Dear R-helpers,
I have a three-column matrix with lots of rows:
xyzs - matrix(rnorm(3*10,0,1),ncol=3)
# And I am multiplying it with some vector V, and summing the rows
(columns after t()) in this way:
V - c(2,3,4)
system.time(vx - apply(t(xyzs) * V, 2 ,sum))
Ok, this does not take long
xyzs - matrix(rnorm(3*10,0,1),ncol=3)
V - c(2,3,4)
system.time(vx - apply(t(xyzs) * V, 2 ,sum))
user system elapsed
1.060.021.08
system.time(wx - as.vector(xyzs %*% V))
user system elapsed
0 0 0
all.equal(vx, wx)
[1] TRUE
?
-Original
Dear group,
Here is the kind of data.frame I obtain every day with my function :
futures -
structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10,
CORN Jul/10, CORN Jul/10, CORN Jul/10, LIVE CATTLE Aug/10,
LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10,
SUGAR NO.11 Jul/10, SUGAR
Brian,
If I do understand correctly, I must use in my function something else than
ddply() if I want to avoid any error each time my df has zero rows?
Am I correct?
TY
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: Tuesday, June 01, 2010 9:47 AM
Dear group,
I have noticed that the MacOS X binary for lme4 is not available on
CRAN at the moment. I am aware that it may be possible to install from
source but I am not very familiar with that procedure and would rather
avoid it. But I need the package for a statistics course next week. So
can
I'm running a long MCMC chain that is generating samples for 22 variables.
I have each run of the chain as a row in a matrix.
So: Chain[,1] is the column with all the samples for variable one.
Chain[,2] is the column with all the samples for variable 2, etc.
I'd like to fit all 22 on a single
thanks for your reply.
I have tried to use rseek.org.But still some problems.
When I add axis(4) and axis(1,at=1:6,labels=gradeinfo$gradenam),the old
tick or labels still
are there as shown in the figure,how could I delete them( the old tick
information in x-axis and left y axis )
My script is
On 31/05/2010, Gabor Grothendieck ggrothendi...@gmail.com wrote:
Use read.csv or read.table(..., sep = ,). Also note that if you
delete the first comma of the header (as in the second example below)
you won't have to specify row.names since it can figure it out from
the fact that there is one
Hi
r-help-boun...@r-project.org napsal dne 01.06.2010 10:20:35:
On 31/05/2010, Gabor Grothendieck ggrothendi...@gmail.com wrote:
Use read.csv or read.table(..., sep = ,). Also note that if you
delete the first comma of the header (as in the second example below)
you won't have to specify
Thank you
Ben Bolker a écrit :
Gildas Mazo gildas.mazo at curie.fr writes:
Is there a simple way to save my figures in png instead of pdf with
Sweave ??
See
http://sites.google.com/site/thibautjombart/r-packages
(scroll to the bottom)
Dear group,
Here is my data frame:
futures -
structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10,
CORN Jul/10, CORN Jul/10, CORN Jul/10, LIVE CATTLE Aug/10,
LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10,
SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10
),
On 2010-06-01 0:16, bill.venab...@csiro.au wrote:
xyzs- matrix(rnorm(3*10,0,1),ncol=3)
V- c(2,3,4)
system.time(vx- apply(t(xyzs) * V, 2 ,sum))
user system elapsed
1.060.021.08
system.time(wx- as.vector(xyzs %*% V))
user system elapsed
0 0 0
On 1 June 2010 11:34, Peter Ehlers ehl...@ucalgary.ca wrote:
Or, for a very slight further reduction in time in
the case of larger matrices/vectors:
as.vector(tcrossprod(V, xyzs))
I mention this merely to remind new users of the
excellent speed of [t]crossprod().
-Peter Ehlers
Thanks,
Hi,
Try with:
par(mfrow=c(11,2))
It should work better!
Remember that mfrow is an argument of the function par(), not a function
itself.
One other tip: think about using pdf, ps, png, or SVG devices, I find it
easier and nicer.
HTH,
Ivan
Le 6/1/2010 10:02, Noah Silverman a écrit :
I'm
As noted in the help file under CSV files:
These wrappers are deliberately inflexible: they are designed to
ensure that the correct conventions are used to write a valid file.
Attempts to change append, col.names, sep, dec or qmethod are ignored,
with a warning.
On Mon, May 31, 2010 at 8:33
It may not be the nicest solution, but my suggestion should work.
Have you tried plot(type=n,...), plotting the axes with axis(), and
plotting the data with lines()?
Ivan
Le 6/1/2010 10:10, Jie TANG a écrit :
thanks for your reply.
I have tried to use rseek.org.But still some problems.
When
On 2010-06-01 1:53, arnaud Gaboury wrote:
Brian,
If I do understand correctly, I must use in my function something else than
ddply() if I want to avoid any error each time my df has zero rows?
Am I correct?
You could define a function to handle the zero-rows case:
f - function(x){
Patrick,
When apply to this following df :
futures -
structure(list(DESCRIPTION = character(0), CREATED.DATE =
structure(numeric(0), class = Date),
QUANTITY = numeric(0), SETTLEMENT = character(0)), .Names =
c(DESCRIPTION,
CREATED.DATE, QUANTITY, SETTLEMENT), row.names = integer(0), class
Is there a reason why append=TRUE should not be used?
I actually use it a lot; that way I have less csv files created. I can
output everything to a single file. Since some statistical tests returns
lists with different number of elements, it is not always possible to
export the output all at
Hi,
You could use melt from the reshape package to create a long format
data.frame. This is more easy to plot with lattice or ggplot2, and you
can then use facetting to arrange several plots on the same page. The
dummy example below produces 10 pages of output with 10 graphs per
page.
On Tue, 1 Jun 2010, Peter Ehlers wrote:
On 2010-06-01 1:53, arnaud Gaboury wrote:
Brian,
If I do understand correctly, I must use in my function something else than
ddply() if I want to avoid any error each time my df has zero rows?
Am I correct?
You could define a function to handle the
On 2010-05-31 18:33, Steven Worthington wrote:
Dear R users,
I have recently begun to reuse some functions I made several months ago. The
scripts write to a .csv file using the 'write.csv' function with the append
option set to TRUE. This used to work fine, albeit with the warning
appending
First, this is not the list to discuss MacOS X binaries: that is
R-sig-mac.
Second, only one person really knows about the situation with package
binaries, the person who produces them (equally true for Windows and
for Mac OS X). And that person does not read R-help, having asked for
such
On May 31, 2010, at 11:40 PM, Nevil Amos wrote:
Is it possible to print page numbers in pdf() with multiple pages?
?mtext
thanks
Nevil Amos
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On Jun 1, 2010, at 12:51 AM, suman dhara wrote:
Sir,
I have n p-dimensional regressor vectors corresponding to n
responses. I
want to plot n regressor vectors in R. Is it possible to plot them
in R?
If yes, please send me the code.
?matplot
Thanks,
Suman Dhara
David Winsemius, MD
Dear All,
I am newbie to R, and I wanted to plot a barplots with R and in such a way
that It will also show me position which I can plot on the bar line.
Here is my code that I am using to plot,
chromosome - c(40.2, 35.6, 36.1, 29.6, 31, 29.6, 31, 29.4, 28.2, 23, 23,
28.2)
barplot (chromosome,
Hi
r-help-boun...@r-project.org napsal dne 01.06.2010 13:01:38:
Dear All,
I am newbie to R, and I wanted to plot a barplots with R and in such a
way
that It will also show me position which I can plot on the bar line.
Here is my code that I am using to plot,
chromosome - c(40.2,
On 06/01/2010 12:44 AM, Jie TANG wrote:
here ,I want to plot two lines in one figure.But I have two problems
1) how to move one of the y-axis to be the right ? I tried to the
commandaxis(2),But I failed.
2) how to add the axis information correctly.Since I have use the cmommand
On 06/01/2010 04:16 AM, Noah Silverman wrote:
Hi,
Working on a report that is going to have a large number of graphs and
summaries. We have 80 groups with 20 variables each.
Ideally, I'd like to produce ONE page for each group. It would have two
columns of 10 graphs and then the 5 number
Hi,
If you want to draw lines on your barchart then
aa = barplot(chromosome, col=purple, xlab=Oryza sativa Chromosomes,
border
= NA, space = 5, ylim = c(0,45))
returns the midpoints of each bar in the vector aa and then you can use
the lines() function to do the drawing.
Martyn
-Original
On 06/01/2010 09:01 PM, khush wrote:
Dear All,
I am newbie to R, and I wanted to plot a barplots with R and in such a way
that It will also show me position which I can plot on the bar line.
Here is my code that I am using to plot,
chromosome- c(40.2, 35.6, 36.1, 29.6, 31, 29.6, 31,
On 2010-06-01 5:01, khush wrote:
Dear All,
I am newbie to R, and I wanted to plot a barplots with R and in such a way
that It will also show me position which I can plot on the bar line.
Here is my code that I am using to plot,
chromosome- c(40.2, 35.6, 36.1, 29.6, 31, 29.6, 31,
Hi all,
Thanks to an offlist e-mail from Thomas (Lumley), I have spent the past few
days getting a bit more edumacated on additional restrictions on the
opportunity for R to appear on the iPhone/iPad. These in fact go well beyond
the GPL issue that I raised in my initial post, which in and of
Noah Silverman noah at smartmediacorp.com writes:
I'm running a long MCMC chain that is generating samples for 22 variables.
I have each run of the chain as a row in a matrix.
So: Chain[,1] is the column with all the samples for variable one.
Chain[,2] is the column with all the
Can anyone tell me how to calculate a mid-p value for a chi-squared test
in R?
Many thanks,
Andrew Wilson
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PLEASE do read the posting guide
Hi All,
I am trying to run a loop that will have varying numbers of rows with each
output.
Previously I have had the same number of rows so I would use (and I
appreciate that this will no doubt achieve some gasps as being thoroughly
inefficient!):
xdfrow-(0)
xdfrow1-(1:32)
xdfrow2-(33:64)
Hi,
I might be a bit tired so that I don't understand everything but I'm a
bit confused.
How would you like your DIST_LOOP matrix to look like?
What do you intend to use xdfrow1...xdfrow7? As I said, I might not be
at the best of my shape!
A sample dataset would really help to see what you
The closest explanation of the term loess I can find is in Local
regression models by Cleveland, W. S.; Grosse, E. Shyu, W. M. (Chapter
8 of Statistical models in S (1992) by Chambers, J. M. Hastie, T. J.)
... local regression models is called loess, which is short for local
regression,
Dear readers of R-help
as most of you will *not* be aware, R-help has continued to work the
way it does, only thanks to a dozen of volunteers,
see https://stat.ethz.ch/mailman/listinfo/r-help .
The volunteers manually moderate e-mails that look like spam (and
sometimes are and sometimes are
My issue relates to adding text to a matrix and finding that the text is
converted to a number.
This is the section of code I'm having trouble with:
# First, I load in a list of names from a .csv file to 'names'
names - read.csv(file(Names.csv))
# Then I define a matrix which will be
Hi Ivan,
Thanks for your help, your initial suggestion did not work, but that is no
doubt down to my lack of making sense!
Here is a short example of my dataset. Basically the loop is set up to match
the ID with the TO column based on DIST = 0. So A1 = 2, A1.1 =1, A2 = 4,
A2.1 = 3. That is
Sir,
Can you suggest some function for discriminant analysis for Binary response
having continuous as well as categorical regressors.
Thanks,
Suman Dhara
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Can any one help it will be very kind, loop statements
I have this table and some more records, I want to reshape it
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
TP53 Dis1 Dis2 Dis3 Dis4 Dis5 Dis6
DCI New1 New2 New3 New4
FDI Hi2 H3 H4
GHD I1 I3 I4 I5 I6 I7 I8
I want my new table or matrix to be some thing
On Tue, Jun 1, 2010 at 5:04 AM, Jessica Queree
j.j.que...@googlemail.com wrote:
My issue relates to adding text to a matrix and finding that the text is
converted to a number.
A matrix can only hold one type of data. Since you started with character,
that's what you get.
A dataframe can hold
Hello,
is there an equivalent to write.csv for saving images to a folder? E.g. if I
have a histogram or piechart I would need to write a command that saves the
resulting image as png or jpg into a certain folder (working directory).
Thanks
--
View this message in context:
Hello!
I have the following problem:
I have a file in R that has in the first row three informations in one
row that I would like to in three different rows.
The first row looks like this:
GenusA_SpeciesC_Tree
GenusA_SpeciesF_Tree
GenusB_SpeciesA_Shrub
...
I tried with strsplit and and
Hi fellow R developers/users,
I've recently revised a package called rtv, and now consider it
reasonably stable.
Description: A package for conveniently representing, manipulating and
visualising time data. Here, time is regarded as a random variable,
and objects are used to represent
There's something very unlogic in your code. You have the whole time the
same datafra
On Tue, Jun 1, 2010 at 1:51 PM, RCulloch ross.cull...@dur.ac.uk wrote:
Hi All,
I am trying to run a loop that will have varying numbers of rows with each
output.
Previously I have had the same number of
You might consider starting your search with
?png
Sarah
On Tue, Jun 1, 2010 at 5:34 AM, thoeb t.hoebin...@gmail.com wrote:
Hello,
is there an equivalent to write.csv for saving images to a folder? E.g. if I
have a histogram or piechart I would need to write a command that saves the
How about this:
testdata - data.frame(sp = c(GenusA_SpeciesC_Tree,
GenusA_SpeciesF_Tree, GenusB_SpeciesA_Shrub),
stringsAsFactors=FALSE)
# for one
unlist(strsplit(testdata[1,1], split=_))
# for all of them
do.call(rbind, sapply(testdata[,1], strsplit, split=_))
Sarah
On Tue, Jun 1, 2010 at
Hi,
I don't know if it would help you since your goal is not really clear to
me, but here are some thoughts:
x - c(GenusA_SpeciesC_Tree, GenusA_SpeciesF_Tree,
GenusB_SpeciesA_Shrub)
test - strsplit(x, _)
test
[[1]]
[1] GenusA SpeciesC Tree
[[2]]
[1] GenusA SpeciesF Tree
[[3]]
[1]
This might not be the most elegant way of doing it, but it should probably
work, given that data is always separated by a _.
string.1-strsplit(c(GenusA_SpeciesC_Tree,GenusA_SpeciesF_Tree,
GenusB_SpeciesA_Shrub),_)
matrix(unlist(string.1),ncol=3,byrow=TRUE)
2010/6/1 Joël Baumann
Hi,
I want to simulate a simple wireless network in R.
There is a source and destination and source sends some data packets to
the destination and some routing table and so on.
Does anyone have any idea and knowledge about this or are there any
packages for this.
The thing that I am going to do
I am trying to understand R arima function. Any pointers would be
appreciated.
Thank you,
Shakira.
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PLEASE do read the
could you just give us the output of dput() for the data you copied in the
mail?
eg dput(seal_dist[,1:100])
and an example of how you want your output. I guess I get what you want to
do, but it's not what your code is doing. And it will be difficult to put
that in a matrix, as you have different
Another option is to use the colsplit() function (in the reshape package):
testdata - data.frame(sp = c(GenusA_SpeciesC_Tree,
GenusA_SpeciesF_Tree, GenusB_SpeciesA_Shrub),
stringsAsFactors=FALSE)
library(reshape)
testdata - cbind(testdata, colsplit(testdata$sp, split=_, names=c(Genus,
Species,
Hi Ross,
I'm still really confused about your question.
Let me ask you some specific questions (maybe someone more experienced
would understand at once, but I'm no expert; I hope I can still help
you! In any case, I would like to understand for myself ;) )
- is seal_dist the name of your
Dear R experts,
I would really appreciate if you had an idea on how to use more
efficiently the aggregate method:
More specifically, I would like to calculate the mean of certain
values on a data frame, grouped by various attributes, and then
create a new column in the data frame that will have
Is this what you're looking for?
seal_list - split(seal_dist,sel)
out - lapply(seal_list,function(x){
indx - subset(x, x$DIST == 0)
x$TO_ID - indx$ID[match(x$TO, indx$TO)]
return(x)
})
output - unsplit(out,sel)
Cheers
Joris
On Tue, Jun 1, 2010 at 3:32 PM, RCulloch
Sorry, forgot to add the sel. This is the first line, then just run the
rest.
sel - as.factor(paste(seal_dist[,10],-,seal_dist[,5],sep=))
cheers
Joris
On Tue, Jun 1, 2010 at 4:56 PM, Joris Meys jorism...@gmail.com wrote:
Is this what you're looking for?
seal_list - split(seal_dist,sel)
It's easiest for us to help if you give us a reproducible example. We
don't have your datasets (ap.dat), so we can't run your code below.
It's easy to create sample data with the random number generators in R,
or use ?dput to give us a sample of your actual data.frame.
I would guess your
It is indeed ddply() from package plyr.
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: Tuesday, June 01, 2010 12:24 PM
To: Peter Ehlers
Cc: arnaud Gaboury; r-help@r-project.org
Subject: Re: [R] data frame manipulation with zero rows
On Tue, 1
Dear R-users,
I'm developing a package that heavily depends on another package
released under the GPL-2 license.
In addition, some few functions depends on other packages released
under the following licences (as described in the corresponding pages
of http://cran.r-project.org/web/packages/):
Dear All,
I'm getting a error while trying to apply the BreastCancer dataset
(package=mlbench) to kknn (package=kknn) that I don't understand as I'm new
to R.
The codes are as follow:
rm = (list = ls())
library(mlbench)
data(BreastCancer)
library(kknn)
BCancer = na.omit(BreastCancer)
d =
Jessica,
Two further comments:
1. When you read your data, your character vectors are
stored as factors; is that what you want?
2. You may be better off using a list.
-Peter Ehlers
On 2010-06-01 7:47, Sarah Goslee wrote:
On Tue, Jun 1, 2010 at 5:04 AM, Jessica Queree
Hi Joris,
Thanks for your help!
The data as requested:
structure(list(FROM = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L),
TO = c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L,
Hi Ivan,
Thanks again for your help! I'll just go through your questions...
I'm still really confused about your question.
-Sorry!!!
Let me ask you some specific questions (maybe someone more experienced
would understand at once, but I'm no expert; I hope I can still help
you! In any case, I
Hello, I have a question about how R can run automatically. Here is the
story:
A file called data.csv will be generated every couple of minutes in a
folder (overwrite itself). What I want R to do is:
1) scan the folder to find the file.
2) if the file is a newly generated file, process the
Esteemed R-forum subscribers,
I'm having a tough time configuring contrasts for my 3-way ANOVA. In short:
I don't know how to configure (all) my contrasts correctly in order to
specify (all) my comparisons of interest.
I succeeded getting my contrasts of interest set up for a simpler 2-way
Take a look at
?split (and unsplit)
eg:
Dur - rnorm(100)
Attr1=rep(c(A,B),each=50)
Attr2=rep(c(A,B),times=50)
ap.dat -data.frame(Attr1,Attr2,Dur)
split.fact - paste(ap.dat$Attr1,ap.dat$Attr2)
ap.list -split(ap.dat,split.fact)
ap.mean -lapply(ap.list,function(x){
?Sys.sleep
On Tue, 1 Jun 2010, zhangted001 wrote:
Hello, I have a question about how R can run automatically. Here is the
story:
A file called data.csv will be generated every couple of minutes in a
folder (overwrite itself). What I want R to do is:
1) scan the folder to find the file.
On 2010-06-01 8:07, zhangted001 wrote:
Hello, I have a question about how R can run automatically. Here is the
story:
A file called data.csv will be generated every couple of minutes in a
folder (overwrite itself). What I want R to do is:
1) scan the folder to find the file.
2) if the file
OK, then I was right. It's exactly what my code does.
Enjoy.
Cheers
On Tue, Jun 1, 2010 at 4:25 PM, RCulloch ross.cull...@dur.ac.uk wrote:
Hi Joris,
Thanks for your help!
The data as requested:
structure(list(FROM = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L,
?difftime
?file.info
file.info(filename)$mtime
Sys.sleep(20)
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Carolina
nikhil.l...@gmail.com
On Jun 1, 2010, at 10:07 AM, zhangted001 wrote:
Hello, I have a question about how R can run automatically. Here is
Dear Erik and R experts,
Thank you for the fast response!
I include an example with the ChickWeight dataset:
ap.dat - ChickWeight
matchMeanEx - function(ind,dataTable,aggrTable)
{
index - which((aggrTable[,1]==dataTable[[Diet]][ind])
(aggrTable[,2]==dataTable[[Chick]][ind]))
Hi Ross,
I think Joris answered your question, but to keep with your own code:
To set up a list, you can use:
DIST_LOOP - list()
Or, if you know the length:
DIST_LOOP - vector(mode=list, length=11)
You would then fill up with (in your for loop):
DIST_LOOP[[i]] - SEL_HR ##Actually, what you had
Mauricio Zambrano wrote:
Dear R-users,
I'm developing a package that heavily depends on another package
released under the GPL-2 license.
Are you including code from that package in yours, or just making use of
it? The former requires that you follow all the GPL rules about your
own.
Hi all,
I also couldn't help but notice that some of my messages are bounced for
following reason:
The message headers matched a filter rule
I included the header of one of the messages below, but neither of these
messages is sent trough Nabble, nor does any mail address has digits in it.
I
Dear group,
Here is my df (obtained with a read.csv2()):
df -
structure(list(DESCRIPTION = c(COTTON NO.2 Jul/10, COTTON NO.2 Jul/10,
PALLADIUM Jun/10, PALLADIUM Jun/10, SUGAR NO.11 Jul/10,
SUGAR NO.11 Jul/10), CREATED.DATE = c(13/05/2010, 13/05/2010,
14/05/2010, 14/05/2010, 10/05/2010,
Please look at the maiz example, the last example in ?MMC in the HH package.
Follow the example all the way to the end. It illustrates a problem and
then the resolution
of the problem.
install.packages(HH) ## if you don't have it yet.
library(HH)
?MMC
Rich
[[alternative HTML version
Where does the problem comes from?? Maybe from my sytem date ??
Simply from not reading the options carefully enough, from ?strptime,
‘%y’ Year without century (00-99). If you use this on input, which
century you get is system-specific. So don't! Most often
values
Hello,
The lowercase 'y' is year without century. This should work for you:
as.Date(x=df$CREATED.DATE, format=%d/%m/%Y)
HTH,
Josh
On Tue, Jun 1, 2010 at 8:57 AM, arnaud Gaboury arnaud.gabo...@gmail.com wrote:
Dear group,
Here is my df (obtained with a read.csv2()):
df -
Change this line to :
pose$CREATED.DATE=as.Date(pose$CREATED.DATE,%d/%m/%Y) # mind the capital
Y
pose
DESCRIPTION CREATED.DATE QUANITY CLOSING.PRICE
1 COTTON NO.2 Jul/10 2010-05-13 1 81.2000
2 COTTON NO.2 Jul/10 2010-05-13 1 81.2000
3 PALLADIUM Jun/10 2010-05-14
Hi Jessica,
this tells me that your text is saved as a factor.
Try :
names - read.csv(file=Names.csv,stringsAsFactors=F)
Cheers
Joris
On Tue, Jun 1, 2010 at 11:04 AM, Jessica Queree
j.j.que...@googlemail.comwrote:
My issue relates to adding text to a matrix and finding that the text is
TY for the tip. The lower case is in fact the culprit.
-Original Message-
From: Erik Iverson [mailto:er...@ccbr.umn.edu]
Sent: Tuesday, June 01, 2010 6:05 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] as.date
Where does the problem comes from?? Maybe from
Stella Pachidi wrote:
Dear Erik and R experts,
Thank you for the fast response!
I include an example with the ChickWeight dataset:
ap.dat - ChickWeight
matchMeanEx - function(ind,dataTable,aggrTable)
{
index - which((aggrTable[,1]==dataTable[[Diet]][ind])
Type in Google
Arima R
Read the first hit, the third, the fifth, and any other that says tutorial
Cheers
Joris
On Tue, Jun 1, 2010 at 4:14 PM, shakira M m.shak...@gmail.com wrote:
I am trying to understand R arima function. Any pointers would be
appreciated.
Thank you,
Shakira.
Dear Erik,
Thank you very much. Indeed ave did the same job amazingly fast! I did not
know the function before.
Many thanks to all R experts who answer to this mailing list, it's amazing
how much help you offer to the newbies :)
Kind regards,
Stella
On Tue, Jun 1, 2010 at 6:11 PM, Erik Iverson
Dear All,
Thank you very much for your kind help and support I got it.
Jeet
On Tue, Jun 1, 2010 at 5:32 PM, Jim Lemon j...@bitwrit.com.au wrote:
On 06/01/2010 09:01 PM, khush wrote:
Dear All,
I am newbie to R, and I wanted to plot a barplots with R and in such a way
that It
Which kind of discriminant analysis?
If you mean LDA, use lda() in package MASS and read the correpsonding
book Modern Applied Statistics with S by Venables and Ripley published
by Springer.
Uwe Ligges
On 01.06.2010 09:24, suman dhara wrote:
Sir,
Can you suggest some function for
On 01.06.2010 02:19, David Winsemius wrote:
On May 31, 2010, at 8:14 PM, jim holtman wrote:
try this:
input - readLines(yourfile.txt)
# determine start
start - grep(\tBegin Main\t, input)[1] # first line if many
Puzzled. I thought backslashes in grepping patterns needed to be
doubled? I
Hi,
I used the term run, as each iteration of the Gibbs sampler produces
22 variables (coefficients for Beta in a regression model)
The example wont work
On 6/1/10 5:54 AM, Ben Bolker wrote:
Noah Silverman noah at smartmediacorp.com writes:
I'm running a long MCMC chain that is
Both pdf() and CairoPDF() produce vector graphics, hence there is no
things such as resolution required here.
Uwe Ligges
On 27.05.2010 11:09, Will Eagle wrote:
Dear all,
how can I set the resolution of embedded plots in PDF using pdf() or
CairoPDF() to a value of e.g. 600 dpi to meet
I guess something strange is loaded or changed in your personal startup
scripts (i.e. through your Renviron or Rprofile settings.
Uwe Ligges
On 27.05.2010 15:34, Rainer Machne wrote:
Hi,
I have a strange package installation problem after update to R 2.11.0
on Fedora Core 12.
A colleague
Hi Nitin,
It can be solved by splitting your data a bit different. You need more
training data than you have evaluation data, eg :
i1 = 1:400
i2=401:d
Then it works on my computer. No clue as to where the error originates from
though.
Cheers
Joris
On Tue, Jun 1, 2010 at 4:27 PM, Nitin
Thanks Jim,
That helps.
Ben Bolker had a nice suggestion on how to get the lattice package to
easily plot all 22 variables in one window.
Ultimately, I'd like to generate a PDF that will print on a standard
(8.5 x 11) page.
A few things I'm still stuck are:
1) How to use the lattice
2010/6/1 Duncan Murdoch murdoch.dun...@gmail.com:
Mauricio Zambrano wrote:
Dear R-users,
I'm developing a package that heavily depends on another package
released under the GPL-2 license.
Are you including code from that package in yours, or just making use of it?
The former requires
Hi Joris,
The matched a filter rule is the principal reason for holding
messages for moderation. Please don't become anxious about the
situation -- one of the reasons we have become concerned about
the situation is that people whose messages get held up do tend
to become worried about it. This is
Hello ,
I can not get apply function to do what I want when doing stuff on rows.
Let's say I want to divide the rows of matrix by their mean, the below show
you get the same result weather you use 1 or 2, i.e. same result for columns
than for means..:(
Thanks a lot for the help,
The mean by col and by row are the same:
colMeans(m) == rowMeans(m)
So:
m / rowMeans(m) # You don't need apply here
m / colMeans(m)
On Tue, Jun 1, 2010 at 2:26 PM, Joachim de Lezardiere
joachim.lez...@gmail.com wrote:
Hello ,
I can not get apply function to do what I want when doing
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