Hi,
Does anyone know how to take a time difference when the format of the time
is as 13:22:23.586? I am trying to take the difference of time between stock
transactions and need to keep the three decimal places for seconds. I have
tried *diff(strptime(x, "%H:%M:%S.000))*, but apperantly that doesn
Not answering your question,
But if you where to ask this regarding plots, the answer would be to use:
par(ask = T)
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (He
Hello dear R community,
*Background:*
Since I see more and more R users who blog about R opening their blogs in
WordPress.com, I contacted one of their workers asking if they could add
Syntax highlight support for R.
He replied to me saying that if there was a "Brush" for R, according to the
forma
On Apr 21, 2010, at 10:17 PM, gallon li wrote:
Dear r-helpers,
I have a very simple question. Suppose my data is like
id=c(rep(1,2),rep(2,2))
b=c(2,3,4,5)
m=cbind(id,b)
m
id b
[1,] 1 2
[2,] 1 3
[3,] 2 4
[4,] 2 5
I wish to select the first observation for each id. That is, I want to
On Apr 21, 2010, at 9:51 PM, zhenjiang xu wrote:
I tried that. It seems the bar width is already maximized, although
there is a lot of space between groups of bars. Thank you anyway.
I apologize. It was reproducible code. I missed the "values"
assignment. There is also a box.width argument
Henrik-
A coding solutions may be
... + (1/(2*stdev*stdev))*sum( ( y-(rev/12)- c(0,y[-n]) *exp(-lap/12) )^2
)
where n is the number of observations in y.
Personally, I would use lm. Your model can be written as a linear function.
Let x=c(0,y[-n]). Then run lm(y~x). The parameter estimat
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of gallon li
> Sent: Wednesday, April 21, 2010 7:18 PM
> To: r-help
> Subject: [R] how to select the first observation only?
>
> Dear r-helpers,
>
> I have a very simple question.
Gallon, your question looks very homeworky. People on this list are not
likely to help you unless you can demonstrate own effort (even if it
failed), and the list is not for homework questions in case it is one. Where
exactly are you stuck?
Daniel
-
cuncta stricte discussu
You set records to NULL perhaps (delete, shift up). Perhaps your system is
susceptible to butterflies on the other side of the world.
Your code may have 'worked' on a small section of data, but the data used
did not include all of the cases needed to fully test your code. So... test
your code!
sc
Try the duplicated() function. As in
m[!duplicated(id), ]
-tgs
On Wed, Apr 21, 2010 at 10:17 PM, gallon li wrote:
> Dear r-helpers,
>
> I have a very simple question. Suppose my data is like
>
> id=c(rep(1,2),rep(2,2))
> b=c(2,3,4,5)
> m=cbind(id,b)
>
> > m
> id b
> [1,] 1 2
> [2,] 1 3
R experts,
Is there anyway to reorder inside each group? In the following example, the
bar of year 1932 is always plotted before the bar of year 1931, may I change
the order inside each groups of bars?
library(lattice)
barchart(yield ~ variety | site,data=barley, groups = year, layout =
c(1,6),au
Actually the memory foot print can be reduced by re-writting the function so
that it makes use of the
kronecker, rather than the outer function.
Note that you do not get 2D map as before ...
> countinstance2<-function(v,p) {
+ sapply(p,function(x,y) sum(kronecker(x,y,"==")),v)
+ }
> v<-sample(
Dear r-helpers,
I have a very simple question. Suppose my data is like
id=c(rep(1,2),rep(2,2))
b=c(2,3,4,5)
m=cbind(id,b)
> m
id b
[1,] 1 2
[2,] 1 3
[3,] 2 4
[4,] 2 5
I wish to select the first observation for each id. That is, I want to
quickly select two rows:
id b
1 2
2 4
only. how
On Wed, 21 Apr 2010, Frank E Harrell Jr wrote:
Thomas Lumley wrote:
On Tue, 20 Apr 2010, Noah Silverman wrote:
> I just read the help page for predict.coxph.
>
> It indicates that the risk score is just exp(lp)
>
> What I'm trying to find, and have seen with some other implementations
>
I tried that. It seems the bar width is already maximized, although there is
a lot of space between groups of bars. Thank you anyway.
On Tue, Apr 20, 2010 at 10:16 AM, David Winsemius wrote:
>
> On Apr 20, 2010, at 9:46 AM, zhenjiang xu wrote:
>
> Dear R users,
>>
>> I am trying to use the follo
Hi Liam,
Unfortunately this currently isn't supported. It's on my to do list:
http://github.com/hadley/ggplot2/issues/issue/94
Hadley
On Tue, Apr 20, 2010 at 7:59 PM, Liam Blanckenberg
wrote:
> Hi all,
>
> I have a question about setting arbitrary breaks/labels when using GGPLOT
> and date/tim
Hello,
The *predict.mvr* function from the *pls* package has the syntax
predict(mvr object, newdata, ncomp = array of integers,
type = c("response", "scores"),...)
When type is *scores*, predicted score values are returned for the
components given in *ncomp*.
What exactly do the score v
Hello,
I am trying to calculate predicted values derived from one dataset into a
hypothetical dataset. I tried this line of code:
graphdata$fmgpredvalues <- predict(Acs250.3.4, graphdata)
and received the following error message:
ERROR: ZXend[1], drop = FALSE] %*%lmeFit$beta
I have made
Hello,
By fixing log_vraisemblance, gradient and hessian function linked to a
family density like exp(theta.xi), I'm looking for some efficients
estimators by PML.
So I've seen optim,nlminb, et maxLik procedure.
But I 'm not sure that the heteroscedacity of my estimators are considered.
Does an
Thank you Thomas.
(a) an embarrassing mistake by me. Of course it should be squared. Thank you
for pointing that out.
(b) Do you possibly have any suggestions on how to solve this issue? I
presume that there is no reason in trying to create a lagged "vector"
manually?
Best Regards
Henrik
--
David, Duncan,
Thanks. I didn't realize the functions in "graphic" also work in "lattice".
I thought they didn't work with each other.
Jun
On Wed, Apr 21, 2010 at 6:51 PM, Duncan Murdoch wrote:
> On 21/04/2010 5:41 PM, Jun Shen wrote:
>
>> Dear all,
>>
>> How do stamp my graphs with date and t
Is it possible to implement the Jonckheere-Terpstra test for ordered
alternatives using the coin package: Conditional Inference Procedures
in a Permutation Test Framework?
I found jonckheere.test{clinfun}, but it uses a normal approximation
when ties are present in the data. To make this concrete
At 4:24 PM -0800 4/20/10, chrisli1223 wrote:
Hi everyone,
I have been searching for answers for the following questions but I don't
have much success. The following questions may actually be quite simple. Any
help would be greatly appreciated.
(1) I have written a script which requires user inp
Dears Gabor and Jim,
Many thanks for the answers: both work fine!
Best,
--
///\\\///\\\///\\\///\\\///\\\///\\\///\\\///\\\
Jose Claudio Faria
Estatistica - prof. Titular
UESC/DCET/Brasil
joseclaudio.fa...@gmail.com
///\\\///\\\///\\\///\\\///\\\///\\\///\\\///\\\
--
View this message in con
See the RandomFields package.
-Don
At 4:56 PM -0700 4/21/10, Laura S. wrote:
Dear all:
Does anyone have any suggestions on how to make a spatially explicit
landscape with spatial autocorrelation in R? In other words, a
landscape where all cells have a spatial reference, and the
environment
Dear all:
Does anyone have any suggestions on how to make a spatially explicit landscape
with spatial autocorrelation in R? In other words, a landscape where all cells
have a spatial reference, and the environment values that are closer in space
are more similar (positive spatial autocorrelatio
On 21/04/2010 7:29 PM, rusers.sh wrote:
Hi all,
Today, i just installed the newest R version 2.10.1 and other necessary
tools for building R package under windows,e.g. Rtools, perl. All are the
newest version.
After the correct configuration under windows, i use it to re-check my old
packag
Hi:
I'm a big fan of the reshape package, but this time I think that the doBy
and plyr
packages may better suit your needs. Since you mentioned wanting to get the
min/mean/max of several variables simultaneously, I took out line54 and
added
some vectors of Gaussian(0, 1) random numbers for testing
On 21/04/2010 5:41 PM, Jun Shen wrote:
Dear all,
How do stamp my graphs with date and time somewhere like left corner of the
graph (not the plotting area). I know date() and Sys.time(), but where to?
Thanks.
Use mtext() to put text onto a plot. For example,
plot(1)
mtext(Sys.time(), side=1, l
Hi all,
Today, i just installed the newest R version 2.10.1 and other necessary
tools for building R package under windows,e.g. Rtools, perl. All are the
newest version.
After the correct configuration under windows, i use it to re-check my old
package. I found the following prolem when check
It may be possible to give a solution without a single for loop.
set.seed(1)
v<-sample(1:10,size=1e6,replace=TRUE)
p<-2:4
countinstance<-function(v,p) {
res<-outer(v,p,FUN="==");
apply(res,2,sum)
}
> system.time(replicate(50,countinstance(v,p)))/50
user system elapsed
0.2146 0.
On Apr 21, 2010, at 6:58 PM, David Winsemius wrote:
Sarkar offers a worked example of taking user input regarding
location for locating a grid viewport outside the plot area.
http://lmdvr.r-forge.r-project.org/figures/figures.html
See Figure 12.1
state <- data.frame(state.x77, state.region
Sarkar offers a worked example of taking user input regarding location
for locating a grid viewport outside the plot area.
http://lmdvr.r-forge.r-project.org/figures/figures.html
See Figure 12.1
state <- data.frame(state.x77, state.region)
trellis.vpname("xlab", prefix = "plot1")
trellis.vpna
Yes, outcome is there.
On Wed, Apr 21, 2010 at 3:47 PM, Steve Lianoglou <
mailinglist.honey...@gmail.com> wrote:
> Hi,
>
> On Wed, Apr 21, 2010 at 5:20 PM, Changbin Du wrote:
> > HI, Dear R community,
> >
> > Last friday, I used the codes, it works, but today, it does not run?
> >
> >
> >> fit.
On Apr 21, 2010, at 6:24 PM, Jon Zadra wrote:
Hi,
I want to add error bars to a plot generated with xyplot. I've
tried both errbar() and plotCI(), but in both cases the points are
not in the same place. It's as if the two functions are using a
different frame of reference for the plott
Hi,
On Wed, Apr 21, 2010 at 5:20 PM, Changbin Du wrote:
> HI, Dear R community,
>
> Last friday, I used the codes, it works, but today, it does not run?
>
>
>> fit.dimer <- rpart(outcome ~., method="class", data=p.df)
> Error in `[.data.frame`(frame, predictors) : undefined columns selected
>
>
>
Hey Henrik
I dont do MLE myself but this recent blog might be helpful.
http://www.johnmyleswhite.com/notebook/2010/04/21/doing-maximum-likelihood-estimation-by-hand-in-r/
-A
On Wed, Apr 21, 2010 at 10:02 AM, Thomas Stewart wrote:
> Two possible problems:
>
> (a) If you're working with a normal
Thanks, David.
mtext is for "graphics". Is there an equivalent function for "lattice"?
Jun
On Wed, Apr 21, 2010 at 4:44 PM, David Winsemius wrote:
>
> On Apr 21, 2010, at 5:41 PM, Jun Shen wrote:
>
> Dear all,
>>
>> How do stamp my graphs with date and time somewhere like left corner of
>> the
Hi,
I want to add error bars to a plot generated with xyplot. I've tried
both errbar() and plotCI(), but in both cases the points are not in the
same place. It's as if the two functions are using a different frame of
reference for the plotting area.
for example:
means <- c(92.5, 92.25, 90.
Now available online is our May-June 2010 R Programming Essentials and
Advanced
courses schedule in San Francisco, Seattle, New York City and other USA
cities
http://www.xlsolutions-corp.com/Rcourses
May - June 2010
*** R/S: Programming Essentials
*** R Fundamentals and Programming Techniques
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Hello all:
Is there a way to set up R such that, when you have a very long output
from a command, it will pause when it has displayed one-screen of
information and ask me to press a button to continue displaying? I
happen to have one such command and the information I need is at the
top of the out
On Apr 21, 2010, at 5:41 PM, Jun Shen wrote:
Dear all,
How do stamp my graphs with date and time somewhere like left corner
of the
graph (not the plotting area). I know date() and Sys.time(), but
where to?
Thanks.
?mtext
Jun
[[alternative HTML version deleted]]
On Apr 21, 2010, at 5:19 PM, William Dunlap wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jeff Brown
Sent: Wednesday, April 21, 2010 8:08 AM
To: r-help@r-project.org
Subject: Re: [R] Count matches of a sequence in a vecto
Dear all,
How do stamp my graphs with date and time somewhere like left corner of the
graph (not the plotting area). I know date() and Sys.time(), but where to?
Thanks.
Jun
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing l
The Machine Learning Task view lists alternative packages you could try.
On Wed, Apr 21, 2010 at 5:20 PM, Changbin Du wrote:
> HI, Dear R community,
>
> Last friday, I used the codes, it works, but today, it does not run?
>
>
>> fit.dimer <- rpart(outcome ~., method="class", data=p.df)
> Error in
HI, Dear R community,
Last friday, I used the codes, it works, but today, it does not run?
> fit.dimer <- rpart(outcome ~., method="class", data=p.df)
Error in `[.data.frame`(frame, predictors) : undefined columns selected
DOEs anyone have comments or suggestions? Thanks in advance!
--
S
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Jeff Brown
> Sent: Wednesday, April 21, 2010 8:08 AM
> To: r-help@r-project.org
> Subject: Re: [R] Count matches of a sequence in a vector?
>
>
> This sort of calculation can't
On Apr 21, 2010, at 11:07 AM, Jeff Brown wrote:
At April 21, 2010 10:16:10 AM EDT mieke posted to Nabble:
Hey there,
I need to count the matches of a sequence seq=c(2,3,4) in a long
vector
v=c(4,2,5,8,9,2,3,5,6,1,7,2,3,4,5,).
With sum(v %in% seq) I only get the sum of sum(v %in% 2),
Dear R-Help,
I am Deniz.
I am currently trying to replicate a semiparametric sample selection paper
and I am working on Klein and Spady estimator. I am using the npindex() and
npindexbw() functions.
The problem is, I need results for single bandwidth and when I set bandwidth
computation to "FALSE
sparseby's FUN argument needs to be a function
that accepts a data.frame.
mean() has a data.frame method that returns a vector
of column means but most of the other summary functions
(e.g., median, min, max, quantile) do not work on data.frames.
You can use aggregate() instead of by() or sparseby
rollapply in the zoo package could be used:
library(zoo)
ix <- rollapply(zoo(v), 3, function(x) all(x == 2:4))
which returns a logical vector or if you want the indexes use this:
which(ix)
On Wed, Apr 21, 2010 at 10:16 AM, mieke wrote:
>
> Hey there,
>
> I need to count the matches o
Hello,
I have attempted to email the author of this package without success,
just wondering if anybody else has experienced this.
I am having an using rpart on 4000 rows of data with 13 attributes.
I can run the same test on 300 rows of the same data with no issue.
When I run on 4000 rows, Rgui.e
I've got a problem with the sparseby command (reshape library), and I have
reached the peak of my R knowledge (it isn't really that high).
I have a small data frame of 23 rows and 15 columns, here is a subset, the
first four columns are factors and the rest are numeric (only one, line54 is
provide
Hi all,
I have a problem when trying to read text tables containing Unicode chars in
R 2.10.1 (WinXP, English locale).
An example file is attached.
The following command is supposed to read the table, but the data is only
read incompletely:
read.table("example_unicode.txt", sep="\t", dec=".", h
This sort of calculation can't be vectorized; you'll have to iterate through
the sequence, e.g. with a "for" loop. I don't know if a routine has already
been written.
--
View this message in context:
http://n4.nabble.com/Count-matches-of-a-sequence-in-a-vector-tp2019018p2019108.html
Sent from t
Hey there,
I need to count the matches of a sequence seq=c(2,3,4) in a long vector
v=c(4,2,5,8,9,2,3,5,6,1,7,2,3,4,5,).
With sum(v %in% seq) I only get the sum of sum(v %in% 2), sum(v %in% 3) and
sum(v %in% 4), but that's not what I need :(
Who can help me?
Thanks a lot!
--
View this messag
Dear R-Help,
my name is Henrik and I am currently trying to solve a Maximum Likelihood
optimization problem in R. Below you can find the output from R, when I use
the "BFGS" method:
The problem is that the parameters that I get are very unreasonable, I would
expect the absolute value of each p
I am trying to test for fixed factor main effects in an unbalanced
mixed effects model but when I fit the reduced model for "mic" factor
effects, the extra degrees of freedom are being allocated to a nested
term rather than the residuals. The model has inc, mic and spp are
independent variab
Thomas Lumley wrote:
On Tue, 20 Apr 2010, Noah Silverman wrote:
I just read the help page for predict.coxph.
It indicates that the risk score is just exp(lp)
What I'm trying to find, and have seen with some other implementations
is the "conditional probability within group". Neither the lp o
Assuming the date as id is the first column followed by 23 values, try
the read.reps function found here:
http://www.mail-archive.com/r-help@r-project.org/msg92123.html
like this:
DF <- read.csv("myfile", as.is = TRUE)
read.reps(DF, 23)
On Wed, Apr 21, 2010 at 2:24 PM, Idgarad wrote:
> I have
Try using contour() instead of levelplot. See the examples
in help('contour') for how to add contour lines to an
existing plot.
-Peter Ehlers
On 2010-04-21 13:08, David Winsemius wrote:
On Apr 21, 2010, at 2:27 PM, Simon Goodman wrote:
I've generated a levelplot showing the density distrib
Wonderful Christian, thank you for the (*very*) helpful reply!
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statis
Hi David,
Thank you so much, that's just what I want!!
Best
Tengfei
On Wed, Apr 21, 2010 at 2:24 PM, David Winsemius wrote:
>
> On Apr 21, 2010, at 3:21 PM, Tengfei Yin wrote:
>
> Hi
>>
>> I have mone quick question I am not quite familiar with, for generic
>> function plot, why some methods
On Apr 21, 2010, at 3:21 PM, Tengfei Yin wrote:
Hi
I have mone quick question I am not quite familiar with, for generic
function plot, why some methods are marked by '*', I think
plot(as.mcmc())
may dispatch the right method, I try to use get to view the
function, but it
seems that get() o
On Apr 21, 2010, at 1:15 PM, David Winsemius wrote:
On Apr 21, 2010, at 12:50 PM, Michael Hosack wrote:
I provided a minimized version of my dataframe at the bottom of
this message containing the results of David's code in variable
('wkoffset') and Jeff Hallman's code in ('WEEK'). Jeff
Hi
I have mone quick question I am not quite familiar with, for generic
function plot, why some methods are marked by '*', I think plot(as.mcmc())
may dispatch the right method, I try to use get to view the function, but it
seems that get() only works for the one with no "*", e.g. get('plot.ecdf')
On Apr 21, 2010, at 2:27 PM, Simon Goodman wrote:
I've generated a levelplot showing the density distribution of a
species
derived from survey transects, with lon, lat co-ordinates.
I'd like to overlay this on a map of the study region specified by:
map('worldHires', xlim = range(mlon), y
Gustaf,
That is correct. Schedule3 does contain all of the Saturdays between April 30
and Nov. 01 for a given year.
Mike
R experts,
>
> How could I extract the week number from a date vector (in Date class)
> such that week numbering (week 1...2...) begins (May 01) and ends
> (October 31
David, there is still a problem. The data in variable 'wkoffset' should
be equivalent to the data in 'WEEK'. May 07 should begin wkoffset=2 and
the preceeding days should all be assigned wkoffset=1. I did make sure to
adjust the year within day.of.week(). Could you please explain what you
me
Slight addition below;
On 2010-04-21 10:51, David Winsemius wrote:
On Apr 21, 2010, at 12:36 PM, David Winsemius wrote:
On Apr 21, 2010, at 12:09 PM, Andrea Bernasconi DG wrote:
Thank you David,
but how to get the value of 0.015939 present in s.npk.aov, and not
given by s.npk.aov$coef["bl
I'm guessing that you are using the words "table" and
"list" to mean "data frame". If that's the case, something
like this might get you started:
dfnew = reshape(Test1,varying=list(paste('Hour',1:23,sep='')),
timevar='Hour',idvar='Date',direction='long')
dfnew = dfnew[order(dfne
On 21.04.2010 20:31, Xiao D wrote:
Dear List members,
I am using R to generate MCMC time series plots.
This is the code I use; however, everytime an error message will come out
saying could not find function "plot.mcmc."
I have coda package and Lattice package installed.
Does anyone know what m
Dear List members,
I am using R to generate MCMC time series plots.
This is the code I use; however, everytime an error message will come out
saying could not find function "plot.mcmc."
I have coda package and Lattice package installed.
Does anyone know what may get wrong here? Thanks so much for y
Phil's algorithm is a good one, unless you're worried about optimizing for
speed. It makes N * M comparisons, where N is the length of the first
vector and M is the length of the second. Explicitly iterating through the
longer vector, you could reduce the number of comparisons to M. As is often
I've generated a levelplot showing the density distribution of a species
derived from survey transects, with lon, lat co-ordinates.
I'd like to overlay this on a map of the study region specified by:
map('worldHires', xlim = range(mlon), ylim = range(mlat)), where mlon, mlat
specifies the study
##I am trying to test for fixed factor main effects in an unbalanced mixed
effects model but when I fit the reduced model for "mic" factor effects, the
extra degrees of freedom are being allocated to a nested term rather than the
residuals. The model has inc, mic and spp are independent varia
I have a series of tables, one for each environment indicating a date (row)
and a sample at each hour of the day (0 to 23)
Test1 Table:
Date,Hour1,Hour2,...Hour23
1/1/10,123,123,...,123
I would like to model this as a time series but how can I translate the
table into a list such that I can get:
> optimize(function(x) -scores(x), interval=c(0.5, 0))
$minimum
[1] 0.1830174
$objective
[1] -11.67820
So the maximum is at x=0.183 with a score of 11.7, numerically.
Uwe Ligges
On 17.04.2010 16:30, Akito Y. Kawahara wrote:
Hi, I am new to R, and have a quick question regarding an R script
You forgot to rebuild or reinstall ?
Uwe Ligges
On 18.04.2010 00:25, BXC (Bendix Carstensen) wrote:
I am updating the Epi package.
I added functions named pc.points and pc.matpoints.
Erroneously I wrote pc.plot and pc.matplot in the NAMESPACE file and of course
got an error from Rcmd check
On 19.04.2010 13:51, Viechtbauer Wolfgang (STAT) wrote:
Dear All,
Suppose I want to write a method for the generic function confint():
args(confint)
function (object, parm, level = 0.95, ...)
So, it looks like the second
really???
> and third argument have been predefined in the generi
Hi list!
I have prepared a custom function (below) in order to calculate separability
indices (Divergence, Bhattacharyya, Jeffries-Matusita, Transformed divergene)
between two samples of (spectral land cover) classes.
I need help to cross-compare results to verify that it works as expected
(si
I've got a problem with the sparseby command (reshape
library), and I have reached the peak of my R knowledge (it isn't really that
high).
I have a small data frame of 23 rows and 15 columns, here
is a subset, the first four columns are factors and the rest are numeric (only
one, line54 is pro
Dear Tal,
I took the definition of the Hubert gamma- and Dunn-index from the Gordon
book. They are actually not about comparing two clusters, at least not in
that reference, and they require dissimilarities.
The adjusted Rand index and Meila's VI, as implemented in
cluster.stats, compare two
On Apr 21, 2010, at 12:50 PM, Michael Hosack wrote:
I provided a minimized version of my dataframe at the bottom of this
message containing the results of David's code in variable
('wkoffset') and Jeff Hallman's code in ('WEEK'). Jeff's code
produced the correct results (thank you Jeff)
You cannot specify a dfunction for z, but need to compute the values in
the matrix yourself as in:
persp(data_for_time, data_for_s,
outer(data_for_time, data_for_s, plot_R_i_3d))
Uwe Ligges
On 20.04.2010 17:35, Jin wrote:
Hello Dear,
I have a function, like z=f(x,y), and try a surfac
Thanks for the fast reply Uwe.
My hope in posting this was to find if anyone had already done work (in R)
in this direction. So far I wasn't able to find any such relevant code, so
I turned to the mailing list.
Regarding new implementations - thanks for offering! - I have already came
around one
On 21.04.2010 16:36, Mario Valle wrote:
n <- 63
a <- 1:n
x <- a-1
y <- outer(x,a)
matplot(x,y,type='l')
Warning message:
In matplot(x, y, type = "l") :
default 'pch' is smaller than number of columns and hence recycled
Why is it complaining if I specifically ask for type="l", so no pch
involve
On Apr 21, 2010, at 12:36 PM, David Winsemius wrote:
On Apr 21, 2010, at 12:09 PM, Andrea Bernasconi DG wrote:
Thank you David,
but how to get the value of 0.015939 present in s.npk.aov, and not
given by s.npk.aov$coef["block","Pr(>F)"] ?
??? That's not a coefficient. It's a p-value.
On 21.04.2010 18:15, Tal Galili wrote:
Hello all,
I would like to compare the similarity of two cluster solutions using a
validation criteria (such as Hubert's gamma coefficient, the Dunn index the
corrected rand index and so on)
I see (from here:http://www.statmethods.net/advstats/cluster.html
I provided a minimized version of my dataframe at the bottom of this message
containing the results of David's code in variable ('wkoffset') and Jeff
Hallman's code in ('WEEK'). Jeff's code produced the correct results (thank you
Jeff) though I have been unable to understand it. David, as you
On 21.04.2010 18:19, Michael Steven Rooney wrote:
The R version is 2.9.2.
Please try with R-2.11.0 RC (release candidate) to be released tomorrow
as well as a recent version of nlme.
If its still fails, please send code (and data, if you do not generate
them in the code) that reproduces t
On Apr 21, 2010, at 12:09 PM, Andrea Bernasconi DG wrote:
Thank you David,
but how to get the value of 0.015939 present in s.npk.aov, and not
given by s.npk.aov$coef["block","Pr(>F)"] ?
??? That's not a coefficient. It's a p-value.
On the other, the procedure to extract coefficients from
The R version is 2.9.2.
nlme version is 3.1-93
The OS is Windows XP.
This is for my work computer but I get the same behavior at home using
Vista.
-- Forwarded message --
From: Sarah Goslee
Date: Wed, Apr 21, 2010 at 11:50 AM
Subject: Re: [R] R crashing oddly
To: Michael Steven Roo
gsubfn is like gsub except instead of a replacement string it uses a
replacement function whose input is the string matched by the regular
expression and whose output replaces the match. The replacement
function can optionally be specified as a formula as we do here. If
there is no left hand sid
Try this:
> set.seed(1)
> x <- c(rnorm(1e3, mean=10, sd=1), 50, 100)
>
> start <- 0
> end <- 110
> h <-10
>
> c1 <- cut(x, br=seq(start, end, h), right=TRUE)
> levels(c1)
[1] "(0,10]""(10,20]" "(20,30]" "(30,40]" "(40,50]" "(50,60]"
"(60,70]" "(70,80]"
[9] "(80,90]" "(90,10
Hello all,
I would like to compare the similarity of two cluster solutions using a
validation criteria (such as Hubert's gamma coefficient, the Dunn index the
corrected rand index and so on)
I see (from here:http://www.statmethods.net/advstats/cluster.html) that
the function cluster.stats() in th
Dear vegan-helpers,
I calculated an NMDS with metaMDS and then displayed the results with ordiplot.
The NMDS consist of 4 axes. I want to plot two diagrams: 1st vs. 2nd and 3rd
vs. 4th axis. I used the ordiplot-command choices = c(1,2) and c(3,4),
respectively. 1st vs. 2nd does not make any prob
Thank you David,
but how to get the value of 0.015939 present in s.npk.aov, and not given by
s.npk.aov$coef["block","Pr(>F)"] ?
On the other, the procedure to extract coefficients from a summary of lm or aov
should be the same.
Andrea
On 21 Apr, 2010, at 3:20 PM, David Winsemius wrote:
>
>
Here is one way of doing it:
> x <- read.table(textConnection("CFISCAFIRMS
YEAR VARVALUE
+ 20345 nike2005EC01 34
+ 20345 nike2006 EC01 45
+ 56779
On Tue, 20 Apr 2010, Noah Silverman wrote:
I just read the help page for predict.coxph.
It indicates that the risk score is just exp(lp)
What I'm trying to find, and have seen with some other implementations
is the "conditional probability within group". Neither the lp or the
risk options see
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