-27 2012-09-27 00:00:01 2012-09-27 23:59:59
#2 2012-09-28 2012-09-28 00:00:01 2012-09-28 23:59:59
#3 2012-07-24 2012-07-24 00:00:01 2012-
I have no idea what your code is doing, nor why you want correlated binary
variables. Correlation makes little or no sense in the context of
binary random
variables --- or more generally in the context of discrete random variables.
Be that as it may, it is an easy calculation to show that if
On Sep 27, 2012, at 2:00 PM, Atte Tenkanen wrote:
> Hi,
>
> There are certainly several ways to test, whether a longer vector includes a
> subvector.
> For instance, c(1,4,6) is included in c(2,1,1,4,6,3). How to test this and
> which would be the fastest way to do it?
>
> all( c(1,4,6) %in
On Thu, Sep 27, 2012 at 5:15 PM, Dr. Alireza Zolfaghari
wrote:
> Hi List,
> Would you please send me a good link to talk me through on how to write a R
> package?
There are many, many, many resources:
http://lmgtfy.com/?q=writing+r+packages+tutorial
Take the first hit.
-steve
--
Steve Lianogl
On Thu, Sep 27, 2012 at 3:05 PM, arun wrote:
> HI,
> For a vector of dates:
> Using Michael's suggestion:
> library(xts)
> library(zoo)
> Dt<-c("2012-09-27","2012-09-28","2012-07-24","2012-06-05","2012-12-03")
> newDt<-strsplit(format(as.yearmon(Dt),"%Y-%m"),split="-")
> StartDt<-do.call(rbind,lap
Hello
This is Elaine.
I am using package lattice to generate boxplots.
Using Richard's code, the display was almost perfect except the outlier
shape.
Based on the following code, the outliers are vertical lines.
However, I want the outliers to be empty circles.
Please kindly help how to modify th
I am trying to Sweave the output of calculating correlations between one
variable and several others. I wanted to print a table where the
odd-numbered rows contain the variable names and the even-numbered rows
contain the correlations. So if VarA is correlated with all the variables in
mydata.df, t
Hi List,
Would you please send me a good link to talk me through on how to write a R
package?
thanks
Alireza
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read t
Hello,
If you really need to trash your disk, why not use seek()?
> fl <- file("Test.txt", open = "wb")
> seek(fl, where = 1024, origin = "start", rw = "write")
[1] 0
> writeChar(character(1), fl, nchars = 1, useBytes = TRUE)
Warning message:
In writeChar(character(1), fl, nchars = 1, useBytes
Hi Dr. Viechtbauer,
I'm doing meta-analysis using your package 'metafor'. I used the 'IRFT' to
transform the incident rate. But when I tried to back-transform the summary
estimates from function rma, I don't know what's the appropriate ti to feed in
function transf.iirft. I searched and found y
Hello, again.
There was another error in the line in question. TRUE does not need
quotes. In fact, with quotes you're comparing to a character string, not
to a logical value.
And the other tip still holds, use as follows in the complete and
corrected line below.
Health2PairsOnly <- PairIDs[
Hello,
That way of refering to variables can be troublesome. Try
PairIDs[, "Pairiddups"]
Hope this helps,
Rui Barradas
Em 27-09-2012 20:46, GradStudentDD escreveu:
Hi,
I have a data set of observations by either one person or a pair of people.
I want to only keep the pair observations, and
Hi R-fellows,
I am trying to simulate a multivariate correlated sample via the Gaussian
copula method. One variable is a binary variable, that should be
autocorrelated. The autocorrelation should be rho = 0.2. Furthermore, the
overall probability to get either outcome of the binary variable sho
Hi R-fellows,
I am trying to simulate a multivariate correlated sample via the Gaussian
copula method. One variable is a binary variable, that should be
autocorrelated. The autocorrelation should be rho = 0.2. Furthermore, the
overall probability to get either outcome of the binary variable sho
HI,
For a vector of dates:
Using Michael's suggestion:
library(xts)
library(zoo)
Dt<-c("2012-09-27","2012-09-28","2012-07-24","2012-06-05","2012-12-03")
newDt<-strsplit(format(as.yearmon(Dt),"%Y-%m"),split="-")
StartDt<-do.call(rbind,lapply(lapply(lapply(newDt,`[`,1:2),function(x)
as.numeric
Hi,
There are certainly several ways to test, whether a longer vector includes a
subvector.
For instance, c(1,4,6) is included in c(2,1,1,4,6,3). How to test this and
which would be the fastest way to do it?
Best,
Atte Tenkanen, FT, MuM
http://users.utu.fi/attenka/
___
Hi,
I am trying to fit an AR model, maximum order =4, order selection
criterion is aic. I wonder why these two give different results:
m1<-ar.ols(x, aic=TRUE, method="ols", order.max=4)
m1<-auto.arima(x,d=0, D=0, max.p=4, max.P=0, max.q=0, max.Q=0, ic="aic")
Could they both use the f
?Distributions
The more general term for percentile is quantile.
help.search("quantile")
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
In R, what is the best way to convert z-scores (Normal distribution) to a
percentiles, and vice-versa.
(I'll be looking to do the same with other distributions eventually.)
===
Peter Petto
Bay Village, OH 440.249.4289
__
R-help@r-project.org maili
#By using cbind in:
PairIDs<-cbind(PairID, PairIDDuplicates)
#You create a numeric matrix (the logical
#vector PairIDDuplicates gets converted
#to numeric - note that your second column
#contains 1s and 0s, not Trues and Falses).
#Matricies are not subsetable using $,
#they are basically a vector
Dear R help,
I am trying solve an MLE convergence problem: I would like to estimate
four parameters, p1, p2, mu1, mu2, which relate to the probabilities,
P1, P2, P3, of a multinomial (trinomial) distribution. I am using the
mle2() function and feeding it a time series dataset composed of four
On 12-09-27 2:53 PM, Anju R wrote:
Sometimes when I try to install certain packages I get a warning message.
For example, I tried to install the package "Imtest" on windows R version
2.15.1 and got the following message:
Warning message:
package ‘Imtest’ is not available (for R version 2.15.1)
You can always roll your own with something like:
> x1 <- c(4, 5, 7)
> x2 <- c(5, 6, 9)
> plot(c(1.5, 2.5), cbind(x1[2], x2[2]), xlab="", ylab="",
xlim=c(1, 3), ylim=c(0, 10), xaxt="n")
> axis(1, at=1:2+.5, labels=c("Group 1", "Group 2"))
> arrows(1.5, x1[2], 1.5, c(x1[1], x1[3]), angle=90,
On 12-09-27 05:34 PM, Bert Gunter wrote:
> Good point, Ben.
>
> I followed up my earlier reply offline with a brief note to Benedikt
> pointing out that "No" was the wrong answer: "maybe, maybe not" would
> have been better.
>
> Nevertheless, the important point here is that even if you do get
>
Good point, Ben.
I followed up my earlier reply offline with a brief note to Benedikt
pointing out that "No" was the wrong answer: "maybe, maybe not" would
have been better.
Nevertheless, the important point here is that even if you do get
convergence, the over-parameterization means that the est
Bert Gunter gene.com> writes:
>
> On Thu, Sep 27, 2012 at 12:43 PM, Benedikt Gehr
> ieu.uzh.ch> wrote:
> > now I feel very silly! I swear I was trying this for a long time and it
> > didn't work. Now that I closed R and restarted it it works also on my
> > machine.
> >
> > So is the only proble
Wouldn't the correct approach be to use the software and give credit
appropriately for the software and libraries you use?
Branding is not an issue... familiarity is. This is open source software...
read the license(s).
---
Hi all,
I would like to run a paired t-test with censored data using the NADA
package. Here is an example data set:
data <- data.frame(group=rep(c('a','b'), each=6),
obs=c(2,2,5,7,10,10,5,5,6,8,5,9),
cen=c(T,T,F,F,F,F,F,F,F,F,F,F))
#A two sample t-test can be executed as follows:
library(NADA
On Thu, Sep 27, 2012 at 12:43 PM, Benedikt Gehr
wrote:
> now I feel very silly! I swear I was trying this for a long time and it
> didn't work. Now that I closed R and restarted it it works also on my
> machine.
>
> So is the only problem that my model is overparametrized with the data I
> have?
P
Hi,
I have a data set of observations by either one person or a pair of people.
I want to only keep the pair observations, and was using the code below
until it gave me the error " $ operator is invalid for atomic vectors". I am
just beginning to learn R, so I apologize if the code is really rough
Hi all,
I need to obtain the LV variance from my PLS-DA analysis.
I tried to read the reference manuale of the package, but I do not found
information about that.
Someone know a way to do it?
Thank you,
Roberto
--
View this message in context:
http://r.789695.n4.nabble.com/PLS-DA-and-LV-varia
Sometimes when I try to install certain packages I get a warning message.
For example, I tried to install the package "Imtest" on windows R version
2.15.1 and got the following message:
Warning message:
package Imtest is not available (for R version 2.15.1)
How can I install the above package?
now I feel very silly! I swear I was trying this for a long time and it
didn't work. Now that I closed R and restarted it it works also on my
machine.
So is the only problem that my model is overparametrized with the data I
have? however shouldn't it be possible to fit an nls to these data?
On 27-09-2012, at 21:15, Benedikt Gehr wrote:
> thanks for your reply
>
> I agree that an lm model would fit just as well, however the expectation from
> a mechanistic point of view would be a non-linear relationship.
>
> Also when I "simulate" data as in
>
> y_val<-115-118*exp(-0.12*(seq(1,
Folks:
Asked this question some time ago, and found what appeared (at first) to be
the best solution, but I'm now finding a new problem. First off, it seemed
like ff as Jens suggested worked:
# outdata_ncells = the number of rows * number of columns * number of bands
in an image:
out<-ff(vmode="
thanks for your reply
I agree that an lm model would fit just as well, however the expectation
from a mechanistic point of view would be a non-linear relationship.
Also when I "simulate" data as in
y_val<-115-118*exp(-0.12*(seq(1,100)+rnorm(100,0,0.8)))
x_val<-seq(1:100)
plot(y_val~x_val)
sum
I was able to puzzle it out with the help of the book "R Graphics" (Murrell).
When par("bg") = "transparenent" one needs to use col="white", otherwise the
old code col=0 works correctly.
The default for pdf and x11, the two I use, is transparent.
Terry Therneau
On 09/27/2012 08:48 AM, PIKAL
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Pradipto Banerjee
> Sent: Thursday, September 27, 2012 9:13 AM
> To: r-help@r-project.org
> Subject: [R] equivalent of Stata "by construct"
>
> I am evaluating a switch from Stata
My guess:
You probably are overfitting your data. A straight line does about as
well as anything except for the 3 high leverage points, which the
minimization is probably having trouble with.
-- Bert
On Thu, Sep 27, 2012 at 10:43 AM, Benedikt Gehr
wrote:
> quantiles<-c(seq(.05,.95,0.05))
> sl
Users forget how much is an OS service. This is OS X not R being slow.
On a recent Linux box it takes about 90s.
But at least you can easily parallelize it: see ?pvec in package
parallel for one way to do this (and one way not to).
If the file contain a high proportion of duplicates, making a
> And as for the thermodynamics analogy, I agree. I'm looking for Alan
> Kay's to help me hide complexity.
Whups, that should read "I'm looking for Alan Kay's daemon to help me hide
complexity."
stupid editor...
cur
--
Curt Seeliger, Data Ranger
Raytheon Information Services - Contractor to O
Hi
I would like to fit a non-linear regression to the follwoing data:
quantiles<-c(seq(.05,.95,0.05))
slopes<-c( 0.00e+00, 1.622074e-04 , 3.103918e-03 , 2.169135e-03 ,
9.585523e-04
,1.412327e-03 , 4.288103e-05, -1.351171e-04 , 2.885810e-04 ,-4.574773e-04
, -2.368968e-03, -3.104634e-03, -5
(Sorry, forgot to cc to the list)
-- Bert
On Thu, Sep 27, 2012 at 10:41 AM, Bert Gunter wrote:
> Perrick:
>
> You have an a extra ")" after yscale =
> so the pushViewport statement is ignored and the rest does what you ask it to.
>
> -- Bert
>
> On Thu, Sep 27, 2012 at 9:39 AM, Pierrick Bruneau
From: "Daniel Nordlund"
> > > > ...
> > > > I'd like to have the code source files from the 'local' git
repository
> > > > without modification, where 'local' could mean c:\yada\ for one
> > person, m:\my documents\wetlands\ for another, ...
>
> each user could set a PROJECT_PATH environment var
On 27/09/2012 12:13 PM, Pradipto Banerjee wrote:
I am evaluating a switch from Stata to R. I don't need to extensive Statistical
methods, but the main reason I am exploring the switch is the coding
flexibility in R (e.g. Stata does not support linear/quadratic programming). I
have been going o
R 2.15.1
OS X.7.4
Colleagues,
I have a large dataset (27773536 records, the file is several GB) that contains
a column of date / time entries in the format:
"2/1/2011 13:25:01"
I need to convert these to numeric values (ideally in seconds; the origin
[e.g., 1970-01-01] is not impor
Hi,
I need some help to manage frailty in Survreg function; in particular I'm
looking for more information about frailty in survreg function applied to a
loglogistic hazard function.
Actually I need to develope a predictor for frailty random variable realization
(similar to the Proportional Haz
I have a stupid problem that is currently driving me crazy...
Let us suppose that I want to draw a big red square in the middle of my
current device (say X11)
I tried the following code :
pushViewport(viewport(xscale=c(0,1), yscale=c(0,1)), just=c("center", "center"))
vp1 <- viewport(x=unit(0.5,
I am evaluating a switch from Stata to R. I don't need to extensive Statistical
methods, but the main reason I am exploring the switch is the coding
flexibility in R (e.g. Stata does not support linear/quadratic programming). I
have been going over the R syntax and I had a quick question:
In St
Hello,
Inline.
Em 27-09-2012 13:52, Krunal Nanavati escreveu:
Hi,
Thanks for all your help. I am stuck again, but with a new problem, on
similar lines.
I have taken the problem to the next step now...i have now added 2 "for"
loops... 1 for the Price variable...and another for the Media variabl
Curt:
Well, if you want to peer a bit more, you might wish to have a look at
http://inlinedocs.r-forge.r-project.org/ . Same philosophy as
Roxygen_x (keep docs together with code, generate Rd files
automatically therefrom) but takes a different approach.
Perhaps worth noting for both cases is tha
On Sep 27, 2012, at 5:59 AM, Alexandra Howe wrote:
> Hello,
>
> I have data which I have arcsin transformed to analyse.
> I want to plot my data with error bars however as my data is
> back-transformed my standard errors are uneven.
> Is there a simple way to draw these asymmetric error bars in
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Curt Seeliger
> Sent: Wednesday, September 26, 2012 5:51 PM
> To: r-help@r-project.org
> Subject: Re: [R] Is there a way to source from a specific Git repository
> without hardcodi
> > I would usually do more than that: I find the R documentation system
> > helpful even when I'm the only user of a package ...
>
> Me too - but I'd never write Rd by hand ;) instead relying on roxygen2
> ...
>
> But my point main was that it's very easy to start a package - and you
> don't ne
Hello,
I am not a Mac user, but I use Eclipse+StatET in both Windows and
Linux, and it is the best IDE (not just text editor) for me. It
supports code syntax, R and Sweave editors, object explorer, document
outline, debugging, ... Besides, Eclipse has other plugins that may be
useful for some tasks
On Thu, Sep 27, 2012 at 1:21 PM, Rantony wrote:
> Hi,
>
> Can anyone please help to get "StartDay" and "End-day of a particular month"
> with time ?
>
> For eg:- Input wil be "2012-09-27"
>
> i need to get output as given below
>
> StartDt <- "2012-09-01 00:00:01"
> EndDt <- "2012-09-30 23:59:59"
Hi,
You can also try this:
df2 <- data.frame(df1, colsplit(df1$x, pattern = "_", names=c("str","name")))
df2list<-list(df2$str,df2$name)
df2[,2:3]<-sapply(df2list,function(x) gsub(".*(\\d)","\\1",x))
df2
# x str name
#1 str1_name2 1 2
#2 str3_name5 3 5
A.K.
- Original Mes
Hi,
Thanks for all your help. I am stuck again, but with a new problem, on
similar lines.
I have taken the problem to the next step now...i have now added 2 "for"
loops... 1 for the Price variable...and another for the Media variable
I have taken 5 price variables...and 2 media variables with th
Hi,
Can anyone please help to get "StartDay" and "End-day of a particular month"
with time ?
For eg:- Input wil be "2012-09-27"
i need to get output as given below
StartDt <- "2012-09-01 00:00:01"
EndDt <- "2012-09-30 23:59:59"
- Thanks in advance
--
View this message in context:
http://r.
Hello,
I have data which I have arcsin transformed to analyse.
I want to plot my data with error bars however as my data is
back-transformed my standard errors are uneven.
Is there a simple way to draw these asymmetric error bars in R?
Thanks for your help.
[[alternative HTML version del
Dear all,
I am new here ,I attempted to use R to estimate the spatial Durbin (mixed)
model,and mydata is a panel data form,and the matrix is generated by geoda
software ,here is my Command and error,really hope your help ,thank you!
#引入gal
library(spdep)
w<- read.gal("E:/splm/zj.GAL",override.id=TR
Hi,
You can also use either of these:
data$date <- as.Date(data$date,format="%m/%d/%y")
data$Days<-as.vector(sapply(lapply(split(data,data$Person),`[`,2),function(x)
difftime(x[,1],x[1,],units="days")))
#or
data$Days<-as.vector(sapply(lapply(split(data,data$Person),`[`,2),function(x)
x[,1]-x[1
Dear Karly,
I don't know if you've resolved your problem but I just came across the
same thing in RStudio and I think I've found out what the problem is. The
problem seems to occur when the plot viewing panel (by default in the bottom
right) is very small. The plot function tries to create a plo
On Sep 27, 2012, at 8:33 AM, Hasan Diwan wrote:
> On 27 September 2012 09:26, peter dalgaard wrote:
>
>> (I gather that .emacs functionality ends up in
>> ~/Library/Preferences/Aquamacs\ Emacs/* . You're not really expected to
>> bypass the menus, though.)
>>
>
> ~/.emacs, same place as on Li
Hi
It seems that it still works.
x<-c(3,7,7,3)
y<-c(4,4,6,6)
par(bg="pink")
plot(1:10,1:10)
polygon(x, y, border=TRUE, col=0)
> version
_
platform i386-pc-mingw32
arch i386
Thanks. Which function I can use for forecasting if I use the function
"auto.arima()"? Should I use the function "forecast" or "predict" or
anything else? Thanks,
Miao
2012/9/27 Jose Iparraguirre
> You'll find this in the forecast package, function auto.arima()
> Regards,
> José
>
> José Iparr
Thank you,
this works perfectly...
best regards,
Johannes
On Thu, Sep 27, 2012 at 1:43 PM, Ista Zahn wrote:
> Hi Johannes,
>
> On Thu, Sep 27, 2012 at 7:25 AM, Johannes Radinger
> wrote:
>> Hi,
>>
>> I am using colsplit (package = reshape) to split all strings
>> in a column according to the s
On 27 September 2012 09:26, peter dalgaard wrote:
> (I gather that .emacs functionality ends up in
> ~/Library/Preferences/Aquamacs\ Emacs/* . You're not really expected to
> bypass the menus, though.)
>
~/.emacs, same place as on Linux.
--
Sent from my mobile device
Envoyait de mon portable
On Sep 27, 2012, at 02:59 , Albyn Jones wrote:
> Have you looked at aquamacs? (emacs for the mac).
> its at aquamacs.org.
Seconded, especially if you also want AUC-TeX features. For pure R purposes, as
others have noted, the built-in editors in R.app and Rstudio are often good
enough and less
I'm updating some (very) old code, and one particular option of its plot method depends on
a once-was-true trick
polygon(x, y, border=TRUE, col=0)
polygon(x, y, border=TRUE, density=0)
would draw the polygon AND erase whatever was underneath it back to background
color.
Is there a reli
On Sep 26, 2012, at 8:09 PM, David Winsemius wrote:
>
> On Sep 26, 2012, at 6:06 PM, David Winsemius wrote:
>
>>
>> On Sep 26, 2012, at 5:48 PM, Steven Wolf wrote:
>>
>>> Hi everyone,
>>>
>>> I've recently moved from using a windows machine to a Mac (some might call
>>> it an upgrade, othe
Thanks everyone for all of your help. This has really helped me filter all of
the noise about text editors on the internet.
-Steve
On Sep 27, 2012, at 5:41 AM, Berend Hasselman wrote:
>
> On 27-09-2012, at 02:48, Steven Wolf wrote:
>
>> Hi everyone,
>>
>> I've recently moved from using a
Hello,
By looking at the output of
pat <- "(str)|(_name)|( name)"
strsplit(c("str1_name2", "str3_name5"), pat)
[[1]]
[1] "" "1" "2"
[[2]]
[1] "" "3" "5"
I could understand why colsplit includes NAs as column 'str' values.
So the hack is to fake we want three coluns and then set the first one
Hi Johannes,
On Thu, Sep 27, 2012 at 7:25 AM, Johannes Radinger
wrote:
> Hi,
>
> I am using colsplit (package = reshape) to split all strings
> in a column according to the same patterns. Here
> an example:
>
> library(reshape2)
>
>
> df1 <- data.frame(x=c("str1_name2", "str3_name5"))
> df2 <- da
Hi,
I am using colsplit (package = reshape) to split all strings
in a column according to the same patterns. Here
an example:
library(reshape2)
df1 <- data.frame(x=c("str1_name2", "str3_name5"))
df2 <- data.frame(df1, colsplit(df1$x, pattern = "_", names=c("str","name")))
This is nearly what I
On Thu, Sep 27, 2012 at 5:49 AM, Barry Rowlingson
wrote:
> On Wed, Sep 26, 2012 at 8:24 PM, baycan global wrote:
>> Hi all,
>>
>> I have seen that R can be switched on to a Broker called IB.
>>
>> There is another one similar to IB that permits to make a "code" and send
>> orders to broker to buy
Hi José,
This sounds like a survey weight. Take a look at the survey package if
you want to calculate statistics incorporating survey weights.
Best,
Ista
On Wed, Sep 26, 2012 at 12:38 PM, Jose Bustos Melo wrote:
> Thank you PIKAL Petr,
>
> This is used when you have a big data base of a nationa
Hello,
Just to add that you can also
lapply(lm.list, coef)
with a different output.
Rui Barradas
Em 27-09-2012 09:24, David Winsemius escreveu:
On Sep 26, 2012, at 10:31 PM, Krunal Nanavati wrote:
Dear Rui,
Thanks for your time.
I have a question though, when I run the 5 regression, whose
We are pleased to announce that the R user conference
useR! 2013
is scheduled for July 10-12, 2013, and will take place at the
University of Castilla-La Mancha, Albacete, Spain.
As for the predecessor conferences, the program will consist of two
parts: invited lectures and user-contributed se
Hi all, I appreciate your help.
Here are a sample of my script. I appreciate any help.
David, I will go ahead and try your suggestion as well.
Thanks.
library(survey)
clust<- svydesign(id=id, weights=wtper, strata=strat, data=data)
summary(cl
Cheers Sarah, Rui, David,
Your effort clarifying my (several) confusions, especially with
examples, most helpful for my understanding.
Not least the value of a fresh global environment _without_ confounding
objects like:
> scl
function(x) { median(x, na.rm=TRUE) }
And proper punctuation termi
> I would usually do more than that: I find the R documentation system
> helpful even when I'm the only user of a package (and there are the prompt*
> functions for quickly creating it, as well as package.skeleton to set things
> up at the beginning). Vignettes are a great way to organize and doc
On Wed, Sep 26, 2012 at 8:24 PM, baycan global wrote:
> Hi all,
>
> I have seen that R can be switched on to a Broker called IB.
>
> There is another one similar to IB that permits to make a "code" and send
> orders to broker to buy or sell stocks?
>
> Can be done trough R, writte in excel and tro
Hi Abraham,
On Wed, Sep 26, 2012 at 6:06 PM, Abraham Mathew wrote:
> I'm trying to teach myself about Bayesian Networks and am working with the
> following data and the bnlearn package.
> I understand the conceptual aspects of BNs, but I'm not sure how to specify
> the response variables in R whe
On 27-09-2012, at 02:48, Steven Wolf wrote:
> Hi everyone,
>
> I've recently moved from using a windows machine to a Mac (some might call it
> an upgrade, others not…I'll let you be the judge). Once I started using
> Notepad ++ on my windows machine, I really began to like it. Unfortunately
On 12-09-27 3:31 AM, Hadley Wickham wrote:
I'd like to have the code source files from the 'local' git repository
without modification, where 'local' could mean c:\yada\ for one
person,
m:\my documents\wetlands\ for another, and
l:\foo\bar\sharedRemote\wet\ to
another user.
...
Yes. Use
l
On 12-09-26 9:21 PM, Bhupendrasinh Thakre wrote:
My vote for R-Studio. Very elegant design and great functionality. However if
coming from languages like Java and others then eclipse is better. R-Studio
have dedicated section for Mac user which you will find useful.
One of the nice features o
Hello,
RSiteSearch('ar BIC')
Regards,
Pascal
Le 27/09/2012 17:43, jpm miao a écrit :
Hello,
Is there a function in R by which one can run AR model with Bayesian
Information Criteria (BIC)? To my knowledge, functions ar and ar.ols could
select the order only by AIC.
Thanks,
Miao
On 27-09-2012, at 07:16, Marcel Curlin wrote:
> Hi
> I have data for events in rows, with columns for person and date. Each
> person may have more than one event;
>
> tC <- textConnection("
> Persondate
> bob 1/1/00
> bob 1/2/00
> bob 1/3/00
> dave 1/7/00
> dave 1/8/00
> dave
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