David Riano ucdavis.edu> writes:
> I have two time series. Both measure the same thing and I would like
> to determine which one is noisier.
> Would it be a good measure of the noise in each time series the
> absolute lag difference?
> Is this a good measure? Any other measure I could use?
You
Hi,
I attached the files I'm using, it may help.
I'm using Windows XP
> sessionInfo()
R version 2.6.0 (2007-10-03)
i386-pc-mingw32
locale: LC_COLLATE=English_New Zealand.1252;LC_CTYPE=English_New
Zealand.1252;LC_MONETARY=English_New
Zealand.1252;LC_NUMERIC=C;LC_TIME=English_New Zealand.1252
attac
I'm using:
"solve(a, b, tol, LINPACK = FALSE, ...)"
Therefore,tol ==.Machine$double.eps
>.Machine$double.eps
[1] 2.220446e-16
It explains why 'solve' works for 'tcp1' but not for 'tcp2':
eigen(tcp1)$values
[1] 5.208856e+09 2.585816e+08 -3.660671e-06
-3.660671e-06/5.208856e+09
[1] -7.027783e
srinivasa raghavan gmail.com> writes:
> I am interested to generate dashboard and reports based on data from MS
> Access. These reports need to be posted on a weekly basis to the web. The
> reporting interface should provide facilities for "what if" scenarios.
>
> Is it possible to interface
Hi folks!
I run the following code to get a CI for a Poisson with lambda=12.73
library(MASS)
set.seed(125)
x <- rpois(100,12.73)
lambda_hat<-fitdistr(x, dpois, list(lambda=12))$estimate
#Confidence Intervals - Normal Approx.
alpha<-c(.05,.025,.01)
for(n in 1:length(alpha)) {
LowerCI<-mean(x
Hi,
I turn to the help list after spending hours on the package help, looking
online, asking colleagues and trying various versions on R itself.
The work I need help on involves a confidentiality contract, so I am limited
in details on the specific model I used. However, my problem is really
simp
Hi!
I have two time series. Both measure the same thing and I would like
to determine which one is noisier.
Would it be a good measure of the noise in each time series the
absolute lag difference?
Is this a good measure? Any other measure I could use?
Thanks for help :)
David Riano
Center for
Dear R-helpers:
I am an entry-level R user and have a question related to overlaying a
barchart and and a xyplot using latticeExtra.
My problem is that when I overlay them I fail to align their x-axes.
I show my problem below through an example.
#the example data frame is provided below
ve
I am seeing different behavior than don Reis on my installation as well:
mtx is the same as his WBtWB
> mtx <- matrix(c(1916061939, 2281366606, 678696067, 2281366606,
3098975504, 1092911209, 678696067, 1092911209, 452399849), ncol=3)
>
> mtx
[,1] [,2] [,3]
[1,] 1916061
This evening I ran into the problem Chuck Clifton referred to in message 86
of Volume 71 Issue 9 of this list. That is, objects created by lmer change
after calling pvals.fnc on that lmer object when using lme4
version 0.999375-16 and 0.999375-28. This is somewhat troublesome. The bug
tracker on
Marlon,
Are you using a current version of R? sessionInfo()?
It would help if you had something we could _fully_ reproduce.
Taking the _printed_ values you have below (WBtWB) and adding or
subtracting what you have printed as the difference of the two visually
equal matrices ( say Delta ) ,
Thanks Nath,
I get the following results:
system.time(tcp1 <- tcrossprod(WB))
user system elapsed
0 0 0
> system.time(tcp2 <- crossprod(t(WB)))
user system elapsed
0 0 0
> all.equal(tcp1, tcp2,check.attributes=FALSE)
[1] TRUE
But, when I try the in
on 01/14/2009 03:32 PM John Lande wrote:
> dear all,
>
> I want to plot a kaplan Meier plot with the following functions, but I fail
> to produce the plot I want:
>
> library(survival)
> tim <- (1:50)/6
> ind <- runif(50)
> ind[ind > 0.5] <- 1; ind[ind < 0.5] <- 0;
> MS <- runif(50)
> pred <- vec
Dear R-Users
if I have a data frame (or zoo data) as follows:
Run Bike Walk Drive
1 2 7 5
5 2 4 2
8 3 2 1
How can I re-order it so that it reads:
Drive Run Bike Walk
5 1 2 7
2 5 2 4
1 8 3 2
Also notice that the q in Aquarium is hidden. Is there a way to make
this not happen?
thanks
Stephen Sefick
On Wed, Jan 14, 2009 at 10:18 PM, stephen sefick wrote:
> #I am putting a test together for an introductory biology class and I
> would like to put different cross hatching inside of each
#I am putting a test together for an introductory biology class and I
would like to put different cross hatching inside of each bar for the
bar plot below
color <- c("Brightly Colored", "Dull", "Neither")
lizards <- c(277, 70, 3)
liz.col <- data.frame(color, lizards)
qplot(color, lizards, data=liz
Thanks for that :)
Gabor Grothendieck wrote:
>
> Its hard coded in but you can circumvent it if you really want like this:
>
> library(zoo)
>
> # make a copy of plot.zoo
> plot.zoo <- zoo:::plot.zoo
> environment(plot.zoo) <- .GlobalEnv
> box <- list
>
> z <- zoo(1:5)
> plot(z, bty = "n")
>
Yes, computing WB.%*%t(WB) may be the problem, by either method.
if the goal is to compute the inverse of WB%*%t(WB), you should
consider computing the singular value or QR decomposition for the
matrix WB.
If WB = Q%*%R, where Q is orthogonal, then WB %*% t(WB) =R %*%t(R),
and the inverse of
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
I've recently been working with cross products etc.
You should try the following function:
tcp1 <- tcrossprod(WB)
or
tcp2 <- crossprod(t(WB))
Both should be the same (check for equality accounting for some floating point
imprecision):
all.equal(tcp1
Dear All,
I'm preparing a simple algorithm for matrix multiplication for a
specific purpose, but I'm getting some unexpected results.
If anyone could give a clue, I would really appreciate.
Basically what I want to do is a simple matrix multiplication:
(WB) %*% t(WB).
The WB is in the disk so I com
dear all,
I want to plot a kaplan Meier plot with the following functions, but I fail
to produce the plot I want:
library(survival)
tim <- (1:50)/6
ind <- runif(50)
ind[ind > 0.5] <- 1; ind[ind < 0.5] <- 0;
MS <- runif(50)
pred <- vector()
pred[MS < 0.3] <- 0; pred[MS >= 0.3] <- 1
df <- as.data.f
I had a similar misunderstanding of using partial in sort() a while
back when I was trying to sort by columns, a la Excel. In this case,
I ended up using the order() function, eg:
require(stats)
swiss[order(rownames(swiss)), ] #sort by location
swiss[order(swiss$fertility), ] #sort by fertility
dear all,
I want to plot a kaplan Meier plot with the following functions, but I fail
to produce the plot I want:
library(survival)
tim <- (1:50)/6
ind <- runif(50)
ind[ind > 0.5] <- 1; ind[ind < 0.5] <- 0;
MS <- runif(50)
pred <- vector()
pred[MS < 0.3] <- 0; pred[MS >= 0.3] <- 1
df <- as.data.f
or this
x[,!(colSums(abs(x)) == 0)]
On Jan 15, 10:00 am, Marc Schwartz wrote:
> Careful:
>
> x <- matrix(c(1, 5, 3, 2, 1, 4, -1, 0, 1),
> ncol = 3, nrow = 3)
>
> > x
>
> [,1] [,2] [,3]
> [1,] 1 2 -1
> [2,] 5 1 0
> [3,] 3 4 1
>
> > x[, colSums(x) != 0]
>
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
I was wondering if anyone could point me in the right direction for reading up
on writing tests in
R. I'm writing some functions for inclusion into a package and would like to
test them to ensure
they're doing what I expect them to do.
Are these app
Its hard coded in but you can circumvent it if you really want like this:
library(zoo)
# make a copy of plot.zoo
plot.zoo <- zoo:::plot.zoo
environment(plot.zoo) <- .GlobalEnv
box <- list
z <- zoo(1:5)
plot(z, bty = "n")
rm(box)
On Wed, Jan 14, 2009 at 7:16 PM, jimdare wrote:
>
> Hi,
>
> I a
Hi,
I apologise for the stupid question but how to you get rid of the box around
a plot in the package zoo? I can't seem to find an equivalent for bty="l"
i.e. just x and y axis.
Cheers
James
--
View this message in context:
http://www.nabble.com/Zoo-Plot-Can%27t-Get-Rid-of-Box-tp21468452p214
I uninstalled and then re-installed R using all the defaults and that seemed
to do the trick. I was able to install the gcl package from the Packages
menu and run the demo.
--
View this message in context:
http://www.nabble.com/Can%27t-find-files-after-install-package-%28Windows%29-tp2146360
__
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.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Duncan Murdoch wrote:
rkevinbur...@charter.net wrote:
This is definitely a newbie question but from the documentation I have not been
able to figure out what the partial sort option on the sort method does. I have
read and re-read the documentation and looked at the examples but for some
r
rkevinbur...@charter.net wrote:
This is definitely a newbie question but from the documentation I have not been
able to figure out what the partial sort option on the sort method does. I have
read and re-read the documentation and looked at the examples but for some
reason it doesn't register.
bcbob43 wrote:
Phil Spector wrote:
Unless you issue the statement
library(gcl)
the gcl package will not be available in the current session.
Using demo does not load the package; only library() does.
I should have mentioned that I did try that and got the following result:
lib
1) Unless you know what you are doing, you don't want to use it since
the speed advantage is small at best.
2) Suppose you want the median of 2n+1 values. Then all you need is a
value with n others to the left and n to the right. That is a
partially sorted vector. Example
set.seed(123)
x
Specifying a partial sort allows you to specify a re-arrangement of the data
into groups where all the values in any group are guaranteed to be at least as
large as any value in the previous group. Within the groups themselves,
though, the values are not sorted.
Here is an example:
> x <- rou
On Jan 14, 2009, at 4:27 PM, Andreas Klein wrote:
Ok... I set up the problem by some code as requested:
Example:
x <- rnorm(100)
mean_x <- mean(x)
mean_boot <- numeric(1000)
for (i in 1:1000) {
mean_boot[i] <- mean(sample(x,100,replace=TRUE))
}
How can I compute the p-Value out of mea
Careful:
x <- matrix(c(1, 5, 3, 2, 1, 4, -1, 0, 1),
ncol = 3, nrow = 3)
> x
[,1] [,2] [,3]
[1,]12 -1
[2,]510
[3,]341
> x[, colSums(x) != 0]
[,1] [,2]
[1,]12
[2,]51
[3,]34
Not quite the result wanted... :-)
Try th
This is definitely a newbie question but from the documentation I have not been
able to figure out what the partial sort option on the sort method does. I have
read and re-read the documentation and looked at the examples but for some
reason it doesn't register. Would someone attempt to explain
Dear rafamoral,
Try this:
x[is.na(x)]<-0
HTH,
Jorge
On Wed, Jan 14, 2009 at 5:32 PM, rafamoral wrote:
>
> I have a dataset which contains some missing values, and I need to replace
> them with zeros. I tried using the following:
>
> x <- matrix(data=rep(c(1,2,3,NA),6), ncol=6, nrow=6)
>
> y <-
Hi Ken,
If your lists are not too large something such as this can work:
> matrixFromList <- function(listX) t(sapply(listX, function(x, n) c(x, rep(NA,
> n))[1:n], n = max(sapply(listX, length
> matrixFromList(list(c(3,2,3),c(6,5)))
[,1] [,2] [,3]
[1,]323
[2,]65
Hello rafamoral,
Try this:
ifelse(is.na(x),0,x)
On Wed, Jan 14, 2009 at 8:32 PM, rafamoral wrote:
>
> I have a dataset which contains some missing values, and I need to replace
> them with zeros. I tried using the following:
>
> x <- matrix(data=rep(c(1,2,3,NA),6), ncol=6, nrow=6)
>
> y <- matr
I have a dataset which contains some missing values, and I need to replace
them with zeros. I tried using the following:
x <- matrix(data=rep(c(1,2,3,NA),6), ncol=6, nrow=6)
y <- matrix(data=0, ncol=ncol(x), nrow=nrow(x))
for(i in 1:nrow(x)) {
for(j in 1:ncol(x)) {
y[i,j] <- ifelse(x[i,j]==NA
Sorry for the double post, but this is probably faster:
x[, colSums(x) != 0]
On Wed, Jan 14, 2009 at 8:22 PM, Gustavo Carvalho
wrote:
> You can also try this:
>
> x[,-(which(colSums(x) == 0))]
>
> Cheers,
>
> Gustavo.
>
> On Wed, Jan 14, 2009 at 8:01 PM, Anthony Dick wrote:
>> Hello-
>>
>> I wo
Phil Spector wrote:
>
> Unless you issue the statement
>
> library(gcl)
>
> the gcl package will not be available in the current session.
> Using demo does not load the package; only library() does.
>
>
I should have mentioned that I did try that and got the following result:
> library(gcl
on 01/14/2009 03:50 PM Lo, Ken wrote:
> Dear list,
>
> I have a list of number sequences. Each number sequence has different
> numbers of elements. Is there a quick way (other than to iterate
> through the entire list) way to coerce list to matrix with NAs filling
> in the short sequences?
>
>
You can also try this:
x[,-(which(colSums(x) == 0))]
Cheers,
Gustavo.
On Wed, Jan 14, 2009 at 8:01 PM, Anthony Dick wrote:
> Hello-
>
> I would like to remove the columns of a matrix that contain all zeros. For
> example, from
> x<-matrix(c(1,5,3,2,1,4,0,0,0), ncol=3,nrow=3)
>
> I would like t
Hi Anthony,
Try this:
x[,apply(x,2,function(x) !all(x==0))]
HTH,
Jorge
On Wed, Jan 14, 2009 at 5:01 PM, Anthony Dick wrote:
> Hello-
>
> I would like to remove the columns of a matrix that contain all zeros. For
> example, from
> x<-matrix(c(1,5,3,2,1,4,0,0,0), ncol=3,nrow=3)
>
> I would l
Unless you issue the statement
library(gcl)
the gcl package will not be available in the current session.
Using demo does not load the package; only library() does.
- Phil Spector
Statistical Computing Facility
Hello-
I would like to remove the columns of a matrix that contain all zeros.
For example, from
x<-matrix(c(1,5,3,2,1,4,0,0,0), ncol=3,nrow=3)
I would like to remove the third column. However, because this is in a
loop I need a way to first determine which columns are all zeros, and
only the
Dear list,
I have a list of number sequences. Each number sequence has different
numbers of elements. Is there a quick way (other than to iterate
through the entire list) way to coerce list to matrix with NAs filling
in the short sequences?
An example of what I mean is this:
A <- list(c(3,2,3)
Thank you very much -- this was very helpful for differentiating among
the aggregating methods!
Matt
On Wed, Jan 14, 2009 at 3:42 PM, Marc Schwartz
wrote:
> on 01/14/2009 02:51 PM Matthew Pettis wrote:
>> I have a specific question and a general question.
>>
>> Specific Question: I want to do an
on 01/14/2009 02:51 PM Matthew Pettis wrote:
> I have a specific question and a general question.
>
> Specific Question: I want to do an analysis on a data frame by 2 or more
> class variables (i.e., use 2 or more columns in a dataframe to do
> statistical classing). Coming from SAS, I'm used to
Gabor Grothendieck wrote:
>
> You should only have to do this from within R:
>
> install.packages("glc", dep = TRUE)
> library(gcl)
> demo("glcdemo", package = "glc")
>
> or in place of the first line you could use the Packages
> menu choosing Install packages...
>
Thanks for your help but I
Hello,
I am not a R coder but I am looking for a R coder to write a small
program (I think it should only take a few hours at most) that will be
used to calculate statistics in a report for the World Health
Organization.
I will pay $250 for fully functional working program. $100 Bonus if
finished
On Wed, Jan 14, 2009 at 1:53 PM, glenn wrote:
> Dear All;
>
> Is it possible to create a list of lists (I am sure it is) along these
> lines;
>
> I have a dataframe data02 that holds a lot of information, and the first
> column is ³date²
>
> I have a list of dates in;
>
> data03<-c(date1,.,dat
> General Question: I assume the answer to the specific question is dependent
> on my understanding list objects and accessing their attributes. Can anyone
> point me to a good, throrough treatment of these R topics? Specifically how
> to read and interpret the output of the str(), and attributes
Ok... I set up the problem by some code as requested:
Example:
x <- rnorm(100)
mean_x <- mean(x)
mean_boot <- numeric(1000)
for (i in 1:1000) {
mean_boot[i] <- mean(sample(x,100,replace=TRUE))
}
How can I compute the p-Value out of mean_boot for the following tests:
1. H0: mean_x = 0 vs
On Jan 14, 2009, at 4:14 PM, David Winsemius wrote:
On Jan 14, 2009, at 3:56 PM, Matthew Pettis wrote:
I have a specific question and a general question.
Specific Question: I want to do an analysis on a data frame by 2 or
more class variables (i.e., use 2 or more columns in a dataframe to
You should only have to do this from within R:
install.packages("glc", dep = TRUE)
library(gcl)
demo("glcdemo", package = "glc")
or in place of the first line you could use the Packages
menu choosing Install packages...
On Wed, Jan 14, 2009 at 3:51 PM, bcbob43 wrote:
>
>
> Tom Backer Johnsen
On Jan 14, 2009, at 3:56 PM, Matthew Pettis wrote:
I have a specific question and a general question.
Specific Question: I want to do an analysis on a data frame by 2 or
more class variables (i.e., use 2 or more columns in a dataframe to do
statistical classing). Coming from SAS, I'm used t
I have a specific question and a general question.
Specific Question: I want to do an analysis on a data frame by 2 or
more class variables (i.e., use 2 or more columns in a dataframe to do
statistical classing). Coming from SAS, I'm used to being able to
take a data set and have the output of
See if one of %in% or match gets your further.
> 1:10 %in% c(1,3,5,9)
[1] TRUE FALSE TRUE FALSE TRUE FALSE FALSE FALSE TRUE FALSE
> match(c(1,3,5,9), 1:10)
[1] 1 3 5 9
> match(c(1,3,5,9), 10:1)
[1] 10 8 6 2
date03 as offered was not a list, but a vector.
date04 <- date02[which(date02$da
Tom Backer Johnsen wrote:
>
> Did you include the statement
>
> library (gcl)
>
> before the call on source () ?
>
> Tom
>
>
Yes. Here is the code I am trying to run. It was provided by the authors of
the gcl package
# library(gcl)
#gcldemo <- function() {
df <-
structure(list(V1 =
I have a specific question and a general question.
Specific Question: I want to do an analysis on a data frame by 2 or more
class variables (i.e., use 2 or more columns in a dataframe to do
statistical classing). Coming from SAS, I'm used to being able to take a
data set and have the output of th
Did you include the statement
library (gcl)
before the call on source () ?
Tom
bcbob43 wrote:
--I was not registered when I first sent this. I registered and it looks
like I have to resend so sorry if this gets sent twice ...
I am a total newbie at R but experienced with computers. If this
Greg Snow imail.org> writes:
>
> My preferred method for this type of thing is to use simulation.
> You have already done the hard parts in
> figuring out what your data is going to look like and how you plan
> to analyze it. Now just write a function
> that will simulate data according to yo
--I was not registered when I first sent this. I registered and it looks
like I have to resend so sorry if this gets sent twice ...
I am a total newbie at R but experienced with computers. If this is not the
right forum for this question, please let me know one that is. I searched in
the R manua
Dear John and Stas,
Thanks so much for your help.
John, I did the correlation on the complete dataset (no missing values). I
tried what you suggested and you were right: hetcor with pd=FALSE gives me
the same result as polychor.
Anyway, thanks to the answers I got in the forum, I understand I sho
on 01/14/2009 11:53 AM Kerpel, John wrote:
> Hi folks! I'm trying to get a histogram legend to give me a filled box
> and a line. The problem is I keep getting both filled boxes and a line.
> How can I get rid of the second box from the code below?
>
>
>
> x<-rnorm(1000,mean=0,sd=1)
>
>
Dear All;
Is it possible to create a list of lists (I am sure it is) along these
lines;
I have a dataframe data02 that holds a lot of information, and the first
column is ³date²
I have a list of dates in;
data03<-c(date1,.,daten)
And would like to create a list;
data04 <- subset(data02, d
Assuming that your dataframe is DF and those are names of columns then
?subset
> DF <- data.frame(posit=runif(20, max=20), tvalues = runif(20, max=
10))
>
> subset(DF, subset= posit > 10 & tvalues > 3.5 )
posit tvalues
4 14.66464 9.527535
6 14.71280 7.231015
9 14.76594 7.626381
10 1
Hi,
new.Match <- Match[10Match$tvalues>3.5,]
Everything before the comma selects rows. You could select columns by
using conditions after the comma.
HTH,
Stephan
emj83 schrieb:
I have a dataframe called Match with two columns: position and tvalue.
I would like to select the parts of the d
If we check the data, we see a mistake in the entry.
> tbl.8.3
mara cig alc count
1 Yes Yes Yes 911
2 No No No 538
3 Yes Yes Yes44
4 No No No 456
5 Yes Yes Yes 3
6 No No No43
7 Yes Yes Yes 2
8 No No No 279
Try:
table8.3 <- read.table(textConnection("
Folks:
NON-R RELATED NATURE article on fundamental statistical inference (and, I
think, more fundamentally, philosophy of science: what constitutes
scientific "validity" ?) issues that I thought might be of general interest
to this list.
http://www.nature.com/news/2009/090113/full/457245a.html
F
Dear all,
I'm using R 2.8.1 under Vista.
I programmed a Simulation with the code enclosed at the end of the eMail.
After the simulation I want to analyse the columns of the single
simulation-runs, i.e. e.g. Simulation[[1]][,1] sth. like that but I
cannot address these columns...
Can anybody
Problem resolved. This error occurred because of a mixture of hardy
and intrepid sources in my sources.list .
Thanks to Phil Spector, Dirk Eddelbuettel and others for their
input.
[[alternative HTML version deleted]]
__
R-help@r-project.org m
I have a dataframe called Match with two columns: position and tvalue.
I would like to select the parts of the dataframe that have a position> 10
but <50 and tvalues >3.5 as a new stand alone dataframe.
Could anyone help me with how to do this?
Thanks Emma
--
View this message in context:
htt
Hi folks! I'm trying to get a histogram legend to give me a filled box
and a line. The problem is I keep getting both filled boxes and a line.
How can I get rid of the second box from the code below?
x<-rnorm(1000,mean=0,sd=1)
hist(x, breaks = 50, main="Histogram of x",freq=FALSE,
xla
FWIW,
Having done some quick checks, the appropriate incantation is
getURL('ftp://ftp.wcc.nrcs.usda.gov/data/snow/snow_course/table/history/idaho/13e19.txt',
ftp.use.epsv = FALSE)
D.
Duncan Temple Lang wrote:
Hi Zack,
You might explore some of the options of libcurl
via the RC
My preferred method for this type of thing is to use simulation. You have
already done the hard parts in figuring out what your data is going to look
like and how you plan to analyze it. Now just write a function that will
simulate data according to your pattern and with the difference(s) that
Could be ...
legend("topright",legend=c("Histogram","Kernel Density
Estimate"),lty=c(NA,1), lwd=c(NA,2), pch = c(15, NA), col =
c("lightblue", "black"), merge=TRUE,inset=.01,cex=.8,adj=0)
A.
John Kerpel wrote:
Hi folks! I'm trying to get a histogram legend to give me a filled box and
a line
On 14 January 2009 at 12:04, Rob James wrote:
| BODY { font-family:Arial, Helvetica, sans-serif;font-size:12px; }I
| can't seem to perform an install from CRAN (stat.sfu.ca, specifically)
| for Ubuntu 64 8.1 (Intrepid) and 2.8.2 of R. Here is the log:
[ There is no R 2.8.2 (yet, and maybe ever
On Wed, 2009-01-14 at 16:38 +, Prof Brian Ripley wrote:
> You need to ask the author (as the posting guide asked you to).
>
> I'm tempted to not help further given the (almost complete) lack of
> cooperation of that author with R's recommendations, but note
> 'ordinal..so' in your log and l
BODY { font-family:Arial, Helvetica, sans-serif;font-size:12px; }I
can't seem to perform an install from CRAN (stat.sfu.ca, specifically)
for Ubuntu 64 8.1 (Intrepid) and 2.8.2 of R. Here is the log:
Looks like some dependencies cannot be satisfied.
Suggestions? Thanks in advance.
sudo apt-g
Hi folks! I'm trying to get a histogram legend to give me a filled box and
a line. The problem is I keep getting both filled boxes and a line. How
can I get rid of the second box from the code below?
x<-rnorm(1000,mean=0,sd=1)
hist(x, breaks = 50, main="Histogram of x",freq=FALSE,
xlab=" x"
PCA is only defined up to a multiplicative constant and different programs use
different constants. Without code or output we cannot tell if this is the
case, or if something more is going on. Try rescaling one of the answers to
see if you can get the other answer, if so, then it is just a dif
The S+ basename() function has an argument called suffix and
it will remove the suffix from the result. This was based on
the Unix basename command, but I missed the special case in the
Unix basename that doesn't remove the suffix if the removal
would result in an empty string. The suffix must in
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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.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
On Wed, 14 Jan 2009, Premal Shah wrote:
Hi all,
I use a mac and was trying to reset my quartz options. However, every time I
restart R, the options are gone and I have to type them again.
How are you setting them?
Any idea as to what's going wrong?
Try asking on R-sig-mac for Mac-specifi
Hi all,
I use a mac and was trying to reset my quartz options. However, every
time I restart R, the options are gone and I have to type them again.
Any idea as to what's going wrong?
Thanks,
Premal
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R-help@r-project.org mailing list
https://stat
You need to ask the author (as the posting guide asked you to).
I'm tempted to not help further given the (almost complete) lack of
cooperation of that author with R's recommendations, but note
'ordinal..so' in your log and look for the obvious fix in
src/Makefile.
On Wed, 14 Jan 2009, Rick
On Wed, 14 Jan 2009, Albrecht, Dr. Stefan (AIM SE) wrote:
Dear R-help list,
I am just trying to increase my memory for R, but for some reason I
cannot really increase the memory size as much as I would like. I
have 3 GB of RAM on a Windows XP (R 2.8.1)
memory.size()
[1] 879.4667
memory
Hi Zack,
You might explore some of the options of libcurl
via the RCurl package for R. Depending on the server,
you may want to send either EPSV or PASV commands at
some point in the dialog.
libcurl, and the functions in RCurl, give you a great
deal of control in cutomizing and controlling th
I'm trying to install the ordinal package
(http://popgen.unimaas.nl/~plindsey/rlibs.html).
I downloaded ordinal03.tgz and untarred it. rmutil was previously
installed (and appears to work ok.) Then I installed ordinal:
[r...@localhost ~]# R CMD INSTALL /home/chippy/Download/ordinal
* Installing t
Professor Ripley and Brian Rowe,
Thank you both for your helpful suggestions. I was unable to solve this
problem. Apparently there is a passive FTP server that is not supported by R.
I'll have to find another programming language, perhaps IDL, that can do this.
Manually retrieving the hundreds
I think we are at the stage where it is your responsibility to provide
some code to set up the problem.
--
David Winsemius
On Jan 14, 2009, at 9:23 AM, Andreas Klein wrote:
Hello.
What I wanted was:
I have a sample of 100 relizations of a random variable and I want a
p-Value for the hypot
Be careful there if the original data contains negative numbers.
As per ?floor:
takes a single numeric argument x and returns a numeric vector
containing the largest integers not greater than the corresponding
elements of x.
Thus:
x <- seq(-2, 2, 0.25)
> x
[1] -2.00 -1.75 -1.50 -1.25 -1.00 -0
Hi
r-help-boun...@r-project.org napsal dne 14.01.2009 14:34:14:
> Dear ALL
> suppose "x=7.5",and i need of only integer part of variable "x" that is
"7"
> only then what command i can use in R.
If you looked at help page for integer you would quickly find links to
trunc, round, floor, ceiling
On Wed, Jan 14, 2009 at 3:07 PM, hadley wickham wrote:
> On Wed, Jan 14, 2009 at 5:51 AM, Gustaf Rydevik
> wrote:
>> Hi all,
>>
>> for some reason I always get stuck spending hours when trying to use
>> reshape or the Reshape package. Heaven knows why.
>> My latest frustration (in 2.7.1, so ignor
Hi all,
I'm new (post #1!) and I hope you'll forgive me if I'm acting like an
idiot...
I have been asked for some power analyses for some mixed-effects models I'm
running using lmer. My studies nearly always contain mixes of
repeated-measures and between-subjects predictor variables.
As an exam
On Wed, 14 Jan 2009, Tony Breyal wrote:
Hi Xin,
I think you can use the read.xls() function from package
xlsReadWrite.
Assuming this is Windows
alternatively, i think the odbcConnectExcel() function in package
RODBC might work too (not sure about excel 2007 files, though there
might be
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