This is very straightforward using the reshape2 package:
library('reshape2')
dc <- dcast(a, name ~ year, value_var = 'amount')
name 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984
1a123456789 10 NA NA NA NA
2b 11 12
Hi:
Here's a Q & D solution that could be improved. It uses the plyr
package. Starting from your data1 data frame,
library('plyr')
dseq <- seq(as.Date('2010-01-01'), as.Date('2011-06-05'), by = '30 days')
# Use the cut() function to create a factor whose levels are demarcated
# by the dates in ds
What build of R can't calculate factorial(150)? I thought the max
double of a 32 bit build would be on the order of 10^308 ~ 2^1024 (the
material below seems to agree with me on this)
But yes, agreed with Ted: it's very helpful to think a bit about what
you are calculating when doing these sorts o
On 19/11/11 05:54, Keith Thompson wrote:
Hi,
I am new to R and looking to do auto-regression / ARIMA type modeling. My
data has both date and time which I need to combine into a single date-Time
value. The time steps are unequal. What package is best for doing the
regression and plotting the
Thanks David and Bill. I don't know how I couldn't think of that - maybe
because I'm working through a brutal flu right now.
Thanks for the help. This list is a fantastic resource.
Trevor
On Fri, Nov 18, 2011 at 4:18 PM, David Winsemius wrote:
>
> On Nov 18, 2011, at 7:04 PM, Trevor Davies wr
On Nov 18, 2011, at 7:04 PM, Trevor Davies wrote:
A late friday afternoon coding question. I'm having a hard time
thinking
of the correct search terms for what I want to do.
If I have a df like this:
a <-
data.frame(name=c(rep('a',10),rep('b',
15)),year=c(1971:1980,1971:1985),amount=1:25)
You can use the reshape() in core R (the stats package)
> reshape(a, timevar="year", idvar="name", direction="wide")
name amount.1971 amount.1972 amount.1973 amount.1974
1 a 1 2 3 4
11b 11 12 13 14
amount.197
Dear colleagues,
I am having issues trying to build a R package I recently wrote.
I am using R 2.14.0 and my package depends on another package called aod.
Running:
R CMD Sweave vignette.Rnw
is perfectly fine and the vignette compiles properly.
but when I use
R CMD build mypackage
to build the ta
Thanks again,
really appreciate that.
Il giorno 18 Nov 2011, alle ore 23:06, Ben Bolker ha scritto:
> [cc'ing back to r-help]
>
> On Fri, Nov 18, 2011 at 4:39 PM, matteo dossena
> wrote:
>> Thanks a lot,
>>
>> just to make sure i got it right,
>>
>> if (using the real dataset) from the LogL
Hi,
I am looking forward to fill the plot using conditions on variables a2 and
a3. Whenever variable(a2) goes above variable(a3) i fill it with some color
.
I am storing the coordinates of a2 and a3 in x and y as well as time where
it is occurring . But it is not producing properly. I must be wro
will this work for you:
> str(x)
'data.frame': 10 obs. of 2 variables:
$ price: chr "0,00" "0,00" "0,02" "0,03" ...
$ mentioned: int 1 1 1 1 1 1 1 1 1 1
> # create vector with range of 'price'
> allPrices <- paste('0,', sprintf("%02d", 0:19), sep = '')
> # find missing prices and add wi
A late friday afternoon coding question. I'm having a hard time thinking
of the correct search terms for what I want to do.
If I have a df like this:
a <-
data.frame(name=c(rep('a',10),rep('b',15)),year=c(1971:1980,1971:1985),amount=1:25)
name year amount
1 a 1971 1
2 a 1972
dear R-listers,
does anyone of you know whether there is any R package that features a
alogarithm to calculate the sum formula from a given exact mass (as
used in mass spectrometry)?
thanks a lot,
lukas
--
Lukas Kohl
Department of Chemical Ecology and Ecosystem Research
http://www.ecosystem-res
I large datset that includes subjects(ID), Dates and events that need to be
counted. Not every date includes an event, and I need to only count one event
per 30days, per subject. So in essence, I need to create a 30-day "black out"
period during which time an event cannot be "counted" for each
[cc'ing back to r-help]
On Fri, Nov 18, 2011 at 4:39 PM, matteo dossena
wrote:
> Thanks a lot,
>
> just to make sure i got it right,
>
> if (using the real dataset) from the LogLikelihood ratio test model1 isn't
> "better" than model,
> means that temporal auto correlation isn't seriously affec
On 11/17/2011 09:46 PM, cgenolin wrote:
Hi the list,
I define a class 'C' that inherit from two classes 'A' and 'B'. 'A' and 'B'
have no slot with similar names.
setClass(
Class="C",
contains=c("A","B")
)
To define the get operator '[' for class "C", I simply use the
You can use Extra Packages for Enterprise Linux (or EPEL) to install your RPM
from.
https://fedoraproject.org/wiki/EPEL
--
View this message in context:
http://r.789695.n4.nabble.com/Problem-Installing-R-2-9-1-1-RHEL-x86-64-binary-tp900310p4084929.html
Sent from the R help mailing list archive
1. We don't do homework on this list.
2. "... which is of the size 10 by 2^10."
No it isn't.
(Hint: Many of the permutations are not distinct).
-- Bert
On Fri, Nov 18, 2011 at 2:02 PM, Gyanendra Pokharel
wrote:
> Hi all,
> I have a set of elements (1, 1, 0,1,1,0,1,0,1,1) with ten elements. I
Hi all,
I have a set of elements (1, 1, 0,1,1,0,1,0,1,1) with ten elements. I have
to construct the permutation matrix of this set which is of the size 10 by
2^10. Can some one help how is this possible? Is there is a particular
function in R or I need to make function?
Best
[[alternativ
On Nov 18, 2011, at 3:52 PM, Roger Coupal wrote:
Hi, I am working with R on a Mac. I need to use a couple of packages
that I found on the Web site. I install them using Packages and
Data: Package Manager. I leave R and come back another time and the
installed packages are not installed. So
Dear Dr. Bolker and R list,
Thanks so much for the help! Apologies for taking so long to write
back, I've been distracted with other work. Between your response and
the suggested section in the EMD book, I was able to figure it out!
I have another question. I ran three optimization runs wi
I don't really know about the second half of your email, but are you sure you
don't just need to reload the packages with the library() command rather than
reinstalling them?
It's also to set up autoload: google Rprofile for instructions.
Michael
On Nov 18, 2011, at 3:52 PM, Roger Coupal wro
Hi,
Two possible routes I can suggest:
1- export the plot in svg format, which supports natively the use of
filling patterns, e.g.
http://www.w3.org/TR/SVG/images/pservers/pattern01.svg
It's possible that the gridSVG package could help you automate the
process of "grid.garnish()-ing" the grobs;
Hi, I am working with R on a Mac. I need to use a couple of packages that I
found on the Web site. I install them using Packages and Data: Package Manager.
I leave R and come back another time and the installed packages are not
installed. So I do it again. Not a big deal, but is there anyway to
Dear Sarah,
RE:
> Hi all,
>
> I'm working with a bunch of large graphs, and stumbled across
> something useful. Probably many of you know this, but I didn't and so
> others might benefit.
>
> Using pch="." speeds up plotting considerably over using symbols.
>
>> x <- runif(100)
>> y <- run
I have built a Kalman Filter model for flu forecasting as shown below.
Y - Target Variable X1 - Predictor1 X2 - Predictor2
While forecasting into the future, I will NOT have data for all three
variables. So, I am predicting X1 and X2 using two Kalman filters. The code
is below
x1.model <- dl
On 18-Nov-11 16:03:44, Gyanendra Pokharel wrote:
> Hi all,
> why factorial(150) shows the error out of range in 'gammafn'?
> I have to calculate the number of subset formed by 150 samples
> taking 10 at a time. How is this possible?
> best
Because factorial(150) is nearly 10^263, which is far grea
That is a function I did not know about, thanks Hadley!
I still don't see the speed increase that you do with the base plot
package, but I'm sticking with ggplot anyway!
> x<-runif(1e6)
> y<-runif(1e6)
> system.time(print(qplot(x,y)))
user system elapsed
42.234 0.520 43.061
> system.time(
You need: system.time(print(qplot(x,y,pch=I('.'
Hadley
On Fri, Nov 18, 2011 at 1:30 PM, Justin Haynes wrote:
> Very cool. Sadly, as far as I can tell, it doesn't work with ggplot though
> :(
>
>
>> x<-runif(1e6)
>> y<-runif(1e6)
>> system.time(plot(x,y,pch='.'))
> user system elapsed
>
I have a couple of Tinn-R problems that I hope someone may be able to make some
suggestions about or point me to a Tinn-R source. So far I have not found
anything particularly relevant by googling on Tinn-R.
NOTE: Just to be on the safe side I uninstalled both R and Tinn-R , ran some
cleanup
Very cool. Sadly, as far as I can tell, it doesn't work with ggplot though
:(
> x<-runif(1e6)
> y<-runif(1e6)
> system.time(plot(x,y,pch='.'))
user system elapsed
0.824 0.012 0.845
> system.time(plot(x,y))
user system elapsed
33.422 0.016 33.545
> system.time(print(qplot(x,y)))
Hello,
I am searching for a method, algorithm or some solution to the following
problem:
I need an estimator for the block length of two series to bootstrap the
correlation coefficient with the stationary bootstrap procedure. The problem
here: the two series have different block lengths!
Here
Hi,
I fit both Poisson and NB (negative binomial) models to some empirical data.
Although models provide me with sensible parameters, in the case of the NB
models i get three inconsistencites:
- First, the total number of occurrences predicted by the model (i.e.
fitted(fit)) is much greater th
As others have said, no need for a loop
Another approach
mydata <- data.frame(A = dpois(x,exp(4.5355343)),
B = dpois(x,exp(4.5355343 + 0.0118638)),
C = dpois(x,exp(4.5355343 -0.0234615)),
D = dpois(x,exp(4.5355343 + 0.0316557)),
E = dpois(x,exp(4.5355343 + 0.0004716)),
F = dpois(x,exp(4.5355343 +
Hello,
This is the code to generate the tree
library(RWeka)
library(randomForest)
library(party)
library(partykit)
if(require(mlbench, quietly = TRUE) && require(party, quietly = TRUE))
m1 <- J48(income2 ~ capital.gain+capital.loss+relationship+marital.status,
data = treino)
#m1 <- J48(in
Hi,
I am looking to build even quintiles for a set of data. I managed to get it
done, but I would like to know if there is a more direct way to write the
data from my loop output x in the bottom of the code into the "empty" matrix
p1, which I filled with NA's. The way I am doing it at the momen
Hi all,
I'm working with a bunch of large graphs, and stumbled across
something useful. Probably many of you know this, but I didn't and so
others might benefit.
Using pch="." speeds up plotting considerably over using symbols.
> x <- runif(100)
> y <- runif(100)
> system.time(plot(x, y,
... and yet another line I left out below! I apologize for this baloney!
On Fri, Nov 18, 2011 at 10:48 AM, Bert Gunter wrote:
> ... I failed to correctly paste the first line of an example:
>
> On Fri, Nov 18, 2011 at 10:44 AM, Bert Gunter wrote:
>> David:
>>
>> As you now realize "\t" etc. is
... I failed to correctly paste the first line of an example:
On Fri, Nov 18, 2011 at 10:44 AM, Bert Gunter wrote:
> David:
>
> As you now realize "\t" etc. is a perfectly legal single tab character.
>
> Now consider:
- left this out --
> gsub("\\","a","\\")
-
David:
As you now realize "\t" etc. is a perfectly legal single tab character.
Now consider:
Error in gsub("\\", "a", "\\") :
invalid regular expression '\', reason 'Trailing backslash'
BUT
> gsub("","a","\\")
[1] "a"
???
The issue is there are two levels of escapes here -- the R parser
Dear Varina,
I think that you may be confusing the Anova() function in the car package
(which takes an idesign argument) with the standard R anova() function.
I hope this helps,
John
John Fox
Senator William McMaster
Professor of Social Statistics
Department o
Hi,
I'm looking the best way to fill irregular polygons with patterns,
Something like the function grid.pattern do, but my case is with
irregular polygons.
Whit this script I can get it, but I'm looking for an "elegant" solution..
library(grid)
grid.polygon(x=c(0.2, 0.8, 0.6, 0.6, 0.8, 0.2),
Le vendredi 18 novembre 2011 à 15:06 +, Mario Giesel a écrit :
> Hello, list,
>
> I've been struggling with this task for a while looking for an efficient way
> to solve it:
> There are two variables 'price' and 'mentioned'.
> I want to 'enlarge' data so that missing price points within the p
I have no idea what you're talking about, and you're probably on the
wrong mailing list. I'm sure there's Mathematica help out there.
Sarah
On Fri, Nov 18, 2011 at 10:53 AM, Somaye Node wrote:
>
> Thanks for your reply,
> i cant open three sites.
> all the best.
>
> --- On Fri, 11/18/11, Sarah G
Hi,
I am new to R and looking to do auto-regression / ARIMA type modeling. My
data has both date and time which I need to combine into a single date-Time
value. The time steps are unequal. What package is best for doing the
regression and plotting the predicted values against the actual data?
You may want to look at the following paper.
Best,
Giovanni
Reference Type: Journal Article
Author: Pinheiro, José C.
Author: Bates, Douglas M.
Primary Title: Unconstrained parametrizations for variance-covariance
matrices
Journal Name: Statistics and Computing
Cover Date: 1996-09-01
Publisher: S
On Fri, 2011-11-18 at 11:03 -0500, Gyanendra Pokharel wrote:
> Hi all,
> why factorial(150) shows the error out of range in 'gammafn'?
> I have to calculate the number of subset formed by 150 samples taking 10 at
> a time. How is this possible?
> best
Do you mean:
> choose(150, 10)
[1] 1.169554e+
S+'s help(apply) says
MARGIN
the subscripts over which the function is to be applied.
For example, if X is a matrix, MARGIN=1 indicates rows
and MARGIN=2 indicates columns. If the dimensions of X
are named, then the names can be used to specify MARGIN.
Note that MARGIN tells which dim
Hi all,
why factorial(150) shows the error out of range in 'gammafn'?
I have to calculate the number of subset formed by 150 samples taking 10 at
a time. How is this possible?
best
[[alternative HTML version deleted]]
__
R-help@r-project.org mai
On Fri, Nov 18, 2011 at 10:26 AM, David Winsemius
wrote:
>
> On Nov 18, 2011, at 9:28 AM, jim holtman wrote:
>
>> It is pretty straightforward in R:
>>
>>> x <-
>>> readLines(textConnection("sadf|asdf|asdf\tqwer|qwer|qwer\tzxcv|zxcv|zxfcgv"))
>>> closeAllConnections()
>>> # convert tabs to newline
Because it's a character vector with only one element, which just happens to be
empty - sort of like why the cardinality of the set { /O } is one. (/O being my
best shot at the empty set symbol in plain text).
You might be thinking of nchar().
Michael
On Nov 18, 2011, at 10:09 AM, "Jaensch,
On Nov 18, 2011, at 10:09 AM, Jaensch, Steffen [TIBBE] wrote:
Hi all,
Can somebody explain why length("") returns 1 and not 0? How do I test
if a given string is the empty string?
character(0) is not the same as ""
--
David Winsemius, MD
West Hartford, CT
__
> nchar("")
>
>
[1] 0
On Friday 18 November 2011 16:09:38 Jaensch, Steffen [TIBBE] wrote:
> Hi all,
>
>
>
> Can s
Le vendredi 18 novembre 2011 à 16:09 +0100, Jaensch, Steffen [TIBBE] a
écrit :
> Hi all,
>
> Can somebody explain why length("") returns 1 and not 0?
Probably because it contains one element, "".
> How do I test if a given string is the empty string?
a <- ""
a == ""
[1] TRUE
Seems to be enough to
Hi all,
Can somebody explain why length("") returns 1 and not 0? How do I test
if a given string is the empty string?
Thanks,
Steffen.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.c
Hi all,
I'm trying to run a repeated measures ANOVA on some univariate ecological
data that was collected over two growing seasons. I ran the test using the
methodology found on this website:
http://rtutorialseries.blogspot.com/2011/02/r-tutorial-series-one-way-repeated.html
Upon running the act
Hello everybody.
I'm trying to send the result of a decision tree for latex, but I do not
get with the package(xtable), there is a package that make this export
Export this for latex
marital.status = Divorced
| educational.num <= 12: <=50K (1795.0/90.0)
| educational.num > 12
| | ho
Hello, list,
I've been struggling with this task for a while looking for an efficient way to
solve it:
There are two variables 'price' and 'mentioned'.
I want to 'enlarge' data so that missing price points within the price range
are added to variable price.
Also variable 'mentioned' is to recei
What kind of object it is? How was it generated?
It doesn't have a plot method?
--
View this message in context:
http://r.789695.n4.nabble.com/Export-Tree-for-latex-tp4083586p4083593.html
Sent from the R help mailing list archive at Nabble.com.
__
R-h
On Nov 18, 2011, at 9:28 AM, jim holtman wrote:
It is pretty straightforward in R:
x <- readLines(textConnection("sadf|asdf|asdf\tqwer|qwer|qwer\tzxcv|
zxcv|zxfcgv"))
closeAllConnections()
# convert tabs to newlines
x <- gsub("\t", "\n", x)
Did the rules get liberalized for escaping patter
The thing to watch out for is if you file is large, 'textConnection'
is very slow at providing the data stream for something like
read.table. It is usually much faster to read in the file with
'readLines', preprocess the data data, write it out to a tempfile and
then read it back in with 'read.tab
On Thu, Nov 17, 2011 at 4:40 PM, Katrina Bennett wrote:
> Hello, I am wondering if someone can help me. I have the following function
> that I derived using nls() SSlogis. I would like to find its derivative. I
> thought I had done this using deriv(), but for some reason this isn't
> working out f
On Nov 18, 2011, at 9:13 AM, Langston, Jim wrote:
Thanks Paul,
That's the path I was marching down, I was hoping for something
a little cleaner, I do the same with Perl or Java.
> tesfil <- "aa|bb|cc\tdd|ee|ff\t"
> read.table(textConnection(gsub("\\\t", "\n", scan(
textConnec
On Fri, 2011-11-18 at 15:17 +0100, René Mayer wrote:
> Thanks a lot Gavin!,
> this was what I was looking for.
> Have I got this right that with no 'cyclic shifts *within* strata' you
> mean that I cannot define a nesting within animal, e.g.,
> animal/year/season (speaking in regression-terms
Hi,
I would like to know what should I garantee about P and GGt in order to have
F = Z %*% P %*% t(Z) + GGt always as a positive definite matrix.
Being more precise:
I am trying to find minimum likelihood parameters by using the function
'optim' to find the lowest value generated by $LogLik fro
Hello,
I have done a stepwise analysis to determine the best model fitting my data.
data <- stepAIC( aov (Group)~ . , data=mydata)
data
> summary(data)
Df Sum Sq Mean Sq F valuePr(>F)
`a` 1 1.9829 1.98290 11.176 0.0011824 **
`b` 1 2.6606 2.66064 14.996 0.0001967 ***
On Nov 18, 2011, at 8:05 AM, wrote:
Hello
How can I apply functions along "layers" of a data matrix?
Example:
daf <- data.frame(
'id' = rep(1:5, 3),
matrix(1:60, nrow=15, dimnames=list( NULL, paste('v', 1:4,
sep='') )),
rep = rep(1:3, each=5)
)
The data frame "daf" contains 3 repetit
It is pretty straightforward in R:
> x <-
> readLines(textConnection("sadf|asdf|asdf\tqwer|qwer|qwer\tzxcv|zxcv|zxfcgv"))
> closeAllConnections()
> # convert tabs to newlines
> x <- gsub("\t", "\n", x)
> # write out to a temp file and then read in as a data frame
> myFile <- tempfile()
> writeLin
On Fri, Nov 18, 2011 at 1:53 PM, Raphael Saldanha
wrote:
> I would like
> to import this dataset direct into R, without saving the file on disk
> (for using the most updated file),
You don't really have much choice - I'm certain that the mdb-tools
won't read from a remote ftp server and I susp
Thanks a lot Gavin!,
this was what I was looking for.
Have I got this right that with no 'cyclic shifts *within* strata' you
mean that I cannot define a nesting within animal, e.g.,
animal/year/season (speaking in regression-terms random-effects for
the animal-specific season and year varia
Thanks Paul,
That's the path I was marching down, I was hoping for something
a little cleaner, I do the same with Perl or Java.
Jim
On 11/18/11 8:35 AM, "Paul Hiemstra" wrote:
>Hi Jim,
>
>You can read the text file using readLines. This puts each line in the
>file into an element of a list. Th
The multiple exponential problem you are attempting has a well-known and long
history.
Lanczos 1956 book showed that changing the 4th decimal in a data set changes the
parameters hugely.
Nevertheless, if you just need a "fit" and not reliable paramters, you could
reparameterize to k1 and k2diff=
I think you need to make an expression. I tried
> nls.fn <- asym/((1+exp((xmid-x.seq)/scal)))
Error: object 'asym' not found
> nls.fn <- expression(asym/((1+exp((xmid-x.seq)/scal
> D(nls.fn,"asym")
1/((1 + exp((xmid - x.seq)/scal)))
>
Does that help? Maybe there are other approaches too.
JN
Dear David,
Thank you. That was very clear. I will try to make minimal examples in the
future.
Best Regards,
Ashim
On Fri, Nov 18, 2011 at 6:50 PM, David Winsemius wrote:
>
> On Nov 18, 2011, at 1:51 AM, Ashim Kapoor wrote:
>
> Dear Dennis,
>>
>> Many thanks.I was wondering if there was a way
I am really sorry for the inconvenience, this will not repeat again.
This message is in plain text and I will try give every information
needed. If I'm not obeying some of the recommendations on the Posting
Guide, please let me know.
The IBGE, a Brazilian foundation of Geography and Statistics, ha
Le vendredi 18 novembre 2011 à 04:45 -0800, Gyanendra Pokharel a écrit :
> Hi all, following is my R -code and shows the error given below
> > n <- 100
> > k<-2
> > x1 <-c(1, 3);x2 <- c(2,5)
> > X <- matrix(c(0,0), nrow = 2, ncol = n)
> > for(i in 1:k)
> + X[i, ] <- mh1.epidemic(n,x1[i],x2[i])
> Er
On 11/18/2011 02:55 AM, Anup Som wrote:
Dear all,
I was trying to normalize a subset of affy data those transcribe
are either P or M (called pm_filter). I am able to normalize pm_filter
subset by using RMA method, however MAS5 is not working.
For RMA method, I used the following com
Hi Jim,
You can read the text file using readLines. This puts each line in the
file into an element of a list. Then you can go through the lines
manually (e.g. using grep, sub, strsplit) and create your data.frame.
cheers,
Paul
On 11/18/2011 12:37 PM, Langston, Jim wrote:
> Hi all,
>
> I've been
...in addition, you might get more answers on the R-sig-mac mailing list.
Paul
On 11/18/2011 12:34 PM, Assa Yeroslaviz wrote:
> Hi,
>
> I have a problem on my mac when trying in R to produce png images.
>
> I am getting this warnings with the ArrayQualityMetrics package:
>
>> arrayQualityMetrics(
On Nov 17, 2011, at 4:40 PM, Katrina Bennett wrote:
Hello, I am wondering if someone can help me. I have the following
function
that I derived using nls() SSlogis. I would like to find its
derivative. I
thought I had done this using deriv(), but for some reason this isn't
working out for me
Hi,
Cross-posting between this list and stackoverflow is discouraged.
cheers,
Paul
On 11/18/2011 12:34 PM, Assa Yeroslaviz wrote:
> Hi,
>
> I have a problem on my mac when trying in R to produce png images.
>
> I am getting this warnings with the ArrayQualityMetrics package:
>
>> arrayQualityMet
On 11/18/2011 01:05 PM, saschav...@gmail.com wrote:
> daf <- data.frame(
> 'id' = rep(1:5, 3),
> matrix(1:60, nrow=15, dimnames=list( NULL, paste('v', 1:4, sep='') )),
> rep = rep(1:3, each=5)
> )
Hi,
This seems like a job for plyr!
library(plyr)
ddply(daf, .(rep), summarise, mn = mean(v1))
Hi:
Here are two ways to do it; further solutions can be found in the doBy
and data.table packages, among others.
library('plyr')
ddply(daf, .(id), colwise(mean, c('v1', 'v2', 'v3', 'v4')))
aggregate(cbind(v1, v2, v3, v4) ~ id, data = daf, FUN = mean)
# Result of each:
id v1 v2 v3 v4
1 1 6
On Nov 18, 2011, at 1:51 AM, Ashim Kapoor wrote:
Dear Dennis,
Many thanks.I was wondering if there was a way to edit the variable
and put
\n's in it. Is there ?
Of course there is. The key however ti to realize that at the moment
there are "\\"'s and "n"'s but not any single characters
On Fri, Nov 18, 2011 at 12:15 PM, Raphael Saldanha
wrote:
> Hi!
>
> I need to import an Access MDB database from this FTP address:
> ftp://geoftp.ibge.gov.br/Organizacao/Localidades/cadastro_localidades_selecionadas.mdb
>
> I tried with Hmisc mdb.get and RODBC without sucess any tip?
>
What
Dear all,
I was trying to normalize a subset of affy data those transcribe
are either P or M (called pm_filter). I am able to normalize pm_filter
subset by using RMA method, however MAS5 is not working.
For RMA method, I used the following commend: est<-rma(affydata,
subset=pm_filter)
Hi all, following is my R -code and shows the error given below
> n <- 100
> k<-2
> x1 <-c(1, 3);x2 <- c(2,5)
> X <- matrix(c(0,0), nrow = 2, ncol = n)
> for(i in 1:k)
+ X[i, ] <- mh1.epidemic(n,x1[i],x2[i])
Error in X[i, ] <- mh1.epidemic(n,x1[i],x2[i]):
number of items to replace is not a mult
Dear All
I am running Sweave with xtable and want to put the caption placement
on top. But this does not work. Any idea what is going wrong?
Here is an example that runs properly with the exception of the
caption placement in the pdf-file.
\documentclass[11pt,a4paper]{article}
\usepackage
Dear all,
I'm trying to use sweave, but running the demo example, I got problem.
First, I'm realluy a beginner with LaTeX and sweave, so sorry if my question
seems weird.
First I've installed mactex distribution :
http://www.tug.org/mactex/2011/
I've set up the "SWEAVE_STYLEPATH_DEFAULT" envi
With John Nash, I am an administrator for the R project's contribution to the
Google Summer of Code 2012. This is a program where Google provides $5000 to a
student to write code for an open source project over a 3 month period in the
summer. We would like to invite anyone interested in being ei
Hello
How can I apply functions along "layers" of a data matrix?
Example:
daf <- data.frame(
'id' = rep(1:5, 3),
matrix(1:60, nrow=15, dimnames=list( NULL, paste('v', 1:4, sep='') )),
rep = rep(1:3, each=5)
)
The data frame "daf" contains 3 repetitions/layers (rep) of 4 variables
of 5 p
On 11/17/2011 06:29 PM, Abraham Mathew wrote:
I have information on two versions of the same site, and I have data
on the number of times people filled out a form on each version
of the site.
Sample data:
Site 1 Site 2
Filled out form
Hi!
I need to import an Access MDB database from this FTP address:
ftp://geoftp.ibge.gov.br/Organizacao/Localidades/cadastro_localidades_selecionadas.mdb
I tried with Hmisc mdb.get and RODBC without sucess any tip?
Regards,
Raphael Saldanha
saldanha.plan...@gmail.com
[[alternative
Hi all,
I've been scratching and poking, but basically, the file I need to read has
two delimiters that I need to contend with. The first is that the file
contains
tabs (\t) , instead of newlines (\n), and the second is that the fields
have
| for the seperators. I can easily do a read if I first
Hi,
I have a problem on my mac when trying in R to produce png images.
I am getting this warnings with the ArrayQualityMetrics package:
> arrayQualityMetrics(rma_fatBody, outdir="normData", force =T)
The report will be written into directory 'normData'.
KernSmooth 2.23 loaded
Copyright M. P. Wan
Le jeudi 17 novembre 2011 à 21:34 -0500, R. Michael Weylandt a écrit :
> Hi Josh,
>
> You're absolutely right. I suppose one could set up some sort of S3
> thing for Henri's problem:
>
> c <- function(..., recursive = FALSE) UseMethod("c")
> c.default <- base::c
> c.corpus <- function(..., recurs
And you always can count with the help of the R-SIG-GEO list:
https://stat.ethz.ch/mailman/listinfo/r-sig-geo
On Fri, Nov 18, 2011 at 4:54 AM, vioravis wrote:
> Thanks a lot for the guidance. I will take a look at these options.
>
> Ravi
>
> --
> View this message in context:
> http://r.789695.n
It works if you separate the print command and put the caption placement in
the print command
, see below:
\documentclass[11pt,a4paper]{article}
\usepackage{Sweave}
\begin{document}
<<>>=
x = runif(100, 1, 10)
y = 2 + 3 * x + rnorm(100)
@
<>=
library(xtable)
p <- (xtable(summary(lm(
On Fri, 2011-11-18 at 10:25 +0100, René Mayer wrote:
> Dear all,
> How can I run a constrained correspondence analysis with
> the following data:
> 15 animals were measured repeatedly month-wise (over to 2 years)
> according to ther diet composition (8 food categories).
>
> our data.frame looks
Dear Renger,
This is occurring because "xtable" divides up the arguments into items related
to the "content" of the table and arguments related to the "layout" of the
table.
The "caption.placement" is an argument to "print.xtable()" rather than to
"xtable()":
print(xtable(summary(lm(y~x)),
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