On Sun, 5 Jun 2011, Abhilash Balakrishnan wrote:
Dear Mr. Murdoch,
I find out that still do not understand why the following does not work:
curve(expression(x))
Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ
As here the input to curve is an expression, as docume
On Sun, 5 Jun 2011, Robert Baruch wrote:
I have libtiff installed from macports, and it is sitting there in
/opt/local/lib:
Well, you haven't told us which build of R you used, but it looks like
it is not the one from MacPorts (maybe the one from CRAN?).
/opt/local is not a standard location
On Mon, 6 Jun 2011, Richard Wang wrote:
Hi,
I would like to get suggestion about parallel computing package on a
multicore windows workstation. I tried doSMP, but it crashes R a lot. I am
wondering if "snow" and "snowfall" can be used on a single workstation (i,e,
not cluster). At suggestio
On 5 June 2011 21:07, Alexander Shenkin wrote:
> Hello Folks,
>
> As some of my old code broke when an updated package changed its
> interface, I started thinking about reproduction of analyses. It's not
> good enough to save our code - we have to save the package versions
> those analyses used a
You might find Tweedie distributions helpful. See packages tweedie and
statmod.
Cheers,
Simon.
On 06/06/11 14:12, siriustar wrote:
Hi, Dear R-help
I know there are some R package to deal with zero-inflated count data. But I
am now looking for R package to deal with zero-inflated continuous da
Hi, Dear R-help
I know there are some R package to deal with zero-inflated count data. But I
am now looking for R package to deal with zero-inflated continuous data.
The response variable (Y) in my dataset contains a larger mount of zero and
the Non-zero response are quite right skewed. Now what i
Dear Mr. Murdoch,
I find out that still do not understand why the following does not work:
> curve(expression(x))
Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ
As here the input to curve is an expression, as documented in the help, and
the expression is simply x.
Dear Mr. Murdoch,
I now understand that curve does not know how to guess what x in curve(x)
means. As you say, it might guess that x means a function, and if there is
a function x, execute it. This is seen in the example:
> x <- function(x) { x }
> curve(x)
I think, if curve guesses I have a f
I have libtiff installed from macports, and it is sitting there in
/opt/local/lib:
$ ls -l /opt/local/lib/*tiff*
-rwxr-xr-x 2 root admin 796684 Jun 4 22:13 /opt/local/lib/libtiff.3.dylib
-rw-r--r-- 2 root admin 990296 Jun 4 22:13 /opt/local/lib/libtiff.a
lrwxr-xr-x 1 root admin 15
Thanks again. FWIW, I tried an even older R version (2.11) + BRugs 0.53 and
0.70 ... but got the same errors as with the iterations reported below. So
I'm giving up on trying to solve the issue.
My workaround is that I'm using the R2WinBUGS package instead. So far that
has worked well -- I'
I accidentally did not respond to the list.
-- Forwarded message --
From: Abhilash Balakrishnan
Date: Sun, Jun 5, 2011 at 8:44 PM
Subject: Re: [R] Question about curve function
To: Matt Shotwell
Dear Mr. Shotwell,
Thank you for the explanation. You seem to be right. In parti
Dear friends.
I am using the command plot (inside the anacor package) to construct one
simple correspondence analysis graphic.
The line that I am using is this:
plot(COL, plot.type = "jointplot", arrows = "FALSE", conf=NULL, xlim =
c(-2.5,1.5), ylim = c(-1.2,3), col.r="BLUE", col.c="BLACK")
An err
Dear friends.
I am using the anacor command to do a simple correspondence analysis.
I got correctly the eingenvalues, the Chi-Square decomposition and the
graphics, but I do not know what is the command to "call" the row and column
coordinates?
Thanks in advance
Claudia
--
Claudia Isabel Rodrígue
Dear R-users
I'm trying to use lasso in lars package for subset regression, I have a
large matrix of size 1000x100 and my aim is to select a subset k of the 100
variables.
Is there any way in lars to fix the number k (i.e. to select the best 10
variables)
library(lars)
aa=lars(X,Y,type="lasso"
Hi,
I would like to get suggestion about parallel computing package on a
multicore windows workstation. I tried doSMP, but it crashes R a lot. I am
wondering if "snow" and "snowfall" can be used on a single workstation (i,e,
not cluster). At suggestion would be appreciated,
Best,
Richard
On 06/06/11 03:01, kfl wrote:
Thank you for your reply.
However, I don't undestand what your are telling me.
I would be pleased if you could give me the synstax in the following
examples:
#Exampel page side 372
trt<-gl(3,4,12,labels=c("T1","T2","T3"))
blk<-gl(4,1,12,labels=c("B1","B2","B3","B4
On 6 June 2011 06:45, Gabor Grothendieck wrote:
> On Sun, Jun 5, 2011 at 3:18 PM, Mark Ebbert wrote:
>> Dear R gurus,
>>
>> Based on my searches I think I know the answer to this question, but I'd
>> like to pose it to you. Is there an easy way to plot the confidence interval
>> around a loess
Hmm, that is a bit tricky. The conversion from a table to a data
frame uses the dimension names, which are always character. To bypass
this, you would need to save the dimension names, convert the ones you
want numeric to numeric (I am assuming everything except Conc, so the
indices would be c(1,
On 11-06-05 4:07 PM, Alexander Shenkin wrote:
Hello Folks,
As some of my old code broke when an updated package changed its
interface, I started thinking about reproduction of analyses. It's not
good enough to save our code - we have to save the package versions
those analyses used as well as t
On Sun, Jun 05, 2011 at 03:07:44PM -0500, Alexander Shenkin wrote:
> Hello Folks,
>
> As some of my old code broke when an updated package changed its
> interface, I started thinking about reproduction of analyses. It's not
> good enough to save our code - we have to save the package versions
> t
On Sun, Jun 5, 2011 at 3:18 PM, Mark Ebbert wrote:
> Dear R gurus,
>
> Based on my searches I think I know the answer to this question, but I'd like
> to pose it to you. Is there an easy way to plot the confidence interval
> around a loess line using lattice? The only thing I've found is ggplot,
On 11-06-05 1:07 PM, Abhilash Balakrishnan wrote:
Dear Sirs,
I am a new user of the R package. When I try to use the curve function it
confuses me.
curve(x^2)
Works fine.
curve(x)
Makes a complaint I don't understand. Why is x^2 valid and x is not?
curve() is a convenience function, an
Dear R gurus,
Based on my searches I think I know the answer to this question, but I'd like
to pose it to you. Is there an easy way to plot the confidence interval around
a loess line using lattice? The only thing I've found is ggplot, but I prefer
to stick with lattice out of preference.
Than
Hello all,
Let's say I have a character string
"Race-ethnicity-coding information"
I want to extract all text before the multiple dashes, including the word
"ethnicity."
I wrote a handy function to extract the first matched text:
grepcut <- function(pattern,x){
start.and.length <- regexpr(p
Hello Folks,
As some of my old code broke when an updated package changed its
interface, I started thinking about reproduction of analyses. It's not
good enough to save our code - we have to save the package versions
those analyses used as well as the R-core. I saw a couple references to
"reprod
Hi Abhilash,
>From ?example, under "arguments":
local: logical: if ‘TRUE’ evaluate locally, if ‘FALSE’ evaluate in
the workspace.
So all you need to do is:
> x <- 0
> example(mean, local=TRUE)
mean> x <- c(0:10, 50)
mean> xm <- mean(x)
mean> c(xm, mean(x, trim = 0.10))
[1] 8.75 5.50
mean> m
I think there is trouble because expr in curve(expr) may be the name of
a function, and it's ambiguous whether 'x' should be interpreted as a
mathematical expression involving x, or the name of a function. Here are
some examples that work:
curve(I(x))
curve(1*x)
On Sun, 2011-06-05 at 12:07 -0500,
Dear Sirs,
I am a new user of the R package. When I try to use the curve function it
confuses me.
> curve(x^2)
Works fine.
> curve(x)
Makes a complaint I don't understand. Why is x^2 valid and x is not?
I check the documentation of curve, and it says the first argument must be
an expression c
Dear Sirs,
I am exploring the R package and its documentation. I find there is the
function example which runs examples from documentation pages. What
confuses me is that running example interferes with the variables I have in
my workspace.
> x <- 0
> example(mean)
> x
Now x is a vector of some
Thank you for your reply.
However, I don't undestand what your are telling me.
I would be pleased if you could give me the synstax in the following
examples:
#Exampel page side 372
trt<-gl(3,4,12,labels=c("T1","T2","T3"))
blk<-gl(4,1,12,labels=c("B1","B2","B3","B4"))
res<-c(13,7,9,3,6,6,3,1,11
Thanks Jeff and Spencer, I will probably set the time zone for my session, but
I had forgotten the possibility of setting the time zone attribute of a POSIXct
object, which would have solved my problem also.
Denis
Le 2011-06-05 à 11:14, Spencer Graves a écrit :
> On 6/5/2011 9:30 AM, Jeff Newmi
On 6/5/2011 9:30 AM, Jeff Newmiller wrote:
Sys.setenv(TZ="Etc/GMT+5")
Or:
x <- as.POSIXct(as.Date('2011-01-15'))
attr(x, 'tzone') <- "Etc/GMT+5"
x
This version works without Sys.setenv, which may not work on some
platforms. Unfortunately, I believe there are some copy operations that
On 05-Jun-11 13:36:18, Lara Poplarski wrote:
> Dear All,
> Could someone please suggest how to find the Kronecker sum
> of two 2x2 matrices,
> i.e. given two matrices:
>
> -A A
> a -a
>
> and
>
> -B B
> b -b
>
> I need:
>
> -A-BA B 0
> a -a-B 0 B
> b
Sys.setenv(TZ="Etc/GMT+5")
Make the timezone you prefer the default for that R session.
FWIW: EST may or may not exist as a valid timezone on your system, but it is an
ambiguous notation anyway.
---
Jeff Newmiller The .
Hi,
For a project I try to keep everything in normal time, not daylight saving
time, to prevent problem when instruments collected data during the nights when
we go from DST to normal time.
But sometimes R tricks me and I do not know how to prevent it.
This is one example:
lights_on = as.POSI
On 11-06-05 9:36 AM, Lara Poplarski wrote:
Dear All,
Could someone please suggest how to find the Kronecker sum of two 2x2
matrices,
i.e. given two matrices:
-A A
a -a
and
-B B
b -b
I need:
-A-BA B 0
a -a-B 0 B
b0 -A-b A
0b
Hi Duncan,
In this case they all had length 1, but I'll be careful at other occasions.
Denis
Le 2011-06-05 à 09:26, Duncan Murdoch a écrit :
> On 11-06-05 8:49 AM, Denis Chabot wrote:
>> Thanks Duncan, I'll go back to if and else!
>
> Be careful, it might not give you the same answer.
>
> I'd
Dear All,
Could someone please suggest how to find the Kronecker sum of two 2x2
matrices,
i.e. given two matrices:
-A A
a -a
and
-B B
b -b
I need:
-A-BA B 0
a -a-B 0 B
b0 -A-b A
0b a-a-b
Many thanks,
Lara
[[alternat
On 11-06-05 8:49 AM, Denis Chabot wrote:
Thanks Duncan, I'll go back to if and else!
Be careful, it might not give you the same answer.
I'd use this variation on the advice from ?ifelse:
new.date <- my.date + x
new.date[is.na(my.date)] <- min(default.date) + x
The thing to watch out for in t
I did not know this function, thanks a lot Gabor.
Denis
Le 2011-06-05 à 08:48, Gabor Grothendieck a écrit :
> On Sun, Jun 5, 2011 at 8:23 AM, Denis Chabot wrote:
>> Hi,
>>
>> I was "losing" my dates in a script and upon inspection, found that my
>> recent switch from separate "if" and "else" t
Thanks Duncan, I'll go back to if and else!
Denis
Le 2011-06-05 à 08:39, Duncan Murdoch a écrit :
> On 11-06-05 8:23 AM, Denis Chabot wrote:
>> Hi,
>>
>> I was "losing" my dates in a script and upon inspection, found that my
>> recent switch from separate "if" and "else" to "ifelse" was the cau
On Sun, Jun 5, 2011 at 8:23 AM, Denis Chabot wrote:
> Hi,
>
> I was "losing" my dates in a script and upon inspection, found that my recent
> switch from separate "if" and "else" to "ifelse" was the cause. But why?
>
> my.date = as.POSIXct("2011-06-04 08:00:00")
> default.date = seq(as.POSIXct("2
On 11-06-05 8:23 AM, Denis Chabot wrote:
Hi,
I was "losing" my dates in a script and upon inspection, found that my recent switch from separate
"if" and "else" to "ifelse" was the cause. But why?
See ?ifelse. The class of the result is the same as the class of the
test, not the classes of t
Hi,
I was "losing" my dates in a script and upon inspection, found that my recent
switch from separate "if" and "else" to "ifelse" was the cause. But why?
my.date = as.POSIXct("2011-06-04 08:00:00")
default.date = seq(as.POSIXct("2011-01-01 08:00:00"), as.POSIXct("2011-09-01
08:00:00"), length=
It seems row.names only shows up separately when it differs from the actual
row numbers. I got them to go away by setting them as such, e.g.
row.names(uniquedf) <- seq(nrow(uniquedf))
--
View this message in context:
http://r.789695.n4.nabble.com/Remove-row-names-column-in-dataframe-tp856647p3574
To drop the burnin period
xx <- x1$x[-seq_len(burnin)]
And it sounds like you want
mean(xx > -1 & xx < 2]
but your code and description differ.
On Sun, 5 Jun 2011, Kehl Dániel wrote:
Dear All,
I have a MCMC result in x1. I was wondering if there is a simpler, more
elegant way of evaluatin
Dear All,
I have a MCMC result in x1. I was wondering if there is a simpler, more
elegant way of evaluating the estimate of an integral then this (I am
pretty sure there is):
Also if I want to count the x's say -1in period.
[code]
z <- -2
burnin <- 2000
int1 <-
length(x1$x[(burnin+1):length(
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