Michael Just gmail.com> writes:
> I am trying to plot multiple histograms with the same scales, etc into one
> plot. The commands below produce a 3 page PDF with each histogram occupying
> the upper right quadrant. And use slightly different scales on the X and Y
> axes.
I suggest that you forge
charter.net> writes:
>
> ff <- complex(length(fs))
> ff[9] <- fs[9]
> ff[5] <- fs[5]
>
> Include the DC component:
>
> ff[1] <- fs[1]
>
> Take the inverse
>
> fi <- fft(ff, inverse=TRUE) / length(ff)
>
> Plot
>
> plot(fi)
>
> Notice that the plot is the Re vs. Im on the x and y axis' res
Hi I believe I have long time solution to the problems with the
read.spss() funtion.
In non-ascii locales the import of spss files into R has consistently
produced errors and sometimes wouldn't allow import at all.
I've made a custom version of the read.spss() function that appear to
solv
Hi,
I have tried this with R-2.7.2 then I got
> alist[['goodbye']]
NULL
> alist[['hi']]
NULL
> alist[['hello']]
[1] 10
The results make sense to me thus you might want to update your R version.
chunhao
B Fox wrote:
>
> This seems odd. When I try to look up a list element which has a space
Thanks, Gabor! This workaround fixes my immediate needs.
--sundar
Gabor Grothendieck said the following on 10/1/2008 10:42 PM:
That may be a bug in R but I think there is another problem on top of that
as I don't think bquote descends into function bodies:
z <- 2
bquote(function(x) {x^.(z)})
That may be a bug in R but I think there is another problem on top of that
as I don't think bquote descends into function bodies:
> z <- 2
> bquote(function(x) {x^.(z)})
function(x) {x^.(z)}
> bquote(function(x, y) { x^.(z) + y})
function(x, y) { x^.(z) + y}
> R.version.string # Vista
[1] "R ver
Hi,
I would like to create a matrix. Say 5*4.
The first column is integer, the second column is date, the third column
is character and the rest columns are integers.
So it's a combination of different types.
I am wondering how can I do that? "matrix" command only allow one type.
**
if expects just one condition (no vectors); see ?ifelse
dataframe$thevector <- ifelse(dataframe$factor=="3", a.mean,
dataframe$thevector)
K
On Wed, Oct 1, 2008 at 12:05 PM, Whitt Kilburn <[EMAIL PROTECTED]> wrote:
> Hello all,
>
> I apologize for a terribly simple question. I'm used to using
Hello,
I am trying to plot multiple histograms with the same scales, etc into one
plot. The commands below produce a 3 page PDF with each histogram occupying
the upper right quadrant. And use slightly different scales on the X and Y
axes.
> s21 <- dat[dat$sc_recov=="21",]
> s21.ED <- subset(s21, s
Hi, R-help,
(sessionInfo at the end)
I'm trying to construct a function using bquote and running into a
strange error message. As an example, what I would like to do is this:
z <- 2
eval(bquote(function(x, y) { x^.(z) + y }))(2, 3)
However, I get the following:
Error in eval(expr, envir, en
Hi Eric,
> data<-list(x1 <- c(0,1,2,3),x2 <- c(7,8),x3 <- c(2,6,6,8), x4 <- c(4,8))
> lapply(X=data, mean)
[[1]]
[1] 1.5
[[2]]
[1] 7.5
[[3]]
[1] 5.5
[[4]]
[1] 6
Hope it helps
Chunhao
eric lee-8 wrote:
>
> Hi. I have a list where each object in the list has multiple parts. I'd
> like to
Dear Duncan,
Thanks for your quick response! I really like the rgl package. First a
clarification. I was not using IMDisplay to try to view the .gif file.
I apologize for that confusion. I attempted to open the file with
Mozilla firefox which works but unfortunately only displays a black box.
Hi everybody!
I was using the function pvals.fnc from package 'languageR' until April.
I do not know which version. Yesterday I updated all my packages
and tried to run my loop again. Now I get the following error message:
error in pvals.fnc(mm, nsim = 1000) :
MCMC sampling is not yet implemen
This seems odd. When I try to look up a list element which has a space in
the name using just the first word (i.e. no spaces), it will sometimes
return the element with a space in the name and sometimes it will return
NULL.
Try this:
alist <- list( 'hello'=10, bye=20, 'hello world'=30, 'goodbye
Suppose p is a vector of paths, e.g., p <- c("~/dir1", "~/dir2/dir3")
Then the following will return the full pathname of the first found
location:
Find(file.exists, file.path(p, "prog.R"))
so you can source that.
On Wed, Oct 1, 2008 at 6:09 PM, Gang Chen <[EMAIL PROTECTED]> wrote:
> Suppose
Hi.
On Sat, Sep 27, 2008 at 5:17 AM, Steele, Dr Douglas <[EMAIL PROTECTED]> wrote:
> Hi
>
> I am using Ubuntu 8.04 64 bit, R as below, Matlab 7.6.0. I would like to
> transfer mat files back and forward between R and Matlab. Whilst I have used
> Matlab for years its been a long time since I ha
Thank you very much!
That's the way to go!
Cheers,
Stefan
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Wednesday, October 01, 2008 10:52 AM
To: Schreiber, Stefan
Cc: r-help@r-project.org; [EMAIL PROTECTED]
Subject: Re: [R] shifting ticks to left or right
I think it is meaningful to ask for a non-trivial Pr (X < x, Y=y) when you
are writing down the likelihood for parameter estimation. This is commonly
the case in likelihood estimation in bivariate failure time models. If one
interprets Pr(Y=y) as the density evaluated y then:
Pr(Xhttp://www.jh
Suppose I have a file prog.R stored in a directory under ~/dirname,
and ~/dirname is set in a shell script file (e.g. .cshrc) as one of
the accessible paths on terminal. On a different directory I could run
prog.R interactively by executing
source("~/dirname/prog.R")
It seems that source() does n
On 2/10/2008, at 11:02 AM, Ravi Varadhan wrote:
I think it is meaningful to ask for a non-trivial Pr (X < x, Y=y)
when you
are writing down the likelihood for parameter estimation. This is
commonly
the case in likelihood estimation in bivariate failure time
models. If one
interprets P
Hi. I have a list where each object in the list has multiple parts. I'd
like to take the mean of just one part of each object. Is it possible to do
this with lapply? If not, can you recommend another function? Thanks.
eric
> x1 <- c(0,1,2,3)
> x2 <- c(7,8)
> x3 <- c(2,6,6,8)
> x4 <- c(4,8)
>
Dear R users,
I am using RandomForest package. While using this package, i got
"Error: cannot allocate vector of size 117.3 Mb"
.message.
I had this problem earlier too but could not manage. Is there any way to solve
this problem or to increase vector size ? My data set
Hello,
I used the aggregate function with success on my data frame "all".
> aggregate(all, list(sc_recov), mean)
I then made a subset of this data frame:
>s.all <-subset(all, select= c(3,4,5,6,7,8,9,10,11,23))
but when I tried to to run the aggregate function on this subset:
> aggregate(s.all,
Hi, Shiva,
The idea of reject inference is very simple. Let's assume a credit card
environment. There are 100 applicants, out of which 50 will be approved and
booked in. Therefore, we can only observe the adverse behavior, such as
default and delinquency, of 50 booked accounts. Again, let's assume
h <- structure(list(V1 = c(-0.351714766, 0.188298251, 0.042951816,
-0.072490327, -0.691885485, -0.816169763, -0.7066502, -0.856286332,
-0.839723411, -0.427242353, -0.372911996, 0.326707494, 0.07847893,
0.687447841, 0.516105863, 0.267076547, 0.727867663, 0.432699191,
0.258610632), V2 = c(0.256636068
R-users
E-mail: r-help@r-project.org
Hi!
>I interprete this as following: the simplest tree with xerror under
>min(xerror) + its own xstd
>Neverthless, in some article I read the following rule:
>the simplest tree with xerror under min(xerror) + xstd corresponding to the
>min(xerror)
>Is this a m
On 2/10/2008, at 4:43 AM, Sasha Pustota wrote:
Package mvtnorm provides dmvnorm, pmvnorm that can be used to compute
Pr(X=x,Y=y) and Pr(X
Yes:
foo <- function(x,y) {
0
}
I'm currently using "integrate" with dmvnorm but it is too slow.
Words fail me
> s1 <- "Hello.World"
> which(strsplit(s1, '')[[1]]=='.')
[1] 6
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of phoebe kong
> Sent: Wednesday, October 01, 2008 2:56 PM
> To: r-help@r-project.org
> Subject: [R] Looking for position of character in
Try this:
gregexpr("\\.", X)
On Wed, Oct 1, 2008 at 3:56 PM, phoebe kong <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I would like to check if a string contains "." in it. If yes, I would like
> to know the position of "." in the string.
>
> Eg: X<-"NM1236.3"
>
> In above example, "." is at position
Hi all,
I would like to check if a string contains "." in it. If yes, I would like
to know the position of "." in the string.
Eg: X<-"NM1236.3"
In above example, "." is at position 7.
I tried functions grep() and match(). It seems they couldn't detect ".".
Thanks in advance for your help,
Sit
P.S. just "pnorm(x, mean=m, sd=s)", not "1-pnorm(x, mean=m, sd=s) +
pnorm(-x, mean=m, sd=s)"
On Wed, Oct 1, 2008 at 2:04 PM, Sasha Pustota <[EMAIL PROTECTED]> wrote:
> Thanks Jay. I realized that I was doing it a silly way shortly after I
> posted and that the answer i was looking for is simply
>
Dear R-help,
I have just been informed that I must not rewrite the 'https' as 'http' as
some web pages may not download (I think I just got lucky on the ones I've
tried thus far). Therefore I would ask if some kind individual could let me
know where I can post questions about the RCurl package (Om
Dear R-Help,
>From reading the help file, it is my understanding the the download.file()
function does not support HTTPS connections. So therefore, understandably,
the follow produces an error:
### R Code
> url <- "https://stat.ethz.ch/pipermail/r-help/2008-October/thread.html";
> destfile <- "//
I have searched the forums but I am having trouble trying accomplish two
specific plots. I am not sure if this is a very basic task but I am having a
lot of trouble Here is my data frame:
dateprod_daily price
1 2003-08-15 12050 4.83
2 2003-08-18 12050
Thanks Jay. I realized that I was doing it a silly way shortly after I
posted and that the answer i was looking for is simply
condXY(y, x, my, mx, r) * dnorm(y, my)
condXY <- function(y, x, my, mx, r) {
m <- mx + r*(y - my)
s <- sqrt(1-r^2)
p <- 1 - pnorm(x, mean=m, sd=s) + pnorm(-x, mean=m
On Tue, Sep 30, 2008 at 8:06 PM, Mike Fugate <[EMAIL PROTECTED]> wrote:
> Good Day,
>
> I'm using the knn function in the package class. With k set to 3, the
> function returns proportions of 1/3, 0.5, 0.6, 2/3, 3/4, and 1.0 for the
> test cases. I don't understand how with k set to 3 the proport
For the sake of anyone else asking this questionas Andy Liaw explained
to me, yes, this is expected because predict() uses all the trees in the
forest, including the ones based on any particular item in the input data
frame.
Rodney
-Original Message-
From: [EMAIL PROTECTED] [mailto:[E
Hello,
Im am trying to fit a GAM (mgcv) using the family gamma and get the followinr
error message:
Error in gamma(link = "identity") :
Non-numeric argument to mathematical function
Does anyone know why do i get this? I tried to check in R-help already but I
couldn't find anything alike.
Hello,
I am trying to produce some graphics to visualize my data. I think I want
histograms. I have a grouping variable that has 48 different groups. I would
like to produce a graphic that contains three of of these groups in the same
device (they are time steps). I would like create graphics for
Hi,
In the help for spectrum it says:
" If available, a confidence interval will be plotted by
'plot.spec': this is asymmetric, and the width of the centre mark
indicates the equivalent bandwidth. "
What you get is a blue bar on the right side of the plot.
I can understand the width o
Hi Bernardo,
Do you have to use logistic regression? If not, try Random Forests... It has
worked for me in past situations when I have to analyze huge datasets.
Some want to understand the DGP with a simple linear equation; others want high
generalization power. It is your call... See, e.g.,
Dear Sasha,
On Wed, Oct 1, 2008 at 11:43 AM, Sasha Pustota <[EMAIL PROTECTED]> wrote:
> Package mvtnorm provides dmvnorm, pmvnorm that can be used to compute
> Pr(X=x,Y=y) and Pr(X
> Are there functions that would compute Pr(X I'm currently using "integrate" with dmvnorm but it is too slow.
Stri
On Wed, 1 Oct 2008, Donald Catanzaro, PhD wrote:
Good Day All,
I have a negative binomial model which I have developed using the MASS
library. I now would like to develop some predictions from it.
Running the predict.glm (stats library) using type="response" gives me a
non-integer value whi
Hi all,
Please, how could I calculate the time that two time segments has in
common? Is there any function to perform this calculation?
For instance, given four POSIXlt objects...
endPeriod<-as.POSIXlt("2008-09-30")
startPeriod<-as.POSIXlt("2007-10-01")
endProject<-as.POSIXlt("2007-05-31")
st
> Does anybody knows a command to centre the tick mark labels exactly
> between the tick points (right shift)?
> And then to exclude the last tick label on the right?
>
> I know one can shift them using the 'hadj' option in par. But I am
> wondering if there is a more convenient command!
A littl
Schreiber, Stefan wrote:
Hey list,
Does anybody knows a command to centre the tick mark labels exactly
between the tick points (right shift)?
And then to exclude the last tick label on the right?
I know one can shift them using the 'hadj' option in par. But I am
wondering if there is a more co
Daniel McGlinn wrote:
Dear list,
I am attempting to utilize the function "movie3d" in the package "rgl"
to create a .gif animation of a 3d graphic. I understand that this
function requires that the program "ImageMagik" is installed, and I have
recently downloaded and installed this program.
Try this also;
xtabs(volume ~ day, data = x)
On Wed, Oct 1, 2008 at 10:58 AM, Max Rausch <[EMAIL PROTECTED]> wrote:
> I have a data frame with the following information
>
> day hourvolume
> 1 2003-07-18 10 836700
> 2 2003-07-18 11 375000
> 3 2003-07-18 12
Daniel Rabczenko wrote:
Hello everybody,
Two - I hope Simple questions about working with workspaces.
Is there a way to force R to start in "clean" workspace / avoid "previously
saved workspace restored"?
Start with --no-restore. (Start with --help for the full list of
command line optio
Schreiber, Stefan wrote:
Hey list,
Does anybody knows a command to centre the tick mark labels exactly
between the tick points (right shift)?
And then to exclude the last tick label on the right?
I know one can shift them using the 'hadj' option in par. But I am
wondering if there is a more co
On Wed, Oct 1, 2008 at 8:58 AM, Max Rausch <[EMAIL PROTECTED]> wrote:
> I have a data frame with the following information
>
> day hourvolume
> 1 2003-07-18 10 836700
> 2 2003-07-18 11 375000
> 3 2003-07-18 12 6
> 4 2003-07-188 102
> 5
Daniel Rabczenko iestat.pl> writes:
>
> Hello everybody,
>
> Two - I hope Simple questions about working with workspaces.
>
> Is there a way to force R to start in "clean" workspace / avoid "previously
> saved workspace restored"?
When you start R, use the option --no-restore. If you are work
Good Day All,
I have a negative binomial model which I have developed using the MASS
library. I now would like to develop some predictions from it.
Running the predict.glm (stats library) using type="response" gives me a
non-integer value which was rather puzzling. I would like to confirm
Thank you, Dieter:
I apologize to all of those who tried to help and who couldn't get my code to
work.
I now know three things: 1) stop working at midnight; 2) always provide the
correct package
name (I meant arm(), not car(); and 3) when testing code make sure you do it in
a new R window.
Die
Two approaches
?aggregate or the reshape package
aggregate(xx$volume, list(day=xx$day), sum)
library(reshape)
names(xx)[3] <- "value"
cast(xx, day ~ ., sum)
--- On Wed, 10/1/08, Max Rausch <[EMAIL PROTECTED]> wrote:
> From: Max Rausch <[EMAIL PROTECTED]>
> Subject: [R] "group by" functionalit
On Wed, Oct 1, 2008 at 7:21 AM, Cézar Freitas <[EMAIL PROTECTED]> wrote:
> Hi. I searched the list and didn't found nothing similar to this. I
> simplified my example like below:
>
> #I need calculate correlation (for example) between 2 columns classified by a
> third one at a data.frame, like be
The first tapply in your question subsets V1 but not V2 so they are
of different length. To subset both tapply over the row names and
perform the subsetting in the function:
tapply(rownames(dataf), dataf$class, function(r) cor(dataf[r, "V1"],
dataf[r, "V2"]))
or
tapply(rownames(dataf), dataf$cl
From: Frank E Harrell Jr
>
> Bernardo Rangel Tura wrote:
> > Em Ter, 2008-09-30 às 18:56 -0500, Frank E Harrell Jr escreveu:
> >> Bernardo Rangel Tura wrote:
> >>> Em Sáb, 2008-09-27 às 10:51 -0700, milicic.marko escreveu:
> I have a huge data set with thousands of variable and one binary
> >
Hey list,
Does anybody knows a command to centre the tick mark labels exactly
between the tick points (right shift)?
And then to exclude the last tick label on the right?
I know one can shift them using the 'hadj' option in par. But I am
wondering if there is a more convenient command!
Thanks a
tapply(data$volume, data$day, sum)
would do I think. But I haven't tried, was to lazy to type in some
data, I'm sorry.
Gabor
On Wed, Oct 1, 2008 at 3:58 PM, Max Rausch <[EMAIL PROTECTED]> wrote:
> I have a data frame with the following information
>
> day hourvolume
> 1 2003
Look at contourplot in the lattice library. There is an example doing
what you want.
Alain Guillet
Mark wrote:
Is it possible to add labelled contour lines to filled.contour plot ?
[[alternative HTML version deleted]]
__
R-help@r-project.o
It would not be possible to answer your original
question until you specify your goal.
Is it to develop a model with external validity
that will generalize to new data? (You are not
likely to succeed, if you are starting with a
"boil the ocean" approach with 44,000+ covariates
and millions o
Try this:
sapply(by(dataf[,c("V1","V2")], dataf$class, cor), '[', 3)
On Wed, Oct 1, 2008 at 9:21 AM, Cézar Freitas <[EMAIL PROTECTED]> wrote:
> Hi. I searched the list and didn't found nothing similar to this. I
> simplified my example like below:
>
> #I need calculate correlation (for example
?aggregate
day hour volume
1 2003-07-18 10 836700
2 2003-07-18 11 375000
3 2003-07-18 12 6
4 2003-07-188 102
5 2003-07-189 39
> aggregate(x$volume, list(x$day), sum)
Group.1 x
1 2003-07-18 2681700
On Wed, Oct 1, 2008 at 9:58 AM, Max Rausch <[EM
If you read the help page for randomForest(), you would expect this
behavior...
In the "Value" section of that help page, it says:
predicted the predicted values of the input data based on
out-of-bag samples.
Andy
From: Rodney Barnett
>
> Is it expected that predict.randomForest() prod
I have a data frame with the following information
day hourvolume
1 2003-07-18 10 836700
2 2003-07-18 11 375000
3 2003-07-18 12 6
4 2003-07-188 102
5 2003-07-189 39
I have been trying create a new data frame with the
Thanks a lot to all of you, it works now
Cheers,
Solène Goy> Subject: Re: [R] How to get the day of the year> From: [EMAIL
PROTECTED]> To: [EMAIL PROTECTED]> CC: r-help@r-project.org> Date: Wed, 1 Oct
2008 13:52:09 +0100> > On Wed, 2008-10-01 at 08:42 +, Solene Goy wrote:> >
Hello,> > I
Dear list,
I am attempting to utilize the function "movie3d" in the package "rgl"
to create a .gif animation of a 3d graphic. I understand that this
function requires that the program "ImageMagik" is installed, and I have
recently downloaded and installed this program. The R console does not
Hi. I searched the list and didn't found nothing similar to this. I simplified
my example like below:
#I need calculate correlation (for example) between 2 columns classified by a
third one at a data.frame, like below:
#number of rows
nr = 10
#the third column is to enforce that I need correla
Package mvtnorm provides dmvnorm, pmvnorm that can be used to compute
Pr(X=x,Y=y) and Pr(Xhttps://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Dear All!
I am looking for a package that contains routines for vertex enumeration
and center of mass computation for convex polytops.
Thanks in advance,
Serguei Kaniovski
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-he
Hi,
I have a data frame containing a column of human judgments, some of which
are missing:
> pr[3]
label
1 4
2 4
3 4
4 4
5 NA
6 3
7 3
8 3
9 3
10NA
11NA
12NA
13 2
14 2
15 2
16NA
17 1
18-1
19-1
20-1
Accompan
Hi. I am using the ripley multivariate function Lcross to caracterise the
spatial interactions of two factors using monte carlo simulation.
envelope(B1, Lcross, nsim=100, rank=1, global=TRUE)->E1
My problem is that My confidence Interval (CI) is not parallel to the X axis
when I plot the re
Hello everybody,
Two - I hope Simple questions about working with workspaces.
Is there a way to force R to start in "clean" workspace / avoid "previously
saved workspace restored"?
When I load workspace"2" working in workspace"1" everything from "1" is
written into "2" is there a way to avoid it
Hi,
I am a R user, with some experience in MacOS, Linux, etc, but I am
having a problem that I cannot solve:
I have a linux server (CentOS 5) and I installed sun jdk1.6. For instance:
$ java -version
Java version "1.6.0_10-rc2"
Java(TM) SE Runtime Environment (build 1.6.0_10-rc2-b32)
Java Hot
Hello, I would like to run a model with nesting design in lmer but I always
got the error message "Matrices must have same number of columns in
rbind2(..1, r)". My model is:
model_3R <-
lmer(N~Bareground+Habitat_type_simple+Presence+(1|Region/Ref_lmer),family =
"poisson")
of course it doesn't sa
Is it possible to add labelled contour lines to filled.contour plot ?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.or
Is it expected that predict.randomForest() produces the response vector when
given the same data frame as provided to randomForest()? See below.
Thanks.
Rodney
> rf <- randomForest (Species ~ ., iris)
> pc <- predict (rf, iris)
> confusion (pc, iris$Species)
true
object setosa
Generate the square wave:
genseq <- function()
{
x <- numeric(4*365)
s <- seq(as.Date("2005-01-01"), as.Date("2008-12-31"), by="month")
ob <- as.vector(s[c(10,22,34,46)] - as.Date("2005-01-01"))
oe <- as.vector(s[c(11,23,35,47)] - as.Date("2005-01-01"))
for(.index in 1:length(ob))
{
I've attached below the output of a program I am running. What I want to do
is, get the equation for each of the lines and determine what the overall
equation is that is applicable to every scenario (I know that each line is
of multiple of the proceeding line; as in the bottom's coefficient is 1x
Dear R users,
Suppose I have a set of 10 candidate models, and these all differ by a
delta-AIC of 5.
Would it be sensible, then, to choose the best-fit model as the one that has *both* minimal AIC
*and* minimal number of d.f.?
Are there any references for such an approach?
Many thanks for y
>I would like to calculate and plot a Weibull distribution (Weibull best-fit
>line, accuracy curves left and right beside the best-fit line) for a simple
>set
>of data points. I need the shape and the scale parameter for the best-fit and
>values like B5, B10, B50, B90, B95.
You can use the sur
> I will post data probably tonight, but here is my problem. I have
> preformed an MDS on a set of data. I have the scores of the four axes
> that
> are the optimal solution. I want to calculate the euclidean distance
> between time steps of the ordination.
See ?dist for a much faster sol
eric lee gmail.com> writes:
>
> Hi,
>
> I'd like to merge the following list and data frame by matching the first
> column in the data frame to the first number in each object of the list.
> I'd also like the merged object to be a list. Any suggestions? Thanks.
>
> eric
>
> a <- list(c(1,11
Actually, help.search("for") finds
Control(base) Control Flow
which is exactly where 'for' is documented. In general, if you want
the manual page of reserved words, then you'll have to quote them:
?"for"
Gabor
On Wed, Oct 1, 2008 at 3:39 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
Hi All,
I have a gTree (x1) produced by ggplot, with the following structure (output
from grid.ls, edited):
viewport[GRID.VP.15]
frame[plot-surrounds]
viewport[GRID.VP.16]
cellGrob[GRID.cellGrob.137]
rect[background]
upViewport[1]
viewport[GRID.VP.17]
cellGrob
One must write ?"for" presumably since for is a reserved word in R.
On Wed, Oct 1, 2008 at 9:39 AM, stephen sefick <[EMAIL PROTECTED]> wrote:
> ?for doesn't return anything help.search("for") doesn't return anything-
> Is the for loop so prevelant in computer programing that the
> documentation is
Use dimnames:
dimnames(Z) <- list(c("X1", "X2"), c("X1", "X2", "X3"))
On Wed, Oct 1, 2008 at 10:38 AM, cruz <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I have an object, Z:
>
>> typeof(Z)
> [1] "list"
>
> It looks like:
>
> [,1] [,2] [,3]
> [1,] 111
> [2,] 111
>
> How can it label
?for doesn't return anything help.search("for") doesn't return anything-
Is the for loop so prevelant in computer programing that the
documentation is implicit or is R paradigm to discourage the use of
the for loop.
I will post data probably tonight, but here is my problem. I have
preformed an MD
Hi,
I have an object, Z:
> typeof(Z)
[1] "list"
It looks like:
[,1] [,2] [,3]
[1,] 111
[2,] 111
How can it label the Rows and Columns with Text? so that it looks like:
Y1 Y2 Y3
X1 111
X2 111
Thanks,
cruz
___
jpmorgan.com> writes:
>
> Dear R Users,
>
> I've been hit with a cryptic error message:
>
> "Error in embed(y, lag) : wrong embedding dimension"
>
> My configuration is:
> - R 2.7.2
> - Windows XP Sp2
>
> Google returns nothing for this. Could someone suggest what this might
> mean ?
>
Plat, H.J. uva.nl> writes:
> There is lots of information about maximum likelihood estimation in R.
> However, I didn't came across anything about maximum likelihood
> with constraints.
> For example, estimation of parameters k(1) to k(20) with
> maximum likelihood, where sum(k(i)) = 0.
If the
Philipp Pagel wzw.tum.de> writes:
> If I only want axis labels on the left and bottom I can set
> alternating=F like in the following example:
>
> Depth <- equal.count(quakes$depth, number=8, overlap=.1)
> xyplot(lat ~ long | Depth, data = quakes, scales=list(alternating=F)
>
> I also would l
One approach would be to smooth the data using `gam' from package `mgcv'.
`vis.gam' will contour the resulting smoothed fit if you select
`plot.type="contour"'. If the results look ok but you want finer control of
what the plot looks like then `predict.gam' can be used to predict from the
fitte
charter.net> writes:
>
> So iis what you are saying n plotting the list of compliex numbers 'R'
recognizes the non-symmery and plots
> it differently (Re vs. Im) rather than versus Time?
Mmm... don't fully understand what you mean, mainly because your program example
did not run and I had do s
On Wed, 2008-10-01 at 08:42 +, Solene Goy wrote:
> Hello,
> I am new to R and I would like to get the day of the year from
> numerous data in the following form: %a %d/%b/%Y %H:%M:%S (for
> example from Tu 10/Jul/2007 21:59:13 I would like to get 191)
>
> Whatever I try, I get NAs.
Tu isn't
Michela,
Thanks for the email. I'm still not completely clear what parts of H you need.
Are you just wanting to extract a model matrix for a smooth (i.e. the matrix
of basis functions evaluated at the covariate values), or do you need other
things, like the smoothing penalty matrix, as well?
Hi R-experts,
There is lots of information about maximum likelihood estimation in R.
However, I didn't came across anything about maximum likelihood with
constraints.
For example, estimation of parameters k(1) to k(20) with maximum likelihood,
where sum(k(i)) = 0.
Is there any standard functi
Dear R Users,
I've been hit with a cryptic error message:
"Error in embed(y, lag) : wrong embedding dimension"
My configuration is:
- R 2.7.2
- Windows XP Sp2
Google returns nothing for this. Could someone suggest what this might
mean ?
Thanks in advance,
Tolga
Generally, this communicatio
If I only want axis labels on the left and bottom I can set
alternating=F like in the following example:
Depth <- equal.count(quakes$depth, number=8, overlap=.1)
xyplot(lat ~ long | Depth, data = quakes, scales=list(alternating=F)
I also would like to get rid of the unlabeled tick marks (top a
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