2008/10/21 Marcioestat <[EMAIL PROTECTED]>:
>
> Hi listers,
> I am working on a program of statistical analysis of simulated data and I've
> been searching the error at the program, but I didn't find it!
> It is something about the WHILE procedure, the error says: Error in while
> (ecart >= d) { :
Call For Posters:
By now, there's a good chance you've heard about USAR2009 - R/R-PLUS/S
Conference.
R-PLUS is the next evolution of R: Interactive, Comprehensive, Highly
Visual.
Join us in Las Vegas, USA for the USAR2009 held at Ceasar's Palace on
April 26-29, 2009.
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leo_wa hotmail.com> writes:
>
> i want to plot the histogram and the curve in the same graph.if i have a
> set of data ,i plot the histogram and also want to see what distribution
> it was.So i want to plot the curve to know what distribution it like.
See the example under histogram/lattice
Scotty Nelson hotmail.com> writes:
> I'm trying to build a for loop, where I estimate a series of models with
> different sets of (time series) data.
> However my for loop doesn't recognize the "i"
>
> # code
> window.1=anomalies.CAK[(positions(anomalies.CAK)
Hello,
I'm trying to build a for loop, where I estimate a series of models with
different sets of (time series) data.
However my for loop doesn't recognize the "i"
# code
window.1=anomalies.CAK[(positions(anomalies.CAK)>=timeDate("1/1/1971") &
positions(anoma
Most likely this is a bug in your pdf viewer: try turning off
anti-aliasing there (or use a better viewer, if that is not an option).
It is a symptom of anti-aliasing of the rectangles used to plot image
pixels.
You haven't told us which 'a pdf device' and there are several for R,
depending on
Greg,
> The rimage package has functions for reading in and plotting
> jpeg files that you could use for displaying the photograph.
> If you then can find 2 points in the image (not on the same
> horizontal or vertical line) for which you know the
> coordinates in the coordinate system that y
Hi listers,
I am working on a program of statistical analysis of simulated data and I've
been searching the error at the program, but I didn't find it!
It is something about the WHILE procedure, the error says: Error in while
(ecart >= d) { : missing value where TRUE/FALSE needed
Thanks in advance
i want to plot the histogram and the curve in the same graph.if i have a set
of data ,i plot the histogram and also want to see what distribution it
was.So i want to plot the curve to know what distribution it like.
--
View this message in context:
http://www.nabble.com/how-to-plot-the-histogram
Try:
as.vect = function(x)
{
t1 = as.matrix(x)
t2 = aperm(t1)
t3 = as.vector(t2)
return(t3)
}
This function converts entire dataframe into a vector.
Alexander Pui
PhD Candidate in Hydroclimatology,
University of New South Wales
Email: [EMAIL PROTECTED]
_
> Well, I am going to type in ever value because the data sheets are of
> counts of insects that I identified, so I should be okay with
> accuracy... I really just need something that allows for more than
> 256 columns as I have encounter over 256 species of insects in even
> small streams. ...
O
How about simply a text editor, typing in your data, separated by commas
or tabs or spaces? One row for each case/subject/observation? R can
read that in easily.
A good, open-source, free data entry program is EpiData.
www.epidata.dk. It is simple to use but probably more than you need for
task
As interest in topics that are often grouped under the catch-all umbrella of
'high performance computing with R' has grown over the last few months, it has
been suggested to create a dedicated mailing list.
Now, thanks to Martin Maechler who tirelessly looks after all R mailing
matters, such a
I nominate the below 2 paragraphs (or a possible shortening of them) as a new
fortune. While not as entertaining as many of the current fortunes, the wisdom
gained and sentiment expressed deserves preservation and easy reference for
future posters who think that Frank is only trying to be funny
Well, I am going to type in ever value because the data sheets are of
counts of insects that I identified, so I should be okay with
accuracy... I really just need something that allows for more than
256 columns as I have encounter over 256 species of insects in even
small streams. I think calc wi
There is a list of free spreadsheets with their row and column limits
at this link:
http://en.wikipedia.org/wiki/OpenOffice.org_Calc
On Mon, Oct 20, 2008 at 3:13 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> sorry excel 2003 with no immediate update in the future.
>
> On Mon, Oct 20, 2008 at 3:1
Hi Stephen,
You don't say what "staging" is - do you mean for data entry or loading a
data file for review, or ... ?
In general, I keep away from Excel for data transfer purposes. It tends to
make "intelligent" decisions on data types leading to strange & bizarre
results (unless you explicitly ty
Dear list members,
I am using Mclust in order to deconvolute a distribution that I
believe is a sum of two gaussians.
First I can make a model:
> my.data.model = Mclust(my.data, modelNames=c("E"), warn=T, G=1:3)
But then, when I try to plot the result, I get the following error:
> mclust1Dplot(
Apologies, just a typo in the first instruction (when translating the
names), the question is still valid
On 21 Oct 2008, at 00:38, Ivan Alves wrote:
Dear all,
I would like to aggregate a data frame (consisting of 2 columns - one
for the bins, say factors, and one for the values) along bins
tIs a bit neater to create list objects for each of
star1 and star2 and then, since the various functions
are not used outside of gravitation stick them right
in gravitation. Also we can use distance for len
as well by giving it a default second arg of 0 and
sqr does not seem to do much so we can
Dear all,
I would like to aggregate a data frame (consisting of 2 columns - one
for the bins, say factors, and one for the values) along bins and
quantiles within the bins.
I have tried
aggregate(data.frame$values, list(bin = data.frame
$bin,Quantile=cut2(data.frame$bin,g=10)),sum)
but the
Hi all,
I have a queston about Categorical response.
i have a data frame containing age, sex, class, success(1=success,
0=non sucess).
age, sex,class are the explantory variables, and sucess is the
response variable. and i can get n (the nunber of times each age
occurs) and r (the number of suce
On 20-Oct-08 21:17:22, Gabor Grothendieck wrote:
> Try this:
> with(data.frame(x = 0:1, times = 3:6), rep(x, times))
>
> or even shorter:
> do.call(rep, data.frame(x = 0:1, times = 3:6))
That is sneaky!
data.frame(x = 0:1, times = 3:6)
# x times
# 1 0 3
# 2 1 4
# 3 0 5
# 4 1
On 20-Oct-08 21:19:21, Stefan Evert wrote:
> On 20 Oct 2008, at 22:57, (Ted Harding) wrote:
>> I'm wondering if there's a compact way to achieve the
>> following. The "dream" is that one could write
>>
>> rep(c(0,1),times=c(3,4,5,6))
>>
>> which would produce
>>
>> # [1] 0 0 0 1 1 1 1 0 0 0 0 0 1
Why not just write your data to a CSV (comma-spaced-variable) or a
tab-spaced variable text file?
You didn't say what software and/or hardware was generating your data,
but most gizmos these days let you dump data to CSV.
No need for Excel at all.
I forget :-( how many rows/columns OpenOffice
Here is one other solution:
x <- 0:1
times <- 3:6
rep(x + 0*times, times)
This solution also works if the length of times is not a whole
number of lengths of x but in that case it does give a warning
which seems reasonable since that is the way recycling in
R works elsewhere too.
On Mon, Oct 20,
On 20 Oct 2008, at 22:57, (Ted Harding) wrote:
I'm wondering if there's a compact way to achieve the
following. The "dream" is that one could write
rep(c(0,1),times=c(3,4,5,6))
which would produce
# [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1
in effect "recycling" x through 'times'.
rep2 <-
Try this:
with(data.frame(x = 0:1, times = 3:6), rep(x, times))
or even shorter:
do.call(rep, data.frame(x = 0:1, times = 3:6))
On Mon, Oct 20, 2008 at 4:57 PM, Ted Harding
<[EMAIL PROTECTED]> wrote:
> Hi Folks,
> I'm wondering if there's a compact way to achieve the
> following. The "dream" i
will this do what you want:
> f.rep <- function(x, times){
+ # make sure the 'x' is long enough
+ x <- head(rep(x, length(times)), length(times))
+ rep(x, times)
+ }
>
> f.rep(c(0,1), c(3,4,5,6,7))
[1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0
>
On Mon, Oct 20, 2008 at 4:57
Hi Folks,
I'm wondering if there's a compact way to achieve the
following. The "dream" is that, by analogy with
rep(c(0,1),times=c(3,4))
# [1] 0 0 0 1 1 1 1
one could write
rep(c(0,1),times=c(3,4,5,6))
which would produce
# [1] 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1
in effect "recycling" x t
Thomas Steiner wrote:
NB: library(fortunes); fortune("WTFM") applies.
This one I didn't get. They help to do it without a loop?
No.
> library(fortunes); fortune("WTFM")
This is all documented in TFM. Those who WTFM don't want to have to WTFM
again
on the mailing list. RTFM.
-- Barry R
Surendra champanerkar gmail.com> writes:
>
> Hallo
> I am going through the paper on Torus randomization by Kyle E. Harms
> torus randomization is an alternative to Chi-sq test of Independence When
> the assumption of independence within the groups is violated
>
> i want to know that Is there a
Define "better".
Really, it depends on what you need to do (are all your data appropriately
represented in a 2D array?) and what resources are available. If all your
data can be represented using a 2D array, then Excel is probably your best
bet for th enear term. If not, you might as well bite
On 10/20/2008 4:34 PM, Jeroen Ooms wrote:
I'm getting a unexpected compiling error when doing make:
gcc -std=gnu99 -I../../src/extra/zlib -I../../src/extra/bzip2
-I../../src/extra/pcre -I. -I../../src/include -I../../src/include
-I/usr/local/include -DHAVE_CONFIG_H -fpic -g -O2 -c platform.c
Brian, thank you for your reply.
> Why do you expect this to be vectorized when the help page clearly says
> otherwise?
Sure, it's in line with the help, no problem there. I just asked if
this feature is somehow possible.
I expected it, because "pos" does take arrays and it "overrides any
adj val
I'm getting a unexpected compiling error when doing make:
gcc -std=gnu99 -I../../src/extra/zlib -I../../src/extra/bzip2
-I../../src/extra/pcre -I. -I../../src/include -I../../src/include
-I/usr/local/include -DHAVE_CONFIG_H -fpic -g -O2 -c platform.c -o
platform.o
platform.c: In function 'd
I'm getting a unexpected compiling error when doing make:
gcc -std=gnu99 -I../../src/extra/zlib -I../../src/extra/bzip2
-I../../src/extra/pcre -I. -I../../src/include -I../../src/include
-I/usr/local/include -DHAVE_CONFIG_H -fpic -g -O2 -c platform.c -o
platform.o
platform.c: In function 'do_
reproducible code please.
On Mon, Oct 20, 2008 at 3:22 PM, phoebe kong <[EMAIL PROTECTED]> wrote:
> Hi All,
>
> I'm going to draw 8 plots in one page. I want the plots to be arranged
> in two rows, each row has 4 plots. So, I set the it par(mfrow=c(2,4)).
>
> However, there might be too many plots
On Mon, 20 Oct 2008, A Van Dyke wrote:
I am trying to plot survival curves using the following code as an example:
rs1799964.coxph<-(coxph(Surv(sassurvmonths,status)~age+stage+rs1799964_TNFA,method="efron"))
plot(rs1799964.coxph,lyt=c(1,3),xlab="Survival in Months",ylab="Proportion
Surviv
Why do you expect this to be vectorized when the help page clearly says
otherwise? See
adj: one or two values in [0,1] which specify the x (and
optionally y) adjustment of the labels. On most devices
values outside that interval will also work.
What is wrong with two
Hi All,
I'm going to draw 8 plots in one page. I want the plots to be arranged
in two rows, each row has 4 plots. So, I set the it par(mfrow=c(2,4)).
However, there might be too many plots in a page, all the 8 plots were
drawn in triangular shape, which makes the x-y coordinate scale not in
same
sorry excel 2003 with no immediate update in the future.
On Mon, Oct 20, 2008 at 3:12 PM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> You didn't say which version of Excel you are using but Excel 2007
> allows 16,384 columns.
>
> On Mon, Oct 20, 2008 at 2:27 PM, stephen sefick <[EMAIL PROTECTE
...and the offset argument can't handle arrays either:
text(c(0,0),c(0,0),c("a","b"),pos=c(1,2),offset=c(1,0.5),col="darkgreen")
The second argument does not have any effect.
Can I pass arrays to text and use diffent locations?
Thomas
PS: I use R version 2.5.0 (2007-04-23) under Win2000.
__
You didn't say which version of Excel you are using but Excel 2007
allows 16,384 columns.
On Mon, Oct 20, 2008 at 2:27 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> I am wondering if there is a better alternative than Excel for data
> storage that does not require database knowledge (I will even
assuming that A1 is the data provided in the first post... it would be
helpful if you used dput() to copy into an email, so that we could
just copy the code and data right into a R session and be off and
running. I hope this helps.
#I used dput() on the object A1
A1 <- (structure(list(s1 = c(0L,
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Johannes Bauer
> Sent: Monday, October 20, 2008 10:42 AM
> To: r-help@r-project.org
> Subject: [R] R Newbie Question
>
> Hello list,
>
> I just started R today and tried something quite simple. I want
I believe that you have to specify which variable is the x and also
the y. I am not familiar with what you are doing, but it looks as
though you need to specify both x and y to get it to plot.
Stephen
On Mon, Oct 20, 2008 at 2:22 PM, A Van Dyke <[EMAIL PROTECTED]> wrote:
>
> I am trying to plot
I am wondering if there is a better alternative than Excel for data
storage that does not require database knowledge (I will eventually
have to learn this, but it is not on my immediate todo list). I need
something that is not limited to 256 columns... I don't need any of
the built in functions in
Hi everyone,
> does any one knows how can I calculate mean for different samples
> i.e. I have a data like this:
>
> s1 s2 s3 s4
> 1 0 0 0 1
> 2 1 0 1 0
> 3 0 0 0 0
> 4 0 0 0 0
> 5 0 1 0 1
> 6 1 0 0 0
> 7 0 0 0 0
> 8 0 0 0 0
> 9 0 0 0 0
> 10 0 0 0
Hello,
I have seen that some papers do this:
1.) Group data by year (e.g. 35 years)
2.) Estimate the mean of the key variable through the distribution that fits
better (some years is a normal distribution , others is a more skewed, gamma
distribution, etc.)
3.) With these estimated means of ea
I am trying to plot survival curves using the following code as an example:
>rs1799964.coxph<-(coxph(Surv(sassurvmonths,status)~age+stage+rs1799964_TNFA,method="efron"))
>plot(rs1799964.coxph,lyt=c(1,3),xlab="Survival in Months",ylab="Proportion
Surviving")
I am gettingthe following error messa
Hi,
text()'s pos does handle more dimensional labels, while adj does not -
or how can I pass an array to text() and use the adj argument (skyblue
text)?
plot(-1:1,-1:1)
text(c(0,0),c(0,0),c("a","b"),pos=c(1,2),col="red")
text(c(0,0),c(0,0),c("a","b"),adj=cbind(c(1,2),c(1,1)),col="skyblue")
Thoma
Hello list,
I just started R today and tried something quite simple. I wanted to
create a colored plot and eventually after hours of fiddling around got
it working. However, my solution seems very suboptimal and I'd really
appreciate your hints on how to improve. I believe that R already offers
ma
It's trivial to do. In particular abs(rnorm(...)) will give you the
random number generator.
The term 'folded normal distribution' is normally only used for mean=0,
when you just double the density, but the general result is
ifelse(x > 0, 1, 0) * (dnorm(x, ...) + dnorm(-x, ...))
On Mon, 20
Dear Andreas,
try:
library(distr)
X <- Norm() ## standard normal distribution
Y <- abs(X)
plot(Y)
Best regards,
Matthias
Andreas Wittmann wrote:
Dear R useRs,
i wanted to ask if the folded normal destribution (Y = abs(X) with X
normal distributed)
with density and random number generator is
I suspect that you need to set the type= in X11(): see its help page.
E.g.
'type = "nbcairo"' is the same device as 'type="cairo"' without
buffering: which is faster will depend on the X11 connection.
Both will be slower than 'type = "Xlib"', especially on a slow X11
connectio
Dear R useRs,
i wanted to ask if the folded normal destribution (Y = abs(X) with X
normal distributed)
with density and random number generator is implemented in R or in any
R-related package
so far? Maybe i can use the non-central chi-square distribution and
rchisq(n, df=1, ncp>0) here?
Tha
Hi,
I often work from home, running R on my office computer and displaying
interactive graphics on my home computer using X11. This was always
more sluggish than sitting at my office computer, obviously, but
recently the graphics transfer has become painfully slow -- so slow,
that I now dive
Hi,,
I want to use the existing cor function in R but with a different way to
compute the correlation method.. basically zero mean correlation.
The forumula I have is
'D' <- function(c1, c2)
sum(c1*c2, na.rm=T)/(sqrt(sum(c1*c1, na.rm=T))*sqrt(sum(c2*c2, na.rm=T)))
I am not sure how i can
I thought I was finished, having gotten everything to work as intended. This
is a model of risk, and the short term forecasts look very good, given the
data collected after the estimates are produced (this model is intended to
be executed daily, to give a continuing picture of our risk). But now
On Mon, Oct 20, 2008 at 11:14 AM, jim holtman <[EMAIL PROTECTED]> wrote:
> yet another way (as it always is in R):
>
>> x<-c(1,0,0,1,0,0,1,0,0,1)
>> y<-c(1,3,2,3,2,1,2,3,2,3)
>> z<-c(1,2,1,2,1,2,1,2,1,2)
>> d<-as.data.frame(cbind(x,y,z))
>> d$myvar <- x*6 + (y-1)*2 + z
>>
>> d
> x y z myvar
> 1
yet another way (as it always is in R):
> x<-c(1,0,0,1,0,0,1,0,0,1)
> y<-c(1,3,2,3,2,1,2,3,2,3)
> z<-c(1,2,1,2,1,2,1,2,1,2)
> d<-as.data.frame(cbind(x,y,z))
> d$myvar <- x*6 + (y-1)*2 + z
>
> d
x y z myvar
1 1 1 1 7
2 0 3 2 6
3 0 2 1 3
4 1 3 212
5 0 2 1 3
6 0 1 2 2
Is this what you want:
> x <- read.table(textConnection(" s1 s2 s3 s4
+ 1 0 0 0 1
+ 2 1 0 1 0
+ 3 0 0 0 0
+ 4 0 0 0 0
+ 5 0 1 0 1
+ 6 1 0 0 0
+ 7 0 0 0 0
+ 8 0 0 0 0
+ 9 0 0 0 0
+ 10 0 0 0 1"), header=TRUE)
> closeAllConnections()
> for (i in 1:5){
Hi everyone,
does any one knows how can I make sample from rows?
i.e. I have a data like this:
s1 s2 s3 s4
1 0 0 0 1
2 1 0 1 0
3 0 0 0 0
4 0 0 0 0
5 0 1 0 1
6 1 0 0 0
7 0 0 0 0
8 0 0 0 0
9 0 0 0 0
10 0 0 0 1
I need to make 5 different sample wit
Thank you all for your answers.
On Mon, Oct 20, 2008 at 6:32 AM, Henrique Dallazuanna <[EMAIL PROTECTED]>wrote:
> Try this:
>
> Using 'envir' argument, in ls and get function:
>
> get.vars.name.prefix<-function(prefix, envir = parent.frame()){
>result<-list()
>len<-nchar(prefix)
>var.
Hi
maybe you can change combination of your columns to factor
new <-factor(paste(x,y,z, sep=""))
and then reorder this factor (see reorder) change it to numeric to get
index
index <- as.numeric(factor)
and then choose from set of 144 data
data[index]
or use your index directly if you need
or if x, y and z are factors :
> dbis <- as.data.frame(apply(d, 2, as.factor)
> dbis$x:dbis$y:dbis$z
[1] 1:1:1 0:3:2 0:2:1 1:3:2 0:2:1 0:1:2 1:2:1 0:3:2 0:2:1 1:3:2
12 Levels: 0:1:1 0:1:2 0:2:1 0:2:2 0:3:1 0:3:2 1:1:1 1:1:2 1:2:1 ... 1:3:2
and for your results :
> as.numeric(dbis$x:dbis$y:dbis$
On Mon, Oct 20, 2008 at 4:10 PM, <[EMAIL PROTECTED]> wrote:
>
> I'm trying to create a new column in my data.frame where subjects are
> categorized depending on values on four other columns. In any other case I
> would just nest a few ifelse statements, however, in this case i have
> 4*6*2*3=14
try this:
x <- c(1,0,0,1,0,0,1,0,0,1)
y <- c(1,3,2,3,2,1,2,3,2,3)
z <- c(1,2,1,2,1,2,1,2,1,2)
d <- data.frame(x, y, z)
ind <- do.call("paste", c(expand.grid(1:2, 1:3, 0:1)[3:1], sep = "\r"))
trg <- do.call("paste", c(d, sep = "\r"))
d$myvar <- match(trg, ind)
I hope it helps.
Best,
Dimitris
I'm trying to create a new column in my data.frame where subjects are
categorized depending on values on four other columns. In any other case I
would just nest a few ifelse statements, however, in this case i have
4*6*2*3=144 combinations and i get weird 'context overflow' errors. So I wonder
Hi,
I was trying to adjust an bayesian network with Deal package.
I know how to adjust when I give the structure and when I don't give the
struture. Now, I want to adjust when I know some variables relationships and
others I don't know.
Anybody knows how can I do that?
Regards,
Atenciosamen
thanks Antony.
regards,
Dhruv
From: Antony Unwin [mailto:[EMAIL PROTECTED]
Sent: Monday, October 20, 2008 7:00 AM
To: r-help@r-project.org
Cc: Sharma, Dhruv
Subject: Re: [R] pairs plots in R
If you want to do efficient exploratory data analysis on this kind o
You can put the code on several lines as shown or on one line. Does
not matter.
Read the info on the dependencies= arg in ?install.packages
for the definitive explanation of how dependencies= works.
On Mon, Oct 20, 2008 at 9:32 AM, Brian Lunergan <[EMAIL PROTECTED]> wrote:
> Gabor Grothendieck w
Thanks Felix.
Regards,
Dhruv
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Felix Andrews
Sent: Sunday, October 19, 2008 11:37 PM
To: Sharma, Dhruv
Cc: r-help@r-project.org
Subject: Re: [R] pairs plots in R
One idea:
if the primary variable of intere
Version 0.3-7 of the gsubfn package is available on CRAN.
Changes to the package are:
- all known bugs have been fixed.
- in gsubfn and strapply the replacement object can be a
list (or a function, formula, character string or proto
object, as before). In the case of a list, regexp matches
yes it is, I meant without the clutter of my previous analyses -
reshape is required by ggplot2 so I had that covered... If I am
understanding you correctly.
thanks for all of the help,
Stephen
On Mon, Oct 20, 2008 at 9:39 AM, Thompson, David (MNR)
<[EMAIL PROTECTED]> wrote:
> Is not melt part o
Is not melt part of reshape package, i.e., require(reshape)?
DaveT.
>-Original Message-
>From: stephen sefick [mailto:[EMAIL PROTECTED]
>Sent: October 19, 2008 12:12 PM
>To: hadley wickham
>Cc: R Help
>Subject: Re: [R] Date classes in ggplot2
>
>well I must have either resolve it without
Gabor Grothendieck wrote:
leaps is not in Depends or Imports for car. It is in Suggests
and those don't get automatically pulled in when dependencies = NA.
What you could do is to replace the builtin menuInstallPkgs with
your own by running this:
assignInNamespace("menuInstallPkgs", function (
> I'd like to avoid looping through an array in order to change values in
> the array as it takes too long.
> I red from an earlier post it can be done by "do.call" but never got it
> to work. The Idea is to change the value of "y" according to values in
> "x". Wherever "x" holds the value 3, the c
On Mon, Oct 20, 2008 at 03:07:45PM +0200, Kåre Edvardsen wrote:
> I'd like to avoid looping through an array in order to change values in
> the array as it takes too long.
> I red from an earlier post it can be done by "do.call" but never got it
> to work. The Idea is to change the value of "y" ac
y <- ifelse(x == 3, 1, y)
-Kaom
On Oct 20, 2008, at 6:07 AM, Kåre Edvardsen wrote:
I'd like to avoid looping through an array in order to change values
in
the array as it takes too long.
I red from an earlier post it can be done by "do.call" but never got
it
to work. The Idea is to change
I'd like to avoid looping through an array in order to change values in
the array as it takes too long.
I red from an earlier post it can be done by "do.call" but never got it
to work. The Idea is to change the value of "y" according to values in
"x". Wherever "x" holds the value 3, the correspondi
Dear all,
I'm a new user of R.
How to manage R to have homogeneous means groups with Duncan test? I've tried
the Duncan test and the results is the difference between means taken in pairs.
Patrice
__
ble contre les messages non sollicités
On 20/10/2008 5:39 AM, A.Noufaily wrote:
Hello,
I am hoping someone can help me with the following:
I am applying the function "mle" on a single data set X n times, each
time using a different set of initial values v[i] (i=1,...,n).
The initial values are all finite. Two cases arise:
A- For som
If you want to do efficient exploratory data analysis on this kind of
dataset, then interactive graphics with parallel coordinate plots
(ipcp in iplots) should help. Of course, it depends what you mean by
large. It might be worth looking at the book "Graphics of Large
Datasets" for some i
Hello,
I am hoping someone can help me with the following:
I am applying the function "mle" on a single data set X n times, each
time using a different set of initial values v[i] (i=1,...,n).
The initial values are all finite. Two cases arise:
A- For some sets of initial values mle is giving par
Hullo all,
[***recap of the question: passing bar=harry fails when bar is among
the ... args.]
callTimes <- function(tom, harry) {
timesDefineInside(foo=tom, bar=harry
}
timesDefineInside <- function(foo, ...) {
foo * bar
}
callTimes(3, 4)
# Error: object "bar" not found
[***end
> > The row names on a data frame should be unique. You can try
> > as.data.frame(xx, row.names=FALSE) to convert zz to be a data frame.
If
> > you need the row name information, add it as a column in the data
frame,
> > e.g. mydataframe$rnames <- rownames(zz). (Note to R-Core: the
> > document
On 20/10/2008 2:06 AM, erwann rogard wrote:
hi,
get.vars.name.prefix<-function(prefix){
result<-list()
len<-nchar(prefix)
var.names<-ls(name=1,pattern=prefix) #name=1 probably wrong option
print(var.names)
for(i in 1:length(var.names)){
name<-var.names[i]
field<-substr(name,len+1,nchar(
Try this:
Using 'envir' argument, in ls and get function:
get.vars.name.prefix<-function(prefix, envir = parent.frame()){
result<-list()
len<-nchar(prefix)
var.names<-ls(envir = envir,pattern=prefix)
print(var.names)
for(i in 1:length(var.names)){
name<-var.names[i]
fie
Dear R-users,
I'm having a small problem while bootstraping data.
What i would like to do, is resmple the data and calulate a function on
this, so i can estimate the measure of reproducability for this data.
The function i wrote works fine, even while bootstraping.
The only problem is that boo
On Mon, 20 Oct 2008, [EMAIL PROTECTED] wrote:
I have a matrix x with repeating row names.
zz<-matrix(0,4,4)
rownames(zz)=c("a","a","b","b")
data.frame(zz) (?)
The row names on a data frame should be unique. You can try
as.data.frame(xx, row.names=FALSE) to convert zz to be a data frame.
I've rolled up R-2.8.0.tar.gz a short while ago. This is a development
release which contains a number of new features.
Also, a number of mostly minor bugs have been fixed. See the full list
of changes below.
You can get it (later today) from
http://cran.r-project.org/src/base/R-2/R-2.8.0.tar.g
> I have a matrix x with repeating row names.
> zz<-matrix(0,4,4)
>
> rownames(zz)=c("a","a","b","b")
>
> data.frame(zz) (?)
The row names on a data frame should be unique. You can try
as.data.frame(xx, row.names=FALSE) to convert zz to be a data frame. If
you need the row name information
That's it.
thanks
Regards/Cordialement
Benoit Boulinguiez
-Message d'origine-
De : Greg Snow [mailto:[EMAIL PROTECTED]
Envoyé : vendredi 17 octobre 2008 19:19
À : Benoit Boulinguiez; r-help@r-project.org
Objet : RE: [R] use of empty space in split graph device
The plot.new functi
"Stella Sim" <[EMAIL PROTECTED]> writes:
> I want to calculate SPE (squared prediction error) in x-space, can
> someone help?
>
> Here are my codes:
>
> fit.pls<-
> plsr(Y~X,data=DAT,ncomp=3,scale=T,method='oscorespls',validation="CV",x=
> T)
> actual<-fit.pls$model$X
(The x = TRUE is not needed
I know, this is a forum about R. But I am so desperate of this problem (BTW,
anyone knows any good Statistics/Math forum to post question like this?):
A and B are both n x n positive definite matrix.
Denote A > B, if A - B is positive definite.
I know this is true: if A > B, then A^{-1} < B^{-1}
Em Qui, 2008-10-16 às 21:11 +1030, Fernando Marmolejo Ramos escreveu:
> Dear all
>
> Recently the "granova" package was launched. I installed but after when I
> invoked it in R it requested for other libraries. They were downloaded and
> install automatically.
>
> I tried to run the example synta
Hallo
I am going through the paper on Torus randomization by Kyle E. Harms
torus randomization is an alternative to Chi-sq test of Independence When
the assumption of independence within the groups is violated
i want to know that Is there any library regarding torus randomization
Someone said me t
> "MJ" == Michael Just <[EMAIL PROTECTED]>
> on Sat, 18 Oct 2008 17:26:02 -0400 writes:
MJ> Henrique,
MJ> This code below from a R-help helper worked.
>> dat <-data.frame(f)
>> dat2<-dat[dat$d<0,]
>> dat.lm <-lm(c~d, data=dat2)
But Henrique's advice, of using 'sub
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