Hello list.
Until last week, I've been using my function without a problem. I can only
imagine what has changed (no package updates), but all of a sudden, a custom
function that uses raster::xyValues stopped working. For some reason, one of
the values is not found, even though it's there. Here is
Please, reply to the r-help and not only to me personally. That way
others can can also help, or perhaps benefit from the answers.
You can use strplit to remove the last part of the strings. strplit
returns a list of character vectors from which you (if I understand
you correctly) only want to sel
Oops, small typo, should be:
barchart_test_heights=sin(c(1:100))
barchart_test_bins=c(c(1:50),c(1:50))
groups=c(rep(1,50),rep(2,50))
# Wish below didn't have spaces!
barchart(barchart_test_bins~barchart_test_heights,groups=groups)
On Wed, Aug 25, 2010 at 4:46 PM, Jonathan Greenberg
wrote:
> I
I apologize, let me add some data to play with:
barchart_test_heights=sin(c(1:100))
barchart_test_bins=c(1:100)
groups=c(rep(1,50),rep(2,50))
I have pre-calculated histogram data and the bins they belong to. I
would like to plot the two histograms superimposed on one-another,
having group 1 bars
On Thu, 2010-08-26 at 06:19 +0800, elaine kuo wrote:
> Yes, I appreciated your answers which well hit my questions. (esp the
> perfect model parts).
>
>
> About the plot part, one more question.
> Is it possible to make the two plots (northern and southern richness)
> sharing the same Y axis (lat
I have the following problem
mydata <- data.frame(a=1:3)
form <- ~ (a>1) - 1
model.matrix(form,mydata)
a > 1FALSE a > 1TRUE
1 1 0
2 0 1
3 0 1
...
However
model.matrix(update(terms(form)[1],~.-1),mydata)
(Intercept) a > 1 - 1TRUE
1
>On Wed, 2010-08-25 at 09:32 +0100, e-letter wrote:
>> On 24/08/2010, David Winsemius wrote:
>> >
>> > On Aug 24, 2010, at 9:37 AM, e-letter wrote:
>> >
>> >> Readers,
>> >>
>> >> According to the documentation for the function 'plotmath' there is no
>> >> apparent possibility to add the percent s
Wow, I was aware of R's lazy evaluation but didn't realize that was the cause
here. If the function, adder(), was not simple as I'd defined, I suppose I
could also force the evaluation of "x" in a wrapping function before passing it
to adder() as an alternative:
> adder <- function(x) function(
>It's possible that my help page is different than yours. Right after
>the syntax/meaning description on mine (which is a Mac OSX system) is
>a paragraph:
>
>"The symbol font uses Adobe Symbol encoding so, for example, a lower
>case mu can be obtained either by the special symbol mu or by
>symbol("
To clarify, when I run the unlist(...) in R_GlobalEnv, I create all the
necessary variables (including effect.distance).
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Hi Philippe, thanks for the suggestion - for my smaller problems I find that
closures are quicker to define and deploy. At larger scales, I've implemented
S4 objects with methods and attributes - though while elegant, I find that OO
(apart from what's built-in) adds significant biolerplate in d
I couldn't resist the temptation.
Here is the example using the R package nleqslv.
library(nleqslv)
npv <- function (irr, cashFlow, times) sum(cashFlow / (1 + irr)^times)
CF <- c(-1000,500,500,500,500,500)
dates <-
c("1/1/2001","2/1/2002","3/1/2003","4/1/2004","5/1/2005","6/1/2006")
cfDate <-
I think you need an I(), i.e.,
form <- ~ I(a > 1) - 1
I hope it helps.
Best,
Dimitris
On 8/26/2010 9:17 AM, Niels Richard Hansen wrote:
I have the following problem
mydata <- data.frame(a=1:3)
form <- ~ (a>1) - 1
model.matrix(form,mydata)
a > 1FALSE a > 1TRUE
1 1 0
2 0 1
3 0 1
...
Howeve
On Thu, 2010-08-26 at 06:11 +0800, elaine kuo wrote:
> Thank you.
> The answers provides the right direction and a code was written accordingly.
>
> One more request, if the label of axis X wants to be drawn from 5 to 1 (left
> to right)
> rather than 1 to 5, is it fine to change axis (4, at = NUL
Hi folks,
Following command only works on Windows
> Test01=read.table(file.choose(), header=TRUE)
It popup open a dialog box for choosing file.
But it doesn't work on Linux (Ubuntu);
> Test01=read.table(file.choose(), header=TRUE)
Enter file name:
Entering file name doesn't work. Please ad
On Wed, 2010-08-25 at 11:30 -0400, RICHARD M. HEIBERGER wrote:
> Kay,
>
> doe this do what you want?
>
>
> dotplot(y1+y2 ~ facs$Treatment|facs$Sites,
> outer=TRUE,
> scales = list(x = list(rot = 90, tck=c(1,0))),
> ylab=c("y1", "y2"),
> xlab=c("Site 1", "Site 2"),
On Thu, 2010-08-26 at 02:30 -0400, David Kane wrote:
> How reliable is R-Forge? http://r-forge.r-project.org/
>
> It is down now (for me). Reporting "R-Forge Could Not Connect to Database: "
>
> I have just started to use it for a project. It has been down for
> several hours (at least) on differ
OK, I admit. It will never win a beauty price, but I won't either so
we go pretty well together.
Seriously, I am aware this is about the worst way to solve such a
problem. But as I said, rewriting the class definitions wasn't an
option any more. I'll definitely take your advice for the next
projec
On Thu, 2010-08-26 at 08:17 +0100, e-letter wrote:
> >On Wed, 2010-08-25 at 09:32 +0100, e-letter wrote:
> >> On 24/08/2010, David Winsemius wrote:
> >> >
> >> > On Aug 24, 2010, at 9:37 AM, e-letter wrote:
> >> >
> >> >> Readers,
> >> >>
> >> >> According to the documentation for the function 'pl
On Thu, Aug 26, 2010 at 5:22 AM, Jonathan Greenberg
wrote:
> Oops, small typo, should be:
>
>
> barchart_test_heights=sin(c(1:100))
> barchart_test_bins=c(c(1:50),c(1:50))
> groups=c(rep(1,50),rep(2,50))
>
> # Wish below didn't have spaces!
> barchart(barchart_test_bins~barchart_test_heights,group
It would presumably help you get a good answer if you explained
what Amazon EC2 was, and what limitations are involved in that
platform. You got a very good answer for standard 64-bit Linux,
because that was the recognizable part of your query. To go beyond
that we need more information.
*Can* you
[I already sent this message to the list almost a day ago using an
other email address, but that message does not seem to get through. In
case it finally does get through, I appologize for the the double
posting]
Dear list,
I got some surprising results when using the svytotal routine from
David Kane kanecap.com> writes:
>
> How reliable is R-Forge? http://r-forge.r-project.org/
>
> It is down now (for me). Reporting "R-Forge Could Not Connect to Database: "
>
> I have just started to use it for a project. It has been down for
> several hours (at least) on different occasions ov
I have data similar to this:
Location Surveyor Result
A1 83
A2 76
A3 45
B1 71
B4 67
C2 23
C5 12
D3 34
E4 75
F4 46
G5 90
et
On 26/08/10 09.30, Dimitris Rizopoulos wrote:
I think you need an I(), i.e.,
form <- ~ I(a > 1) - 1
Yes, it solves the concrete problem, but does not really answer
the question. Let me rephrase. Should the use of terms like (a>1)
be discouraged in R and replaced by I(a>1) systematically, or
Hello useRs and guRus,
I was trying to add the support of str() in the ascii package, and I
realized that str() does not have a print method. It uses cat() from inside
the function and returns nothing.
Since it is not usual in R, I wonder why? Is there a particular reason?
Thank you very much fo
Dear users,
***I have a function f to simulate data from a model (example below used
only to show my problems)
f<-function(n,mean1){
a<-matrix(rnorm(n, mean1 , sd = 1),ncol=5)
b<-matrix(runif(n),ncol=5)
data<-rbind(a,b)
out<-data
out}
*I want to simulate 1000 datasets (here only 5)
On 08/26/2010 01:06 AM, Neta wrote:
I have a collection of results. I use R to get the linearization presented.
how can I get R to show the equation and R^2 value on the plot area next to
the graph?
Hi Neta,
You can use functions like textbox or boxed.labels (plotrix package) to
display arbit
On 08/26/2010 06:36 AM, Dennis Murphy wrote:
...
It looks like he's using barchart() in lattice, which does have a box.ratio
argument.
That's why I didn't suggest using barp with width=0.5
Jim
__
R-help@r-project.org mailing list
https://stat.ethz.c
Thankyou!!! I knew this was so simple and I had tried paste but with the
wrong syntax so I moved on. This opens a whole new world to me!!
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Dear all,
I have a list that contains 3 sublists( x1, x2, x3)
mylist<-list(x1=c("A","A","A","B","C","Z","Y"),x2=c("D","D","E","E","F","Z","X"),x3=c("A","A","A","B","Y","Z"))
mylist
$x1
[1] "A" "A" "A" "B" "C" "Z" "Y"
$x2
[1] "D" "D" "E" "E" "F" "Z" "X"
$x3
[1] "A" "A" "A" "B" "Y" "Z"
I also ha
The answer is indeed a LINUX cluster with more memory. Thanks, Josquin
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R-help@r-p
one way is the following:
mylist <- list(x1 = c("A","A","A","B","C","Z","Y"),
x2 = c("D","D","E","E","F","Z","X"),
x3 = c("A","A","A","B","Y","Z"))
newlist <- c("A","B","C","D","E","F")
tab <- t(sapply(mylist, function (x)
table(factor(x, levels = newlist
tab[tab == 0] <- NA
tab
Dear all
I want to save several ggplots in one pdf document. I tried this
for (i in names(iris)[2:4]) {
p<-ggplot(iris, aes(x=Sepal.Length, y=iris[,i], colour=Species))
p+geom_point(aes(size=3))
}
with different variations of y input but was not successful. In past I
used qplot in similar fashi
You haven't wrapped p in the print command, which is one of the ways to
make sure the plot gets printed when we need it.
print(p+geom_point(aes(size=3))) does the trick
On 08/26/2010 06:08 AM, Petr PIKAL wrote:
Dear all
I want to save several ggplots in one pdf document. I tried this
for (i
Answers below.
On Thu, Aug 26, 2010 at 11:20 AM, Evgenia wrote:
>
> Dear users,
>
> ***I have a function f to simulate data from a model (example below used
> only to show my problems)
>
> f<-function(n,mean1){
> a<-matrix(rnorm(n, mean1 , sd = 1),ncol=5)
> b<-matrix(runif(n),ncol=5)
> data<-
Thanks Dimitris,
It works nicely!
Regards,
Olga
On Thu, 2010-08-26 at 11:55 +0200, Dimitris Rizopoulos wrote:
> one way is the following:
>
> mylist <- list(x1 = c("A","A","A","B","C","Z","Y"),
> x2 = c("D","D","E","E","F","Z","X"),
> x3 = c("A","A","A","B","Y","Z"))
> newlist <- c("
Thanks. I knew I forgot something obvious.
Petr
r-help-boun...@r-project.org napsal dne 26.08.2010 12:29:52:
> You haven't wrapped p in the print command, which is one of the ways to
> make sure the plot gets printed when we need it.
> print(p+geom_point(aes(size=3))) does the trick
> On 08/
Dear all,
I want to equalize a symbolic derivative (of a function with two
variables) with zero and convert it to a variable, e.g. x.
I'm computing the derivative by: (found it in the archive)
library(Ryacas)
f <- function(x,y) (100-x-y)*x-10*x
yacas(f) # register f with yacas
Df <- f
body(Df) <
I am trying to use Rcmdr 1.5_4 with R-2.11.1 (order to run to run the new
version of misdist-0.5.3 which is built under R-2.11.1). however relimp is
required for Rcmdr and the version of relimp_1.0.1 downloaded from CRAN will
not work with the latest version of Rcmdr (I get error message telling
> grep("f[0-9]+=", "f1=5,f22=3,", value = T)
[1] "f1=5,f22=3,"
How do I make the line output c("f1", "f22") instead? (Actually, c(1,22)
would be even better).
Thank you.
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On 08/26/2010 08:11 AM, elaine kuo wrote:
...
One more request, if the label of axis X wants to be drawn from 5 to 1
(left to right)
rather than 1 to 5, is it fine to change axis (4, at = NULL) ?
If so, which value should be input ?
Hi Elaine,
There are a couple of ways to do this. One is to u
Hi
Coming a bit late to the thread another way to reduce the space between the
panels may be along the lines of Deepayan's reply with the layout.widths in
http://finzi.psych.upenn.edu/R/Rhelp02/archive/43626.html
I regularly use this and the latticeExtra addition will be welcome - I
have no
Dear John,
Thank a lot for your answer. Indeed, i misunderstood the place of
indicator variables in the path diagram, i thank they were considered
as exogeneous variables. Thanks for your answer, i changed variances
of latente variables to NA, and remove indicator variables from
fixed.x a
It isn't entirely clear what you are trying to do - your grep() statement
simply returns the entire string you started with.
But to turn that string into a vector, you will need some combination
of gsub(), strsplit(), and as.numeric() with the exact values depending
on the exact form of the output
I too am unsure of what is required but another way is
gsub(",$","",gsub("[f53=]+","",a))
[1] "1,22"
Regards
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
ARMIDALE NSW 2351
Email home: mac...@northnet.com.au
At 21:47 26/08/2010, you wrote:
It isn't
Hi
I want to extract the random slopes from a lmer (I am doing a random
regression), but are the answers obtained from ranef or coef?
My model is: mod1<-lmer(B~ A +(A|bird), family=quasibinomial)
And I want to obtain a slope for each individual bird but am not sure which
output I need and can'
On Thu, Aug 26, 2010 at 6:16 AM, Dimitri Shvorob
wrote:
>
>> grep("f[0-9]+=", "f1=5,f22=3,", value = T)
> [1] "f1=5,f22=3,"
>
> How do I make the line output c("f1", "f22") instead? (Actually, c(1,22)
> would be even better).
>
strapply in gsubfn extracts matches based on content rather than
deli
On Thu, Aug 26, 2010 at 4:42 AM, David Hajage wrote:
> Hello useRs and guRus,
>
> I was trying to add the support of str() in the ascii package, and I
> realized that str() does not have a print method. It uses cat() from inside
> the function and returns nothing.
>
> Since it is not usual in R, I
Hello Again Gurus and Lurkers:
I¹m trying to build a very user-friendly function which does aov without
having the user type in a formula (which would be tedious in this case).
The idea is to take the response from a PCA score matrix, and the factors
from a list. A simple example is the function
Hi John,
I'm having some trouble sorting out your request, for the following reasons:
1) I don't know what misdist is (can't find it on CRAN or google) or why you
need Rcmdr to run it. Presumably you meant mixdist?
2) relimp is not required by Rcmdr, merely suggested. Rcmdr will work fine
without
On Thu, Aug 26, 2010 at 6:57 AM, Immanuel Seeger wrote:
> Dear all,
> I want to equalize a symbolic derivative (of a function with two
> variables) with zero and convert it to a variable, e.g. x.
>
> I'm computing the derivative by: (found it in the archive)
> library(Ryacas)
> f <- function(x,y)
On Aug 26, 2010, at 10:35 AM, Niels Richard Hansen wrote:
>
>
> On 26/08/10 09.30, Dimitris Rizopoulos wrote:
>> I think you need an I(), i.e.,
>>
>> form <- ~ I(a > 1) - 1
>
> Yes, it solves the concrete problem, but does not really answer
> the question. Let me rephrase. Should the use of t
Hi Greg,
thanks for the suggestion:
I have attached some small dataset that can be used to reproduce the
odd behavior of the approxfun-function.
If it gets stripped off my email, it can also be downloaded at:
http://bioinf.gen.tcd.ie/approx.data.Rdata
Strangely, the problem seems specific to the
Dear all,
This problem came up initially while debugging a function, but it
seems to be a more general problem of R. I hope I'm wrong, but I can't
find another explanation. Let me illustrate with the raster package.
For an object "RasterLayer" (which inherits from Raster), there is a
method xyVal
On Aug 26, 2010, at 8:15 AM, Gabor Grothendieck wrote:
On Thu, Aug 26, 2010 at 6:16 AM, Dimitri Shvorob
wrote:
grep("f[0-9]+=", "f1=5,f22=3,", value = T)
[1] "f1=5,f22=3,"
How do I make the line output c("f1", "f22") instead? (Actually,
c(1,22)
would be even better).
strapply in gsu
Hi,
First, some toy data would have helped a lot, and that can explain why
you got so few answers...
Second, I have maybe some thoughts about it, not sure if this can help.
The error tells you that the data passed to aov is not a matrix. In your
function, you don't specify explicitly the dat
On Thu, Aug 26, 2010 at 9:45 AM, Immanuel Seeger wrote:
> Thanks, but my problem is that I have my formulas in functions.
> How can I do that for functions or e.g. for the body(Df) below?
Try this:
> library(Ryacas)
> Solve(deparse(body(f)), "x")
[1] "Starting Yacas!"
expression(list(x == -((10
On Aug 26, 2010, at 8:47 AM, Bryan Hanson wrote:
Hello Again Gurus and Lurkers:
I’m trying to build a very user-friendly function which does aov
without
having the user type in a formula (which would be tedious in this
case).
The idea is to take the response from a PCA score matrix, and th
Hi,
I'm trying to apply the function daisy() to a data.frame 1x10 but I have
not enough space (error message: cannot allocate vector of length
1476173280).
I didn't imagine I was not able to work with a matrix of just 1
observations... I have setted in Rgui --max-mem-size=2G (I'm not abl
Hi folks,
Following command prints 2 graphs side-by-side:-
layout(matrix(1:2, nrow=1))
plot(Date,Input_No.)
plot(Test01$Date, Test01$Input_No.)
However each is a square graph I need a rectangular layout. Pls advise how to
make it. TIA
B.R.
satimis
_
On Aug 26, 2010, at 10:44 AM, Stephen Liu wrote:
Hi folks,
Following command prints 2 graphs side-by-side:-
layout(matrix(1:2, nrow=1))
?layout
#Read the details for the widths and heights arguments more carefully
and (as it suggests) see the examples.
plot(Date,Input_No.)
plot(Test01$
Dear Anne,
> -Original Message-
> From: Anne Mimet [mailto:ami...@mnhn.fr]
> Sent: August-26-10 7:45 AM
> To: John Fox
> Cc: r-help@r-project.org
> Subject: RE: [R] SEM : Warning : Could not compute QR decomposition of
> Hessian
>
> Dear John,
>
> Thank a lot for your answer. Indeed, i m
Hi!
I didn't see the toy data, my bad.
I guess David's solution fixed the problem.
Ivan
Le 8/26/2010 16:10, Bryan Hanson a écrit :
> Hi Ivan, there is toy data given in the original post.
>
> The object passed to aov is a matrix, it is called scores, and it is passed
> via the formula that is
coef(mod1)$bird will give you a matrix with two columns. The first
column is the intercept for each bird and the second column is the
slope for each bird.
ranef(mod1) will also give you a matrix of two columns. These
represent the random effects. That is, how much the intercept (or
slope) is shift
A big thanks David! Eliminating the env = parent.frame() seems to fix
everything. I hate it when I'm that close... I included that because from
?as.formula it seemed to be the default, but on re-read, I guess it doesn't
really say that. In addition, I had included it even though it was the
defau
On Thu, 26 Aug 2010, Gavin Simpson wrote:
On Thu, 2010-08-26 at 02:30 -0400, David Kane wrote:
How reliable is R-Forge? http://r-forge.r-project.org/
It is down now (for me). Reporting "R-Forge Could Not Connect to Database: "
late to chime in, so had tossed the first piece. As this
relate
Hello,
Is there a simple way to get the class type for each column of a
data.frame? I am in the situation where I would like to get all the
columns of a data.frame that are factors.
I have tried:
apply(df,2,class)
but all the columns come back as class "character".
Thanks
Dan
--
On Aug 26, 2010, at 11:31 AM, Daniel Brewer wrote:
Hello,
Is there a simple way to get the class type for each column of a
data.frame? I am in the situation where I would like to get all the
columns of a data.frame that are factors.
I have tried:
apply(df,2,class)
lapply(df, class)
but a
That's because apply works on arrays/matrices, not data.frames.
It therefore coerces your data.frame to a matrix of type
character, since you have factors, thus the result.
You want sapply or lapply, since a data.frame is actually a
list.
sapply(df, class)
and then to get what you want:
df[sa
On Aug 26, 2010, at 11:31 AM, Daniel Brewer wrote:
Hello,
Is there a simple way to get the class type for each column of a
data.frame? I am in the situation where I would like to get all the
columns of a data.frame that are factors.
I have tried:
apply(df,2,class)
but all the columns come ba
Since a data.frame is a list, you should use lapply() or sapply():
> df <- data.frame(a=1:5, b=LETTERS[1:5])
> lapply(df, class)
$a
[1] "integer"
$b
[1] "factor"
> sapply(df, class)
a b
"integer" "factor"
HTH,
Ivan
Le 8/26/2010 17:31, Daniel Brewer a écrit :
> Hello,
>
> I
On Thu, 2010-08-26 at 10:42 +0200, David Hajage wrote:
> Hello useRs and guRus,
>
> I was trying to add the support of str() in the ascii package, and I
> realized that str() does not have a print method. It uses cat() from inside
> the function and returns nothing.
>
> Since it is not usual in R
On Thu, Aug 26, 2010 at 4:31 PM, Daniel Brewer wrote:
> Hello,
>
> Is there a simple way to get the class type for each column of a
> data.frame? I am in the situation where I would like to get all the
> columns of a data.frame that are factors.
>
> I have tried:
> apply(df,2,class)
> but all the
Joris,
I looked at the problem. Here is a minimal example reproducing
the error
setGeneric("myplus",function(x,y,...) standardGeneric("myplus"))
setMethod("myplus",c(x="numeric",y="numeric"),
function(x,y,z=0) x+y+z
)
setMethod("myplus",c(x="numeric",y="list"),
func
On Aug 26, 2010, at 10:36 AM, David Winsemius wrote:
>
> On Aug 26, 2010, at 11:31 AM, Daniel Brewer wrote:
>
>> Hello,
>>
>> Is there a simple way to get the class type for each column of a
>> data.frame? I am in the situation where I would like to get all the
>> columns of a data.frame that
On Thu, 2010-08-26 at 07:35 -0700, abanero wrote:
> Hi,
>
> I'm trying to apply the function daisy() to a data.frame 1x10 but I have
> not enough space (error message: cannot allocate vector of length
> 1476173280).
>
> I didn't imagine I was not able to work with a matrix of just 1
> obs
Hi,
Gavin Simpson wrote:
> What do you want to do with the dissimilarities?
Clustering.
Gavin Simpson wrote:
> If clustering, try the clara() function
I can't because the variables are mixed (numeric and categorical) and (I
suppose..) I should use "gower" distance in advance, shouldn't I?
Hi
Thanks for your help - however if I have a model of:
mod1<-lmer(B~ A+C+(A|bird), family=quasibinomial)
coef then gives me an individual slope for factors A and C. However the random
effect is only nested within factor - so I am only trying to allow the slope to
vary in relation to effect
On 26 August 2010 at 11:28, R P Herrold wrote:
| On Thu, 26 Aug 2010, Gavin Simpson wrote:
|
| > On Thu, 2010-08-26 at 02:30 -0400, David Kane wrote:
| >> How reliable is R-Forge? http://r-forge.r-project.org/
| >>
| >> It is down now (for me). Reporting "R-Forge Could Not Connect to Database:
"
On Thu, Aug 26, 2010 at 11:31 AM, Daniel Brewer wrote:
> Hello,
>
> Is there a simple way to get the class type for each column of a
> data.frame? I am in the situation where I would like to get all the
> columns of a data.frame that are factors.
>
> I have tried:
> apply(df,2,class)
> but all th
Many thanks!
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Thanks, but my problem is that I have my formulas in functions.
How can I do that for functions or e.g. for the body(Df) below?
And: How can I refer to the first solution of expression?
Thanks a lot!
-Ursprüngliche Nachricht-
Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Gesend
Hello,
I was wondering if I could be taken off the list to get emails?
Thanks
Sonia
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PLEASE do read the posting guide http://
Ben Holt bio.ku.dk> writes:
> I have data similar to this:
>
> Location Surveyor Result
> A1 83
> A2 76
> A3 45
> B1 71
> B4 67
> C2 23
> C5 12
> D3 34
> E4
???
You were provided exactly what you requested. I think you either need
to read up on what random effects models mean or more clearly
communicate what YOU mean. (It's unclear to me, anyway).
--
Bert Gunter
Genentech Nonclinical Statistics
On Thu, Aug 26, 2010 at 8:57 AM, Samantha Patrick
wrot
Certainly. The link at the bottom of each and every email to this list has
the web address at which you may do so.
For your convenience:
https://stat.ethz.ch/mailman/listinfo/r-help
Sarah
On Thu, Aug 26, 2010 at 9:05 AM, Sonia Spirling
wrote:
> Hello,
>
> I was wondering if I could be taken off
On Thu, Aug 26, 2010 at 4:44 PM, Marc Schwartz wrote:
>> sapply(iris, class)
> Sepal.Length Sepal.Width Petal.Length Petal.Width Species
> "numeric" "numeric" "numeric" "numeric" "factor"
Note that comparing the result of class(foo) is a bad way of telling
if something is
try
lapply(df, class)
Steve E
On Thu, Aug 26, 2010 at 4:31 PM, Daniel Brewer
wrote:
> Hello,
>
> Is there a simple way to get the class type for each column of a
> data.frame? I am in the situation where I would like to get all the
> columns of a data.frame that are factors.
>
> I have tried:
On Thu, 26 Aug 2010, Dirk Eddelbuettel wrote:
On 26 August 2010 at 11:28, R P Herrold wrote:
| Is anyone aware of explanations, other than a release process
| that does not require unique versioning of differing content?
| [it seems pretty basic to me that a 'receiver' of new content
| could
On 8/26/2010 8:43 AM, Niels Richard Hansen wrote:
setGeneric("myplus",function(x,y,...) standardGeneric("myplus"))
setMethod("myplus",c(x="numeric",y="numeric"),
function(x,y,z=0) x+y+z
)
setMethod("myplus",c(x="numeric",y="list"),
function(x,y,...) callGeneric(x,un
I'm sure this has appeared before on this list, but the biggest
help to me has been: "Gelman and Hill", Data Analysis and
Regression Using Multilevel/Hierarchical Models, which contains
clear explanations and the R code to go along. Also check out
http://lme4.r-forge.r-project.org/book/
Looking
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Dimitri Shvorob
> Sent: Thursday, August 26, 2010 3:16 AM
> To: r-help@r-project.org
> Subject: [R] Quick GREP challeng
On Aug 26, 2010, at 11:01 AM, Dirk Eddelbuettel wrote:
>
> On 26 August 2010 at 11:28, R P Herrold wrote:
> | On Thu, 26 Aug 2010, Gavin Simpson wrote:
> |
> | > On Thu, 2010-08-26 at 02:30 -0400, David Kane wrote:
> | >> How reliable is R-Forge? http://r-forge.r-project.org/
> | >>
> | >> It is
Did anyone manage to have a look at this? I have now tried foreach with doMC
backen and multicore, but still no luck. If someone can help that would be
much appreciated!
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Greetings Gabor,
is it possible to open a channel to a data frame in the default environment?
there are cases when using a sql update statement is the simplest
alternative, so instead of dumping the df and then updating and then
reimporting it I would like to update the df directly in R.
Thank y
Dear All,
I'm looking to perform an ANOVA between two nested panel fixed effects models.
I tried with anova, as well as with waldtest. anova tells me there is no method
available for plm objects, while waldtest tells me my models are not nested. I
think they are instead. The difference between
On Thu, Aug 26, 2010 at 1:18 PM, stephenb wrote:
> is it possible to open a channel to a data frame in the default environment?
> there are cases when using a sql update statement is the simplest
> alternative, so instead of dumping the df and then updating and then
> reimporting it I would like t
I need the parameters estimated for a non-linear equation, an example of the
data is below.
rm(list=ls())
Time<-c( 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4,
4, 4, 5, 5, 5, 5, 5, 8, 8, 8, 8, 8)
Level<-c( 100, 110, 90, 95, 87, 60, 65, 61, 55, 57, 40, 41, 50,
47,
44, 44, 42,
Look for a vectorized solution such as ?table or ?aggregate to obtain a list
of combination counts, followed by logical indexing or ?subset to get a set of
valid combinations, and then use ?sample to get your random selections of
locations/surveyors and then process only those combinations from
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