Hi Yunlei,
Is your problem constrained or not?
If it is unconstrained and your matrix is not positive definite, the minimum is
unbounded (unless you are extremely lucky and the matrix is positive
semi-definite and the vector which multiplies the unknowns is exactly
perpendicular to all the eige
Hi Chunhao,
>> I google the website and I found that there are three ways to perform
>> repeated measure ANOVA: aov, lme and lmer.
It's also a good idea to search through the archives.
>> I use the example that is provided in the above link and I try
>> > tt<-aov(p.pa~group*time+Error(subject
Hi R users,
I google the website and I found that there are three ways to perform
repeated measure ANOVA: aov, lme and lmer.
http://www.mail-archive.com/[EMAIL PROTECTED]/msg58502.html
But the questions are which one is good to use and how to do post-hoc test?
I use the example that is provide
Regarding dubuggers, RSiteSearch("debug", "fun") produced 183
hits for me just now. The first two reference a "debug" package, while
the third describes the "debug" function in the "base" package.
The "debug{base}" function is great, but not as good as the Matlab
debugger. The "de
Is this what you want:
> "var3" %in% names(ipf)
[1] FALSE
> "var1" %in% names(ipf)
[1] TRUE
>
On Tue, Jul 29, 2008 at 6:20 PM, <[EMAIL PROTECTED]> wrote:
>
> Hello R mailing list
>
> Is there a better way than this to see if an element exists *within* a
> list object :
>
> #generate "file.txt"
?findInterval
> x <- c(1.0,1.2,1.4,1.6,1.8,2.0)
> findInterval(1.4,x)
[1] 3
> findInterval(1.5,x)
[1] 3
> findInterval(10,x)
[1] 6
On Tue, Jul 29, 2008 at 8:32 PM, Jörg Groß <[EMAIL PROTECTED]> wrote:
> I am searching for a way to get nearest position of a number in a vector
> from a search-valu
Thank you to all who applied. These all seem to work the way I want them to.
The outer function seems really useful, I probably could use that for a lot
of my work.
Thanks!
Rolf Turner-3 wrote:
>
>
> On 30/07/2008, at 6:12 AM, dxc13 wrote:
>
>>
>> useR's,
>>
>> I am trying trying to find out
Read the last line of every message to r-help.
On Tue, Jul 29, 2008 at 6:15 PM, rcoder <[EMAIL PROTECTED]> wrote:
>
> Hi everyone,
>
> I am trying to apply linear regression to adjacent columns in a matrix (i.e.
> col1~col2; col3~col4; etc.). The columns in my matrix come with identifiers
> at the
methods(rollapply)
shows which classes have rollapply methods.
On Tue, Jul 29, 2008 at 5:48 PM, rcoder <[EMAIL PROTECTED]> wrote:
>
> Hi Gabor,
>
> Thanks for your reply. Assuming I have a time series that is ready made
> (i.e. not constructed it using a zoo function) will the procedure below
> st
I am searching for a way to get nearest position of a number in a
vector from a search-value.
So perhaps if I have this vector:
1.0 1.2 1.4 1.6 1.8 2.0
I want to get the position of a special number.
with match(1.4 ,x), I get "3" for position three.
But now I want the nearest numbe
On 29-Jul-08 16:28:26, Edna Bell wrote:
> 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
Gad and David - would you be so kind as to include me in any off-line
discussion of this vital topic of R and GPL?
Many thanks.
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: [EMAIL PROTECTED
Hello Gad:
On Jul 29, 2008, at 4:41 PM, Gad Abraham wrote:
If your legal department is still concerned, have them look at the
license in Java. It most likely is GPL (unless you have a special
commercial version of Java installed) and thus makes any Java
program you create subject to the G
Gad Abraham wrote:
If your legal department is still concerned, have them look at the
license in Java. It most likely is GPL (unless you have a special
commercial version of Java installed) and thus makes any Java program
you create subject to the GPL in Java due to the JIT compiler used in
J
On 27/07/2008 3:10 PM, Stephen Tucker wrote:
Hi list,
I was using Sweave and was wondering if anyone has had any luck changing the
font colors of the code chunks. For instance, in my .Rnw preample I tried
including:
===
\usepackage[usenames]{colors}
\definecolor{darkred}{rgb}{0.545,0,0}
\defi
If your legal department is still concerned, have them look at the
license in Java. It most likely is GPL (unless you have a special
commercial version of Java installed) and thus makes any Java program
you create subject to the GPL in Java due to the JIT compiler used in
Java. If your legal
Hi everyone,
I am trying to apply linear regression to adjacent columns in a matrix (i.e.
col1~col2; col3~col4; etc.). The columns in my matrix come with identifiers
at the top of each column, but when I try to use these identifiers to
reference the columns in the regression function using rollap
Hi Gabor,
Thanks for your reply. Assuming I have a time series that is ready made
(i.e. not constructed it using a zoo function) will the procedure below
still retain the dates in the matrix?
Thanks,
rcoder
quote author="Gabor Grothendieck">
rollapply along an index:
library(zoo)
z <- zoo(mat
Hi everyone,
I'm having trouble applying the Cor() function to two matrices, both of
which contain NAs. I am doing the following:
a<-cor(m1, m2, use="complete.obs")
... and I get the following error message:
Error in cor(m1, m2, use = "complete.obs") :
no complete element pairs
Does anyone
On Jul 29, 2008, at 6:47 PM, Jörg Groß wrote:
legend("topleft",c("Your text here"),pch=1)
Thanks!
But is it possible to get rid of the legend-box and the box-shadow?
I don't find an argument in the manual for that.
Hi,
The documentation for the argument is right there - but as always i
On 30/07/2008, at 10:47 AM, Jörg Groß wrote:
legend("topleft",c("Your text here"),pch=1)
Thanks!
But is it possible to get rid of the legend-box and the box-shadow?
I don't find an argument in the manual for that.
You explicitly said that you *wanted* a box!!!
But be tha
If v is your vector of sample variances (and assuming that their distribution
is chi-square) you can define
f(df) <- sum(dchisq(v,df,log=TRUE))
and now you need to maximize f, which can be done using any optimization
function (like optim).
--- On Sat, 26/7/08, Julio Rojas <[EMAIL PROTECTED]> wr
In general there is no relation. The output of density gives you
something you can plot, essentially. So the x-values are simply a
series of points covering the range of the density and the y values are
the ordinates at those abscissae.
try
plot(density(rnorm(1000)^2)))
for example.
Bill Ven
Hi All:
I know this has been discussed at length already, but 1) I get R-Help
in digest and didn't see this until 3am Pacific time this morning, and
2) We, REvolution, have been discussing this as of late. I thought I
would pass on some of the knowledge we have recently obtained vis a
vi
> legend("topleft",c("Your text here"),pch=1)
Thanks!
But is it possible to get rid of the legend-box and the box-shadow?
I don't find an argument in the manual for that.
Am 30.07.2008 um 00:08 schrieb Jorge Ivan Velez:
>
> Dear Jörg,
>
> Perhaps,
>
> plot(1:10)
> legend("topleft",c("Your t
Hi all,
How do I operate R from within Excel?
I mean, highlight a bunch of cells, and send to R, and do some statistics in
R, and return back the numbers and do some plots in R.
For example, I have some parameters for Gaussian Random variable, and I
would find the most convenient way to send th
Prof Brian Ripley 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 Unixen.
>
> On Tue, 29 Jul 2008, losemind wro
Try
abs(outer(xk,x,"-"))
(see ?outer)
--- On Wed, 30/7/08, dxc13 <[EMAIL PROTECTED]> wrote:
> From: dxc13 <[EMAIL PROTECTED]>
> Subject: [R] finding a faster way to do an iterative computation
> To: r-help@r-project.org
> Received: Wednesday, 30 July, 2008, 4:12 AM
> useR's,
>
> I am trying
Hello R mailing list
Is there a better way than this to see if an element exists *within* a
list object :
#generate "file.txt" using current routine
cat("var1=33\nvar2=TRUE",file="file.txt")
#load file to a list called "ipf"
f <- function(.file){source(.file,local=TRUE);as.list(environment())}
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
Is there an easy way to add a text into an R-plot and place it in the
upper left corner?
Like that:
| (a) |
| |
| |
|
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
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 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 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
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
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]
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
"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
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
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
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
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
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
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,
# 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
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
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
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
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.
_
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
*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
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
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
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
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 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
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
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
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
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
> > 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
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
$statistic,V,73
Hi still yet again!
I have the following code:
> try(log(rnorm(25)),silent=TRUE)
[1] -0.26396185 NaN NaN -0.13078069 -2.44997193
-2.15603971 NaN 0.94917495 0.07244544 NaN
[11] -1.06341127 -0.42293099 -0.53769569 0.95134763 0.93403340
NaN -0.10502078
plot(replace(day, is.list(index(lin)), coredata(lin)))
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide
Hello
I have created a graph using the following commands:
<<<
startBReP3O1T <- diffs$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,
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/
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
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
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
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
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.
> 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
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,
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
-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 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
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
(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
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 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 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
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
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
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
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
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
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.
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
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
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
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]
?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
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
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
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
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
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 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
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;
>
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