Hi,
Try the format from this example:
mp <- barplot(VADeaths)
tot <- colSums(VADeaths)
text(mp, tot+6, format(tot), xpd = TRUE, col = "red")
text(mp, c("A","B","C","D"), xpd = TRUE, col = "blue")
I hope it helps.
A.K.
- Original Message -
From: Manish Gupta
To: r-help@r-pr
Hi,
Sorry about that. My mistake.
Try this:
barplot(VADeaths, names=rep("", 4), col=gray(1:3/4+0.01))
legend("bottom", legend=expression(scriptstyle("Strongly
disagree/disagree"),scriptstyle("Neutral"),scriptstyle("Strongly
agree/agree")), fill=gray(1:3/4+0.01), inset=-0.2, bty='n', horiz=
I'm plotting a lattice bwplot from a data.frame and would like to add some
horizontal lines indicating some reference values (preferably with text
annotation) as well as a grid of regular log tics. The following illustrates
the plot:
data(ToothGrowth)
require("lattice")
p = bwplot(exp(len) ~ su
Here Text message is kind of error message for no data available.
regards
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that makes sense, thanks for the reply
On 13 June 2012 05:53, Uwe Ligges wrote:
>
>
> On 11.06.2012 11:45, Ceci Tam wrote:
>
>> diag(n) is alright when n = 5e3, it took 0.7 sec on my machine for
>> diag(5e3). However, it's slow when n = 23000, diag(23000) took 15 sec
>>
>
>
> 1. As others have s
Hello,
There's a shorter one, using merge().
DF2 <- merge(data.frame(x=1:8), DF1, all.x = TRUE)
DF2$y[is.na(DF2$y)] <- 0
Rui Barradas
Em 13-06-2012 02:02, arun escreveu:
Hi,
I got confused by your heading when I sent the previous reply. There is
another way to reach the same conclusion.
This does not meet my need as i need to write text as well.
On Wed, Jun 13, 2012 at 9:41 AM, Manish Gupta wrote:
> Hi,
>
> I am working on R plot but i need to write some text for the bars where the
> value is = 0
>
> > counts
>
> 3 4 5
> 0 12 * 0* 4
> 1 3 *0* 1
>
> http://r.789695
On Tue, 12 Jun 2012, Paul J Ossenbruggen wrote:
I have the same question. This paper may help you.
http://eeecon.uibk.ac.at/wopec2/repec/inn/wpaper/2012-10.pdf
Thanks for posting this, Paul. The paper above describes the current state
of affairs in "R2BayesX". But we are also working on exte
On Wed, Jun 6, 2012 at 2:43 AM, bestbird wrote:
> hi,
> I met some problems in R, plz help me.
> 1. How to do a intersect operation among several groups in one list, without
> a loop statement? (I think It may be a list)
> create data:
> myData <- data.frame(product = c(1,2,3,1,2,3,1,2,2),
Hi,
I am working on R plot but i need to write some text for the bars where the
value is = 0
> counts
3 4 5
0 12 * 0* 4
1 3 *0* 1
http://r.789695.n4.nabble.com/file/n4633189/Screenshot-2.png
My aim is to write some comment in plot.
http://r.789695.n4.nabble.com/file/n4633
? text
Michael
On Tue, Jun 12, 2012 at 8:49 PM, Manish Gupta wrote:
> Hi,
>
> I am working on R plot but i need to write some text for the bars where the
> value is = 0
>
>> counts
>
> 3 4 5
> 0 12 0 4
> 1 3 0 1
> http://r.789695.n4.nabble.com/file/n4633190/1.png
>
> My aim is to wr
As noted below:
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
For more on how to do that see:
i) https://github.com/hadley/devtools/wiki/Reproducibility
ii)
http://stackoverflow.com/questions/59632
Another alternative is to use
?subset
HTH,
Jorge.-
On Tue, Jun 12, 2012 at 6:06 PM, Rui Barradas <> wrote:
> Hello,
>
> The trick is to use index vectors. They allow us to do without loops.
>
> Try the following.
>
>
> muscle <- read.table(text='
>
> "ID""adj.P.Val" "logFC"
Thanks for the reply.
The one you provide is the vertical legend not for horizontal legend.
x.intersp= simply spreads out the space between symbols and legend text.
using the existing data in R for example,
barplot(VADeaths, names=rep("", 4), col=gray(1:3/4+0.01)) #five levels in
this data, but j
Hi -- I am wondering if there is a suite of R functions or an R package for the
following types of purpose:
Suppose I had a nxpxp array (say, A) and I wanted to invert each pxp "layer" in
the array (i.e., solve(A[i,,]) for i in 1:n). Or, suppose I had a nxpxq array
(A) and a nxqxr array (B) an
Hi,
Try this.
plot(0)
legend("topright", inset=c(0.2,0),legend=c("Strongly
disagree/disagree","Neutral","Strongly agree/agree"), lty=1:3, pch=1:3,
x.intersp=2)
A.K.
- Original Message -
From: Jianyun Wu
To: r-help@r-project.org
Cc:
Sent: Tuesday, June 12, 2012 10:45 PM
Subject:
Thanks it works
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__
R-help@r-project.org mailing list
https://st
Hi,
I am working on R plot but i need to write some text for the bars where the
value is = 0
> counts
3 4 5
0 12 0 4
1 3 0 1
http://r.789695.n4.nabble.com/file/n4633190/1.png
My aim is to write some comment in plot.
http://r.789695.n4.nabble.com/file/n4633190/2.png
How ca
Hi,
I got confused by your heading when I sent the previous reply. There is
another way to reach the same conclusion.
Try this:
DF1<-data.frame(x=c(2,3,4,6,7),y=c(0.12,0.25,0.11,0.16,0.20))
DF2<-data.frame(x=1:8,y=rep(NA,8))
df3<-merge(DF1,DF2,all=TRUE)
df3[is.na(df3)]<-0
g<-rle(df3[,1])
gV
Hi,
Try this:
dat1<-data.frame(x=1:8,y=c(NA,0.12,0.25,0.11,NA,0.16,0.20,NA))
dat1[is.na(dat1)]<-0
dat1
x y
1 1 0.00
2 2 0.12
3 3 0.25
4 4 0.11
5 5 0.00
6 6 0.16
7 7 0.20
8 8 0.00
A.K.
- Original Message -
From: Silvia Lucato
To: r-help@r-project.org
Cc:
Sent: Tuesday, June
Tena koe Jacob
Possibly your best option is to simulate.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jacob Warren (RIT Student)
Sent: Wednesday, 13 June 2012 4:40 a.m.
To: r-help@r-project.org
Subject:
can anyone help me?
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R-help@r-project.org mailing l
Dear R-help,
I have an xts data set that i have subset by date.
now it contains a date-time-stamp, and two columns (price and volume
traded): my objective is to create tables of volume traded at a price - and
i've been successfully using aggregate to do so in interactive use.
say the data looks
thanks. i think i understand: the difference is that the first command
converts my 'searched-for' date to a number and matches it, but the second
does not?
On 13 June 2012 12:58, Joshua Ulrich wrote:
> On Tue, Jun 12, 2012 at 9:48 PM, Matthew Johnson
> wrote:
> > Dear R experts,
> >
> > I am le
On Tue, Jun 12, 2012 at 9:48 PM, Matthew Johnson wrote:
> Dear R experts,
>
> I am learning the very useful XTS package, but cannot figure out the
> purpose of some commands.
>
> in particular, the .indexDate() command does not work as expected.
>
> say:
>
> x <- timeBasedSeq('2010-01-01/2010-01-0
Dear R experts,
I am learning the very useful XTS package, but cannot figure out the
purpose of some commands.
in particular, the .indexDate() command does not work as expected.
say:
x <- timeBasedSeq('2010-01-01/2010-01-02 12:00')
x <- xts(1:length(x), x)
then i can subset on date as follows:
Hi All,
I produced a barplot and made a horizontal legend below the graph.
Because the results are from a survey, there are three levels, namely
strongly disagree/disagree, neutral and strongly agree/agree.
> rownames(survey)[1] "Strongly disagree/disagree" "Neutral"
> "Stron
Hello,
The trick is to use index vectors. They allow us to do without loops.
Try the following.
muscle <- read.table(text='
"ID" "adj.P.Val" "logFC" "Gene.symbol"
"1419156_at" "5.32e-12""2.6462565" "Sox4"
"1433575_at" "5.32e-12""3.9417089" "Sox4"
"
I'm noticing a huge performance hit when using foreach with doSNOW. Any ideas
on why the parallel loop runs ~8 times slower than the same foreach run
sequentially?
> require(doSNOW)
Loading required package: doSNOW
Loading required package: foreach
foreach: simple, scalable parallel programmin
On 11.06.2012 11:45, Ceci Tam wrote:
diag(n) is alright when n = 5e3, it took 0.7 sec on my machine for
diag(5e3). However, it's slow when n = 23000, diag(23000) took 15 sec
1. As others have said already, for huge data, the Matrix implementation
for diagonals is the right idea, since it ac
A simplistic approach which works for the given example is as
follows: Let your data frame be "X". Do:
x <- 1:8
y <- numeric(8)
y[X$x] <- X$y
Y <- data.frame(x=x,y=y)
Then "Y" is the desired result.
cheers,
Rolf Turner
On 13/06/12 09:55, Silvia Lucato wrote:
Dear R users,
Hello,
(You forgot to unlist() the result of strsplit.)
Another way, saving some space between characters is to use strheight().
rotate90 <- function(x, y, label){
label <- unlist(strsplit(label, ''))
y <- y - (seq_along(label) - 1)*strheight(label)
list(x=x, y=y, label
> Dear R users,
>
> I have a very basic query, but was unable to find a proper anwser.
>
> I have the following data.frame
>
> x y
> 2 0.12
> 3 0.25
> 4 0.11
> 6 0.16
> 7 0.20
>
> and, due to further calculations, I need the data to be stored as
>
> x y
> 1
Read through this page on float (figure) placement:
http://en.wikibooks.org/wiki/LaTeX/Floats,_Figures_and_Captions
In particular, trying including \usepackage{float} in your TeX preamble
and then place figures using the big [H] option:
\begin{figure}[H]
That may accomplish what you want in term
Thank you, it works.
by the way, sometime I guess *apply functions act like loop, because
when I apply a complicated function, they run slowly and one by one.
On Tue, Jun 12, 2012 at 3:39 PM, Rui Barradas wrote:
> Hello,
>
> Try
>
> sapply(seq_len(ncol(mt2)), function(j) mt1[mt2[, j], j])
>
>
> H
Hello,
Try
sapply(seq_len(ncol(mt2)), function(j) mt1[mt2[, j], j])
Hope this helps,
Rui Barradas
Em 12-06-2012 17:19, Zhenjiang Lan escreveu:
mt1:
[,1] [,2] [,3]
[1,] "G" "A" "A"
[2,] "C" "T" "C"
mt2:
[,1] [,2] [,3]
[1,]122
[2,]212
[3,]22
Dear R users,
I have a very basic query, but was unable to find a proper anwser.
I have the following data.frame
x y
2 0.12
3 0.25
4 0.11
6 0.16
7 0.20
and, due to further calculations, I need the data to be stored as
x y
1
On Jun 12, 2012, at 4:02 PM, Greg Snow wrote:
If you insert `\n` between each letter, then it should do what you
want with a single call to text or mtext. To expand on David's
example:
plot(1:10)
text(5,5, paste( strsplit(txvec, ''), collapse='\n') )
In fact it could be made even easier:
Greetings,
I have a very, very, simple research question. I want to predict one
dichotomous variable using another dichotomous variable. Straightforward,
right? The issue is that the dataset has two issues causing some
complications for me.
1) The subjects are not independent -- they are sibli
On 11.06.2012 17:47, Duarte Viana wrote:
Hello all,
I am running the function "rda" of the vegan library, like I did many
times without having troubles, and I get the following error message:
"Error in if (rank) { : argument is not interpretable as logical".
I am just testing a response matri
I'm trying to analyze the following data set (sample):
"ID""adj.P.Val" "logFC""Gene.symbol"
"1419156_at""5.32e-12" "2.6462565" "Sox4"
"1433575_at""5.32e-12" "3.9417089" "Sox4"
"1428942_at""2.64e-11" "3.9163618" "Mt2"
"1
[1]Click here to unsubscribe if you no longer wish to receive our emails
Hello,
We have an excellent consultant, Pritesh Agrawal - SAP ABAP . Below is
his/her quick profile for your review.
Pritesh Agrawal - SAP ABAP -
Rate: Open Total Exp: 9+ Available: immediate
I just tested your code and I _think_ you have a misconception about
dbWriteTable().
Your code has some oddities so I am only guessing; for example, what is
"zz" and why is it in this snippet?
In the absence of information on the database TABLE, it is even harder
to guess what you are doing,
Try this,
rotate = function(x) paste(strsplit(x,"")[[1]],collapse="\n")
t <- "this is a text"
plot.new()
text(1/2,1/2,t)
par(lheight=0.8)
text(1/2,1/2,rotate(t))
HTH,
b.
On 13 June 2012 01:49, Stuart Rosen wrote:
> For labelling a plot, I am trying to rotate a character string using text()
>
Or use 'fixed=TRUE' as an argument to grepl to avoid the regular
expression matching (but learning regular expressions will be a useful
tool in the long run).
On Tue, Jun 12, 2012 at 9:15 AM, Jeff Newmiller
wrote:
> ?grepl
>
> Note that this function uses regular expressions, in which certain cha
If you insert `\n` between each letter, then it should do what you
want with a single call to text or mtext. To expand on David's
example:
plot(1:10)
text(5,5, paste( strsplit(txvec, ''), collapse='\n') )
On Tue, Jun 12, 2012 at 10:16 AM, David Winsemius
wrote:
>
> On Jun 12, 2012, at 9:49 AM
TTR::stockSymbols
Michael
On Tue, Jun 12, 2012 at 1:12 PM, Vikram Bahure
wrote:
> Dear R users,
>
> I was working on get.hist.quote to download data, but for this I need to
> know yahoo quotes (for stocks) from the beginning. Is there any simple way
> to get all symbols for particular market or
Hi,
I am building a website with using R and bioconductor packages. I am wondering
where I can find a good quality hosting service which provides servers running
R and allows me to install various bioconductor packages.
Does anyone have such experience?
Thank you.
-Xiaokuan
[[alternati
Hi JEssica
I have tried but for the moment I´m getting crazyso I let for the
future to understand for instance why the viewer doesn´t goes, and so on.
Actually I´m using "Tex Maker" and I attach code using
funtion \begin{lstlisting} so on
The only "inconvenience" in Latex is always is
Dear R users,
I was working on get.hist.quote to download data, but for this I need to
know yahoo quotes (for stocks) from the beginning. Is there any simple way
to get all symbols for particular market or I have to go on the website and
get the quotes.
Thanks for your help in advance.
Regards
V
I have the same question. This paper may help you.
http://eeecon.uibk.ac.at/wopec2/repec/inn/wpaper/2012-10.pdf
Best of luck.
Paul
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On Jun 12, 2012, at 1:58 PM, David Winsemius wrote:
On Jun 12, 2012, at 12:44 PM, Michael wrote:
Hi all,
Lets say I have a matrix A which is m x n.
I also have a mask matrix MASK which is m x n with values in T/F,
where T
values make a sub-matrix in regutangular shape...
I applied B=A[
On Jun 12, 2012, at 12:44 PM, Michael wrote:
Hi all,
Lets say I have a matrix A which is m x n.
I also have a mask matrix MASK which is m x n with values in T/F,
where T
values make a sub-matrix in regutangular shape...
I applied B=A[MASK] and it didn't work as expected...
Perhaps: (no
Thanks I have just subscribed.
John Kane
Kingston ON Canada
> -Original Message-
> From: nord...@dshs.wa.gov
> Sent: Tue, 12 Jun 2012 17:26:02 +
> To: jrkrid...@inbox.com, r-help@r-project.org
> Subject: RE: [R] Attempting to update from R 2.14 to 2.15
>
>> -Original Message
I have the same question. This paper may help you.
http://eeecon.uibk.ac.at/wopec2/repec/inn/wpaper/2012-10.pdf
Best of luck.
Paul
--
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Sent from the R help mailing list archive at
On Tue, Jun 12, 2012 at 11:44:51AM -0500, Michael wrote:
> Hi all,
>
> Lets say I have a matrix A which is m x n.
>
> I also have a mask matrix MASK which is m x n with values in T/F, where T
> values make a sub-matrix in regutangular shape...
>
> I applied B=A[MASK] and it didn't work as expect
The unreleased 0.6 version of igraph has a much faster implementation
for this, you can download a nightly build source R package from here:
https://code.google.com/p/igraph/downloads/list
Make sure that you download the right file, there are C, Python
packages there as well. I can send you a wind
Hello,
On Tue, Jun 12, 2012 at 12:44 PM, Michael wrote:
> Hi all,
>
> Lets say I have a matrix A which is m x n.
>
> I also have a mask matrix MASK which is m x n with values in T/F, where T
> values make a sub-matrix in regutangular shape...
>
> I applied B=A[MASK] and it didn't work as exp
> -Original Message-
> From: John Kane [mailto:jrkrid...@inbox.com]
> Sent: Tuesday, June 12, 2012 10:07 AM
> To: Nordlund, Dan (DSHS/RDA); r-help@r-project.org
> Subject: RE: [R] Attempting to update from R 2.14 to 2.15
>
> The problem is that the Ubuntu site still has 2.14 and won't upda
HI,
Maybe this link will be useful.
(http://ubuntuforums.org/showthread.php?p=11876430).
I also updated to R2.15 from 2.14 in Ubuntu 11.10 (Now I upgraded to 12.04, so
the update command is disabled). I added one line in /etc/apt/sources.list
deb http://cran.mtu.edu/bin/linux/ubuntu oneiric/
Providing a reproducible example is a good idea, but what about this:
> A <- matrix(1:20, nrow=4)
> MASK <- matrix(FALSE, nrow=4, ncol=5)
> MASK[2:3, 1:3] <- TRUE
> MASK
[,1] [,2] [,3] [,4] [,5]
[1,] FALSE FALSE FALSE FALSE FALSE
[2,] TRUE TRUE TRUE FALSE FALSE
[3,] TRUE TRUE TRUE
> -Original Message-
> From: John Kane [mailto:jrkrid...@inbox.com]
> Sent: Tuesday, June 12, 2012 9:42 AM
> To: Nordlund, Dan (DSHS/RDA); r-help@r-project.org
> Subject: RE: [R] Attempting to update from R 2.14 to 2.15
>
> Thanks Dan.
>
> Results
> sudo apt-key adv --keyserver keyserver.
The problem is that the Ubuntu site still has 2.14 and won't update til
something like Ubuntu 12.10 or whatever the next release is.
Thanks for all the help.
John Kane
Kingston ON Canada
> -Original Message-
> From: nord...@dshs.wa.gov
> Sent: Tue, 12 Jun 2012 16:58:27 +
> To: jr
Thanks arun,
Thanks arun . It looked good but no luck.
I have updated my /etc/apt/sources.list, apparently installed the gpg code etc
thanks to Dan N's pointers by no go. Still the same error message when doing
sudo ap-get update.
John Kane
Kingston ON Canada
> -Original Message---
I have an unbalanced design I would like to run a power analysis on.
What I have been able to find has pointed me to using the pwr.f2.test
function as described below. My problem is that I don't know how to
appropriately define the numerator and demoninator df.
If someone can help here is some mor
Dear R users,
I'm applying a correlation structure in a mixed model (nmle function) to
control for spatial correlation between land parcels that are adjacent to
each other. I generated X,Y coordinates in ArcGIS for each land parcel and
used them in the correlation form like this:
test.exp<-corExp
Hi all,
Lets say I have a matrix A which is m x n.
I also have a mask matrix MASK which is m x n with values in T/F, where T
values make a sub-matrix in regutangular shape...
I applied B=A[MASK] and it didn't work as expected...
Any thoughts?
[[alternative HTML version deleted]]
_
Thanks Dan.
Results
sudo apt-key adv --keyserver keyserver.ubuntu.com --recv-keys E084DAB9
[sudo] password for john:
Executing: gpg --ignore-time-conflict --no-options --no-default-keyring
--secret-keyring /tmp/tmp.28YKFHKFht --trustdb-name /etc/apt/trustdb.gpg
--keyring /etc/apt/trusted.gpg -
Hello,
On Tue, Jun 12, 2012 at 11:13 AM, R. Michael Weylandt
wrote:
> Long answer: use a closure if for some bizarre reason you just can't
> use a loop.
Do you mean use recursion?
Take care
Oliver
--
Oliver Ruebenacker
Bioinformatics Consultant (http://www.knowomics.com/wiki/
On Jun 12, 2012, at 11:45 AM, Maaike_020 wrote:
Hi all,
I just started using the rgl package in R, and I need to know how I
connect
the dots I created. A simple line from one coördinate (x0,y0,z0) to
another
(x1,y1,z1) will do. I tried using the code rgl.lines, but it doesn't
accept
lin
I am using the igraph package and I would like to obtain only 1
largest independent set from a sparse graph with aprox. 600 vertices.
When I used the largest.independent.vertex.sets command I cannot
obtain all the sets in a decent time, even if I need only 1. Any ideas
on how to solve this?
Use a 2-column matrix of integer subscripts, the first column
being the row indexes and the the second the column indexes:
> mt<-matrix(1:12,ncol=6)
> vt<-c(1,2,2,2,1,2)
> mt[cbind(vt,1:6)]
[1] 1 4 6 8 9 12
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Mess
I think you are looking for
mt[cbind(vt, seq_along(vt))]
which uses some trickines known as matrix indexing (basically, the two
columns created by cbind() are considered row & col indices -- then
seq_along() gives the col1,col2, bit as needed.
HTH,
Michael
On Tue, Jun 12, 2012 at 10:28 AM, Zhe
Hello,
On Tue, Jun 12, 2012 at 10:25 AM, Rantony wrote:
> Hi,
>
> here i have some code...
>
>
> a <-1
> b <- b+1
> c <- b+a/20
> if c >5
> { d<- 1 }
> else { d<- 0 }
>
> i want to repeat this code from the 2nd line [b <- b+1] without using loop.
> Actulay i want to use "*Label *" and cal
Hello,
Yes, there is. See ?grepl or help('grepl').
Hope this helps,
Rui Barradas
Em 12-06-2012 14:51, anjali escreveu:
Hi ,
Is there any inbuilt functions to check whether a substring is present in a
string and give the result as boolean
Thanks
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View this message in context:
http://r.78
Hello,
You're right, to put lists or vectors as elements of data frames is not
the best practice.
Note, however, that the opposite is not true, it's common and good
practice to have data frames and other objects as list elements,
especially if they are in some way related. If, for instance,
Dear John,
R-sig-mixed-models is a better list for this kind of questions.
It looks like the model finds no evidence for a random slope. Notice the very
small variance of the random slope. In the model without random intercept, the
random slope tries to mimic the effect of a random intercept.
I thought I solved my question but actually not...
What I really want is to translate a matrix:
mt1:
[,1] [,2] [,3]
[1,] "G" "A" "A"
[2,] "C" "T" "C"
mt2:
[,1] [,2] [,3]
[1,]122
[2,]212
[3,]222
[4,]111
I want to translate all the "1"
On Jun 12, 2012, at 9:49 AM, Stuart Rosen wrote:
For labelling a plot, I am trying to rotate a character string using
text() so that characters are upright and reading down, for
example, ...
L
i
k
e
t
h
i
s
.
It appears that par crt does not work with text. Does anyone have
any other s
?grepl
Note that this function uses regular expressions, in which certain characters
have special meanings, so depending on what string you are looking for you may
have to know something about regex patterns to get it to work.
-
Probably Spector
(http://www.amazon.com/Data-Manipulation-R-Use/dp/0387747303) if you
want a dead-tree book. Otherwise, just read the archives of this list
and the R-bloggers aggregator and you should be able to pick up most
everything you need.
Michael
On Tue, Jun 12, 2012 at 9:58 AM, XINLI LI
Is this homework? If it is, please read the posting guide.
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54
Hi,
I work in a kitchen production factory and I try to predict sales of kitchen
for an horizon of 12 weeks, and I have to turn into account promotions. My
sales are express in number of command, my promotion are express with dummy
variables. (1 if promotion, 0 else).
The first problem is that my
Hi all,
I just started using the rgl package in R, and I need to know how I connect
the dots I created. A simple line from one coördinate (x0,y0,z0) to another
(x1,y1,z1) will do. I tried using the code rgl.lines, but it doesn't accept
lines that don't stem from O (0,0,0). Please help!
Thanks!
Dear R User's
Please help me to debug this issue. I am trying to write some data ( i=
6) to PostgreSQL database, but it not writing.
Is there any issue in the way I use "dbWriteTable"?
++ Source Code
library("DBI")
library("RPostgreSQL")
drv1 <- dbDriv
See if the SECURE APT section of the following link helps.
http://cran.r-project.org/bin/linux/ubuntu/README
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
>
I am very new to Linux so I probably am doing something stupid but I cannot
seem to update to R 2.15
Using Ubuntu 12.02 Precise Penguin
I realise that debian packages are not updated regularly so I tried to follow
the insructions at the R-site
So far, I have modified /etc/apt/sources.list to r
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Zhenjiang Lan
> Sent: Tuesday, June 12, 2012 8:28 AM
> To: r-help@r-project.org
> Subject: [R] How to index a matrix with different row-number for each
> column?
>
> here's my qu
Thank you very much! that's exactly what I want.
On Tue, Jun 12, 2012 at 10:32 AM, R. Michael Weylandt
wrote:
> I think you are looking for
>
> mt[cbind(vt, seq_along(vt))]
>
> which uses some trickines known as matrix indexing (basically, the two
> columns created by cbind() are considered row &
here's my question: suppose I have a matrix:
mt<-matrix(1:12,ncol=6)
now I have a vector
vt<-c(1,2,2,2,1,2)
which means I want to get:
the 1st row for column1;
the 2nd row for column2;
the 2nd row for column3;
the 2nd row for column4;
...
that what I want is this vector:
1,4,6,8,9,12
Thierry,
Thank you for your thoughts. I agree with your analysis, but am still surprised
that the results are not approximately, but exactly the same to the limit of
the precision of the printed results. The exact comparability of the results
makes me wonder if something else is going on that I
Short answer: no.
Long answer: use a closure if for some bizarre reason you just can't
use a loop.
Bonus tip: change the last three lines to:
d <- c > 5
for speed and clarity!
Michael
On Tue, Jun 12, 2012 at 9:25 AM, Rantony wrote:
> Hi,
>
> here i have some code...
>
>
> a <-1
> b <- b+1
>
Hi all
I have searched for a solution to the following problem and even tried
emailing the original poster, but have not had any luck in finding an
answer.
Magnus Torfason-2 wrote
>
> I'm getting the following warnings inside a loop:
>
> Warning messages:
> 1: In calc.measures(g, 1, i) :
>
grepl
Michael
On Tue, Jun 12, 2012 at 8:51 AM, anjali wrote:
> Hi ,
> Is there any inbuilt functions to check whether a substring is present in a
> string and give the result as boolean
> Thanks
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/String-Manipulation-in-R-t
Dear Group:
What is the best reference book on data manipulation with R in the
market now?
Thanks,
XING
__
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/post
R 2.15.0
Windows XP
Can someone help me understand why a random intercept model gives the same
results as the random intercept and slope models?
I am rather surprised by the results I am getting from lme. I am running three
models
(1) random intercept
fitRI <- lme(echogen~time,rando
Your problem is that n.keep is less than 0. Why that is I cannot tell
without appropriate context.
E.g.,
x <- 1:10
x[2:5]
x[0:1]
x[(-3):1]
Michael
On Tue, Jun 12, 2012 at 6:01 AM, sanshine wrote:
> Hello everyone,
>
> I`m trying to normalize and analize an illumina SNP array.
> But when i`m
Hi,
here i have some code...
a <-1
b <- b+1
c <- b+a/20
if c >5
{ d<- 1 }
else { d<- 0 }
i want to repeat this code from the 2nd line [b <- b+1] without using loop.
Actulay i want to use "*Label *" and call *GoTo* method like in other
coding Languages.
For eg:- like this,
Rui Barradas wrote
>
> Hello,
>
> From the help page for readBin:
>
> "readBin and writeBin read and write C-style zero-terminated character
> strings."
>
>
> Your description of the string doesn't match this one. To read any sort
> of character strings, use readChar.
>
> readChar(to.read
Thanks for the suggestion, Thierry.
Nevertheless, in this example I'm not considering "shoe" as a random,
nuisance factor with zero mean. I'm considering three specific shoe
models, and I'm interested in modelling how the output changes between
the different shoes for those grounds, given that the
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