Dear R users
I plot the trellis graphics by using the lattice package. Everything is
OK. Now, I want set
some parameters of the trellis graphics.
1. The tick label site. By default, only two tick labels had been output in
x-axis of my plot.
I want output four or five tick labels. In the tradi
Assuming that the number of rows is even and that your matrix is A,
element-wise product of pairs of rows can be calculated as
A[seq(1,nrow(A),by=2),]*A{seq(2,nrow(A),by=2),]
--- On Mon, 28/7/08, rcoder <[EMAIL PROTECTED]> wrote:
> From: rcoder <[EMAIL PROTECTED]>
> Subject: [R] product of su
OK. Thank you for pointing out my mistake.
I still have my original question. How does the output relate to estimating the
parameters of a given density? I read that for a gausian kernal:
bw.nrd0 implements a rule-of-thumb for choosing the bandwidth of a Gaussian
kernel density estimator. It de
Sorry, poor example. I started with normal deviates and jumped without thinking
to Poisson. The main crux of the question is how does the output of density
relate to the parameters that describe some of the standard distributions (mean
and std for normal, lambda for Poisson, n and p for Binomial
Hi Kevin,
>> I still have my original question. How does the output relate to
>> estimating the parameters
>> of a given density? I read that for a gausian kernal:
This isn't the place for such questions: you need to do some _basic_ reading
on the subject so that you begin to understand somethi
Sorry I tried WikiPedia and only found:
Wikipedia does not have an article with this exact name.
I will try to find some other sources of information.
Kevin
Mark Difford <[EMAIL PROTECTED]> wrote:
>
Hi Kevin,
>> I still have my original question. How does the output relate to
>> estimat
Hi Kevin,
Clicking on the link I sent gets me there (?), though things are pretty slow
at the moment. Perhaps try this related link, and from it get back to the
first one:
http://en.wikipedia.org/wiki/Density_estimation
You can also get to this via histogram, so search for that in Wiki, and
the
On Mon, 2008-07-28 at 06:32 -0700, rlearner309 wrote:
> Hi,
> I have a simple graph:
> x <- c(1,2,3)
> plot(x, pch=16,type="b")
>
> I would like to add some notes just beside these 3 dots, and the notes are
> stored in a vector:
>
> a <- c(12,54,84)
>
> So the result will be: there should be
Hi All:
I have a seemingly typical SPSS data file with 219 rows and 486
variables. When I attempt to read this file into R with read.spss() in
the foreign package, I consistently get a "crash". This is the sequence
of events:
> library(foreign)
> sessionInfo()
R version 2.7.1 Patched (200
On Mon, 2008-07-28 at 12:15 -0700, Arthur Roberts wrote:
> Hi, all,
>
> Does anyone now of a way to put multiple plots on gap.plot?
>
> Much appreciated,
Hi Art,
You must have read my mind. In solving the problem you had with
gap.plot, I considered including an "add" argument that would allow th
Hi R Gurus!
When you build a package, you need to put in keywords in the Rd files.
Where would you find the list of keywords, please?
TIA,
Edna Bell
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the p
Is this what you are after: this comes from the example of wilcox.test
and shows the structure and how to reference the p-value:
> z <- wilcox.test(x, y, paired = TRUE, alternative = "greater")
> str(z)
List of 7
$ statistic : Named num 40
..- attr(*, "names")= chr "V"
$ parameter : NULL
$
In the previous versions of R (2.6.1), when a vector of NA was given to the
functions 'sd' or 'var' with parameter na.rm = TRUE, it used to return NA. Now
(2.7.1) it returns an ERROR :
Example in 2.6.1:
> sd(c(NA, NA, NA, NA), na.rm = TRUE)
[1] NA
Example in 2.7.1:
> sd(c(NA, NA, NA,
It works for me, on Windows with that version of foreign. I also tried
under Linux with valgrind, and nothing untoward was reported. The
warnings are
Warning messages:
1: In read.spss("mySPSSfile.sav") :
mySPSSfile.sav: File-indicated character representation code (1252)
looks like a Windo
Hello everyone,
I have a vector of p-values and I am trying to find the max for the vector and
add it to a list.I am using a loop to loop through 50 times.I have used the
following code but it does not pick out the max and add to the list.Any help
would be appreciated.
w<-c(v[[1]][3],v[[2]][3]
A funny thing happened when I wanted a student of mine to install Brugs.
Using the InstallPackages in the windows version, firts gives an erro, but
trying again works flawlessly.
R version is 2.7.0 on WinXP.
Any explanation?
Bendix Carstensen
__
Bend
BXC (Bendix Carstensen) wrote:
A funny thing happened when I wanted a student of mine to install Brugs.
Using the InstallPackages in the windows version, firts gives an erro, but
trying again works flawlessly.
R version is 2.7.0 on WinXP.
Any explanation?
No, particularly if it worked the
Dear list,
I was attempted to install 2.7.1 in my mac only recently because of
the fantastic quartz display support.
I had no problem in building from source.
But after installation, I always got font warnings as following when
using quartz:
Warning messages:
1: In axis(side = side, at = a
Hi rcoder,
Assuming that the number of rows of your matrix x is even, try also:
x <- matrix(1:72,12)
apply(x,2, tapply, rep(1:(nrow(x)/2),each=2),prod)
# or using a function which argument "x" is your matrix
prod.mat=function(x) {
k=nrow(x)
g=rep(1:(k/2),each=2)
apply(x,2, tapply, g,prod)
}
pr
Dear all,
I am using kernlab package in R, and I have amino acid sequences with different
lenghts as input for a SVM and I need to go through this sequences using
windows (sliding or fixed) of size X.
Does anyone has any suggestions about which function I should use?
I thought I could use strin
Hi all,
I´m using the arima() function to study a time series but it gives me
the following error:
Error en optim(init[mask], armafn, method = "BFGS", hessian = TRUE,
control = optim.control, :
non-finite finite-difference value [3]
I know that I can change the method of the arima() t
Dear Rusers,
I am still an unexperienced builder of functions and loops, so my question is
very basic: Is it possible to introduce a second variable (j) into my loop.
To examplify:
# This works fine:
fn <- function (x) {if (x>46 & x<52) 1 else 0}
res <-NULL
for (i in 40:60) res <-c(res,fn(i))
Hi Art,
Ignore the last email, I realized that I had already written the "add"
bit and it will be in plotrix 2.4-5. As far as I can see, both the
"type" and "xlim" problems are solved.
Jim
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailma
On 7/29/2008 7:55 AM, Oehler, Friderike (AGPP) wrote:
Dear Rusers,
I am still an unexperienced builder of functions and loops, so my question is
very basic: Is it possible to introduce a second variable (j) into my loop.
To examplify:
# This works fine:
fn <- function (x) {if (x>46 & x<52) 1 els
Dear Frederike,
#Both your functions are vectorized. So you don't need loops. Working
with vectorized functions is much faster than looping.
fn <- function (x) {
ifelse(x>46 & x<52, 1, 0)
}
res <- fn(40:60)
fn <- function (x,y) {
ifelse(x>46 & x<52 & y<12, 1, 0)
}
datagrid <- exp
On 29.Jul.2008, at 14:13, ONKELINX, Thierry wrote:
Dear Frederike,
#Both your functions are vectorized. So you don't need loops. Working
with vectorized functions is much faster than looping.
fn <- function (x,y) {
ifelse(x>46 & x<52 & y<12, 1, 0)
}
datagrid <- expand.grid(i = 40:60, j = 0
On Tue, 29 Jul 2008, BXC (Bendix Carstensen) wrote:
A funny thing happened when I wanted a student of mine to install Brugs.
Using the InstallPackages in the windows version, firts gives an erro, but
trying again works flawlessly.
There was a problem with the http connection. It happens
There was a bug in 2.6.1 which has since been corrected: there is no need
to report corrected bugs in obsolete versions. Two different ways to
compute the sd of a zero-length vector gave different answers.
This is covered by the following NEWS item for 2.7.0 (version from
R-patched)
o
Hi ,
I am looking to making a panel of pie charts fo some of my dritribution data
. I was wondering if there is a way in any R package to write a small script to
do so.
Thanks,
Amin
[[alternative HTML version deleted]]
__
R-help@r-proj
jcarmichael gmail.com> writes:
>
>
> Hello.
>
> I am attempting to duplicate a negative binomial regression in R. SAS uses
> generalized estimating equations for model fitting in the GENMOD procedure.
>
> proc genmod data=mydata (where=(gender='F'));
> by agegroup;
> class id gender type;
>
On Tue, Jul 29, 2008 at 5:37 PM, G.H. Zuo <[EMAIL PROTECTED]> wrote:
> Dear R users
>
> I plot the trellis graphics by using the lattice package. Everything is
> OK. Now, I want set
> some parameters of the trellis graphics.
>
> 1. The tick label site. By default, only two tick labels had been o
day<-structure(c(7.7, 7.7, 7.7, 7.7, 7.7, 7.71, 7.7, 7.71, 7.71, 7.7,
7.7, 7.7, 7.7, 7.69, 7.68, 7.68, 7.67, 7.67, 7.67, 7.66, 7.65,
7.65, 7.65, 7.64, 7.64, 7.63, 7.63, 7.63, 7.62, 7.62, 7.62, 7.62,
7.63, 7.63, 7.63, 7.63, 7.63, 7.64, 7.64, 7.65, 7.65, 7.65, 7.66,
7.66, 7.67, 7.67, 7.67, 7.68, 7.68
On Jul 29, 2008, at 5:24 AM, Edna Bell wrote:
Hi R Gurus!
When you build a package, you need to put in keywords in the Rd files.
Where would you find the list of keywords, please?
Simplest way is to google for "r keywords". First hit is:
http://www.stat.ucl.ac.be/ISdidactique/Rhelp/doc/keyw
Try:
window(d, time(lin)) <- coredata(lin)
On Tue, Jul 29, 2008 at 9:01 AM, stephen sefick <[EMAIL PROTECTED]> wrote:
> day<-structure(c(7.7, 7.7, 7.7, 7.7, 7.7, 7.71, 7.7, 7.71, 7.71, 7.7,
> 7.7, 7.7, 7.7, 7.69, 7.68, 7.68, 7.67, 7.67, 7.67, 7.66, 7.65,
> 7.65, 7.65, 7.64, 7.64, 7.63, 7.63, 7.63
Have a look at mfcol in ?par.
--- On Tue, 7/29/08, Amin Momin <[EMAIL PROTECTED]> wrote:
> From: Amin Momin <[EMAIL PROTECTED]>
> Subject: [R] Panel of pie charts
> To: r-help@r-project.org
> Received: Tuesday, July 29, 2008, 7:34 AM
> Hi ,
> I am looking to making a panel of pie charts fo som
when I plot this it gives me the coredata(lin) (Ithink). I would like
d<- c(day[1:17,], lin, day[84:96,])
to be the result
On Tue, Jul 29, 2008 at 9:18 AM, Gabor Grothendieck <[EMAIL PROTECTED]
> wrote:
> Try:
>
> window(d, time(lin)) <- coredata(lin)
>
> On Tue, Jul 29, 2008 at 9:01 AM, stephe
?png
?write.table
--- On Mon, 7/28/08, Rajasekaramya <[EMAIL PROTECTED]> wrote:
> From: Rajasekaramya <[EMAIL PROTECTED]>
> Subject: [R] writing the plots
> To: r-help@r-project.org
> Received: Monday, July 28, 2008, 10:54 AM
> hi there,
>
> I want to write the plots in the pdfs and the detail
Hi,
Suppose I have the following vector (data points):
> x
[1] 36.0 57.3 73.3 92.0 300.4 80.9 19.8 31.4 85.8 44.9 24.6 48.0
[13] 28.0 38.3 85.2 103.6 154.4 128.5 38.3 72.4 122.7 123.1 41.8 21.7
[25] 143.6 120.2 46.6 29.2 44.8 25.0 57.3 96.4 29.4 62.9 66.4 30.0
[37]
Hello R users
It's some time I am playing with a dataset to do some cluster analysis. The
data set consists of 14 columns being geographical coordinates and monthly
temperatures in annual files
latitutde - longitude - temperature 1 -. - temperature 12
I have some missing values in some cases
Unfortunately, when I get to the 'myCuts' line, I receive the following error:
Error: evaluation nested too deeply: infinite recursion / options(expressions=)?
...and I also receive warnings about memory allocation being reached (even
though I've already used memory.limit() to maximise the memo
Hi Paco,
I got the same problem with you before. Thus, I just impute the missing values
For example:
newdata<-as.matrix(impute(olddata, fun="random"))
then I believe that you could analyze your data.
Hopefully it helps.
Chunhao
Quoting pacomet <[EMAIL PROTECTED]>:
Hello R users
It's some ti
Dear Paco,
in order to use the methods in the cluster package (including pam), look up
the help page of daisy, which is able to compute dissimilarity matrices
handling missing values appropriately (in most situations).
A good reference is the Kaufman and Rousseeuw book cited on that help page.
The code I gave replaces that portion of d that
overlaps with lin with lin. Is that not what you
wanted? (please try to use minimal examples
as shown here)
> library(zoo)
> d <- zoo(1:10) + 100
> lin <- - head(d, 4)
> window(d, time(lin)) <- coredata(lin)
> d
1234567
A quick comment on this: imputation is an option to make things
technically work, but it is not
necessarily good. Imputation always introduces some noise, ie, it fakes
information that is not really there.
Whether it is good depends strongly on the data, the situation and the
imputation metho
Hi list,
is there a package or function to compute the frequencies of pairs of chars in
a variable across a grouping variable? Eg:
d <- data.frame(ID=gl(2,3), F=c("A","B","C","A","C","D"))
> d
ID F
1 1 A
2 1 B
3 1 C
4 2 A
5 2 C
6 2 D
Now I want to summarize the frequencies of all pair
On Tue, 29 Jul 2008, Ben Bolker wrote:
jcarmichael gmail.com> writes:
Hello.
I am attempting to duplicate a negative binomial regression in R. SAS uses
generalized estimating equations for model fitting in the GENMOD procedure.
proc genmod data=mydata (where=(gender='F'));
by agegroup;
c
on 07/29/2008 09:51 AM [EMAIL PROTECTED] wrote:
Hi list,
is there a package or function to compute the frequencies of pairs of
chars in a variable across a grouping variable? Eg:
d <- data.frame(ID=gl(2,3), F=c("A","B","C","A","C","D"))
d
ID F 1 1 A 2 1 B 3 1 C 4 2 A 5 2 C 6 2 D
Now
Dear all,
may I suggest to include this quotation of Patrick Burns in the fortunes
package? :-)
Best,
Roland
Patrick Burns wrote:
A good reason to use '&&' rather than '&' is if evaluating
whatever is on the right will create an error if what is on
the left is FALSE. '&&' and '||' stop if
(mean.std.s2n.loss.gain[[1]])
Probe.Set.ID rho_prime r ho_prime_sdpom Expr1
matchinggenes Mean std_dev
29 SNP_A-190846347 2.47 0.75 0 PRKCZ -
0.34560. 1344 30 SNP_A-213137044 2.61
Dear friends - I have a plot of simulated data to compare to the
observed and want the simulated to appear accumulated so that a darker
grey corresponds to more lines of simulated data, but on top of that I
want the measured values (Na) to be very visible. The code below meets
only few of the d
(mean.std.s2n.loss.gain[[1]])
Probe.Set.ID rho_prime rho_prime_sd pom Expr1 matchinggenes
Meanstd_dev
29 SNP_A-190846347 2.47 0.75 0 PRKCZ
-0.345616170.13443676
30 SNP_A-213137044 2.61 0.58 0
Colleagues,
(Running R 2.7.0)
I have a script that I want to delete as it completes execution. The
penultimate line of the script (before the quit command) is:
file.remove("Scriptname")
The script is executed as:
R --no-save < Scriptname
In OS X and Linux this is successful an
Dear all,
I need to compute tensor product of B-spline defined over equi-spaced
break-points.
I wrote my own program (it works in a 2-dimensional setting)
library(splines)
# set the break-points
Knots = seq(-1,1,length=10)
# number of splines
M = (length(Knots)-4)^2
# short cut to splineDe
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Prof Brian Ripley wrote:
| On Tue, 29 Jul 2008, Ben Bolker wrote:
|
|> jcarmichael gmail.com> writes:
|>
|>>
|>>
|>> Hello.
|>>
|>> I am attempting to duplicate a negative binomial regression in R.
|>> SAS uses
|>> generalized estimating equations fo
Dear R-Users,
I am resending this message just to reminder my question regarding the
calculation of a bootstrap confidence intervals for a GAM plot.
I am trying to apply a bootstrap to a GAM in order to calculate the 95%
confidence intervals for a smooth curve obtained by the “plot.gam”
fun
Hi R Gurus:
Here is some code that I was experimenting with, please:
> f1 <- function(x) {
+ e1 <- new.env(parent=.GlobalEnv)
+ environment(e1)
+ print(environment())
+ return(mean(x))
+ }
> f1(1:15)
[1] 8
>
My question: why isn't the environment within the function set to e1, please?
Thanks,
> I am looking to making a panel of pie charts fo some of my
> dritribution data . I was wondering if there is a way in any R
> package to write a small script to do so.
pie() will do you a one-off pie chart, but there is no equivalent using
grid/ lattice graphics. You could write a panel.pie
e1 <- ...
creates a new environment e1
environment(e1)
does nothing
print(environment(e1))
print environment e1
By the way, if you are doing a lot of manipulations of environments
you might want to look at the proto package which reframes the
whole thing in terms of object oriented programming.
Hi Everyone,
I'm apologize for asking this in the R-general list, but I'm unsure of
where else to ask this burning question of mine:
Where do statisticians talk on the internet about professional/ career
developement/ issues? I've found many a list (much like this one) that
specailize in tal
Hi again!
Suppose I have the following:
> xy <- round(rexp(20),1)
> xy
[1] 0.1 3.4 1.6 0.4 1.0 1.4 0.2 0.3 1.6 0.2 0.0 0.1 0.1 1.0 2.0 0.9
2.5 0.1 1.5 0.4
> table(xy)
xy
0 0.1 0.2 0.3 0.4 0.9 1 1.4 1.5 1.6 2 2.5 3.4
1 4 2 1 2 1 2 1 1 2 1 1 1
>
Is there a way to s
Is there a way to set the environment within a function,, please?
On Tue, Jul 29, 2008 at 11:25 AM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> e1 <- ...
> creates a new environment e1
>
> environment(e1)
> does nothing
>
> print(environment(e1))
> print environment e1
>
> By the way, if you
No but look at proto since I suspect the creation
of proto objects is basically what you are trying
to do through the back door. Home page:
http://r-proto.googlecode.com
On Tue, Jul 29, 2008 at 12:29 PM, Edna Bell <[EMAIL PROTECTED]> wrote:
> Is there a way to set the environment within a functi
Hi again!
I put in ls() to check the objects in my workspace.
Is there a limit on how many objects I can have, please? Or does it
depend on the memory, please?
TIA,
Edna Bell
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/
Hello
I have created a graph using the following commands:
<<<
startBReP3O1T <- diffs$BReP3O1T - diffs$diff_BReP3O1T
endBReP3O1T <- diffs$BReP3O1T
x <- seq(47,89, length = 10)
ymin <- min(min(startBReP3O1T), min(endBReP3O1T))
ymax <- max(max(startBReP3O1T), max(endBReP3O1T))
y <- seq(ymin, ymax,
plot(replace(day, is.list(index(lin)), coredata(lin)))
[[alternative HTML version deleted]]
__
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
Hi still yet again!
I have the following code:
> try(log(rnorm(25)),silent=TRUE)
[1] -0.26396185 NaN NaN -0.13078069 -2.44997193
-2.15603971 NaN 0.94917495 0.07244544 NaN
[11] -1.06341127 -0.42293099 -0.53769569 0.95134763 0.93403340
NaN -0.10502078
Hello everyone,
I have a list which I am trying to calculate a max value.I have the list as
w<-c(v[[1]][1],...v[[100]][1]). The problem I am getting is that the function
max is saying the list is an "invalid
type (list) of argument".When I show the element v[[1]][1] it shows as
$statistic,V,73
> > try(log(rnorm(25)),silent=TRUE)
> [1] -0.26396185 NaN NaN -0.13078069 -2.44997193
> -2.15603971 NaN 0.94917495 0.07244544 NaN
> [11] -1.06341127 -0.42293099 -0.53769569 0.95134763 0.93403340
> NaN -0.10502078 NaN 0.30283262 NaN
> [21] -0.1
On Tue, 29 Jul 2008, Dennis Fisher wrote:
Colleagues,
(Running R 2.7.0)
I have a script that I want to delete as it completes execution. The
penultimate line of the script (before the quit command) is:
file.remove("Scriptname")
The script is executed as:
R --no-save < Scriptna
On Tue, 2008-07-29 at 17:23 +0100, [EMAIL PROTECTED] wrote:
> > I am looking to making a panel of pie charts fo some of my
> > dritribution data . I was wondering if there is a way in any R
> > package to write a small script to do so.
> pie() will do you a one-off pie chart, but there is no eq
You can do it very easily using subsetting and a bit of paste().
I assumed you didn't need startdata and enddata after the plot
has been made, but if you do you can change the last line of
the function to return them.
Sarah
myfunction <- function(dataname)
{
# where dataname is a string, eg myf
What's v? And w? And what exactly do you want to do?
A small reproducible example would be very helpful. Without
knowing what your data look like, it's hard to make helpful
suggestions.
Sarah
On Tue, Jul 29, 2008 at 12:56 PM, Paul Adams <[EMAIL PROTECTED]> wrote:
>
> Hello everyone,
> I have a
on 07/29/2008 11:38 AM Peter Flom wrote:
Hello
I have created a graph using the following commands:
<<<
startBReP3O1T <- diffs$BReP3O1T - diffs$diff_BReP3O1T
endBReP3O1T <- diffs$BReP3O1T
x <- seq(47,89, length = 10)
ymin <- min(min(startBReP3O1T), min(endBReP3O1T))
ymax <- max(max(startBReP3O
On Tue, 29 Jul 2008, Ben Bolker wrote:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Prof Brian Ripley wrote:
| On Tue, 29 Jul 2008, Ben Bolker wrote:
|
|> jcarmichael gmail.com> writes:
|>>
|>> Hello.
|>>
|>> I am attempting to duplicate a negative binomial regression in R.
|>> SAS uses
|>> g
On Tue, 29 Jul 2008, Edna Bell wrote:
Hi still yet again!
I have the following code:
try(log(rnorm(25)),silent=TRUE)
[1] -0.26396185 NaN NaN -0.13078069 -2.44997193
-2.15603971 NaN 0.94917495 0.07244544 NaN
[11] -1.06341127 -0.42293099 -0.53769569 0.9513476
Hello -
Paul Adams wrote:
Hello everyone, I have a list which I am trying to calculate a max
value.I have the list as w<-c(v[[1]][1],...v[[100]][1]). The problem
I am getting is that the function max is saying the list is an
"invalid type (list) of argument".When I show the element v[[1]][1]
it
Prof Brian Ripley wrote:
'poisson' _family_, I presume?
oops, yes.
__
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 commente
*R we Wishing on a cloud-*
I would like to Test R as an cloud analytical tool.
remote submit csv files to the web for data mining ,
get data crunching with scalable processing power and disk storage ,
display Silverlight graphs of summarization,
downloadable in pdf's.
data mining on demand, o
One clarification:
I did the dirty and completely unprofessional thing, and assumed that this
function would only be run in the workspace with all your data. I use functions
like this frequently for something I need to do multiple times for a particular
project, but will never do anywhere else, bu
hi ,
Is there anyways to delet the list names once created.
i tried
rm(names(mylist))
i didnt work
kindly help me
Ramya
--
View this message in context:
http://www.nabble.com/List-names-help-tp18717742p18717742.html
Sent from the R help mailing list archive at Nabble.com.
_
hi,
i am seeking for a solution to create a biplot that shows:
a) data points instead of labels and that
b) shows two different groups (field: site with two factor levels:
forest/corridor) in different colours
i tried to include "pch=19" to show points instead of labels in the syntax
below but
Hey All,
I am a PhD student in Forestry science and I wish use kriging to spatial
LiDAR cloud points. The shape of my semivariogram is like "hole (wave)
model", but unfortunately I don't know the code of flexible model or hole
model to fit my semivariogram in R. I Found it in google or in R he
Sure, just set them to NULL.
> mylist <- c(a=1, b=2, c=3)
> mylist
a b c
1 2 3
> names(mylist) <- NULL
> mylist
[1] 1 2 3
Sarah
On Tue, Jul 29, 2008 at 1:45 PM, Rajasekaramya <[EMAIL PROTECTED]> wrote:
>
> hi ,
>
> Is there anyways to delet the list names once created.
>
> i tried
> rm(names(my
# is that what you want?
table(cut(xy,seq(0,max(xy)+.4,by=.4)))
# or this
table(cut(xy,hist(xy)$breaks)) # not the same
regards,
PF
+-
| Patrizio Frederic
| Research associate in Statistics,
| Department of Economics,
| University of Modena and
Just a quick follow up to my own post here. In offline exchanges, Peter
noted a typo in the call to plot(), where I was missing the final ")".
Also, I had the order of the unnamed arguments to segments() off.
Here is the corrected code, with some additional teaks:
plotdiffs <- function(x, y,
My apologies it worked perfectly.
On Tue, Jul 29, 2008 at 12:40 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> plot(replace(day, is.list(index(lin)), coredata(lin)))
>
--
Let's not spend our time and resources thinking about things that are so
little or so large that all they really do for u
It is memory. You should probably not have 40% of your RAM allocated
to R so that if you do create some large objects, you will have room
for them. How much physical memory do you have, can you put some
"numbers" on how many objects you need, if you have a lot of them,
have you considered using l
Hi all,
In R GUI window, if you use "up" and "down" key, you will be able to recall
the previous and next command that has been used and stored in the command
history cache.
But there is one inconvenience:
In Matlab, you can type the first a few characters of the command that you
previously use
I assume that you are doing this on one column of the matrix which
should only have 2160 entries in it. can you send the actual code you
are using. I tried it with 10,000 samples and it works fine. So I
need to understand the data structure you are using. Also the
infinite recursion sounds stra
A patch to do this was posted on 2007-09-29 by Glenn Davis. Some people
not addicted to Matlab find the behaviour very inconvenient and prefer the
getline/readline behaviour (triggered by ^R/^S) of Rterm and R on Unixen.
On Tue, 29 Jul 2008, losemind wrote:
Hi all,
In R GUI window, if you
"Edna Bell" <[EMAIL PROTECTED]> wrote:
> Hi still yet again!
>
> I have the following code:
>
> > try(log(rnorm(25)),silent=TRUE)
> [1] -0.26396185 NaN NaN -0.13078069 -2.44997193
> -2.15603971 NaN 0.94917495 0.07244544 NaN
> [11] -1.06341127 -0.42293099 -0.537
Dear R users,
I´m trying to optimize simultaneously two binomials inequalities used to
acceptance sampling, which are nonlinear solution, so there is no simple
direct solution. Please, let me explain shortly the the problem and the
question as following.
The objective is to obtain the smallest v
useR's,
I am trying trying to find out if there is a faster way to do a certain
computation. I have successfully used FOR loops and the apply function to
do this, but it can take some time to fully compute, but I was wondering if
anyone may know of a different function or way to do this:
> x
[1]
matrix(rep(x, each=13) - xk, nrow=13)
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of dxc13
> Sent: Tuesday, July 29, 2008 2:13 PM
> To: r-help@r-project.org
> Subject: [R] finding a faster way to do an iterative computation
>
>
> useR's,
>
> I
On 30/07/2008, at 6:12 AM, dxc13 wrote:
useR's,
I am trying trying to find out if there is a faster way to do a
certain
computation. I have successfully used FOR loops and the apply
function to
do this, but it can take some time to fully compute, but I was
wondering if
anyone may know
On Tue, Jul 29, 2008 at 2:45 PM, Prof Brian Ripley
<[EMAIL PROTECTED]> wrote:
> A patch to do this was posted on 2007-09-29 by Glenn Davis. Some people not
> addicted to Matlab find the behaviour very inconvenient and prefer the
> getline/readline behaviour (triggered by ^R/^S) of Rterm and R on U
On 30/07/2008, at 9:16 AM, hadley wickham wrote:
On Tue, Jul 29, 2008 at 2:45 PM, Prof Brian Ripley
<[EMAIL PROTECTED]> wrote:
A patch to do this was posted on 2007-09-29 by Glenn Davis. Some
people not
addicted to Matlab find the behaviour very inconvenient and prefer
the
getline/readline
On Tue, Jul 29, 2008 at 4:39 PM, Rolf Turner <[EMAIL PROTECTED]> wrote:
>
> On 30/07/2008, at 9:16 AM, hadley wickham wrote:
>
>> On Tue, Jul 29, 2008 at 2:45 PM, Prof Brian Ripley
>> <[EMAIL PROTECTED]> wrote:
>>>
>>> A patch to do this was posted on 2007-09-29 by Glenn Davis. Some people
>>> not
Is there an easy way to add a text into an R-plot and place it in the
upper left corner?
Like that:
| (a) |
| |
| |
|
Dear Jörg,
Perhaps,
plot(1:10)
legend("topleft",c("Your text here"),pch=1)
See ?legend and ?text for more information.
HTH,
Jorge
On Tue, Jul 29, 2008 at 6:04 PM, Jörg Groß <[EMAIL PROTECTED]> wrote:
> Is there an easy way to add a text into an R-plot and place it in the upper
> left corner
1 - 100 of 129 matches
Mail list logo
