Diego,
Start with Bob Muenchen's site: http://www.rforsasandspssusers.com/
HTH,
Jim Porzak
TGN.com
San Francisco, CA
www.linkedin.com/in/jimporzak
use R! Group SF: www.meetup.com/R-Users/
On Mon, Jun 8, 2009 at 5:47 PM, DIEGO CHAVEZ wrote:
> Sent: Sunday, June 07, 2009 5:43 PM
> Subject: Ques
Perhaps you should try http://www.rforge.net/pgfSweave/
On Mon, Jun 8, 2009 at 5:38 PM, maiya wrote:
>
> Wow! Thank you for that Ted, a wonderfully comprehensive explanation and
> now
> everything makes perfect sense!!
>
> Regarding your last point, I would love to hear other people's experienc
There is a technical report on comparison of R and other statistical
software
(http://www.ats.ucla.edu/stat/technicalreports/Number1/R_relative_statpack.pdf).
You can have an overall view from this document.
You may have better chance to get answers if you specify what you want
to do and if R can
Sent: Sunday, June 07, 2009 5:43 PM
Subject: Question about R an SPSS
Dears Sirs:
Venables, Smith and R-Development Core Team
I am reading about the functional features of R statistical software, because
I want to compare these progarm with the basic module of SPSS software.
I woul
I was feeling pretty silly when I saw there was actually a locations
parameter in stars, as well as axes etc.
But now the problem is that the x and y axes in stars must be on the same
scale! Which unfortunately makes my data occupy only a very narrow band of
the plot. I guess one option would be
Wow! Thank you for that Ted, a wonderfully comprehensive explanation and now
everything makes perfect sense!!
Regarding your last point, I would love to hear other people's experience. I
myself, as a complete newbie in both R and LaTeX, am perhaps not the best
judge... But there are several graph
Hi jim,
I am caliculating these caliculation on quaring on 4 tables and preparing
subsets of those and from there I am getting the final results
that is
"034P91"|"d947468f-95ff-4844-8d04-36619ed2cced"|1|"Control"|9686|"brain"|9301|7912|10746|10742|9710|12591|7933|12612|"Gram"|2.014|1.721|1.8902|0
Dear list members,
i am currently want to install Rpy2 in a linux box which has R 2.4.0
installed
RPy requries R 2.7.0 or above
but i have no root previlleges
so my question is how to install R 2.7.0 on my own directory?
and replace the system installed R 2.4.0 when i input R command from the
bash
hi,
new to R and using the car package to do some scatterplots with ellipses
hoping to add the area and center points of each ellipse to the legend?
looking for some direction / ideas here is the script, the data is
where
golf shots end up by club.
x (dispersion), y (distance), g
On Mon, Jun 8, 2009 at 8:56 PM, Mao Jianfeng wrote:
> Dear Ruser's
>
> I ask for helps on how to substitute missing values (NAs) by mean of the
> group it is belonging to.
>
> my dummy dataframe is:
>
>> df
> group traits
> 1 BSPy01-10 NA
> 2 BSPy01-10 7.3
> 3 BSPy01-10 7.3
> 4
I posted a screen capture at
http://groups.google.com/group/r-help-archive/web/TinnRplayswithR.png
On Jun 8, 5:07 pm, "Karin & Martijn" wrote:
> Dear R-users,
>
> I have installed the latest version of Tinn R (Version 2.2.0.2).
> With the older versions it was possible to run Tinn R next to R an
When I run Tinn-R I am able to activate the GUI or the Rterm. I am not
quite sure what you mean.
Options - Applications - R - Path normally allows one to set it up so
that you want to work with it the way you describe.
I will try post a screen capture. Alternatively you can post a video
screencapt
Dear Mao,
Here is another way:
yourdata$traits2 <- with(yourdata, do.call(c, tapply(traits, group,
function(y){
ym <-
mean(y,na.rm=TRUE)
y[is.na(y)]<- ym
On Jun 8, 2009, at 9:56 PM, Mao Jianfeng wrote:
Dear Ruser's
I ask for helps on how to substitute missing values (NAs) by mean of
the
group it is belonging to.
my dummy dataframe is:
df
group traits
1 BSPy01-10 NA
2 BSPy01-107.3
3 BSPy01-107.3
4 BSPy01-115.3
5
Here are four ways:
# using lapplyBy - DF is already sorted by group
library(doBy)
f <- function(x) {
x$traits[is.na(x$traits)] <- mean(x$traits, na.rm = TRUE)
x
}
do.call(rbind, lapplyBy(~ group, DF, f))
# using by - same f as before
do.call(rbind, by(DF, DF$group, f))
# using
Try this:
d$traits[is.na(d$traits)] <- ave(d$traits,
d$group,
FUN=function(x)mean(x,
na.rm = T))[is.na(d$traits)]
On 6/8/09, Mao Jianfeng wrote:
> Dear Ruser's
>
> I ask for helps on how to substitute missing
Dear Ruser's
I want to substitute each "NA" by the group mean of which the "NA" is
belonging to. For example, substitute the first record of traits "NA" by the
mean of "BSPy01-10" in the dummy dataframe.
I have ever tried to solve this problem by using doBy package. But, I
failed. I ask for the a
Dear Ruser's
I ask for helps on how to substitute missing values (NAs) by mean of the
group it is belonging to.
my dummy dataframe is:
> df
group traits
1 BSPy01-10 NA
2 BSPy01-107.3
3 BSPy01-107.3
4 BSPy01-115.3
5 BSPy01-115.4
6 BSPy01-115.6
7 BSPy01-11
How about this:
> "%==%" <- function(x, y) {
if (length(x) > 1) {
sapply(x, function(z) isTRUE(all.equal(z, y)));
} else {
sapply(y, function(z) isTRUE(all.equal(z, x)));
}
}
> seq(0, 1, by=0.1) %==% 0.1
[1] FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALS
On Jun 8, 2009, at 5:27 PM, Barry Rowlingson wrote:
On Mon, Jun 8, 2009 at 10:40 PM, Tan, Richard
wrote:
Hi,
This is not exactly an R question but I am trying to use gsub to
replace
a string that contains 5-9 alpha-numeric characters, at least one of
which is a number. Is there a good wa
On Mon, Jun 8, 2009 at 7:18 PM, Wacek
Kusnierczyk wrote:
> Gabor Grothendieck wrote:
>> Try this. See ?regex for more.
>>
>>
>>> x <- 'This happened in the 21. century." (the dot behind 21 is'
>>> regexpr("(?![0-9]+)[.]", x, perl = TRUE)
>>>
>> [1] 24
>> attr(,"match.length")
>> [1] 1
>>
>
> yes,
Gabor Grothendieck wrote:
> Try this. See ?regex for more.
>
>
>> x <- 'This happened in the 21. century." (the dot behind 21 is'
>> regexpr("(?![0-9]+)[.]", x, perl = TRUE)
>>
> [1] 24
> attr(,"match.length")
> [1] 1
>
yes, but
gregexpr('(?![0-9]+)[.]', 'a. 1. a1.', perl=TRUE)
Sweave does something clever with warnings, which I have so far been
unable to figure out. There are a couple of threads on the list about
this, but the best in here is a hack to redirect all the output and
stick it back in.
http://www.nabble.com/-R--Sweave-and-warning-messages-td7759353.html#
On Mon, Jun 8, 2009 at 10:40 PM, Tan, Richard wrote:
> Hi,
>
> This is not exactly an R question but I am trying to use gsub to replace
> a string that contains 5-9 alpha-numeric characters, at least one of
> which is a number. Is there a good way to write it in a one line regex?
The only way I
Hi,
This is not exactly an R question but I am trying to use gsub to replace
a string that contains 5-9 alpha-numeric characters, at least one of
which is a number. Is there a good way to write it in a one line regex?
Thanks,
Richard
__
R-help@r-proj
Dear R-users,
I have installed the latest version of Tinn R (Version 2.2.0.2).
With the older versions it was possible to run Tinn R next to R and then select
code that one could send to R.
Now it seems that one has to start the R gui in Tinn R and then it is possible
to select en run code of R
On Mon, Jun 8, 2009 at 3:36 PM, Don MacQueen wrote:
Though I do agree that the way you've written the general case with any/
is.na and sum/na.rm is cleaner and clearer because more general, I don't
agree at all with what you say about nested ifelse's vs. a series of
assignments:
> In my opinion
Thanks very much for that, most useful...
Ted.Harding-2 wrote:
>
> On 08-Jun-09 13:11:03, sedm1000 wrote:
>> I've looked long and hard for this, but maybe I am missing something...
>>
>> There is a nice module that displays histograms reflected in the
>> y axis, i.e.
>> http://addictedtor.fre
I've gotten to the point wih an R script where I would like to encapsulate
several blocks of codes in R functions.
In order to keep the top level script simple I would like to put them in a
separate file. This should help the readability of the top level main script.
Is source(...) the best w
After doing your plot command, type
par()$usr
This will give you the x and y axis ranges. Make sure that the "at"
values you give are within the x-axis range.
-Don
At 2:59 PM -0400 6/8/09, Graves, Gregory wrote:
My x axis is a series of daily dates (e.g., 01/01/2000, 01/02/2000,
etc.) fro
I think one of the other good suggestions might have had a typo in it.
And, I would like to append an alternate approach
that can be generalized to more columns.
In my opinion, nested ifelse() expressions are
difficult to read and understand, and therefore
difficult to get right.
Easier to wr
The axis will use the internal representation of the dates. See R News 4/1
for more on that.
On Mon, Jun 8, 2009 at 2:59 PM, Graves, Gregory wrote:
> My x axis is a series of daily dates (e.g., 01/01/2000, 01/02/2000,
> etc.) from 2000 to end of 2008. The default only gives me 4 ticks. I
> want
Dear all,
I need to change the default of "r" when plotting the
output of the envelope function of spatstat package.
I can do this manually for the Lest output:
## Example from spatstat Kest help
data(cells)
L <- Lest(cells, correction="isotropic", r=seq(from=0,to=0.5, by=0.05))
x11(1000,400)
On Mon, Jun 8, 2009 at 1:48 PM, Cecilia Carmo wrote:
> I have the following dataframe:
> firm<-c(rep(1:3,4))
> year<-c(rep(2001:2003,4))
> X1<-rep(c(10,NA),6)
> X2<-rep(c(5,NA,2),4)
> data<-data.frame(firm, year,X1,X2)
> data
>
> So I want to obtain the same dataframe with a variable X3 that is:
Ask yourself:
a) whether anyone could possibly see what you are seeing with the data
you have offered, and ...
(Just a guess that this is the problem)
b) what POSIXct dates are represented by 1:9? Would they even be in
the range 2001-2009?
> ddd <- as.POSIXct(strptime(c("01/01/2000", "01
My x axis is a series of daily dates (e.g., 01/01/2000, 01/02/2000,
etc.) from 2000 to end of 2008. The default only gives me 4 ticks. I
want more. Why doesn't this work?
sdate<-as.POSIXct(strptime(date,format="%m/%d/%Y"))
plot(ppt~sdate,type="l",ylim=c(0,47),col=1,lwd=1,pch=16,ylab="S
On Jun 8, 2009, at 1:48 PM, Cecilia Carmo wrote:
Hi R-helpers!
I have the following dataframe:
firm<-c(rep(1:3,4))
year<-c(rep(2001:2003,4))
X1<-rep(c(10,NA),6)
X2<-rep(c(5,NA,2),4)
data<-data.frame(firm, year,X1,X2)
data
So I want to obtain the same dataframe with a variable X3 that is:
X1,
Am Montag, den 08.06.2009, 13:33 -0400 schrieb David Winsemius:
> On Jun 8, 2009, at 12:42 PM, Juergen Rose wrote:
>
> > Am Montag, den 08.06.2009, 18:33 +0200 schrieb Juergen Rose:
> >> Am Montag, den 08.06.2009, 12:30 -0400 schrieb milton ruser:
> >>> Dear Rose,
> >>>
> >>> no attached file came
On 6/8/2009 1:48 PM, Cecilia Carmo wrote:
> Hi R-helpers!
>
> I have the following dataframe:
> firm<-c(rep(1:3,4))
> year<-c(rep(2001:2003,4))
> X1<-rep(c(10,NA),6)
> X2<-rep(c(5,NA,2),4)
> data<-data.frame(firm, year,X1,X2)
> data
>
> So I want to obtain the same dataframe with a variable X3 th
Hi R-helpers!
I have the following dataframe:
firm<-c(rep(1:3,4))
year<-c(rep(2001:2003,4))
X1<-rep(c(10,NA),6)
X2<-rep(c(5,NA,2),4)
data<-data.frame(firm, year,X1,X2)
data
So I want to obtain the same dataframe with a variable X3
that is:
X1, if X2=NA
X2, if X1=NA
X1+X2 if X1 and X2 are not N
The help page for extractPredictions suggests and testing confirms
that the function expects a _list_ of models. The predict function
is suggested as the method to get predictions from a single model.
Giving the argument as a list does work with a single model, however:
> predict(glmmat)
On Jun 8, 2009, at 12:34 PM, Marc Schwartz wrote:
On Jun 8, 2009, at 9:15 AM, Mark Heckmann wrote:
Hi,
i need to recognize itemization structures in strings which follow
the
format: "digit-digit-dot" like e.g.
1.
2.
19.
211.
Given the string " This happened in the 21. century."
On Jun 8, 2009, at 12:42 PM, Juergen Rose wrote:
Am Montag, den 08.06.2009, 18:33 +0200 schrieb Juergen Rose:
Am Montag, den 08.06.2009, 12:30 -0400 schrieb milton ruser:
Dear Rose,
no attached file came with the message.
bests
milton
I try once more to attache the file. I hope that 89 KB
Try this. See ?regex for more.
> x <- 'This happened in the 21. century." (the dot behind 21 is'
> regexpr("(?![0-9]+)[.]", x, perl = TRUE)
[1] 24
attr(,"match.length")
[1] 1
On Mon, Jun 8, 2009 at 10:15 AM, Mark Heckmann wrote:
> Hi,
>
>
>
> i need to recognize itemization structures in string
On Jun 8, 2009, at 9:15 AM, Mark Heckmann wrote:
Hi,
i need to recognize itemization structures in strings which follow the
format: "digit-digit-dot" like e.g.
1.
2.
19.
211.
Given the string " This happened in the 21. century." (the dot
behind 21 is
used in German instead of 21st
On Jun 8, 2009, at 12:42 PM, Juergen Rose wrote:
Am Montag, den 08.06.2009, 18:33 +0200 schrieb Juergen Rose:
Am Montag, den 08.06.2009, 12:30 -0400 schrieb milton ruser:
Dear Rose,
no attached file came with the message.
bests
milton
I try once more to attache the file. I hope that 89 KB
Try this:
x <- "This happened in the 21. century."
gregexpr("[[:digit:]]\\.", x)
This returns the position of the digit-dot in the string.
On Mon, Jun 8, 2009 at 11:15 AM, Mark Heckmann wrote:
> Hi,
>
>
>
> i need to recognize itemization structures in strings which follow the
> format: "digit
I am using survreg from the survival package to run a left censored tobit
model on non-survival data. I have to four questions that I hope someone
can help me with:
1) Is there anything I should take into consideration when using frailty()
to estimate random intercepts?
2) Is there anyway o
Hi Michael,
with res.uc$conf you'll get the single configurations for each rater.
You can use these to produce the plot you want to have.
Best,
Patrick
r-help-requ...@r-project.org wrote:
Send R-help mailing list submissions to
r-help@r-project.org
To subscribe or unsubscribe via t
Hi,
i need to recognize itemization structures in strings which follow the
format: "digit-digit-dot" like e.g.
1.
2.
19.
211.
Given the string " This happened in the 21. century." (the dot behind 21 is
used in German instead of 21st) I want know where the dots are but I do not
want t
Dear Sunny Vic,
I am forwarding it to the list, to help the helpers :-)
bests..
milton
On Mon, Jun 8, 2009 at 12:41 PM, sunny vic wrote:
> Hi Milton,
> here you go
>
> X1=rnorm(11, 50, 10)
> X2=rnorm(11, 20, 10)
> X3=rnorm(11, 50, 60)
> X4=rnorm(11, 10, 2)
> X5=rnorm(11, 5, 22)
>
> x<-cbind(
Here are 2 useful paths (it's up to you to decide if either is the right path).
The my.symbols function in the TeachingDemos package allows you to create your
own functions to create the symbols.
But in this case, you can just use the locations argument to the stars function:
> stars(cbind(1,sq
Am Montag, den 08.06.2009, 18:33 +0200 schrieb Juergen Rose:
> Am Montag, den 08.06.2009, 12:30 -0400 schrieb milton ruser:
> > Dear Rose,
> >
> > no attached file came with the message.
> >
> > bests
> > milton
>
> I try once more to attache the file. I hope that 89 KB is not to large
> for t
Dear Terry,
Thanks a lot for your reply. The result of survreg reads like:
Call:
survreg(formula = Surv(end, status) ~ fico_demean, data = raw,
dist = "loglogistic", model = TRUE)
Value Std. Error z p
(Intercept) 2.98365 1.34e-03 2233.2 0.00e+00
fico_demean
AFAICS, the problem is that you have not (carefully?) read the docs and are
approaching R as a C programmer and not taking a whole object point of view.
The loop is completely unnecessary! PLEASE READ AN INTRO TO R
(again,perhaps) before posting.
So, unless I misunderstand (and my apologies if I d
Am Montag, den 08.06.2009, 12:30 -0400 schrieb milton ruser:
> Dear Rose,
>
> no attached file came with the message.
>
> bests
> milton
I try once more to attache the file. I hope that 89 KB is not to large
for the mailing list.
> On Mon, Jun 8, 2009 at 12:20 PM, Juergen Rose
> wrote:
>
Am Montag, den 08.06.2009, 18:20 +0200 schrieb Juergen Rose:
> Hello,
>
> In the attached file training.csv (I apologize for the large file) I
> have 238 objects belonging to 13 classes, which are described by 183
> properties. I would like to find a svm model for these objects.
>
> I tried the f
Hello,
In the attached file training.csv (I apologize for the large file) I
have 238 objects belonging to 13 classes, which are described by 183
properties. I would like to find a svm model for these objects.
I tried the following R statements.
library('e1071')
datatraining <- read.csv("training
On Sunday 07 June 2009, Dirk Eddelbuettel wrote:
> On 7 June 2009 at 06:40, Neil Tiffin wrote:
> | I am adding your note to google code issues
> | (http://code.google.com/p/rpostgresql/issues/list ), issue Number 1.
> | Normally I monitor R-SIG-DB
> | (https://stat.ethz.ch/mailman/listinfo/r-sig-d
You can always open a connection and write the rows out as you have
calculated them. It would be nice if you had included at least a subset of
the calculations that you are doing so that we can understand the problem
you are trying to solve.
On Mon, Jun 8, 2009 at 11:43 AM, venkata kirankumar
wro
Dear Amit,
The following should get you started:
# Some data
set.seed(123)
X <- matrix(rnorm(20*10), ncol=10)
X
# Group of replicates
g <- rep(1:(ncol(X)/2), each=2)
g
# Mean of replicate variables
t(apply(X, 1, tapply, g, mean, na.rm = TRUE))
I created a grouping variable (g) and then calculat
Hi Sonny Vic,
how about you send a reproducible code?
cheers
milton
On Mon, Jun 8, 2009 at 11:25 AM, sunny vic wrote:
> Hi all
> I am using the caret package and having difficulty in obtaining the
> results
> using regression, I used the glmnet to model and trying to get the
> coefficients an
Hi
i am using a script which involves the following loop. It attempts to reduce a
data frame(zz) of 95000 * 41 down to a data frame (averagedreplicates) of 95000
* 21 by averaging the replicate values as you can see in the script below. This
script however is very slow (2days). Any suggestions
The as-good-as final scientific programme has been posted on the
conference website http://www.r-project.org/dsc-2009
The absolutely final date for regular registration is July 5, but
earlier registration is preferred and certainly recommended if you wish
to book hotels via the conference bureau.
On Mon, Jun 8, 2009 at 10:29 AM, Herbert
Jägle wrote:
> Hi,
>
> i do have a dataframe representing data from a repeated experiment. PID is a
> subject identifier, Time are timepoints in an experiment which was repeated
> twice. For each subject and all three timepoints there are 2 sets of four
> va
Hi all,
I am trying to create a "index.csv" with caliculating different
types of caliculations .
In that i have to caliculate on 10,000 studies and have to insert many no of
rows more than 500,000
for that right now I am inserting every row after caliculating and doing
data.frame
but its
Hi,
i do have a dataframe representing data from a repeated experiment. PID
is a subject identifier, Time are timepoints in an experiment which was
repeated twice. For each subject and all three timepoints there are 2
sets of four values.
df <- data.frame(PID = c(rep("A", 12), rep("B", 12),
Hi all
I am using the caret package and having difficulty in obtaining the results
using regression, I used the glmnet to model and trying to get the
coefficients and the model parameters I am trying to use the
extractPrediction to obtain a confusion matrix and it seems to be giving me
errors.
Dear all,
I use at least two pc to perform my data analysis. Both are powered
with R updated at the latest release (currently 2.9.0 under windows xp
pro). I bring .Rdata on my portable drive and use them on any of my pc
(this worked also under (k)ubuntu equipped machines). Please notice
that not al
At 08:38 08/06/2009, carol white wrote:
Hi,
It is known that the area of square plotted by forestplot is
proportional to the studies' weights. But since there is not weight
parameter in forestplot function, how does this function proportion
the area of each square based on the related study's
On 08-Jun-09 13:11:03, sedm1000 wrote:
> I've looked long and hard for this, but maybe I am missing something...
>
> There is a nice module that displays histograms reflected in the
> y axis, i.e.
> http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=136
>
> but is it possible to reflec
I've looked long and hard for this, but maybe I am missing something...
There is a nice module that displays histograms reflected in the y axis,
i.e. http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=136
but is it possible to reflect in the x-axis, so to have two datasets, one
pointi
---begin included message --
I am trying to use R to do loglogistic hazard estimation. My plan is to
generate a loglogistic hazard sample data and then use survreg to estimate
it. If everything is correct, survreg should return the parameters I have
used to generate the sample data.
I have w
Dear Alex,
Take a look at the examples in
# Option 1
?heatmap
# Option 2
require(gplots)
?heatmap.2
HTH,
Jorge
On Mon, Jun 8, 2009 at 9:32 AM, Alex Roy wrote:
> Hello Group,
>How can I draw heatmap with variable names in the plot?
>
> Thanks
>
> Alex
>
>[[alterna
It actually means that the MSE (0.04605) is 130.42% of var(y), thus the
model had not provided any better explanatory power than predicting by
mean(y). The pseudo R^2 is just 100% - 130.42% = -30.42%. Remember
that this is not the resubstituttion estimate because it is computed
from the OOB estim
Allan Engelhardt wrote:
> See
> http://wiki.r-project.org/rwiki/doku.php?id=misc:r_accuracy:decimal_numbers#sequences_of_decimal_numbers
>
>
as usual, be careful about what is advertised in r docs and related
texts. on the r_accuracy page, you'll read:
"For further information, see the digit
Couldn't you get that just by giving heatmap() the transpose of your
data?
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Alex Roy
> Sent: Monday, June 08, 2009 9:32 AM
> To: r-help@r-project.org
> Subject: [R] Heatmap
>
>
On 08-Jun-09 06:23:04, Peter Dalgaard wrote:
> maiya wrote:
>> OK, this is really weird!
>>
>> here's an example code:
>>
>> t1<-c(1,2,3,4)
>> t2<-c(4,2,4,2)
>> plot(t1~t2, xlab="exp1", ylab="exp2")
>> dev.copy2eps(file="test.eps")
>>
>> that all seems fine...
>>
>> until you look at the eps fi
Hello Group,
How can I draw heatmap with variable names in the plot?
Thanks
Alex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the post
See
http://wiki.r-project.org/rwiki/doku.php?id=misc:r_accuracy:decimal_numbers#sequences_of_decimal_numbers
and also
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
GrzeÅ wrote:
> Do you heve any idea why I get after this instruction everywh
http://wiki.r-project.org/rwiki/doku.php?id=misc:r_accuracy:decimal_numbers
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
GrzeÅ wrote:
> Do you heve any idea why I get after this instruction everywhere false?
>
>> seq (0, 1, by=0.1) == 0.3
Hi
r-help-boun...@r-project.org napsal dne 08.06.2009 10:45:15:
>
> Do you heve any idea why I get after this instruction everywhere false?
> > seq (0, 1, by=0.1) == 0.3
> [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
>
> But after different step it's ok:
> > seq(0, 1,
Hi
"Burke, Robin" napsal dne 08.06.2009 11:28:46:
> Thanks for the quick response. Sorry for being unclear with my example.
Here
> is something more concrete:
>
> user <- c(1, 2, 1, 2, 3, 1, 3, 4, 2, 3, 4, 1);
> time <- c(100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, 1100,
1200);
>
Solution!!
Peter, that seems to do the trick!
dev.copy2eps(file="test.eps", useKerning=FALSE)
correctly places the labels without splitting them!
the same also works with postscript() of course.
I also found another thread where this was solved
http://www.nabble.com/postscript-printer-breakin
Dear list, this might be a silly question, but I am a bit lost
I have bootstrapped the C-index of a logistic regression model, 500 times.
> results <- boot(data=data,statistic=cindex,R=500,formula = myformula)
When I plot the results I get a histogram of the bootstrapped where the Y axis
is
Hi,
With the help of Dirk Eddelbuettel, I was able to fix a few little
problems that prevented the rimage (0.5-7) package from working under
(at least) R 2.8 or 2.9 and g++ 4.2 or 4.3. I have uploaded a new
release, 0.5-8, that fixes these problems.
It have attempted to contact the package'
Thanks for the quick response. Sorry for being unclear with my example. Here is
something more concrete:
user <- c(1, 2, 1, 2, 3, 1, 3, 4, 2, 3, 4, 1);
time <- c(100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, 1100, 1200);
userCount <- c(1, 1, 2, 2, 1, 3, 2, 1, 3, 3, 2, 4);
period <- 10
Do you heve any idea why I get after this instruction everywhere false?
> seq (0, 1, by=0.1) == 0.3
[1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
But after different step it's ok:
> seq(0, 1, by=0.1) == 0.4
[1] FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE
On Mon, Jun 8, 2009 at 3:04 PM, baptiste auguie
wrote:
> Marie Sivertsen wrote:
>
>> Dear list,
>>
>> I have a vector of elements which I want to combined each with each, but
>> none with itself. For example,
>>
>>
>>
>>> v <- c("a", "b", "c")
>>>
>>>
>>
>> and I need a function 'combine' such tha
Thank you both Dimitiris and Jorge Ivan! I have just found it myself that
'combn' does what I need:
> combn(v, 2, simplify=TRUE)
is exactly what I need.
Mvh.,
Marie
On Mon, Jun 8, 2009 at 3:03 PM, Dimitris Rizopoulos <
d.rizopou...@erasmusmc.nl> wrote:
> one way is:
>
> combn(c("a", "b",
Marie Sivertsen wrote:
Dear list,
I have a vector of elements which I want to combined each with each, but
none with itself. For example,
v <- c("a", "b", "c")
and I need a function 'combine' such that
combine(v)
[[1]]
[1] "a" "b"
[[2]]
[1] "a" "b"
[[3]]
[1] "b" "c"
I a
one way is:
combn(c("a", "b", "c"), 2)
I hope it helps.
Best,
Dimitris
Marie Sivertsen wrote:
Dear list,
I have a vector of elements which I want to combined each with each, but
none with itself. For example,
v <- c("a", "b", "c")
and I need a function 'combine' such that
combine(v)
Many thanks
--- milton ruser schrieb am Mo, 8.6.2009:
Von: milton ruser
Betreff: Re: [R] mean
An: "amor Gandhi"
CC: r-h...@stat.math.ethz.ch
Datum: Montag, 8. Juni 2009, 14:27
oops.
If x1 is the individual try
x2.mean<-aggregate(data[2:2], list(x1), mean)
x2.mean
you can change "mean" by
Dear Marie,
Try this:
combn(v,2,list)
HTH,
Jorge
On Mon, Jun 8, 2009 at 8:56 AM, Marie Sivertsen wrote:
> Dear list,
>
> I have a vector of elements which I want to combined each with each, but
> none with itself. For example,
>
> > v <- c("a", "b", "c")
>
> and I need a function 'combine' su
Dear list,
I have a vector of elements which I want to combined each with each, but
none with itself. For example,
> v <- c("a", "b", "c")
and I need a function 'combine' such that
> combine(v)
[[1]]
[1] "a" "b"
[[2]]
[1] "a" "b"
[[3]]
[1] "b" "c"
I am not very interested in the orders of th
Ben Bolker wrote:
amor Gandhi wrote:
Hi,
I have gote the following data
x1 <- c(rep(1,6),rep(4,7),rep(6,10))
x2 <- rnorm(length(x1),6,1)
data <- data.frame(x1,x2)
and I would like to compute the mean of the x2 for each individual of x1,
i. e. x1=1,4 and 6?
You'll probably get sev
oops.
If x1 is the individual try
x2.mean<-aggregate(data[2:2], list(x1), mean)
x2.mean
you can change "mean" by any function.
cheers
milton
On Mon, Jun 8, 2009 at 8:20 AM, milton ruser wrote:
> Hi Amor,
> I think you forgot to include the individual ID.
> ?aggregate
>
> cheers
> milton
> br
Hi Amor,
I think you forgot to include the individual ID.
?aggregate
cheers
milton
brazil=toronto
On Mon, Jun 8, 2009 at 8:11 AM, amor Gandhi wrote:
> Hi,
>
> I have gote the following data
>
> x1 <- c(rep(1,6),rep(4,7),rep(6,10))
> x2 <- rnorm(length(x1),6,1)
> data <- data.frame(x1,x2)
>
> an
amor Gandhi wrote:
>
> Hi,
>
> I have gote the following data
>
> x1 <- c(rep(1,6),rep(4,7),rep(6,10))
> x2 <- rnorm(length(x1),6,1)
> data <- data.frame(x1,x2)
>
> and I would like to compute the mean of the x2 for each individual of x1,
> i. e. x1=1,4 and 6?
>
You'll probably get seven
sorry, that should be
tapply(x2, x1, mean)
Best,
Dimitris
Dimitris Rizopoulos wrote:
have a look at ?tapply(), e.g.,
x1 <- c(rep(1,6),rep(4,7),rep(6,10))
x2 <- rnorm(length(x1),6,1)
tapply(x1, x1, mean)
I hope it helps.
Best,
Dimitris
amor Gandhi wrote:
Hi,
I have gote the following
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