What OS are you using?
What device are you using to produce 'bitmap output'?
'Arial' (sic) is not a freely available font, but if you have it on your
system you may be able to make use of it by adding a 'family'. The
details depens on details you failed to give: please see the footer of
this
On Mon, 11 Aug 2008, Hyunchul Kim wrote:
Hi, all
I need to specify a font (for example, type=helvetica and size=10) in bitmap
output (for example, dev.print(bitmap, 'test.png').
How can I do this?
You would do better to use dev2bitmap(). In any case, please read the
help for bitmap:
point
On Mon, 11 Aug 2008, Rodrigo Gazaffi wrote:
dears R-users,
I'm interested in model selection problem, and i have faced some problems
that i would like to ask for help.
well,
this is a very small example with 4 variable (just one var. is the response
- z) with 100 individuals
i would like to do
On Mon, 11 Aug 2008, Kenneth Roy Cabrera Torres wrote:
Hi Carlos:
I think you got a encoding problem.
Maybe is esier to convert it.
I don't know how to convert in Mac OS, but
in linux you can use "iconv" that converts many codes
to other.
Well, R has an iconv() command even on Mac OS X, and
Hello all,
am a newby in R,
am trying the knn function, and am doing just a stupid test :
knn(c(1,2,3,4,5,6), c(3), k=4 ,prob=TRUE,factor(c(1:6)))
the result is unstable !! i have each time different result :
[1] 5
attr(,"prob")
[1] 0.167
Levels: 1 2 3 4 5 6
[1] 4
attr(,"prob")
[1] 0.1
The change is in pchisq:
2.4.1:
pchisq(3, 0, lower=F)
[1] NaN
2.7.1:
pchisq(3, 0, lower=F)
[1] 0
And that is correct according to the definition.
On Mon, 11 Aug 2008, Lutz Ph. Breitling wrote:
Dear List,
recently tried to reproduce the results of some custom model selection
function after
Hello all,
am a newby in R,
am trying the knn function, and am doing just a stupid test :
knn(c(1,2,3,4,5,6), c(3), k=4 ,prob=TRUE,factor(c(1:6)))
the result is unstable !! i have each time different result :
[1] 5
attr(,"prob")
[1] 0.167
Levels: 1 2 3 4 5 6
[1] 4
attr(,"prob")
[1] 0.166
David Stoffer wrote:
>
> I stand corrected. I thought I checked this a long time ago, but
> apparently not. tsdiag.Arima DOES NOT use the fact that the series it is
> testing (or diagnosing, if you will) are residuals from an ARIMA fit.
>
> I keep a list of R time series bloopers here:
> h
Hi,
I want to run a logistic regression on summary data and I was not able to
find the adequate R function to do so. The data is summarized daily and
instead of the binary y_t I have n_d and m_d, where is the number of
instances in which a choice could have been made and m_d is the number of
insta
Vibhanshu Abhishek wrote:
> Hi,
>
> I want to run a logistic regression on summary data and I was not able to
> find the adequate R function to do so. The data is summarized daily and
> instead of the binary y_t I have n_d and m_d, where is the number of
> instances in which a choice could have bee
Try reading the help page (and crediting the authors of package class).
Description:
k-nearest neighbour classification for test set from training set.
For each row of the test set, the 'k' nearest (in Euclidean
distance) training set vectors are found, and the classification
rcoder wrote:
Hi Gary,
Thanks for your reply. This works fine. Is there any way to send the ACF
data to a matrix, so I could analyse the data in excel for e.g.?
The coefficients are returned in the acf element of the result, as
mentioned in the documentation (?acf). The side affect of plotti
On Sun, 2008-08-10 at 18:09 +0200, Jörg Groß wrote:
...
> No, what I want is to plot the mean for two different groups.
> So that I get group 1 and 2 on the x-axis and also label it like that.
>
Hi Jorg,
This is a wild guess, but are 1 and 2 anywhere on the x axis of your
plot? That is, if the plo
Em Sex, 2008-08-08 às 13:27 -0400, Michael Gormley escreveu:
> In trying to install the pbatR package, I was greeted with the error
>
> Error: package 'tcltk' does not have a name space
> Execution halted
>
> Directly installing the package tcltk2 returned the following error:
>
> Loading requir
Hello,
I have a matrix and I want to sample 20 rows that are the the percentiles of
0-100 in 0.05 increments. I have a vector of my sequence
(0, 0.05, 0.10, 0.15,1.0) and also
a normalised vector of rownumbers. That is, there are 234 rows (for
example) so I do
perc<-c(1:234/234)
which loo
Gareth Campbell wrote:
Hello,
I have a matrix and I want to sample 20 rows that are the the percentiles of
0-100 in 0.05 increments. I have a vector of my sequence
(0, 0.05, 0.10, 0.15,1.0) and also
That's actually 21 entries, but that probabl doesn't matter...
a normalised vector of
We've seen a few reports on problems in R 2.7.[01] on Windows in Korean.
It does work on my laptop, but I've managed to reproduce this on a WinXP
desktop: it seems the problem lies in the encoding of the translations of
the menus, and might be a problem in the updated iconv used.
Workarounds:
Hi
I'm having a problem using the zoo library and I can't see what I'm
doing wrong. For example setting up the data
> t1 = zoo (matrix (1:12, nrow = 3), order.by = as.Date
(c("2008-08-01","2008-08-02","2008-08-03")))
> colnames (t1) = c ("A", "B", "C", "D")
> t2 = zoo (matrix (1:12, nrow = 3), o
Hei,
is it possible to create from R a series of files with unique(random) names?
I have tryed with file.create and dir.create but did not manage to do it.
Thanks a lot for help
Sara
[[alternative HTML version deleted]]
__
R-help@r-project.org m
How about this?
name.length = 10
name = paste(sample(letters,name.length,T),collapse="")
Then you can use file.exists(), to check if the file already exists.
On Mon, Aug 11, 2008 at 12:28 PM, sara martino <[EMAIL PROTECTED]> wrote:
> Hei,
> is it possible to create from R a series of files w
I am using Linux (Ubuntu 8.04).
I tried
> dev.print(bitmap, file='test.png', family='arial')
but it gave error message of
"Error in postscript(fild=cmd, width=width, height=height,
pointsize=pointsize, : Unknown family 'arial' "
What should I do?
Thanks a lot in advance.
Hyunchul
On Mon, Aug
On Mon, 11 Aug 2008, Hyunchul Kim wrote:
I am using Linux (Ubuntu 8.04).
I tried
> dev.print(bitmap, file='test.png', family='arial')
but it gave error message of
"Error in postscript(fild=cmd, width=width, height=height,
pointsize=pointsize, : Unknown family 'arial' "
What should I do?
Crea
See ?tempfile
On Mon, 11 Aug 2008, sara martino wrote:
Hei,
is it possible to create from R a series of files with unique(random) names?
I have tryed with file.create and dir.create but did not manage to do it.
Thanks a lot for help
Sara
[[alternative HTML version deleted]]
__
I'm not sure I understand, but is this what you want ?
quant <- seq(0,1,0.05)
perc<-c(1:234/234)
rows <- sapply(quant, function(x){which.min(abs(x - perc))})
perc[rows]
2008/8/11 Duncan Murdoch <[EMAIL PROTECTED]>
> Gareth Campbell wrote:
>
>> Hello,
>> I have a matrix and I want to sample 20
On Sun, Aug 10, 2008 at 10:08 PM, R_Learner <[EMAIL PROTECTED]> wrote:
>
> I'm trying to control the number of labels in the x-axis of my plot. My code
> is the following:
>
> graph1<-barplot(total_skew)
> axis(1,at = graph1,labels=raw_date[1:length(total_skew)], las=2)
>
> however, the length of m
seq_along
On Mon, Aug 11, 2008 at 7:56 AM, stephen sefick <[EMAIL PROTECTED]> wrote:
> On Sun, Aug 10, 2008 at 10:08 PM, R_Learner <[EMAIL PROTECTED]> wrote:
>>
>> I'm trying to control the number of labels in the x-axis of my plot. My code
>> is the following:
>>
>> graph1<-barplot(total_skew)
>>
Even though there is a "[.zoo", there is no "[.zoo<-" in zoo.
For assignments it simply defaults to the ordinary [ for
vectors and matrices. Any of these work:
window(t1, both.dates)[, both.cols] <- t2[both.dates, both.cols]
window(t1, both.dates)[, both.cols] <- window(t2, both.dates)[, both.c
If I save a dataframe with a character-typed last column to a relational
database with dbWriteTable, the values in the last column of the
resulting table in the database will have a '\r' (carriage return)
appended. If I read back the dataframe with dbReadTable the last column
in the resulting
Dear R users,
I plot data with
points(my_data[x]~x, col = x, type = "o", lwd="4")
where x is an integer running from 1 to 10, I get points drawn at the plot.
When want to do a legend to this I try
legend(leg.txt[x], col = x, text.col = 1, pch = 1, bty = "n")
where leg.txt contains the names of
OK, sorry, I misread the help page: pt.lwd does what was requested below.
Zroutik
On Mon, Aug 11, 2008 at 2:16 PM, Zroutik Zroutik <[EMAIL PROTECTED]> wrote:
> Dear R users,
>
> I plot data with
>
> points(my_data[x]~x, col = x, type = "o", lwd="4")
>
> where x is an integer running from 1 to 10,
Which package is this? (Clearly DBI, not mentioned, but which backend,
and what OS? I was unaware that there was a DBI package that works with
(Microsoft) SQL Server.)
Please note the 'at a minimum' information the R posting guide asked you
for.
Such questions are perhaps best asked on R-s
Hi,
I'm looking for a manual (we based or pdf) which would explain in detail
with graphical examples what all the option can do in plot and par. Does
anybody now anything like this? A couple of manuals to R I went through do
have plot parameters mentioned, but sometimes it is hard to understand or
An introduction to R has some of these things. The way that I am
learning is by doing- I need something and then I figure it out with
help from the list, manuals, and stubbornness.
best
Stephen Sefick
On Mon, Aug 11, 2008 at 8:38 AM, Zroutik Zroutik <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I'm looki
On Aug 11, 2008, at 8:38 AM, Zroutik Zroutik wrote:
Hi,
I'm looking for a manual (we based or pdf) which would explain in
detail
with graphical examples what all the option can do in plot and par.
Does
anybody now anything like this? A couple of manuals to R I went
through do
have plot p
I recommend the graphics in the Rattle GUI install with dependency all
=true and use the explore options..check log for any iterative learning...
On Mon, Aug 11, 2008 at 6:08 PM, Zroutik Zroutik <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I'm looking for a manual (we based or pdf) which would explain in
Yes! This is a type of manual I was looking for. Thank you so much. I see
I'll spend the whole night "out-sourcing" ideas and possibilities :D
On Mon, Aug 11, 2008 at 2:53 PM, Ben Tupper <[EMAIL PROTECTED]>wrote:
>
> On Aug 11, 2008, at 8:38 AM, Zroutik Zroutik wrote:
>
> Hi,
>>
>> I'm looking f
Calculating predicted survival curves in the case of a time dependent variable
is a thorny problem; not computationally but conceptually. It is not always
clear exactly what you WANT to plot. My book has a whole section on the
subject,
which is only to say that it is too long a subject to sum
stephen sefick gmail.com> writes:
>
> e1 <- function(x,b,t){
> d<-(x)*(b^t)
> plot(d)
> }
>
> e1(2, 2,seq(from=0, to=6, by=1))
>
> Is there a way to do this with a change in time. I would like to use
> differential equations.
I'm not sure what you mean by "do this with a chan
population# = (initialvalue @ t) - (death rate) + (fecundtity per individual) dt
to be able to vary the values of the inputs death rate and fecundity
with time using initial value @ t from the previous time step to
evaluate the function at the present time step.
On Mon, Aug 11, 2008 at 9:43 AM, Be
Hi,
I am trying to run the following code and it seems to be very
temperamental. After a few successful runs I started receiving the error
message from running test = VAR(y,). Could you please advise why the
code would work several times and then start sending back the error
message or pro
Dear R-help,
I have a longitudinal dataset with about 5,000 subjects. To avoid overly
cluttered plot of the individual profile, I want the plot to be shadowed and to
randomly select say 100 subjects which will be depicted using darker lines on
the plot.
Kindly guide me through on how to do
Dear R-users,
I am currently writing an R package that contains three C++ files in the
src/ source directory
When i issue a R CMD check pkge_name, all the *.cpp files in the src/
directory will be compiled through something like (seen in 00install.out
file):
* Installing *source* package 'p
Hello,
I'm fitting a Poisson GLM with the glm( ) function and I would like to know how
to obtain the confidence intervals for predictions (fitted values)...
I mean like in function lm( ):
prediction.matrix=exp(predict(model1.lm,interval="prediction")
(where model1.
No, the df are H-(p+q), against what your intuition tells you. The problem
is that you're thinking of the residuals as losing a df for each parameter,
but the asymptotics of the problem involve the sample autocorrelations of
the residuals. You can read the details in the original paper by Box an
timately, you'll come down to the truth"/
__ Information from ESET Mail Security, version of virus signature
database 3345 (20080811) __
The message was checked by ESET Mail Security.
http://www.eset.com
part000.txt - is OK
_
Hi,
How can I make a t-test if I haven't got all data set but I know the
numbers of subjects for each group, the mean and di standard deviation
for each group?
Thanks
Angelo Scozzarella
__
R-help@r-project.org mailing list
https://stat.ethz.ch
I have a slightly different question. Rather than build the list all at once I
would like to know how to append. Essentially ending up with a list of lists.
Something like:
x <- c(31,2,3)
a <- list(list(x))
x <- c(42,3,5)
a <- append(a, list(x))
Can it be done this way? So now the length(a) is
2008/8/11 Lafaye de Micheaux Pierre
<[EMAIL PROTECTED]>:
> I would like to be able to choose which *.cpp files will be compiled and to
> exclude some *.cpp files from the compilation.
> This is because some of my *.cpp files contain #include instructions calling
> some other *.cpp file.
> For exa
Greetings,
Please forward Web links for the most recent R GUI and a list of add-ons
that will run in the graphic interface. I need either the R GUI for (MS)
Windows Vista, or a Web-based GUI that will run longitudinal models with
covariates (e.g., multilevel modeling, latent growth curves, and
you've a couple of options- maybe they will fit your needs maybe they
won't. Look at the GUI page on CRAN
http://www.sciviews.org/_rgui/
you could google this and find out on your own what may or may not
suit your needs. Most everybody on this list uses the command line
interface
stephen sefick
timately, you'll come down to the truth"/
__ Information from ESET Mail Security, version of virus signature
database 3345 (20080811) __
The message was checked by ESET Mail Security.
http://www.eset.com
part000.txt - is OK
_
Hi,
I'd like to take difference for a sequence a between a_i and a_i-2, for
instance,
a<-c(2,3,4,8,1)
I need (2, 5, -3) as a result. If not using a for loop, can anyone help me?
Thanks a lot.
Dot
--
View this message in context:
http://www.nabble.com/R-help--how-to-take-difference-in-next-two-
Is the FastICA R implementation as good as the MatLab Implementation ?
I would appreciate talking to someone who has used FastICA for R.
Thank you very much.
--
Maura E.M
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing li
How do I change the font size of the label in ggplot. For example
x=c("abcdefghijklmn", "opqrstuvwxyz")
y=c(1, 2)
df=data.frame(name=x, value=y)
p=ggplot(df, aes(name, y))+geom_point()
Now the name of "abcdefghijklmn" and "opqrstuvwxyz" are too long. How can I
change the font size of them. And
On 8/11/2008 11:26 AM, dott wrote:
Hi,
I'd like to take difference for a sequence a between a_i and a_i-2, for
instance,
a<-c(2,3,4,8,1)
I need (2, 5, -3) as a result. If not using a for loop, can anyone help me?
Thanks a lot.
Dot
a <- c(2,3,4,8,1)
diff(a, lag=2)
[1] 2 5 -3
?diff
--
Chuc
Here are a couple of ways:
> tail(a, -2) - head(a, -2)
[1] 2 5 -3
> c(ts(a) - lag(ts(a), -2))
[1] 2 5 -3
On Mon, Aug 11, 2008 at 11:26 AM, dott <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> I'd like to take difference for a sequence a between a_i and a_i-2, for
> instance,
> a<-c(2,3,4,8,1)
> I nee
I have used Rattle ( R Analytical tool to learn easily )and R
Commander and found them to be quite good. I don't use the command
line interface because I find it too time consuming from other
languages like SAS,SPSS. I am not sure on Vista compatibility though.
The Rattle log file auto generates t
Confidence intervals for what? For a Poisson glm() you can predict the
mean response or the linear predictor, and they are not the same (unlike a
linear model).
I suggest you predict the linear predictor (the default), use the s.e.s to
for a confidence interval and then if desired transform (
Hi,
I have a graph and I would like to write some values inside the legend that
were saved in a variable. Please revise the code below in which I've wrote 2
different legends, but I am not happy with either of them. What I want is a
legend with tile "Legend" and underneath a line with a name a
I can't reproduced this (function aes), but have you looked at
?par
cex
cex.axis (and so on)
On Mon, Aug 11, 2008 at 11:55 AM, dhzdhz <[EMAIL PROTECTED]> wrote:
>
> How do I change the font size of the label in ggplot. For example
>
> x=c("abcdefghijklmn", "opqrstuvwxyz")
> y=c(1, 2)
> df=data.fra
Felipe Carrillo yahoo.com> writes:
>
> Hello all:
> Is there a way to detect in R if a dataset is normally distributed or skewed
without graphically seeing it?
> The reason I want to be able to do this is because I have developed and
application with Visual Basic where
> Word,Access and Excel "t
#Code begin:
# ---
a = 2 # result of a function for example
plot(1:10, a*(1:10))
#here you go
legend(1, 20, paste("value", "=" , a, sep=" "), title="Legend" ,bty = 'n')
val <- substitute("value" ==.(a), list(a = a))
legend(6, 5, do.call("expression",list("Legend", val)), bty = '
Hi dhzdhz,
There's currently no particularly easy way to do this. It is possible
to use grid to edit these labels directly (see
http://had.co.nz/ggplot2/book/grid.pdf), but the next version of
ggplot2 will have a more convenient theming system that will make it
much easier to make these sorts of
Charles C. Berry wrote:
Should someone start a pool on when SAS will offer 'R language support' ??
At least Biostatistics Denmark developed something in this direction.
Too bad that Rolf Poalis made only this announcement:
http://tolstoy.newcastle.edu.au/R/help/04/04/0009.html
in the beginning
Hi,
Thanks so much, i even don't know why i felt the compulsion to complicate
myself with the legend and bquote ;-) Fancy code is not necessarily the best -
i may say!
Again, thanks,
Monica
> Date: Mon, 11 Aug 2008 12:49:04 -0400
> From: [EMAIL PROTECTED]
> To: [EMAIL PROTECTED]
> Subject: R
Perhaps you want
val <- substitute("value" == a, list(a = a))
legend(6, 5, do.call("expression", list("Legend", val)), bty = 'n')
?
Your first attempt is a character vector and your second fixed up quote
and substitute. But more likely would be
legend(6, 20, val, bty="n", title="Legend")
entities must not be multiplied beyond necessity
On Mon, Aug 11, 2008 at 1:05 PM, Monica Pisica <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> Thanks so much, i even don't know why i felt the compulsion to complicate
> myself with the legend and bquote ;-) Fancy code is not necessarily the best
> - i ma
Just to add a couple more thougths to this thread. If you go the xfig driver
route, there are also the programs jfig and winfig that edit fig files and may
be easier to install on some platforms (winfig may require a payment to get
full use, but I believe jfig is free).
But you may want to thi
The problem is that the coordinate system after the call to filled.contour is
not the same as used to create the contour plot. There are a couple of ways to
do what you want.
1. use the plot.axes argument (see the help page and the examples) and have
that plot the arrows as part of the call.
2008/8/11 Lafaye de Micheaux Pierre
<[EMAIL PROTECTED]>:
> Thank you Barry,
>
> In fact, these are only C files (not C++) and i use the extern "C"
> directive.
>
> I would prefer not to rename the files because many of these files serve
> also in other projects where they should have the .cpp exten
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
stephen sefick wrote:
| population# = (initialvalue @ t) - (death rate) + (fecundtity per
individual) dt
| to be able to vary the values of the inputs death rate and fecundity
| with time using initial value @ t from the previous time step to
| eval
Thanks, this certainly works as well, especially the "second" legend syntax.
Monica
> Date: Mon, 11 Aug 2008 18:08:02 +0100
> From: [EMAIL PROTECTED]
> To: [EMAIL PROTECTED]
> CC: r-help@r-project.org
> Subject: Re: [R] : bquote inside legend()
>
> Perhaps you want
>
> val <- substitute("value"
Hi,
I'm completely new to R and am struggling with producing residuals from a
lowess curve. I have looked at the pdf help pages and searched online and have
tried the following code:
> l.agebl <- lines(lowess(agebl, f=.5))
> resid(l.agebl)
> resid(lowess(agebl, f=.5))
NULL
> residuals(lowess
Hello,
I can`t figure out how can increase the velocity of the fitting data by nls.
I have a long data .csv
I want to read evry time the first colunm to the other colunm and analisy with
thata tools
setwd("C:/dati")
a<-read.table("Normalizzazione.csv", sep=",", dec=".", header=F)
for (i in 1:di
Hello,
I can`t figure out how can increase the velocity of the fitting data by nls.
I have a long data .csv
I want to read evry time the first colunm to the other colunm and analisy with
thata tools
setwd("C:/dati")
a<-read.table("Normalizzazione.csv", sep=",", dec=".", header=F)
for (i in 1:di
Thanks Jim and Ben for your replies, Reading further about data normalization
found shapiro.test. I understand that if the p-value is smaller than 0.05 then
the data isn't normal, I just don't understand what the "W" means.
Hi Felipe,
Here's one way:
library(nortest)
if(sf.test(fishlength)$p
On Mon, Aug 11, 2008 at 9:27 AM, <[EMAIL PROTECTED]> wrote:
> I have used Rattle ( R Analytical tool to learn easily )and R
> Commander and found them to be quite good. I don't use the command
> line interface because I find it too time consuming from other
> languages like SAS,SPSS. I am not sure
error in isoda(y = 1, times = seq(0, 1, by = 0.02), func = e1B, parms
= c(b = 0.9)) :
Model function must return a list
this is the error message I get when I try and paste in the code.
Stephen Sefick
On Mon, Aug 11, 2008 at 1:36 PM, Ben Bolker <[EMAIL PROTECTED]> wrote:
> -BEGIN PGP SIGNE
Hello,
I can`t figure out how can increase the velocity of the fitting data by nls.
I have a long data .csv
I want to read evry time the first colunm to the other colunm and analisy with
thata tools
setwd("C:/dati")
a<-read.table("Normalizzazione.csv", sep=",", dec=".", header=F)
for (i in 1:di
Hello,
I can`t figure out how can increase the velocity of the fitting data by nls.
I have a long data .csv
I want to read evry time the first colunm to the other colunm and analisy with
thata tools
setwd("C:/dati")
a<-read.table("Normalizzazione.csv", sep=",", dec=".", header=F)
for (i in 1:di
thanks
On Mon, Aug 11, 2008 at 2:18 PM, Ben Bolker <[EMAIL PROTECTED]> wrote:
>
> There was a spurious tilde in the code coming
> from my digital signature -- oops. ignore the
> tilde in the function:
>
> library(odesolve)
>
> ## in general x, b are state and parameter VECTORS
> e1B <- function(
There was a spurious tilde in the code coming
from my digital signature -- oops. ignore the
tilde in the function:
library(odesolve)
## in general x, b are state and parameter VECTORS
e1B <- function(t,x,b) {
list(x*b^t,NULL)
}
L1 = lsoda(y=1,times=seq(0,1,by=0.02),func=e1B,parms=c(b=0.
Hi,
stephen sefick wrote:
e1 <- function(x,b,t){
d<-(x)*(b^t)
plot(d)
}
e1(2, 2,seq(from=0, to=6, by=1))
Is there a way to do this with a change in time. I would like to use
differential equations. I am trying to model
a population with an initial value, fecundity per time st
Is anyone aware of a way to check whether multinom has converged by
checking a component of the output ? I'm
not familar with nnet but, since multinom calls nnet , maybe there is an
extra argument once can send to multinom to capture this
information. Thanks.
___
Please use R's help resources before posting.
help.search("difference") will give you your answer (?diff).
-- Bert Gunter
Genentech
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of dott
Sent: Monday, August 11, 2008 8:27 AM
To: r-help@r-project.org
Subje
You might also have a look at Maindonald and Braun (2007) Data
Analysis and Graphics Using R, 2nd edition, Cambridge. The web site
for the book has R code for the illustrations in the book.
Best Regards
John
2008/8/11 Zroutik Zroutik <[EMAIL PROTECTED]>:
> Yes! This is a type of manual I was lo
Matlab has .*, so that
t(x1,x2).*t(x3,x4).* t(x5,x6)
but what is .* in R?
Thanks.
Dot.
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View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
if you mean matrix multiplication, it's %*% but i'm not sure if that's
what you meant by below ?
On Mon, Aug 11, 2008 at 3:44 PM, dott wrote:
Matlab has .*, so that
t(x1,x2).*t(x3,x4).* t(x5,x6)
but what is .* in R?
Thanks.
Dot.
--
View this message in context:
http://www.nabble.com/how-c
dott gmail.com> writes:
>
>
> Matlab has .*, so that
> t(x1,x2).*t(x3,x4).* t(x5,x6)
>
> but what is .* in R?
>
> Thanks.
> Dot.
>
Element-wise multiplication is just * in R
(do you really mean "sum" or is that a general expression
for "calculation"?)
See also
http://wiki.r-projec
Hi,
How can I generate a random signal that's correlated with a given signal at
a given correlation (say 0.7)?
I've been looking at rmvnorm etc but don't seem to figure it out. Thanks
-
Yasir H. Kaheil
Columbia University
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View this message in context:
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Hi:
2008/8/11 Zroutik Zroutik <[EMAIL PROTECTED]>:
Yes! This is a type of manual I was looking for. Thank you so
much. I see
I'll spend the whole night "out-sourcing" ideas and possibilities :D
On Mon, Aug 11, 2008 at 2:53 PM, Ben Tupper
<[EMAIL PROTECTED]>wrote:
Look at:
http://a
Barry Rowlingson wrote:
2008/8/11 Lafaye de Micheaux Pierre
<[EMAIL PROTECTED]>:
Thank you Barry,
In fact, these are only C files (not C++) and i use the extern "C"
directive.
I would prefer not to rename the files because many of these files serve
also in other projects where they should h
I have a list
List(Sku=" ", Shape=1, Scale=3, DayOfYear=daylist)
Note: picture daylist as c(2,3,4,3) it is a list with variable length.
Then I have a list of lists
al <- c(al, List(List(Sku=" ", Shape=1, Scale=3, DayOfYear=daylist))
Note: same comment on daylist as above.
So far this creates a
Hi,
I want to calculate the T-Test from means and sd.
How can I do it?
Thanks
Angelo Scozzarella
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PLEASE do read the posting guide http://www.R-project.org/posting-guid
On 12/08/2008, at 3:11 AM, Angelo Scozzarella wrote:
How can I make a t-test if I haven't got all data set but I know
the numbers
of subjects for each group, the mean and di standard deviation for
each group?
Use your head. I.e. do the same thing you'd do if you weren't using
a compute
On Sat, 09-Aug-2008 at 04:06PM -0700, dusa.adrian wrote:
[...]
|> Warning in structure(.External("dotTclObjv", objv, PACKAGE = "tcltk"), class
|> = "tclObj") :
|> X11 protocol error: BadWindow (invalid Window parameter)
|>
|> > R.version
|>_
|> platform i486-pc-linux-gnu
Just plug your values into the t-test formula, you don't need R for
this, you can use a calculator. If you want a p-value then use the pt()
function in R after getting the t statistic.
Angelo Scozzarella wrote:
Hi,
I want to calculate the T-Test from means and sd.
How can I do it?
Thanks
Hi Kevin,
check
?dump
set.seed(1234)
list1 <- list(a1=rnorm(100), a2=rnorm(10))
list2 <- list(a1=rnorm(50), a2=rnorm(25))
mylist <- list(top1=list1, top2=list2)
dump(list="mylist", file="mydumpedlist.r")
rm(mylist)
# please note that you have to quote the name
# of the object(s) you want to du
Hi, I wonder if there is any function which gives the number of times which
an element is repeated in the matrix. Let say there is a matrix, I would
like to find out how many times number 2 has been repeated in the matrix, or
in other words, how many elements of the matrix are equal 2.
Thanks for y
What do you want to do with the data? Are you just storing it to read
in it later? Have you looked at 'save/load'? If you want a character
representation, try 'dput'; this can be read back in with 'source'.
So it all depends on what you are planning to do with it.
On Mon, Aug 11, 2008 at 4:14 P
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