On Mon, 3 Mar 2008, Nordlund, Dan (DSHS/RDA) wrote:
[..., quoting Hadley Wickham]
gc()
>>> used (Mb) gc trigger (Mb) max used (Mb)
>>> Ncells 133095 3.6 35 9.4 35 9.4
>>> Vcells 87049 0.7 786432 6.0 478831 3.7
a <- runif(1e7)
gc()
>>>
The axis labels go in the margins -- you left no space for the margins
when you asked for mar = c(0,0,0,0) and hence the labels were clipped to
invisibility.
It you want to write in the outer margins you have to use mtext() or
title() with outer=TRUE.
Looking at the pictures in 'An Introductio
On Tue, 4 Mar 2008, Thomas Allen wrote:
Hi
I came across this strange bug the other day, I'm not sure how to solve it
and I wonder if anyone can even replicate it.
Using OS Ubuntu 7.10
Step 1) Make an R package using the package.skeleton() command with
only these two functions:
error <- func
I'm searching for a statistical regression technique to be applied to data
relating to the proportion of area.
Specifically the data i'm trying to analyse relates to how much fuel was burnt
in response to fire and recorded as a percentage consumed.
Any ideas or suggestions would be greatly apprec
(Ted Harding) <[EMAIL PROTECTED]> wrote in
news:[EMAIL PROTECTED]:
> I just realised I made a bad mistaqke (see below)
>
> On 21-Feb-08 21:39:56, Ted Harding wrote:
>> On 21-Feb-08 20:58:25, Evgenia wrote:
>>>
>>> Dear R users,
>>> I would like to sample from a mixture distribution
>>> p1*f1+p
Hello list,
I am using R 2.4.1 on MacOSx. I've been exploring the seas package (0.3-5)
and am having a problem with plot.interarrival. I have modified the
example below by using width = "mon". This works out well except for the
tick mark labels on the x-axis. The x-axis should be labeled from 1-
Dear all,
Though labeling the x and y axis in the plot command seems to be straight
forward, I can not get it to work if I do the following:
## Creating example data
edata <- c(1,2,1,2)
edata <- matrix(edata, 2, 2, byrow = T)
colnames(edata) <- c("a", "b")
edata <- data.frame(edata)
## p
Hi Ista,
That's a bug, and will be fixed in the next version (hopefully to be
released in the next couple of days)
Hadley
On Mon, Mar 3, 2008 at 7:10 PM, Ista Zahn <[EMAIL PROTECTED]> wrote:
> Hi,
> I've run into a difficulty with qplot function (in the ggplot2
> package). I can facet histogra
Thanks for your reply. For each condition, I will have a matrix or data
frames of 1000 rows and 4 columns. I also have a total of 64 conditions for
now. So, in total, I will have 64 matrices or data frames of 1000 rows and 4
columns. The format of data I would like to store would be data frames or
What is the format of the data you are storing (single value,
multivalued vector, matrix, dataframe, ...)? This will help formulate
a solution. What do you plan to do with the data? Are you going to
do further analysis, write it to flat files, store it in a data base,
etc.? How big are the data
Hi, I would first look into the many manuals that you can get in the
"Manuals" section of the cran-project page. Click the link contributed
documentations and explore from there. There are also quite a few websites
that give more insight into using R which you may google. If you have a
better idea
G'day Thomas,
On Tue, 4 Mar 2008 14:40:35 +1300
"Thomas Allen" <[EMAIL PROTECTED]> wrote:
> I came across this strange bug the other day, I'm not sure how to
> solve it and I wonder if anyone can even replicate it.
> Step 1) Make an R package using the package.skeleton() command with
> only thes
DateTime,Temp,SpCond,DOConc,Depth,pH,ORP,Turbidity+,Chlorophyll,Battery,Cond,DO%,Salinity,TDS
01/13/2006 17:01,10.87,84,9.36,0.664,7.3,132,28.8,3.1,11.5,0.062,84.6,0.04,0.055
01/13/2006 17:16,10.9,84,9.36,0.66,7.31,133,28.7,2.9,11.5,0.062,84.7,0.04,0.055
01/13/2006 17:31,10.92,84,9.36,0.655,7.3,132
Hi
I came across this strange bug the other day, I'm not sure how to solve it
and I wonder if anyone can even replicate it.
Using OS Ubuntu 7.10
Step 1) Make an R package using the package.skeleton() command with
only these two functions:
error <- function(){
cmd <- paste(" -a ",1," -a ",1,"
Try this:
> a <- 1:2
> b <- 1:3
> g <- array(do.call(c, apply(expand.grid(a = a, b = b), 1, list)),
+dim = c(length(a), length(b)))
> g
[,1] [,2] [,3]
[1,] Integer,2 Integer,2 Integer,2
[2,] Integer,2 Integer,2 Integer,2
> g[[2,3]]
a b
2 3
On Mon, Mar 3, 2008 at 8:21 PM, Vadim
Hi Flo,
Ckeck ?agrep
HTH.
Jorge
On 3/3/08, Flo <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> How can I solve this problem?
> I have to find in a list all the words which have the same letters, but
> one must be different.
> Ex "pain": rain, pine...
>
> I hope you will understand my poor english! Tha
Davood,
I developed an MC simulation model for wind
hazard analysis last year. I found three important
issues to increase efficiency:
1) Reuse most variables in each loop
2) Write results (1000 stats) to external files, perhaps
one file for each condition.
(a good ID for each file can be implement
Dear R Users,
Given two vectors, say a = seq(2) and b = seq(3), I want to make an 2*3 array,
where (i,j) element is list(a=a[i], b=b[j]). I tried the outer() function but
it generates an error message that I don't understand, see below.
What do I do wrong?
The expan.grid function is not
This is a nice list:
http://www.amazon.com/Use-R/lm/RNFBA3UHW2M73/ref=cm_lmt_srch_f_1_rsrsrs0
Tom
kayj wrote:
>
> Hi All,
>
> I am a new user in R and I would like to buy a book that teaches me how
> to use R. In addition, I may nees to do some advanced statistical
> analysis. Does anyone
Hi,
I've run into a difficulty with qplot function (in the ggplot2
package). I can facet histograms even when the faceting variable
contains missing values, but only so long as the faceting variable is
not a factor.
Example:
y1 <- rnorm(10)
x1 <- c(rep(1,5), rep(2,4), NA)
x2 <- factor(c(rep
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Rolf Turner
> Sent: Monday, March 03, 2008 4:37 PM
> To: hadley wickham
> Cc: r-help; [EMAIL PROTECTED]
> Subject: Re: [R] renaming objects
>
>
> On 4/03/2008, at 11:48 AM, hadley wickham wrote:
>
>
Use
mat <- eig[["vectors"]]
If x is a list, ``x[i]'' gives you a (sub-) list of length 1,
whereas ``x[[i]]'' gives you the single element of that list.
A possibly subtle but important distinction.
cheers,
Rolf Turner
On 4/03/2008, at 12:42 PM, SS-R wrote:
>
> H
On 4/03/2008, at 11:48 AM, hadley wickham wrote:
> On Mon, Mar 3, 2008 at 4:37 PM, Rolf Turner
> <[EMAIL PROTECTED]> wrote:
>>
>> On 4/03/2008, at 10:38 AM, Ericka Lundström wrote:
>>
>>> On 03/03/2008, at 22.20, [EMAIL PROTECTED] wrote:
Is there a way to rename R objects?
I am looki
It could be done like this:
> a <- if (TRUE) . ~ . + freeny.x[,4] else . ~ .
> fo <- update(freey.y ~ freeny.x[, -4], a)
> fo
freey.y ~ freeny.x[, -4] + freeny.x[, 4]
On Mon, Mar 3, 2008 at 6:24 PM, Rolf Turner <[EMAIL PROTECTED]> wrote:
>
> Matthieu wants to manipulate the *formula* --- i.e. co
Hi
I need to extract the data returned by Eigen to plot the eigenvectors.
However, when I try and eigv = eigen(covariance); it returns an object with
the matrices containing eigenvalues and vectors.. how can I extract the
eigenvector matrix from this??
When I try mat = eig["vectors"] it returns a
Dear All,
I am running a Monte Carlo simulation study and have some questions on how
to manage data storage efficiently at the end of each 1000 replication loop.
I have three conditions coded using the FOR {} loops and a FOR loop that
generates data for each condition, performs analysis, and compu
On Mar 3, 2008, at 4:59 PM, Peter Dalgaard wrote:
> Andrew Robinson wrote:
>> On Mon, Mar 03, 2008 at 10:22:41PM +0100, Peter Dalgaard wrote:
>>
>>> Patrick Burns wrote:
>>>
Douglas Bates wrote:
> On Mon, Mar 3, 2008 at 8:25 AM, Duncan Murdoch <[EMAIL PROTECTED]
> > w
Here is how you can get the proportions from your data frame:
> prop.table(table(paste(x$testdate, x$testtime), x$Behavior),margin=1)
EA FL HO MA OS
PE SI
28Mar96 1014 0. 0. 0.5000 0. 0.
0.
Matthieu wants to manipulate the *formula* --- i.e. control
what predictors go into the formula --- rather than to add something
to one of the predictors (or not, as the case maybe).
I.e. what is wanted is that if ``condition'' is TRUE then the
formula should be
freeny.y ~ freeny.x[,-4]
Terry Therneau wrote:
>
> It is easier to get survival curves using the predict function. Here is a
> simple example:
>> tfit <- survreg(Surv(time, status) ~ factor(ph.ecog), data=lung)
>> tdata <- data.frame(ph.ecog=factor(0:3))
>> qpred <- predict(tfit, newdata= tdata, type='quantile', p=1:99/
or, in other terms:
> set.seed(1)
> x <- runif(1e7)
> x.add <- tracemem(x)
> y <- x
> y.add <- tracemem(y)
> identical(x.add, y.add)
[1] TRUE
ie, both objects have the same memory address - therefore, not a copy:
> x.add
[1] "<0x200>"
> y.add
[1] "<0x200>"
now, observe what happens whe
This is interesting!
> gc()
used (Mb) gc trigger (Mb) max used (Mb)
Ncells 151051 4.1 35 9.4 35 9.4
Vcells 94969 0.8 20380703 155.5 30094956 229.7
> a <- runif(1e07)
> gc()
used (Mb) gc trigger (Mb) max used (Mb)
Ncells 151046 4.1 35 9.4
This does not really answer your question but the following
lets you refer to an object by a second name although the object
still has its original name as well.
> a <- 3
> makeActiveBinding("b", function() a, .GlobalEnv)
NULL
> b
[1] 3
> a <- 4
> b
[1] 4
On Mon, Mar 3, 2008 at 4:20 PM, <[EMAIL
Andrew Robinson wrote:
> On Mon, Mar 03, 2008 at 10:22:41PM +0100, Peter Dalgaard wrote:
>
>> Patrick Burns wrote:
>>
>>> Douglas Bates wrote:
>>>
>>>
>>>
On Mon, Mar 3, 2008 at 8:25 AM, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 3/
Dear R,
I am using stepwise multiple glm to select a subset of variables using the step
command. My question is how do you calculate the percentage variance for each
parameter in the reduced model. An example of my code thus far is,
fit<-glm(Abalone~offset(Area)+Sessile invertebrates+Bare rock+
On Mon, Mar 3, 2008 at 4:37 PM, Rolf Turner <[EMAIL PROTECTED]> wrote:
>
> On 4/03/2008, at 10:38 AM, Ericka Lundström wrote:
>
> > On 03/03/2008, at 22.20, [EMAIL PROTECTED] wrote:
> >> Is there a way to rename R objects?
> >> I am looking for a way to rename objects
> >> without making new o
On Mon, Mar 03, 2008 at 10:22:41PM +0100, Peter Dalgaard wrote:
> Patrick Burns wrote:
> > Douglas Bates wrote:
> >
> >
> >> On Mon, Mar 3, 2008 at 8:25 AM, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> >>
> >>
> >>
> >>> On 3/3/2008 9:10 AM, Rogers, James A [PGRD Groton] wrote:
> >>>
All,
I have a workspace containing only data frame objects. I would like to
loop though each one and clean-up text columns in them. How can I have
R loop through the list? I have tried to find an answer in R help but
the closest solution I can find is to make a static list of data
frames, as illus
On 4/03/2008, at 10:38 AM, Ericka Lundström wrote:
> On 03/03/2008, at 22.20, [EMAIL PROTECTED] wrote:
>> Is there a way to rename R objects?
>> I am looking for a way to rename objects
>> without making new objects.
>>
>> #For example:
>> x = c(1:40)
>> # I wish to use a function to rename x, al
>From: Matthieu Stigler <[EMAIL PROTECTED]>
>Date: 2008/03/03 Mon PM 04:07:09 CST
>To: r-help@r-project.org
>Subject: [R] How to include an externally defined NULL value in lm
maybe you should a<-0 unless there's special
behavior of NULL that's unknownst to me ?
>Hello!
>
>I would love to be abl
UseRs,
I am working on a dataset (see small example below) where individuals
were followed on a specific date-time combo and multiple repeated
measurements were taken (e.g., height in meters, behavior class in 2
letter code). Observation numbers varied between individual (ranging
from 1 obser
See ?write.zoo
e.g.
write.zoo(z1, file = "") # display on console
On Mon, Mar 3, 2008 at 3:55 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> # chron
> library(chron)
> fmt.chron <- function(x) {
> chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x))
> }
> z1 <- read.zoo(SC3.csv, sep = ","
Hello!
I would love to be able to include an external variable to a lm call, I
mean something:
if(TRUE)
a<-freeny.x[,4]
else
a<-NULL
lm(freeny.y~freeny.x[,-4] +a)
but it does not work with a<-NULL, whereas
lm(freeny.y~freeny.x[,-4] +NULL)
I don't understand why and did not find an a
On 03/03/2008, at 22.20, [EMAIL PROTECTED] wrote:
> Is there a way to rename R objects?
> I am looking for a way to rename objects
> without making new objects.
>
> #For example:
> x = c(1:40)
> # I wish to use a function to rename x, already created, to y,
> perhaps by
> obj.rename(x,y)
> # or
>
Hello Kayj,
There are very good tutorials at the R website. See here:
http://cran.r-project.org/other-docs.html
Julian
kayj wrote:
> Hi All,
>
> I am a new user in R and I would like to buy a book that teaches me how to
> use R. In addition, I may nees to do some advanced statistical analysi
Is there a way to rename R objects?
I am looking for a way to rename objects
without making new objects.
#For example:
x = c(1:40)
# I wish to use a function to rename x, already created, to y, perhaps by
obj.rename(x,y)
# or
obj.rename("x","y")
There are numerous examples of renaming
variables
Patrick Burns wrote:
> Douglas Bates wrote:
>
>
>> On Mon, Mar 3, 2008 at 8:25 AM, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
>>
>>
>>
>>> On 3/3/2008 9:10 AM, Rogers, James A [PGRD Groton] wrote:
>>>
As someone of partly French heritage, I would also ask how this
distribu
It works.
Thank you so much.
Abou
==
AbouEl-Makarim Aboueissa, Ph.D.
Assistant Professor of Statistics
Department of Mathematics & Statistics
University of Southern Maine
96 Falmouth Street
P.O. Box 9300
Portland, ME 04104-9300
Tel: (207) 228-8389
Fax: (207) 780-5607
E
# chron
library(chron)
fmt.chron <- function(x) {
chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x))
}
z1 <- read.zoo(SC3.csv, sep = ",", header = TRUE, FUN = fmt.chron)
z2 <- read.zoo(SC2.csv, sep = ",", header = TRUE, FUN = fmt.chron)
z3<-c(z1, z2)
write.table(z3, sep="," , "SC.csv")
Ho
Begin forwarded message:
> Subject: R: Studdy Missing Data, differentiate between a percent
> with in the valid answers and with in the different missing answers
> take a look at the Website from Dirk Enzmann:
> http://www2.jura.uni-hamburg.de/instkrim/kriminologie/Mitarbeiter/
> Enzmann/Softwa
Thanks. I found the issue. I was starting R in a terminal with the
command 'sudo R', help.start() was not able to communicate Firefox
when already running. Works fine when R is started as user. --Dale
On Mon, Mar 3, 2008 at 3:09 PM, Marc Schwartz <[EMAIL PROTECTED]> wrote:
> Patrick Connolly w
I guess you are looking for the number of rows satisfying the following
condition.
Assuming that you have a cutoff k = 1, 2, 5, 10, 15
1) x[i,1] < k regardless of the value of x[i,2], OR
2) x[i,1] == k and x[i,2] == 1
Here is my take on this problem. It is not elegant but it seems to do the
job.
Patrick Connolly wrote:
> On Sun, 02-Mar-2008 at 05:32PM -0500, Dale Steele wrote:
>
> |> Using linux fedora 8 (x86_64) I get the following when firefox is
> |> already running. Is there a way to adjust settings in either R or
> |> firefox to open a new tab when help.start() is invoked? Thanks.
On Mon, Mar 3, 2008 at 11:01 AM, **linda** <[EMAIL PROTECTED]> wrote:
>
> Hello,
> I am trying to make a barplot with nested count data which is build like
> this: first there are several birds (n)laying 3 clutches composed of 2 eggs
> half of the second and third clutch received treatment and
Using t.test() in this case will likely be very slow. A faster
alternative would be to use rowttests() from the genefilter package of
Bioconductor.
Best,
Jim
Henrique Dallazuanna wrote:
> You can try this:
>
> cbind(data.sub, p.value=apply(data.sub, 1, function(x)t.test(x)$p.value))
>
> On 0
On Sun, 02-Mar-2008 at 05:32PM -0500, Dale Steele wrote:
|> Using linux fedora 8 (x86_64) I get the following when firefox is
|> already running. Is there a way to adjust settings in either R or
|> firefox to open a new tab when help.start() is invoked? Thanks.
Using FC6, it works as it says, e
On 3/3/2008 2:27 PM, kayj wrote:
> Hi All,
>
> I am a new user in R and I would like to buy a book that teaches me how to
> use R. In addition, I may nees to do some advanced statistical analysis.
> Does anyone recommend some books or websites where I can learn R.
I would start with "An Intr
Hi All,
I am a new user in R and I would like to buy a book that teaches me how to
use R. In addition, I may nees to do some advanced statistical analysis.
Does anyone recommend some books or websites where I can learn R.
Thanks
--
View this message in context:
http://www.nabble.com/I-ne
Hi,
How can I solve this problem?
I have to find in a list all the words which have the same letters, but one
must be different.
Ex "pain": rain, pine...
I hope you will understand my poor english! Thank you, Flo
-
Douglas Bates wrote:
>On Mon, Mar 3, 2008 at 8:25 AM, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
>
>
>>On 3/3/2008 9:10 AM, Rogers, James A [PGRD Groton] wrote:
>> > As someone of partly French heritage, I would also ask how this
>> > distribution came to be called "Gaussian". It seems very unfai
There was some discussion a while back of an R pgf driver
for latex. Is anyone working on that?
On Mon, Mar 3, 2008 at 2:09 PM, Greg Snow <[EMAIL PROTECTED]> wrote:
> The gp.plot function has a type option that can either be 'p' for points
> or 'l' for lines, the function is not that complicated,
Thanks for all response. Here is more closer to my question,
p<-0.05
f<-0.05
ld<-sqrt(f* (1 - f) * p * (1 - p))
D <- (p * (1 - f) - ld)/p
d <- ((1 - p) * f - ld)/(1 - p)
haplo<-data.frame(D,d)
> haplo
D d
1 -1.387779e-16 -7.304099e-18
cond1<-pmin(haplo[,1],haplo[,
The gp.plot function has a type option that can either be 'p' for points
or 'l' for lines, the function is not that complicated, you could easily
add additional options.
I don't know why your code is not creating the pdf file, you could try
plotting to the screen first (don't change the output and
--- John Fox <[EMAIL PROTECTED]> wrote:
> Dear Doug,
>
> As I recall, according to Stigler, yes -- he wasn't
> the first to
> formulate Stigler's law of eponymy (but I don't
> recall to whom he
> attributed it).
Possibly a disgruntles M. de Moivre?
>
> Regards,
> John
>
> On Mon, 3 Mar 2
On 03/03/08 12:52, Xuejun Qin wrote:
> Hi, there,
> I cannot get accurate value for calculation.
> for example:
> ld<-sqrt(1*0.05*0.95*0.05*0.95)
> 0.05*0.95-ld=-6.938894e-18
> 0.05*0.95-ld==0 is False.
>
> I met this problem in my program, how can I handle it. Thanks.
I think what you are ex
Hi,
Is there any package which provides the functions of create one
dimensional and/or Two dimensional classifiers?
Thanks much.
--
Waverley @ Palo Alto
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
Hi R,
I am getting this error while trying to use 'lrm' function with nine
independent variables:
> res =
lrm(y1994~WC08301+WC08376+WC08316+WC08311+WC01001+WC08221+WC08106+WC0810
1+WC08231,data=y)
singular information matrix in lrm.fit (rank= 8 ). Offending
variable(s):
WC08101 WC0822
Dear Doug,
As I recall, according to Stigler, yes -- he wasn't the first to
formulate Stigler's law of eponymy (but I don't recall to whom he
attributed it).
Regards,
John
On Mon, 3 Mar 2008 12:17:59 -0600
"Douglas Bates" <[EMAIL PROTECTED]> wrote:
> On Mon, Mar 3, 2008 at 8:25 AM, Duncan Mur
On Mon, Mar 3, 2008 at 9:57 AM, Charilaos Skiadas <[EMAIL PROTECTED]> wrote:
> Nothing's wrong. It just means that the package or one of its
> dependencies, has its own xtabs function, which "hides" the default
> xtabs function, since it comes first in the search path. So when you
> next write x
Yes, the sociologist Robert Merton.
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Mar 3, 2008, at 12
apparently you want to check the genefilter package...
it defines functions like:
rowttests
colttests
rowFtests
colFtests
rowVars
rowSds
moreover, a quick look at Biobase is recommended...
that would save you lots of time as you wouldn't have to reinvent the
wheel.
b
On Mar 3, 2008, at
Hey Flo,
I'm not entirely sure what you're problem is -- due to the language thing,
probably -- but I think you want a function that will give you (or computes
in an intermediate step) the "edit distance" between strings. There is the
"sdists" function in the cba package that might help (it can g
If I understand you correctly what you want to do is
do t-test (mu=0) for each column of the data.
Treating the data as a data.frame rather than a matrix
you can do something like this and then pick out the
p-values but with 140 t-tests I don't know what you'll
get in terms of anything meaninful.
On Mon, Mar 3, 2008 at 8:25 AM, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 3/3/2008 9:10 AM, Rogers, James A [PGRD Groton] wrote:
> > As someone of partly French heritage, I would also ask how this
> > distribution came to be called "Gaussian". It seems very unfair to de
> > Moivre, who disc
R FAQ 7.31. G.
On Mon, Mar 03, 2008 at 12:52:43PM -0500, Xuejun Qin wrote:
> Hi, there,
> I cannot get accurate value for calculation.
> for example:
> ld<-sqrt(1*0.05*0.95*0.05*0.95)
> 0.05*0.95-ld=-6.938894e-18
> 0.05*0.95-ld==0 is False.
>
> I met this problem in my program, how can I handl
Xuejun Qin wrote:
> Hi, there,
> I cannot get accurate value for calculation.
> for example:
> ld<-sqrt(1*0.05*0.95*0.05*0.95)
> 0.05*0.95-ld=-6.938894e-18
> 0.05*0.95-ld==0 is False.
>
> I met this problem in my program, how can I handle it. Thanks.
>
>
> xj.
>
> ___
On Mon, 2008-03-03 at 12:52 -0500, Xuejun Qin wrote:
> Hi, there,
> I cannot get accurate value for calculation.
> for example:
> ld<-sqrt(1*0.05*0.95*0.05*0.95)
> 0.05*0.95-ld=-6.938894e-18
> 0.05*0.95-ld==0 is False.
>
> I met this problem in my program, how can I handle it. Thanks.
Answer 1
FAQ 7.31 (You need to understand what floating point numbers are)
On 3/3/08, Xuejun Qin <[EMAIL PROTECTED]> wrote:
> Hi, there,
> I cannot get accurate value for calculation.
> for example:
> ld<-sqrt(1*0.05*0.95*0.05*0.95)
> 0.05*0.95-ld=-6.938894e-18
> 0.05*0.95-ld==0 is False.
>
> I met this
ICANN 2008EXTENDED SUBMISSION DEADLINE MARCH 10 2008
Dear Colleague
Due to numerous requests we decided to extend the submission deadline
for
HI, Thanks for your help previously. I can go around R and scilab now, except
import my research data.
This is an example dataset (below) of radon radiation levels. How can I
import this txt file using Rcmdr?
How do I import .xls files? My alpha guard radon monitors output .dvd files
that can b
Hi, there,
I cannot get accurate value for calculation.
for example:
ld<-sqrt(1*0.05*0.95*0.05*0.95)
0.05*0.95-ld=-6.938894e-18
0.05*0.95-ld==0 is False.
I met this problem in my program, how can I handle it. Thanks.
xj.
__
R-help@r-project.org mai
You can try this:
cbind(data.sub, p.value=apply(data.sub, 1, function(x)t.test(x)$p.value))
On 03/03/2008, Keizer_71 <[EMAIL PROTECTED]> wrote:
>
> Hi Everyone,
>
> I need some simple help.
>
> Here are my codes
>
> ##will give me 1 probesets
> data.sub = data
G'day Carlos,
On Mon, Mar 3, 2008 at 11:52 AM
Carlos Alzola <[EMAIL PROTECTED]> wrote:
> I am trying to get information on how to fit a linear regression
> with constrained parameters. Specifically, I have 8 predictors ,
> their coeffiecients should all be non-negative and add up to 1. I
> unde
Hi Christophe --
This is a variant of the problem that Jim Regetz is having in a thread
in R-devel. Here's where the trouble is
> as(c, "A")
Error in .local(.Object, ...) :
argument "value" is missing, with no default
By default, 'as(c, "A")' will create a new instance of it's second argument
I believe I wrote too hastily and that what you want is sum(X < 1) which
will sum the indicator (T/F) function.
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL
perhaps you need something like sum(X[X < 1])
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of AbouEl-Makarim Aboueissa
Sent: Monday, M
Hi Everyone,
I need some simple help.
Here are my codes
##will give me 1 probesets
data.sub = data.matrix[order(variableprobe,decreasing=TRUE),][1:1,]
dim(data.sub)
data_output<-write.table(data.sub, file = "c://data_output.csv", sep = ",",
col.names = NA)
Dear ALL:
Please see below. I hope this will make it more clear.
[1,]11
[2,]11 number of all observations less than 1 with
indicator 1 (including those 1 with indicator 1 but not 1 with
indicator 0)=2
[3,]10
[4,]10
[5,]10
[6,]10
[7,
Hello,
I am trying to make a barplot with nested count data which is build like
this: first there are several birds (n)laying 3 clutches composed of 2 eggs
half of the second and third clutch received treatment and this treatment
was tested to influence sex of offspring. I want a barplot showing c
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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-guide.html
and provide commented, minimal, self-contained, reproducible code.
One must specify that there is a header if there is one, each % item
should only
be listed once, and format is currently only used for "Date" class so
you want to
use the FUN= argument -- although in the development version of zoo format is
availailable more generally. Here are several ways to do t
For me is not very clear, but if I understand:
sapply(sort(unique(data[data[,2]==1,1])),
function(x)sum(data[data[,2]==1 & data[,1] <= x, 1]))
But the output is:
2 6 31 71 86
On 03/03/2008, AbouEl-Makarim Aboueissa <[EMAIL PROTECTED]> wrote:
> Dear Ellison:
>
> it did not do i
Ciao a tutti!
ho un problema con un esercizio in R.
In una lista devo trovare tutte le parole che si differenziano per una
lettera da un'altra parola.
Esempio: per "casa": cosa, rosa...
C'è una formula generale da applicare a tutte le parole?
Grazie dell'aiuto, Flo
-
> Workshop on user interfaces and interactive graphics in R
> ==
> Tuesday, 1st April 10.00-16.30 Postgraduate Statistics Centre,
> Department of Maths and Statistics, Lancaster University, UK
> =
> This one-day work
On Mar 3, 2008, at 11:12 AM, stephen sefick wrote:
> x<-read.zoo("SC2.csv", sep="," , format="%m%m/%d%d/%y%y%y%y %h%h:%m%
> m")
>
> #Error in read.zoo("SC2.csv", sep = ",", format =
> "%m%m/%d%d/%y%y%y%y %h%h:%m%m") :
> index contains NAs Error message
You need header=TRUE in th
Dear Ellison:
it did not do it.
I edited my previous email to make my question more clear.
The out put should be: (2,11,33,43,46)
For example:
number of all observations less than 1 with indicator 1 (including those 1 with
indicator 1 but not 1 with indicator 0) =2
num
Nothing's wrong. It just means that the package or one of its
dependencies, has its own xtabs function, which "hides" the default
xtabs function, since it comes first in the search path. So when you
next write xtabs(...), it is this new xtabs that is being loaded. If
you want to call the or
Dear all,
I've been trying to install the lme4 package from
http://r-forge.r-project.org/projects/lme4/. However, when wanting to load
the package, I get an message saying that "x-tabs" are masked (see pasted
code). Can anyone point to what has gone wrong?
Kind regards,
Andreas Nord
Sweden
>
> "BDR" == Prof Brian Ripley <[EMAIL PROTECTED]>
> on Mon, 3 Mar 2008 14:16:15 + (GMT) writes:
[]
BDR> In a later message Louise mentioned the desire to use TeX fonts for
BDR> annotation, to match a LaTeX document. Paul Murrell has pointed out
his
BDR>
Dear Carlos,
One approach is to use structural equation modeling (SEM). Some SEM
packages, such as LISREL, Mplus and Mx, allow inequality and nonlinear
constraints. Phantom variables (Rindskopf, 1984) may be used to impose
inequality constraints. Your model is basically:
y = b0 + b1*b1*x1 + b2*b2*
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