Hi al,
I have a dataset (see attached), which basically involves 4 treatments for a
chemotherapy drug. Samples were taken from 2 biopsy locations, and biopsy were
taken at 2 time points. So each subject has 4 data points (from 2 biopsy
locations and 2 time points). The objective is to study tre
Hi,
## Better way to make your matrix
A <- rbind(c(16,10,2,4,8,7),
c(16,10,12,14,8,7),
c(16,10,13,15,19,17),
c(16,9,13,15,9,7))
## Create logical matrices for conditions 1 & 2
B1 <- rowSums(A[, 2:4] < 10) > 0
B2 <- rowSums(A[, 5:6] < 10) > 0
## initialize X at the "catch all"
X <- rep(3L, nrow
On Aug 8, 2011, at 12:18 AM, Lei Liu wrote:
Hi there,
I have two questions on matrix manipulation. For the first one, I
want to calculate the product of each column of a matrix (say A)
with another vector (say b). That is, if A has 5 columns (a1, a2,
a3, a4, a5), I want to obtain a matri
Suppose I have a matrix like
A=matrix(0,4,6)
A[1,]=c(16,10,2,4,8,7)
A[2,]=c(16,10,12,14,8,7)
A[3,]=c(16,10,13,15,19,17)
A[4,]=c(16,9,13,15,9,7)
> A
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 16 102487
[2,] 16 10 12 1487
[3,] 16 10 13 151917
[4,]
Hi Lei,
Here are some examples. Look at the documentation for ?"%*%" ?"%o%"
and ?"[" with the drop = FALSE argument (in case you ever do select
just one column/row of a matrix).
## your first A matrix
(A <- matrix(1:55, ncol = 5))
## you were rather unclear what
## your b vectors dimensions wer
Hi there,
I have two questions on matrix manipulation. For the first one, I
want to calculate the product of each column of a matrix (say A) with
another vector (say b). That is, if A has 5 columns (a1, a2, a3, a4,
a5), I want to obtain a matrix with columns (a1*b, a2*b, aA3*b, a4*b,
a5*b). C
On 11-07-25 4:50 PM, John Hangen wrote:
Hello,
Today I installed R for Windows version 2.13.1, but realized that it's not
something that I need right now.I attempted to uninstall it, but am getting
the following error message:
A brute force uninstall is fairly easy: just delete the direc
Hi Michael,
The coefficients of ridge regression are given by:
\beta^* = (X'X + k I)^{-1} X' y, (1)
where k > 0 is the penalty parameter and I is the identity matrix.
The ridge estimates are related to OLS estimates \beta as follows:
\beta^* = Z \beta,
On 08/08/11 13:20, Peter Ehlers wrote:
On 2011-08-07 16:57, Rolf Turner wrote:
Neither Ted Harding's post, nor Prof. Ripley's, really appear to
address the OP's follow-up question which was
*why* did he get the ``weird error'' (non-finite finite difference, NaNs
produced) when he applied fitd
Hi BJ,
Suppose that your data set is called 'Data1', please copy and paste
the output created by typing:
dput(Data1[1:15, ])
and post that to the list ***in plain text*** (or upload a text file
on some file hosting service). The email you sent is nearly
impossible to parse because this is a pla
On 2011-08-07 16:57, Rolf Turner wrote:
Neither Ted Harding's post, nor Prof. Ripley's, really appear to
address the OP's follow-up question which was
*why* did he get the ``weird error'' (non-finite finite difference, NaNs
produced) when he applied fitdistr(),
as he was advised to do by Jorge
Dear help list: I am trying to reshape a data frame from long to wide format
and with a reduced variable list using reshape2. The original data frame
format is: Site Obs_no LengthSite 1 Obs 1 10Site 1 Obs 2
13Site 1 Obs 3 14.Site 2 Obs 1 5Site 2 O
On 11-08-07 8:09 PM, Alexandre Aguiar wrote:
Hi.
Need some advice. I have to load a library during R_init_X() with
dlopen(). The library file is 23 megabytes long and loads several other
libraries. Everything happens during dlopen() execution, out of control,
and takes around 1 second on fir
I've tried to upgrade to latest (?)
http://cran.mtu.edu/bin/windows/contrib/2.12/RODBC_1.3-3.zip , but no luck.
It looks like nothing is being escaped as I tried to use character class,
but I got same error message.
Mikhail
> -Original Message-
> From: r-help-boun...@r-project.org [mailto
I apologize for the first e-mail as I didn't use plain text. Here is the
full message.
--8<
-->8--
Hello all!
Can someone confirm whether there is a bug or not?
I was trying to use sqlUpdate in place of sqlSave as data s
On Aug 7, 2011, at 5:51 PM, peter dalgaard wrote:
On Aug 7, 2011, at 21:31 , David Winsemius wrote:
On Aug 7, 2011, at 2:37 PM, peter dalgaard wrote:
On Aug 7, 2011, at 20:27 , David Winsemius wrote:
On Aug 7, 2011, at 2:08 PM, Bansal, Vikas wrote:
Dear all,
I was working on number
Hello all!
Can someone confirm whether there is a bug or not?
I was trying to use sqlUpdate in place of sqlSave as data set I import has
duplications. However I get errors while using fast=FALSE argument to safely
update/ignore duplicates:
Error while executing the query[RODBC] ERROR:
Try
library(lattice)
x = seq(from = as.Date("2011-01-01"), by = "1 month", length = 10)
y = rnorm(10)
xyplot(y~x)
?xyplot
?seq
Regards
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
ARMIDALE NSW 2351
Email: home mac...@northnet.com.au
At 10:00 08/08/20
Hi.
Need some advice. I have to load a library during R_init_X() with
dlopen(). The library file is 23 megabytes long and loads several other
libraries. Everything happens during dlopen() execution, out of control,
and takes around 1 second on first load in a quadcore. Besides, due to
secu
Hi Caitlin,
Here are a couple of options for you:
plot(x = seq(from = as.Date("2011-05-01"), to = as.Date("2011-05-30"),
by = "day"),
y = cumsum(runif(30)))
## using a bit of a shortcut
plot(x = as.Date("2011-05-01") + 0:29,
y = cumsum(runif(30)))
Does that do what you want?
Josh
On
Hi all.
I need to plot 30 values and I'd like the x-axis to have 30 dates values,
e.g., 5/1/2011, 5/2/2011, etc. so an observer could see how the values
correspond as the month progresses.
Is there a convenient function I could use? I considered creating two
vectors, x and y, then simply plotting
Neither Ted Harding's post, nor Prof. Ripley's, really appear to
address the OP's follow-up question which was
*why* did he get the ``weird error'' (non-finite finite difference, NaNs
produced) when he applied fitdistr(),
as he was advised to do by Jorge Ivan Velez. The ``weird error'' did
The statconn-RExcel project (RAndFriends is available as service from this
project) has its own mailing list.
Subscription is possible at the web site you mentioned.
If you think your problem is caused by the RAndFriendsSetup, please subscribe
to the mailing list
and report your problems there.
On Fri, Aug 5, 2011 at 3:54 PM, khadeeja ismail wrote:
> other attached packages:
> [1] Rcmdr_1.6-2
>
I bet that this is what generates the lack of error messages. Check
Rcmdr documentation on how to turn it off.
Liviu
__
R-help@r-project.org mailing li
On Aug 7, 2011, at 21:31 , David Winsemius wrote:
>
> On Aug 7, 2011, at 2:37 PM, peter dalgaard wrote:
>
>>
>> On Aug 7, 2011, at 20:27 , David Winsemius wrote:
>>
>>>
>>> On Aug 7, 2011, at 2:08 PM, Bansal, Vikas wrote:
>>>
Dear all,
I was working on number of files and at
On Aug 7, 2011, at 21:25 , Carl Witthoft wrote:
> Just chiming in here...
>
> Granted some of the help files are a bit cryptic until you get used to them
> (but still much easier to understand than the Unix man pages :-) ). This
> sort of thing does provide a chance to learn about the power o
Thanks to all who emailed me regarding my problem.
options(show.error.messages=TRUE) seems to work now, so as one of you pointed
out I suspect a problem with my system.
Regrds,
Hajja
Hi
r-help-boun...@r-project.org napsal dne 05.08.2011 15:21:34:
> khadeeja ismail
> Odeslal: r-help-boun...@r
Hi Matt,
You have not provided us nearly enough information to provide
meaningful feedback. You say your computer got, 'hosed'. Have you
managed to reinstall R? If so, what version are you running? Can you
install other R packages? What is your current OS? Still linux?
What flavour? How wer
Hi Mark,
I do not have any XP or 32 bit systems lying around to try out
anymore, but I have R 2.13.1 running just fine on two Win 7 x64
systems. Have you tried downloading it again and reinstalling? My
antivirus has an unwelcomed predilection to remove R.dll which causes
havoc unless I set a man
Can you provide some sample data and show what you expect?
On Sun, Aug 7, 2011 at 4:32 PM, Bruce Miller wrote:
> I need to reformat long data into a wide format (matrix for another program)
> but it does not seem straight forward.
> Can not seem to get it to work since some data elements are stri
Hello
I had the same problem today.
Error Message: "This Installation can only be uninstalled on 64-bit
Windows."
Computer System: Windows 7 Enterprise SP1 32-bit
R Version attempted: R 2.13.1 (installed as part of the R and Friends
package from http://rcom.univie.ac.at/download.html)
I have
On 2011-08-06 15:14, Alexander Engelhardt wrote:
Hey,
I have a set of income data which I'd like to fit to a gamma
distribution. How can I estimate the two parameters of the gamma
distribution for a vector, e.g.
c(2039L, 2088L, 5966L, 2353L, 1966L, 2312L, 3305L, 2013L, 3376L,
3363L, 3567L, 4798L
I need to reformat long data into a wide format (matrix for another
program) but it does not seem straight forward.
Can not seem to get it to work since some data elements are strings.
Anyone have time to look at the issue?
Tnx
Bruce
--
Bruce W. Miller, Ph.D.
Conservation Ecologist
Neotropica
Greetings all,
I have been working with RStudio and R only for a little while. I
came across a package called 'reshape' that helped me 'rename'
columns. Unfortunately, my computer got hosed (too much playing with
linux too late at nite) and I had to re-install everything, BUT when I
tried to rein
Hi Vikas,
You're overworking yourself here, gsub is vectorized!
df$V5 <- gsub("[^AaCcGgTt\\.,]", "", df$V5)
This will be *substantially* faster than looping (using apply) over
every row of your data frame, since you just care about the 5th column
anyways. Also, I switched your regexp for one th
Dear all,
The 5th column of my data frame is like this-
.$.$.$.$.$,$,$...,.,,.,,....,,,,,T...,,,.,,,...,,
,..,,,...,,,..,,..,,,,,,.,,...G....,,.,,
,t.,,c,,.a.,,,.A,,,...,..,,,.,,,,...,,,$
.
Dear all,
Thanks to everyone.I tried both the methods and it is working.
Thanking you,
Warm Regards
Vikas Bansal
Msc Bioinformatics
Kings College London
From: David Winsemius [dwinsem...@comcast.net]
Sent: Sunday, August 07, 2011 8:31 PM
To: peter dalgaard
On Aug 7, 2011, at 2:37 PM, peter dalgaard wrote:
On Aug 7, 2011, at 20:27 , David Winsemius wrote:
On Aug 7, 2011, at 2:08 PM, Bansal, Vikas wrote:
Dear all,
I was working on number of files and at the end I got a data frame
with approx. million rows.To prin this data frame in output,
Just chiming in here...
Granted some of the help files are a bit cryptic until you get used to
them (but still much easier to understand than the Unix man pages :-)
). This sort of thing does provide a chance to learn about the power of R.
From Rolf's example, suppose that integrate() did
On Aug 7, 2011, at 20:27 , David Winsemius wrote:
>
> On Aug 7, 2011, at 2:08 PM, Bansal, Vikas wrote:
>
>> Dear all,
>>
>> I was working on number of files and at the end I got a data frame with
>> approx. million rows.To prin this data frame in output, I used
>>
>> capture.output(print.dat
Many thanks Peter and David,
Jorge Ivan Velez gave me just what Peter suggests: a resampling solutionÂ… which
is perfect.
I was surprised this is not available as a packaged and documented test:
perhaps an opportunity for a helpful paper for someone enterprising.
Best, tim
On Aug 7, 2011, at 7
?write.table
HTH,
Jorge
On Sun, Aug 7, 2011 at 2:08 PM, Bansal, Vikas <> wrote:
> Dear all,
>
> I was working on number of files and at the end I got a data frame with
> approx. million rows.To prin this data frame in output, I used
>
> capture.output(print.data.frame(end,row.names=F), file = "
Hi Vikas,
See ?print.default and ?options Your options are to set max.print in
options higher or manually pass a max value to print. I'm not sure if
it will let you print a million rows---print is typically used for
output meant for the user and (again typically) users do not read
through a mill
On Aug 7, 2011, at 2:08 PM, Bansal, Vikas wrote:
Dear all,
I was working on number of files and at the end I got a data frame
with approx. million rows.To prin this data frame in output, I used
capture.output(print.data.frame(end,row.names=F), file = "summary",
append = FALSE)
where en
On Aug 7, 2011, at 20:05 , David Winsemius wrote:
>
> On Aug 6, 2011, at 1:19 PM, Timothy Bates wrote:
>
>> Dear R-users,
>> I am comparing differences in variance, skew, and kurtosis between two
>> groups.
>>
>> For variance the comparison is easy: just
>>
>> var.test(group1, group2)
>>
>>
Dear all,
I was working on number of files and at the end I got a data frame with approx.
million rows.To prin this data frame in output, I used
capture.output(print.data.frame(end,row.names=F), file = "summary", append =
FALSE)
where end is the name of my data frame and summary is the name of
On Aug 6, 2011, at 1:19 PM, Timothy Bates wrote:
Dear R-users,
I am comparing differences in variance, skew, and kurtosis between
two groups.
For variance the comparison is easy: just
var.test(group1, group2)
I am using agostino.test() for skew, and anscombe.test() for
kurtosis. However
Hi Richard,
Sorry for the delay, I had stopped tracking this conversation (the
more interesting question, though is what was I searching for in my
gmail archives that lead me to see this and notice your followup
email...). Anyway, Michael is correct about what is going on. One
aspect that might
Dear R and S-Plus Community:
The 2012 R User Conference - useR!2012 - will be held in Nashville
Tennessee USA, June 12-15, 2012 on the campus of Vanderbilt University.
We would like to begin estimating the number of attendees, their area of
interest, and the number seeking hotel vs. lower-
On Aug 7, 2011, at 8:24 AM, Christopher DiBari wrote:
Hello List:
Does anyone know if there's a way to annotate the plots produced by
the
plot.xmean.ordinaly function in the rms package. I'd like to produce
publication quality figures, but first need to annotate the axis
labels. Is
the
Chris for my purposes I need the output tied to the model's variable names,
so I'm not sure I want to add new naming options to the function. You might
just add a line after "nam <- nam[-1]" to specify your vector of names,
starting with the response variable, e.g., nam <- c('velocity
m/s','age','
Hello List:
Does anyone know if there's a way to annotate the plots produced by the
plot.xmean.ordinaly function in the rms package. I'd like to produce
publication quality figures, but first need to annotate the axis labels. Is
there a way to do that?
I appreciate any advice.
Chris
[
On Sat, Aug 06, 2011 at 09:37:22PM +0200, Petr Savicky wrote:
> On Fri, Aug 05, 2011 at 11:47:49PM +0530, Ron Michael wrote:
> > Hi all, I have happened to work on MS .NET for sometime now, and I found
> > that this language offers RNG what is called as Donald E. Knuth's
> > subtractive random nu
Hello
I think you are right. Includegraphics outside od R chunck seems to avoid
unnecessary float space.
Many thanks
Ed.
On Sat, Aug 6, 2011 at 10:18 PM, Duncan Mackay wrote:
> Hi Eduardo
>
> I have just realized that I did not my send previous message to r-help
>
> You do not need to put it
I'm a new user of R and a novice user in copula R package.
I want to fit 3-dimensional t copula for my trivariate data. So I used the
command
t.cop <- tCopula(c(0.785,0.283,0.613),dim=3,dispstr="un",df=6,df.fixed =
TRUE)
where c(0.785,0.283,0.613) is the correlation pattern of my data with 0.785
Hi Sarah,
sorry for the lack of details..hope now my dilemma will be clearer:
so, I have my dataset, represented by a matrix:100 subjects and 20
variables, expressed as 1(presence) and 0 (absence)
I then tried to obtain a dissimilarity m. using the jaccard() command.
To perform a NMDS, I used the
Well, 'Alexander Engelhardt' failed to follow the posting guide, and
so did not get a reply from some knowledgeable people.
(1) When using fitdistr you need to roughly scale your data: e.g.
divide by 1000 here.
(2) fitdistr does 'go down the maximum likeilhood route' and it does
find reasona
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