Great!
Thanks to both of you!
Sincerely,
Erin
On Wed, Apr 15, 2009 at 12:54 AM, Coen van Hasselt
coenvanhass...@gmail.com wrote:
Alternatively you could also drop the column like this:
xx$x2-NULL
On Wed, Apr 15, 2009 at 15:51, Peter Alspach
palsp...@hortresearch.co.nz wrote:
Tena koe
Dear Members,
I have been searching for a package in R which can handle multiple seasonality
suggested by taylor(2003).
It will be great help if anybody has used this on R before (i.e. which package).
Thanks in Advance.
Best Regards
Atul Malik
[[alternative HTML version deleted]]
Dear Members,
I have been searching for a package in R which can handle multiple
seasonality suggested by taylor(2003).
It will be great help if anybody has used this on R before (i.e. which
package).
Thanks in Advance.
Best Regards
Atul Malik
-
Atul Malik (Consultant)
DecisionCraft
Dieter Menne wrote:
I do not know if this a problem with me, my data or cph/survest in package
design. The example below works with a standard data set, but not with my
data, but I cannot locate the problem.
Dieter Menne found out after hours, that in one case and explicit cast is
I've been using the ave function to compute some statistics on a
data frame. After a while I noticed that, for some reason, it was
returning numerical statistics as strings instead of numbers. I delved
into the code of the functions and traced the problem to the following
fact:
Gábor Csárdi schrieb:
The would prefer two parallel arrows one for each direction.
You can set 'curved' to a value close to zero and then the arrows will
be only a bit curved.
No I am lost ... do you mean?
...
E(g)$curved - 0.5
...
plot.igraph(g, layout=layout.kamada.kawai,
Joseph Voelkel wrote:
Hi, all,
Using RGui, is it possible to create a graphics window that can be moved
outside of the RGui window? (This can be done--in fact must be
done--using Rterm, but I wish to use RGui.)
My interest for this is to use two monitors: in my private monitor I
wish to
Hi all,
A colleague of mine tried to install the package EMV, which had been
removed from CRAN.
she ran into some kind of trouble, R locked up, and she closed the program.
Now when she starts R, utils can't be loaded which of course create
an unworkable environment.
Below I've copy-pasted the
On Wed, Apr 15, 2009 at 9:35 AM, Knut Krueger r...@krueger-family.de wrote:
Gábor Csárdi schrieb:
The would prefer two parallel arrows one for each direction.
You can set 'curved' to a value close to zero and then the arrows will
be only a bit curved.
No I am lost ... do you mean?
...
Dear List,
I'm trying to interpret the results of the Kruskal's Non-metric
Multidimensional Scaling algorithm (isoMDS, MASS package).
The 'goodness of fit' is reported as The final stress achieved (in percent).
What does this mean exactly? I've tried to google for an answer but I've not
come
(FICB[,temp])
[1] 0.30 0.55 0.45 2.30 0.45 0.30 0.25 0.30 0.30
1.05 1.00 1.00
[13] 0.30 0.30 0.30 0.55 0.30 0.30 0.30 0.25 1.00
0.30 0.30 0.45
[25] 0.30 1.30 0.30 0.30 0.45 0.30 0.30 0.30 NA
NA NA NA
[37] 0.30 NA 0.30 0.30 0.30 0.30 NA NA 0.35
NA 0.35 0.30
[49] 0.30 0.40 NA 0.40
try this:
string - c(0.30, 0.55, 0.45, 2.30, NA, NA, NA,
0.50, 0.30, 0.30, 0.45, 0.30)
splt - strsplit(string, \\.)
sapply(splt, function (x) if(length(x) == 2) x[1] else as.character(NA))
sapply(splt, function (x) if(length(x) == 2) x[2] else as.character(NA))
I hope it helps.
Best,
On 4/15/2009 1:38 AM, Erin Hodgess wrote:
Dear R People:
Suppose I have the following data frame:
x1 x2 x3
1 -0.1582116 0.06635783 1.765448
2 -1.1407422 0.47235664 0.615931
3 0.8702362 2.32301341 2.653805
str(xx)
'data.frame': 3 obs. of 3 variables:
$ x1:
I have created two functions to compute geometric means. Method 1 can
handle even number of negative values but not large number, vice versa
for method 2. How can I merge both functions so that both large number
and negative values can be handled ?
geometric.mean1 - function(x)
Hello R users,
I've used the following help:
Comparing two regression line slopes
I knew the method based on the following statement :
t = (b1 - b2) / sb1,b2
where b1 and b2 are the two slope coefficients and sb1,b2 the pooled
standard error of the slope (b)
which can be calculated in R
Knut, I think you have an older version of igraph that does not
support curved edges. Again, please read
http://lists.gnu.org/archive/html/igraph-help/2009-04/msg00104.html
and install version 0.5.2, it is on CRAN now (except for OSX).
Gabor
On Wed, Apr 15, 2009 at 11:30 AM, Knut Krueger
I'm not being able to control all parameters of a xyplot.
dataset example:
tabp[1:10,]
time n X id name
1 1 95 0.00 1 Coral reef
2 1 93 0.00 2 Coral reef
3 1 92 0.00 3 Coral reef
4 1 90 0.00 4 Coral reef
5 1 87 8.994321 5 Coral reef
Benedikt,
Is this homework? Let's see, the same question as last week, when Eik
Vetorazzi showed you how to interpret the test value.
The significance of this test depends on your pre-specified alpha. Don't
rely on an arbitrary value of 0.05. It would be a good idea to do some
reading (any
Hello Harsh,
I found useful the fgui package (
http://www.people.fas.harvard.edu/~tjhoffm/fgui.html ).
Regards,
Gabriele Franzini
ICT Applications Manager
Nerviano Medical Sciences SRL
Nerviano Italy
-Original Message-
From: Barry Rowlingson [mailto:b.rowling...@lancaster.ac.uk]
For each variable x, y and z I would like to run the same set of
commands. I have tried
for (n in FICB[,calct30], FICB[, calct60]) {
FICB[,temp] - format([n],digits=2)
etc
}
How do I do this correctly?
__
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On 15-Apr-09 09:26:55, richard.cot...@hsl.gov.uk wrote:
I have created two functions to compute geometric means. Method 1 can
handle even number of negative values but not large number, vice versa
for method 2. How can I merge both functions so that both large number
and negative values can be
Gábor Csárdi schrieb:
Hi Gabor, it seems that anybody doesnt not want us to find a solution ;-)
and install version 0.5.2, it is on CRAN now (except for OSX).
I got the 0.5.1 at 06.April.2009
I just tried the Munich Mirror got 0.5.1
there is also only 0.5.1 on
Hello benedikt,
You say the slopes differ significantly if the p-value is less than a
given threshold, most of the time 0.05.
Please, note that fitting a linear regression through three points is
senseless...
Regards,
Alain
Benedikt Niesterok wrote:
Hello R users,
I've used the
I have created two functions to compute geometric means. Method 1 can
handle even number of negative values but not large number, vice versa
for method 2. How can I merge both functions so that both large number
and negative values can be handled ?
geometric.mean1 - function(x)
Gábor Csárdi schrieb:
Dear Gabor, I am very sorry but i am not able to reproduce your example.
there is no change, i am using r 2.8.0
library(igraph)
g - graph.ring(3, dir=TRUE, mut=TRUE)
g$layout - layout.circle
E(g)$curved - 0.5
plot(g)
E(g)$curved - 0.1
plot(g)
It is a good idea to
geometric.mean1 - function(x) prod(x)^(1/length(x))
geometric.mean2 - function(x) exp(mean(log(x)))
geometric.mean1(c(-5,-4,4,5))
[1] 4.472136
geometric.mean2(c(-5,-4,4,5))
[1] NaN
Warning message:
In log(x) : NaNs produced
comp.x - as.complex(c(-5,-4,4,5))
Gustaf Rydevik wrote:
Hi all,
A colleague of mine tried to install the package EMV, which had been
removed from CRAN.
she ran into some kind of trouble, R locked up, and she closed the program.
Now when she starts R, utils can't be loaded which of course create
an unworkable environment.
Below
Hi,
I am running a simulation and have to perform ANOVA to determine the rank of
factors. Used the aov() function and it works great for full factorial design.
1. For a massive set of data, I tried using biglm, while it can create the
linear model, all the residuals (for assumption
Is there a way to specify a *vector* of statements, as an argument to a
function. What I'm trying to do is to provide the ability to add elements
to a series of panels drawn by a function. (As documented, split.screen
does not provide this capability.) My idea is to mimic the 'plot.axes'
Knut Krueger schrieb:
http://cneurocvs.rmki.kfki.hu/igraph/download/igraph_0.5.2.zip
Server is now available again
I got the file.
Thanks Knut
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PLEASE do read the
Thanks!
That's a really fast soltution. Now the R process takes a few seconds
instead of a couple of hours with my loop.
greets.
jholtman wrote:
try this:
x
V1 V2 V3
1 500 320 0
2 510 310 0
3 520 310 0
4 520 320 0
y
V1
On 15/04/2009 6:31 AM, Dan Kelley wrote:
Is there a way to specify a *vector* of statements, as an argument to a
function.
There are several ways. expression() converts its arguments into a
vector of expressions, e.g.
e - expression(cat(a\n), cat(b\n))
e[[1]]
cat(a\n)
eval(e[[1]])
a
That works perfectly. Thanks very much!!
Duncan Murdoch-2 wrote:
On 15/04/2009 6:31 AM, Dan Kelley wrote:
Is there a way to specify a *vector* of statements, as an argument to a
function.
There are several ways. expression() converts its arguments into a
vector of expressions, e.g.
Yes it works, but I still have a problem.
The thing is that I know the dimensions of my plot but in the R code, not in
the latex code. So I tried to do :
\SweaveOpts{prefix.string=proj1}
label=fig1,fig=T,include=F,width=\Sexpr{wid}, height=\Sexpr{hei}=
plot(1:10)
@
\begin{figure}
On Wed, Apr 15, 2009 at 12:20 PM, Duncan Murdoch murd...@stats.uwo.ca wrote:
Gustaf Rydevik wrote:
Hi all,
A colleague of mine tried to install the package EMV, which had been
removed from CRAN.
she ran into some kind of trouble, R locked up, and she closed the
program.
Now when she starts
Dear Lore,
The easiest thing to do is to write a function that saves your plot to a file
and generates the latex code.
label=fig1, fig=FALSE, result = tex=
pdf(fig1.pdf, width = wid, heigth = hei)
plot(1:10)
dev.off()
cat(\begin{figure}\)
cat(\includegraphics[width = , wid, , height = ,
Dear all,
I have been looking for possibilities to read and write Excel 2007-files
in and from R.
The 'reading' part is ok through odbcConnectExcel2007 and sqlFetch(RODBC).
For 'writing' I thought of using sqlSave in the same package, but it
does not work (I think because this function
Using string from another responder's post here are two
solutions:
1. The first converts to numeric and manipulates that:
cbind(part1 = floor(as.numeric(string)), part2 = 100 * as.numeric(string) %% 1)
2. The second uses strapply from the gsubfn package.
It matches from the beginning ( ^ ) a
I have 3 layouts differents each others and I need make a layout with these 3
layouts, someone can me help???
--
View this message in context:
http://www.nabble.com/composition-of-layouts-tp23057309p23057309.html
Sent from the R help mailing list archive at Nabble.com.
I want to fit the model y=a*x^b using nls; where a should be different for
each level of a factor.
What is the easiest way to fit it? Can i do it with nls?
I've looked the help pages and the MASS example in page 249 but the formula
is different and I don't know how to specify it for my model.
Dieter Menne wrote:
Dieter Menne wrote:
I do not know if this a problem with me, my data or cph/survest in package
design. The example below works with a standard data set, but not with my
data, but I cannot locate the problem.
Dieter Menne found out after hours, that in one case and
Hi,
Algorithm question: I have two sets of intervals, where an interval is
an ordered pair [a,b] of two numbers. Is there an efficient way in R to
generate the intersection of two lists of same?
For concreteness: I'm representing a set of intervals with a data.frame:
list1 =
Check out:
http://tolstoy.newcastle.edu.au/R/help/05/07/8950.html
but use
wkbk$SaveAs(\\test.xlsx)
instead of the corresponding line there.
You must have Excel 2007 on your machine for this to work.
On Wed, Apr 15, 2009 at 8:43 AM, Katrien Baert katrien.ba...@ugent.be wrote:
Dear all,
I
Adding
par.settings = list(grid.pars = list(lwd = 3))
to my code which uses simpleKey gives exactly what I want.
However, if I use auto.key, I lose all the formatting and get
default colors and symbols.
I am indebted to Deepayan for once again providing the key
(pun intended) to a solution.
Dear Gábor,
thank you for your help, I tried the 0.5.2 Version.
It works fine.
The team is very satisfied with the curved graph.
Regards Knut
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PLEASE do read the posting
On Tue, Apr 14, 2009 at 1:23 AM, Harsh singhal...@gmail.com wrote:
HI R users,
I would appreciate information/examples/suggestions on building GUIs
for R applications.
I am currently working on a project that would require the following
functionalities :
1) Display a window to the user.
On 4/15/2009 8:59 AM, Thomas Meyer wrote:
Hi,
Algorithm question: I have two sets of intervals, where an interval is
an ordered pair [a,b] of two numbers. Is there an efficient way in R to
generate the intersection of two lists of same?
For concreteness: I'm representing a set of intervals
one way is:
list1 - as.data.frame(list(open=c(1,5), close=c(2,10)))
list2 - as.data.frame(list(open=c(1.5,3), close=c(2.5,10)))
data.frame(
open = pmax(list1$open, list2$open),
close = pmin(list1$close, list2$close)
)
I hope it helps.
Best,
Dimitris
Thomas Meyer wrote:
Hi,
Not of the self but still not complicated:
list1 - data.frame(open=c(1,5), close=c(2,10))
list2 - data.frame(open=c(1.5,3), close=c(2.5,10))
Intersec - data.frame(Open = pmax(list1$open, list2$open), Close =
pmin(list1$close, list2$close))
Intersec[Intersec$Open Intersec$Close, ] - NA
Intersec
On Wed, Apr 15, 2009 at 1:27 PM, Manuel Gutierrez
manuelgutierrezlo...@gmail.com wrote:
I want to fit the model y=a*x^b using nls; where a should be different for
each level of a factor.
What is the easiest way to fit it? Can i do it with nls?
I've looked the help pages and the MASS example in
There is a very nice intervals package in CRAN. It is impressively
efficient even for intersections of many millions of intervals. If I
remember correctly, it is purely in-core, so on a 32-bit R you'll be
limited to something like 100 million intervals. Is that enough for
your application?
On Wed, Apr 15, 2009 at 2:33 PM, Paul Smith phh...@gmail.com wrote:
On Wed, Apr 15, 2009 at 1:27 PM, Manuel Gutierrez
manuelgutierrezlo...@gmail.com wrote:
I want to fit the model y=a*x^b using nls; where a should be different for
each level of a factor.
What is the easiest way to fit it? Can
After few corrections, it does work. But I have several plots to include in my
document and, because of those commands, they're all on the same line even if
there are 20 plots. I mean that Latex doesn't car about textwidth anymore and I
get an overfull box (too wide).
What could I do to
Hi Dieter,
I'll take a shot at this.
As I understand it, the stress is telling you how the ordination distances
compare with original dissimilarities that you calculated.
It is a measure how well your ordination has done in representing the
relationship of your sites. Note that the stress
I would like to randomly shuffle a distance object, such as the one
created by ade4{dist.binary} below. My first attempt, using
sample(jc.dist) creates a shuffled vector, losing the lower triangular
structure of the distance object. How can I Ishuffle the lower
triangular part of a distance
Hi all,
I am dealing with a big data frame. When printing something like
allData[[3]]
1 625.364 38.223 21.014 0.216 1.241411V 1050o 58.38065 -0.06178768
2 383.709 55.811 21.435 0.296 1.241411V 1050o 58.38308 -0.03328282
3 434.669 58.597 21.207 0.233 1.241411V 1050o 58.38334
Thanks, Uwe. This is exactly what I wanted.
Joe
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Wednesday, April 15, 2009 3:42 AM
To: Joseph Voelkel
Cc: r-help@r-project.org
Subject: Re: [R] Creating a graphics window (in Windows,with RGui) that
is
Hi!
First, pardon me if this is a faq. I think I should be using some sort
of apply, but I am not managing to figure those out.
I have a data frame similar to this:
d - data.frame(x = LETTERS[1:5], y = rnorm(5), z = rnorm(5))
d
x y z
1 A 0.1605464 -0.2719820
2 B
Have a look at the width argument in ?options
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg /
Hello alltogheter,
I have a little problem regarding merging to zoo series.
I want to merge two zoo series to reduce the timegaps between the stamps.
I use the following code:
data.test -
as.POSIXct(seq(data.input01[1,1],data.input01[nrow(data.input01),1],900),tz=GMT)
data.troughput01 -
Hi,
Here is one-way to do it (the following code shows a simulation example):
n - 200
set.seed(123)
x - runif(n)
f - gl(n=2, k=n/2) # a two-level factor
x1 - x * (f == 1)
x2 - x * (f == 2)
a - c(rep(2, n/2), rep(5, n/2))
b - 0.5
nsim - 100
nls.coef - matrix(NA, nsim, 3)
for (i in
Hi Karin,
I'm not sure I understand... Is this what you want ?
d$y - mean(d$y)/sd(d$y)
2009/4/15 Karin Lagesen karin...@ifi.uio.no
Hi!
First, pardon me if this is a faq. I think I should be using some sort
of apply, but I am not managing to figure those out.
I have a data frame similar
You don't say what your intent is, but for most applications it's
important to preserve the pairwise matches. Here's one way
to do that.
library(ecodist) # for the convenient functions lower() and full()
library(ade4)
x1 - c(rep(0,4),1)
x2 - c(rep(0,2),rep(1,3))
x3 - c(rep(1,3), rep(0,2))
X -
Dear all,
I have about 20,000 binary images, on the sense of habitat / non-habitat.
I am running labcon() on these images (512x512pix), but when the
number of patches is very large, the labcon stop, without error, and
never end.
I am running on a Linux machine, with 6Gb ram (memory is not
This seems more a LaTeX problem than an R problem. But can you provide us (an
sample example of) the LaTeX code the yields the overfull box.
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research
pmax/pmin did the trick nicely -- the right-size tool I was hoping for.
Thanks to all,
-tom
On 4/15/2009 9:14 AM, ONKELINX, Thierry wrote:
Not of the self but still not complicated:
list1 - data.frame(open=c(1,5), close=c(2,10))
list2 - data.frame(open=c(1.5,3), close=c(2.5,10))
Intersec -
Dear All,
Is it possible to run a GEE analysis with a custom link function in R?
In particular I'm wanting to use a mafc.logit() generated link function
(using the package psyphy). I can use this with a GLM, but it looks like
the gee package only accepts predefined link functions.
Thanks,
Mark
In general, how can I increase a vector of length m ( n) to length n
by padding it with m - n missing values, without losing attributes?
The two approaches I've tried, using length- and adding missings with
c, do not work in general:
a - as.Date(2008-01-01)
c(a, NA)
[1] 2008-01-01 NA
length(a)
You're right, I'm sorry to disturb you with my Latex problem.
If anyone has an idea, here is the Latex code that I get :
\begin{figure}
\multido{\i=1+1}{7}{\includegraphics[page=\i,width=1.5in, height =
1.5in]{image.pdf} \\ }
\end{figure}
I tried to break the line with \\ or with \linebreak
Dear R-Help List,
I am using optim, with method=L-BFGS-B, to maximize a likelihood inside
a large simulation exercise. This runs fine for most simulated data
sets, but for some reason, about 1 out of 100 times, optim will just hang.
Using a dumb approach to the problem (i.e. printing the
I just want to add that I didn't have any overful box before I use :
label=fig1, fig=FALSE, results = tex=
pdf(fig1.pdf,
width = wid, heigth = hei)
plot(1:10)
plot(1:10)
dev.off()
cat(\\begin{figure}[h])
cat(\\centering)
cat(\\multido{\i=1+1}{7}{\includegraphics[page=\i,width=1.5in, height =
If I am right then you must get the seasonal factor etc (if any) out before
fitting ant ARIMA (or statistical model) i.e. fit ARIMA on residual series
not original series.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of djhurio
A more compact way to code factors in nls is to use the syntax factor[].
Here's an example using a simplified version of Ravi's example:
n - 200
set.seed(123)
x - runif(n)
a - gl(n=2, k=n/2) # a two-level factor
eps - rnorm(n, sd=0.5)
y - as.numeric(a) * x^.5 + eps
nls(y ~ a[]*x^b,
On Wed, Apr 15, 2009 at 3:57 PM, Tiago Marques ti...@mcs.st-and.ac.uk wrote:
I am using optim, with method=L-BFGS-B, to maximize a likelihood inside a
large simulation exercise. This runs fine for most simulated data sets, but
for some reason, about 1 out of 100 times, optim will just hang.
How does one indicate that a particular survival time is right censored in
the Survreg routine?
[[alternative HTML version deleted]]
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Oops! I made a mistake. Corrected below.
On Wed, 2009-04-15 at 11:05 -0400, Manuel Morales wrote:
A more compact way to code factors in nls is to use the syntax factor[].
Here's an example using a simplified version of Ravi's example:
n - 200
set.seed(123)
x - runif(n)
a - gl(n=2,
Dear R Helpers,
I have noticed that when I use lmer to analyse data, the summary function
gives different values for the AIC, BIC and log-likelihood compared with the
anova function.
Here is a sample program
#make some data
set.seed(1);
In general, how can I increase a vector of length m ( n) to length n
by padding it with m - n missing values, without losing attributes?
The two approaches I've tried, using length- and adding missings with
c, do not work in general:
a - as.Date(2008-01-01)
c(a, NA)
[1] 2008-01-01 NA
What are the R equivalents to the Stata command egen?
egen temp = anycount(t0vas t30vas t60vas t120vas t240vas t360vas),
values(0,1,2,3,4,5,6,7,8,9,10)
egen temp2 = rowtotal(t0vas t30vas t60vas t120vas t240vas t360vas)
__
R-help@r-project.org mailing
Also consider the View function for looking at the dataframe
On Wed, Apr 15, 2009 at 10:13 AM, Vladan Arsenijevic
arsen...@ist.utl.pt wrote:
Hi all,
I am dealing with a big data frame. When printing something like
allData[[3]]
1 625.364 38.223 21.014 0.216 1.241411 V 1050o 58.38065
Hi Tiago,
It is hard for me to speculate without knowing more about your problem.
Here is what I would suggest, assuming your problem specification and its
computer implementation are correct:
(1) You may try to terminate the algorithm by specifying a different
stopping criterion than the
Duncan
I tried writeLines. But I need to print about 23 lines and it is
really slow.
Thanks
Aparna
-Original Message-
From: Duncan Murdoch [mailto:murd...@stats.uwo.ca]
Sent: Tuesday, April 14, 2009 4:34 PM
To: Vemuri, Aparna
Cc: r-help@r-project.org
Subject: Re: [R] Controlling
How about:
xx[,-match(x2,names(xx))]
or
xx[,names(xx) != x2]
etc.
Bert Gunter
Genentech Nonclinical Biostatistics
650-467-7374
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Erin Hodgess
Sent: Tuesday, April 14, 2009 10:39
This may not be everything you would like but perhaps its sufficient:
structure(length-(a, 2), class = class(a))
[1] 2008-01-01 NA
Your second example looks ok as is. Can't tell what you don't
like about it.
On Wed, Apr 15, 2009 at 10:55 AM, hadley wickham h.wick...@gmail.com wrote:
In
Hello,
I've run 7 candidate models using mixed-effects logistic regression with
the lmer function from the lme4 package, and I'm getting the following
error for 5 of those models: Warning message: In mer_finalize(ans : false
convergence (8). The candidate models all run with the same data,
Hi All,
I have a data set which I need to plot and show the values of one of the
variables as a second x-axis.
library(lattice)
year-c(2001,2002,2003,2004,2005,2006)
fac-c(arts,arts,arts,sci,sci,sci)
staff-c(95,98,99,32,31,36)
part-c(32,31,33,15,16,14)
df1-data.frame(year,fac,staff,part)
Manuel == Manuel Morales manuel.a.mora...@williams.edu writes:
nls(y ~ a[fac]*x^b, start=list(a=c(1,1), b=0.25))
Did you mean a[f]?
nls(y ~ a[f]*x^b, start=list(a=c(1,1), b=0.25))
Mike
__
R-help@r-project.org mailing list
Hi all
I have developed a zero-inflated negative binomial model using the
zeroinfl function from the pscl package, which I have carried out model
selection based on AIC and have used likelihood ratio tests (lrtest from
the lmtest package) to compare the nested models [My end model contains
2
Stavros, you are quite correct -- I discovered that the hard way a
little while ago when testing my two-line solution. Use of pmin/pmax
don't handle, for instance, cases where more than one interval in one
set is wholly contained by an interval in the other. (I have a
mis-posted msg awaiting
On Tue, 14 Apr 2009, jimm-pa...@gmx.de wrote:
Hi all,
I'm fitting a line to my dataset. Later I want to predict missing values that exceed the
[min,max] interval of my empirical data, therefore I choose surface=direct
for extrapolation.
Hello,
I am using the dunn metric, but something is wrong and I dont understand
what or what that this error mean. Please can you help me with this?
The instructions are:
#Indice de Dunn
disbupa=dist(bupa[,1:6])
a=hclust(disbupa)
cluster.stats(disbupa,a,bupa[,7])$dunn
And the error is:
David S. Schwarz wrote:
How does one indicate that a particular survival time is right censored in
the Survreg routine?
Read the documentation? :-)
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt
Here is one way to find the overlaps:
l1 - rbind(c(1,3), c(5,10), c(13,24))
l2 - rbind(c(2,4), c(7,14), c(20,30))
l1
[,1] [,2]
[1,]13
[2,]5 10
[3,] 13 24
l2
[,1] [,2]
[1,]24
[2,]7 14
[3,] 20 30
# create matrix for overlaps
start -
On 4/15/2009 11:45 AM, Vemuri, Aparna wrote:
Duncan
I tried writeLines. But I need to print about 23 lines and it is
really slow.
This took about 1 second here:
writeLines(as.character(1:23), C:/temp/test.txt)
I can't see how to make it much faster than that.
Duncan Murdoch
Hi, All
Forgive me if this is a stupid newbie question. I'm having no luck
googling an answer to this, probably because I don't know the right R
terminology to frame my question. I want to know how to run an R
function on each combination of the values of 2 or more variables. In
SAS-speak this
On Tue, 31 Mar 2009 13:02:23 -0700 (PDT) dfermin dfer...@umich.edu
wrote:
D Has anyone else got this problem? If so do you have a work around or
D a solution?
D
D I'm using R version 2.8.1 installed from the Fedora 10 repositories
D if that helps.
I have Fedora 10 and R 2.8.1 as well and have
Hi Michael,
Thanks for the reply.
I understand that the stress is a measure of how good the algorithm managed to represent the ordinal distances between items. And I also see why it's dependent on the number of dimensions.
I was hoping someone could tell me exactly what the formula for the
I think you want to have a look at the plyr or doBy packages.
It would be easier to give a precise answer with a minimal example.
HTH,
baptiste
On 15 Apr 2009, at 18:03, Lane, Jim wrote:
Hi, All
Forgive me if this is a stupid newbie question. I'm having no luck
googling an answer to this,
Check out plyr:
http://had.co.nz/plyr/
On Wed, Apr 15, 2009 at 2:03 PM, Lane, Jim jim.l...@rbc.com wrote:
Hi, All
Forgive me if this is a stupid newbie question. I'm having no luck
googling an answer to this, probably because I don't know the right R
terminology to frame my question. I want
That does it perfectly -- and it's pretty much the same technique as
used in the intervals pkg.
-tom
On 4/15/2009 12:43 PM, jim holtman wrote:
Here is one way to find the overlaps:
l1 - rbind(c(1,3), c(5,10), c(13,24))
l2 - rbind(c(2,4), c(7,14), c(20,30))
l1
[,1] [,2]
[1,]13
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