> I thought all the values are false, all none of them, because there
> aren't any that are true:
>
> any(logical(0))
> [1] FALSE
>
This is for the same reason why a product over an empty set of
factors is 1, and a sum over an empty set of terms is 0.
GP
--
> > Is a list O(1) for setting and getting?
>
> Can you elaborate? R is a vector language, and normally you create a list
> in one pass, and you can retrieve multiple elements at once.
When you use a hash table you expect it to be O(1) (on average) for
getting and setting values (conditional on h
> On 1/29/06, context grey <[EMAIL PROTECTED]> wrote:
>>
>> Hi,
>>
>> Is there something like a hashtable or (python)
>> dictionary in R/Splus?
On 29 Jan 2006, [EMAIL PROTECTED] wrote:
> use a 'list':
Most of the time, a list will be what you want, but it has some
important differences from a Pyt
Do you mean they are in the clipboard in the format shown?
If that is the case then do this:
as.matrix(read.table("clipboard"))
On 1/30/06, Michael <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> Suppose I have the following matrix which is a constant matrix I've copied
> from some other document:
>
On 29 Jan 2006, [EMAIL PROTECTED] wrote:
> On Sun, 29 Jan 2006, Elizabeth Purdom wrote:
>
>> I came across the following behavior, which seems illogical to me.
>
> What did you expect and why?
>
>> I don't know if it is a bug or if I'm missing something:
>>
>>> all(logical(0))
>> [1] TRUE
>
> All
On Mon, 30 Jan 2006, Jacques VESLOT wrote:
> sorry if it has already been discussed but i can't understand this:
>
> > seq(0.1,1,by=0.1)
> [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
> > match(0.1,seq(0.1,1,by=0.1))
> [1] 1
> > match(0.2,seq(0.1,1,by=0.1))
> [1] 2
> > match(0.3,seq(0.1,1,by=
On Mon, 30 Jan 2006, [iso-8859-2] Alea }iberna wrote:
> Dear expeRts!
>
> I'm thinking of buying a new computer and am considering dual-core
> processors, such as AMD Athlon64 X2. Since I'm not a computer expert, pleas
> forgive me if some of my questions are silly.
>
> First, am I correct that us
the problem is in
0.3 == seq(0.1,1,by=0.1)[3]
versus
all.equal(0.3, seq(0.1,1,by=0.1)[3])
look at ?Comparison for more info.
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnen
Hope this is clear:
> x <- seq(0.1, 1, by=0.1)
> 0.3 == x[3]
[1] FALSE
> abs(0.3 - x[3])
[1] 5.551115e-17
Andy
From: Jacques VESLOT
>
> sorry if it has already been discussed but i can't understand this:
>
> > seq(0.1,1,by=0.1)
> [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
> > match(0.1,se
Thanks Charles.
The error is occuring in a fit of a glm using the fixed-effects terms
only. This fit provides the starting estimates for the fixed effects
in the GLMM model. I changed the way that this fit is performed and
apparently didn't do it correctly.
I think it would be better if you and
sorry if it has already been discussed but i can't understand this:
> seq(0.1,1,by=0.1)
[1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
> match(0.1,seq(0.1,1,by=0.1))
[1] 1
> match(0.2,seq(0.1,1,by=0.1))
[1] 2
> match(0.3,seq(0.1,1,by=0.1))
[1] NA
> match(0.4,seq(0.1,1,by=0.1))
[1] 4
> R.versi
This might be a bit closer to what Pryseley wanted:
st <- function(dat, x, y) {
f <- formula(substitute(y ~ x), env=environment(dat))
plot(f, dat)
abline(lm(f, dat))
}
Note that the variable names in the plot when tested on `dats' as Pryseley
created.
Andy
From: Jacques VESLOT
>
>
I hope I'm not making your life unnecessarily difficult. As I will
demonstrate below my signature, my original straight application of
lme4_0.995-2/Matrix_0.995-4 is failing without providing any
optimization information. For reference, I've provided optimization
output from lme4_0.995-1/Matrix_0.9
Hi
On 29 Jan 2006 at 17:28, Gabor Grothendieck wrote:
Date sent: Sun, 29 Jan 2006 17:28:29 -0500
From: Gabor Grothendieck <[EMAIL PROTECTED]>
To: Michael <[EMAIL PROTECTED]>
Copies to: R-help@stat.math.ethz.ch
Subject:
st <- function(data, x, y){
attach(data)
rcc <- coef(lm(y~x))
plot(x,y)
abline(rcc)
detach(data)}
st(data=dats, x=visual24, y=visual52)
Pryseley Assam a écrit :
>Hello R-users
>
> I am new to R and trying to write some functions. I have problems writing
> functions tha
# Here it is (using the formula interface):
dats <- data.frame(visual24 = rnorm(30, 3, 5),
visual52 = rt(30, 7))
st <- function(formula, data, ...) {
# Just use the formula to specify which variables to use
rcc <- coef(lm(formula, data))
# Make sure to k
What is wrong with your first solution:
st <-function(x,y){
## y ... Response
## x ... terms
rcc<-coef(lm(y ~ x))
plot(x,y)
abline(rcc[1],rcc[2])
}
st(dats$visual24,dats$visual52)
Or use attach:
st <-function(data,x,y){
attach(data)
rcc<-coef(lm(x~y))
plot(x,y)
abline(rcc[
Hi
Well, colClesses can be used for what you want to do. See
> test<-read.table("c:/temp/test.txt",
colClasses=c("character","numeric", "character", "factor"))
> str(test)
`data.frame': 10 obs. of 3 variables:
$ doba : num 189 256 286 105 272 45 29.5 43 68.5 99
$ otac : chr "0.6" "0.6" "0
You didn't say if you did the selection from the menu or the command line...
In any case, the menu `Packages' -> `Set CRAN Mirror...' runs the command
chooseCRANmirror(), which looks like
> chooseCRANmirror
function (graphics = TRUE)
{
if (!interactive())
stop("cannot choose a CRAN
Hello R-users
I am new to R and trying to write some functions. I have problems writing
functions that takes a data set as an arguement and uses variables in the data.
I illustrate my problem with a small example below:
sample data #--
visual24<-rnorm(30,3,5)
v
On Mon, 30 Jan 2006, Ole Edsberg wrote:
> Hello,
>
> I have a data set on which I run the sammon algorithm as follows:
>
> library(MASS)
> data = read.table('problemforr.dat')
Hmm. This is a data frame of 387 rows and 387 columns and Euclidean
distance is used. Squeezing 387 dims (and PCA show
Yes, you are right, and you found some relevant posts.
Uwe Ligges
Aleš Žiberna wrote:
> Dear expeRts!
>
> I'm thinking of buying a new computer and am considering dual-core
> processors, such as AMD Athlon64 X2. Since I'm not a computer expert, pleas
> forgive me if some of my questions are si
Dear expeRts!
I'm thinking of buying a new computer and am considering dual-core
processors, such as AMD Athlon64 X2. Since I'm not a computer expert, pleas
forgive me if some of my questions are silly.
First, am I correct that using a dual-core processor is (for R point of
view) the same as usin
Hello,
Not exactly the same. By the way, why do you use do.call()? Couldn't you
do simply:
expand.grid(split(t(replicate(3, c(0, 1, NA))), 1:3))
Best,
Philippe Grosjean
Jacques VESLOT wrote:
> this looks similar:
> do.call(expand.grid,split(t(replicate(3,c(0,1,NA))),1:3))
>
>
> Adrian DUSA
See
?par
Look for arguments
mfrow and mfcol
Hope this is what you are looking for,
Ales Ziberna
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of R Granell, Medicine
Sent: Monday, January 30, 2006 12:35 PM
To: r-help@stat.math.ethz.ch
Subject: [R] multip
this looks similar:
do.call(expand.grid,split(t(replicate(3,c(0,1,NA))),1:3))
Adrian DUSA a écrit :
>Dear R-helpers,
>
>I'm trying to develop a function which specifies all possible expressions that
>can be formed using a certain number of variables. For example, with three
>variables A, B and
Adrian DUSA roda.ro> writes:
>
> I'm trying to develop a function [...snip...]
Sorry for the traffic, I forgot to say that I'm using
library(combinat)
for the "combn" function...
Thank you,
Adrian
__
R-help@stat.math.ethz.ch mailing list
https://stat
Hello,
Here comes a simple question:
Is there a way of obtaining more than one histogram in the graph-window so
you can edit all the plots simultaneously?
and how about scatter plots?
Thanks in advance
__
R-help@stat.math.ethz.ch mailing list
https:
Dear R-helpers,
I'm trying to develop a function which specifies all possible expressions that
can be formed using a certain number of variables. For example, with three
variables A, B and C we can have
- presence/absence of A; B and C
- presence/absence of combinations of two of them
- presence
Hi List,
I have a time series of 122 values, actualy it is a time series of daily
indian monsoon rainfall. now i want to filter this time series for a particular
oscilation say 10 to 20days oscilation. i want to find out what amount of
variance is explained by this mode. Which package is
> Suppose I have the following matrix which is a constant matrix I've copied
> from some other document:
>
> 1.2 3.4 1.4 ...
> 2.3 3.7 2.6 ...
> ...
> How do I make it into a matrix or array in R?
> What is the fastest way of initializing a constant matrix with this
> copy/pasted values?
you can
Sorry Michael, but I don't understand your question.
If you want to intialize a constant matrix (there is not such thing in
R, just create a numerical matrix and use it without changing its
values), you just use matrix(). For help and arguments of the function,
type:
> ?matrix
Best,
Philippe
Hello,
I have a data set on which I run the sammon algorithm as follows:
library(MASS)
data = read.table('problemforr.dat')
y = cmdscale(data, add=TRUE)
s = sammon(data, y$points)
(In case it should be relevant, I make the data available at
http://idi.ntnu.no/~edsberg/problemforr.dat)
With R 2.
Hi all,
Suppose I have the following matrix which is a constant matrix I've copied
from some other document:
1.2 3.4 1.4 ...
2.3 3.7 2.6 ...
...
How do I make it into a matrix or array in R?
What is the fastest way of initializing a constant matrix with this
copy/pasted values?
Thanks a lot!
Hi!!
I've had some problems with the new version of textplot, included in the
gplots package. (some kind of postscript problem)
I wonder if any of you have used this function or if there is any other R
funcion to show the text in a R object as an image.
Thanks in advance,
A
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