Thank you very much for very valuable comments.
They are very informative.
Bests,
Niklas
2012/9/16 Ted Harding ted.hard...@wlandres.net
[See at end]
On 15-Sep-2012 20:36:49 Niklas Fischer wrote:
Dear R users,
I have a reproducible data and try to create new variable clo is 1 if
know
On Fri, 14-Sep-2012 at 02:03PM -0400, Earl Brown wrote:
| Hello R-helpers.
|
| I've tried to recreate a parallel version of tapply() and table()
| using a combination of the parallel functions mclapply() and pvec()
| and papply(), but haven't been successful. In the end, I'm trying
| to get a
On 16-Sep-2012 05:22:47 David Winsemius wrote:
On Sep 15, 2012, at 7:17 PM, mcg wrote:
Dear moderator;
I'm on the R-Mailing list with the same (giepe...@gmail.com) email address,
still I get the Post by non-member... message. Am I not a member than?
It appears you are currently
Here is my code, which plots three lines with errorbar. How could I add an
extra line without errorbar to the plot? Thank you very much.
beta.data - data.frame (
method = rep(c(Wrong, Correct, Full Bayes), each = T_obs),
mean.beta = c(mean.beta1, mean.beta2, mean.beta3),
t = rep(points, 3),
Hello,
I'm working with a dataset that has 2 columns and 1000 entries. Column 1 has
either value 0 or 1, column 2 has values between 0 and 10. I would like to
count how often Column 1 has the value 1, while Column 2 has a value greater
5.
This is my attempt, which works but doesn't seem to be
On Sep 16, 2012, at 07:48 , David Winsemius wrote:
On Sep 15, 2012, at 7:15 PM, mcg wrote:
Hello R-users,
I would like to use subscript in chemical formulas for the different
treatments in a boxplot.
Fot title, xlab and ylab sub- and superscript is no problem, but for the
different
On 16.09.2012 12:41, SirRon wrote:
Hello,
I'm working with a dataset that has 2 columns and 1000 entries. Column 1 has
either value 0 or 1, column 2 has values between 0 and 10. I would like to
count how often Column 1 has the value 1, while Column 2 has a value greater
5.
This is my attempt,
Hello,
Since logical values F/T are coded as integers 0/1, you can use this:
set.seed(5712) # make it reproducible
n - 1e3
x - data.frame(A = sample(0:1, n, TRUE), B = sample(0:10, n, TRUE))
count - sum(x$A == 1 x$B 5) # 207
Hope this helps,
Rui Barradas
Em 16-09-2012 11:41, SirRon
Hello,
Relative to the op's request for rectangls, I'm not understanding them.
In your plot using geom_bar, the levels of as.factor(start) are sorted
ascending. If both
as.factor(mydata$start)
[1] 5291000 10988025 11767950 11840900 12267450 12276675
Levels: 5291000 10988025 11767950
Hi Niklas,
I like A.K.'s method. Here's another way to do what I think is the same
thing you're asking for (this is how I did it before I knew ifelse()
existed!)
rep_data$clo - 0
rep_data[ rep_data$know %in% c(very well, fairly well)
rep_data$getalong %in% c(4,5),]$clo - 1
Best,
Steve
On Sun, Sep 16, 2012 at 9:06 AM, Dennis dcfl...@gmail.com wrote:
Here is my code, which plots three lines with errorbar. How could I add an
extra line without errorbar to the plot? Thank you very much.
beta.data - data.frame (
method = rep(c(Wrong, Correct, Full Bayes), each = T_obs),
Hello,
Here's another one.
logic.result - with(rep_data, know %in% c(very well, fairly well)
getalong %in% c(4,5))
rep_data$clo - 1*logic.result # coerce to numeric
Rui Barradas
Em 16-09-2012 13:29, Stephen Politzer-Ahles escreveu:
Hi Niklas,
I like A.K.'s method. Here's another way to
Thank you for the many replies on this issue. I turns out qplot is not
suited to multiple annotations, so the best suggestions were to use
ggplot. The following worked for making an annotated stacked bar plot:
ggplot(algaedata) +
geom_bar(aes(x = year, y = cellsperml, colour = DIV, group =
On Sep 16, 2012, at 4:40 AM, peter dalgaard wrote:
On Sep 16, 2012, at 07:48 , David Winsemius wrote:
On Sep 15, 2012, at 7:15 PM, mcg wrote:
Hello R-users,
I would like to use subscript in chemical formulas for the different
treatments in a boxplot.
Fot title, xlab and ylab
HI,
Try this:
set.seed(1)
dat1-data.frame(col1=sample(0:1,1000,replace=TRUE),col2=sample(0:10,1000,replace=TRUE))
count(dat1$col1==1 dat1$col25)[2,2]
#[1] 209
A.K.
- Original Message -
From: SirRon thechrist...@gmx.at
To: r-help@r-project.org
Cc:
Sent: Sunday, September 16, 2012
Dear all,
I want to manipulate a character string such as
ex-cbind(data$response1,data$response2)
in R in two ways:
1) extracting the response1 portion of ex
2) replacing $ with .
I am wondering that is it possible efficiently doing these in R?
Best
Ozgur
--
View this message in
trim is for calculating trimmed mean such that
the fraction (0 to 0.5) of observations to be trimmed from each end of x
before the mean is computed
from ?help(mean)
Ozgur
--
View this message in context:
http://r.789695.n4.nabble.com/Usage-of-trim-in-mean-tp4643281p4643293.html
Sent from
On Sun, Sep 16, 2012 at 3:35 PM, Özgür Asar oa...@metu.edu.tr wrote:
Dear all,
I want to manipulate a character string such as
ex-cbind(data$response1,data$response2)
in R in two ways:
1) extracting the response1 portion of ex
I'm not sure what you mean by portion -- if you just want
Hello,
Try the following.
1)
pattern - response.
m - regexpr(pattern, ex) #gregexpr to get all response
regmatches(ex, m)
2)
gsub(\\$, \\., ex)
Hope this helps,
Rui Barradas
Em 16-09-2012 15:35, Özgür Asar escreveu:
Dear all,
I want to manipulate a character string such as
-Original Message-
From: ruipbarra...@sapo.pt
Sent: Sun, 16 Sep 2012 13:13:47 +0100
To: jrkrid...@inbox.com
Subject: Re: [R] qplot: plotting precipitation data
Hello,
Relative to the op's request for rectangls, I'm not understanding them.
Neither am I really, I just googled a
Maybe a bug in ggplot2::geom_rect?
I'm Cceing this to Hadley Wickham, maybe he has an answer.
Rui Barradas
Em 16-09-2012 17:04, John Kane escreveu:
-Original Message-
From: ruipbarra...@sapo.pt
Sent: Sun, 16 Sep 2012 13:13:47 +0100
To: jrkrid...@inbox.com
Subject: Re: [R] qplot:
I have a data frame with columns which draw on the same underlying
universe, so I want them to be factors with the same level set:
--8---cut here---start-8---
z - data.frame(a=c(a,b,c),b=c(b,c,d),stringsAsFactors=FALSE)
str(z)
'data.frame': 3 obs. of 2
Dear Rui Barradas and Michael Weylandt,
Many thanks for your replies.
My second question is solved now.
But I think I did not expressed my first wish in a clear way
Indeed,
in ex-cbind(data$response1,data$response2),
I want to extract the variable name between $ and , (corresponds to
Hello,
The obvious simplification is to call union() only once. With 10M rows
it should save time.
Then I've asked myself whether unique() wouldn't be faster.
f1 - function(x){
x[[1]] - factor(x[[1]], levels = union(x[[1]], x[[2]]))
x[[2]] - factor(x[[2]], levels = union(x[[1]],
Thanks Rui and Stephen,
They look very interesting. I am glad there are many ways to do so.
All the bests,
2012/9/16 Rui Barradas ruipbarra...@sapo.pt
Hello,
Here's another one.
logic.result - with(rep_data, know %in% c(very well, fairly well)
getalong %in% c(4,5))
rep_data$clo -
Hello,
This should do it. You can collapse the first two instructions, but I've
left it like this for clarity.
s - unlist(strsplit(ex, [,)[:blank:]]))
s - gsub(^.*\\$, , s)
s[nchar(s) 0]
Rui Barradas
Em 16-09-2012 17:26, Özgür Asar escreveu:
Dear Rui Barradas and Michael Weylandt,
Many
Hey all,
Virgin post to this list - hope I've got it right ;o)
I've been learning R intensively the last two weeks and gone from newbie
status to *reasonably* comfortable with it.
Here's an issue I just cannot solve however as it appears to be some kind
of bug in R itself. But I won't claim
Just found a typo elsewhere in the code which looks like it's the culprit.
I'm not sure if the report below is still relevant. Will advise if so.
On Sun, Sep 16, 2012 at 6:59 PM, Bit Rocker bitracket...@gmail.com wrote:
Hey all,
Virgin post to this list - hope I've got it right ;o)
I've
Hi,
Try this:
ex-cbind(data$response1,data$response2)
gsub(.*\\(.*\\$(.*)\\,.*\\$.*\\),\\1,ex)
#[1] response1
unlist(strsplit(gsub(.*\\(.*\\$(.*)\\,.*\\$(.*)\\),\\1 \\2,ex), ))
#[1] response1 response2
A.K.
- Original Message -
From: Özgür Asar oa...@metu.edu.tr
To:
Hi Davide,
I had some time this afternoon and I wonder if this approach is llkely to get
the results you want? As before it is not complete but I think it holds
promise.
On the other hand Rui is a much better programer than I am so he may have a
much cleaner solution. My way still looks
Does anyone have any guidance on swap and memory configuration when
running R v2.15.1 on UNIX/LINUX? Through some benchmarking across
multiple hardware (UNIX, LINUX, SPARC, x86, Windows, physical, virtual)
it seems that the smaller memory machines have an advantage.
Typically my organization
Is it possible to violate the identity sum(table(v)) == length(v) ??
I see no way to do that and it holds in my small examples,
but it is violated in the huge set I have:
system.time(z - unique(data.frame(u=U,s=S)))
tab1 - table(z$u)
tab1 - tab1[tab10] # S is factor so some counts were 0
tab2 -
On Sun, Sep 16, 2012 at 7:34 PM, Sam Steingold s...@gnu.org wrote:
Is it possible to violate the identity sum(table(v)) == length(v) ??
Quite easily:
x - c(1:5, NA)
sum(table(x)) # 5
length(x) # 6
Perhaps look at the exclude= argument.
Cheers,
Michael
I see no way to do that and it holds
I have a question about how one can modify or override the compilers
that R uses for package installations? Or if perhaps this configuration
is in some editable file somewhere.
Initially I built the version of R 2.15.1 on Solaris SPARC (virtual T4),
but found out the build was done as 32 bit.
On 16-09-2012, at 20:47, Eberle, Anthony wrote:
I have a question about how one can modify or override the compilers
that R uses for package installations? Or if perhaps this configuration
is in some editable file somewhere.
Initially I built the version of R 2.15.1 on Solaris SPARC
Le mercredi 12 septembre 2012 à 07:08 -0700, Tim Hesterberg a écrit :
One approach is to bootstrap the vector 1:n, where n is the number
of individuals, with a function that does:
f - function(vectorOfIndices, theTable) {
(1) create a new table with the same dimensions, but with the counts
On 16 September 2012 at 13:47, Eberle, Anthony wrote:
| I have a question about how one can modify or override the compilers
| that R uses for package installations? Or if perhaps this configuration
| is in some editable file somewhere.
You have several choices:
a) system-wide:
On 16 September 2012 at 13:30, Eberle, Anthony wrote:
| Does anyone have any guidance on swap and memory configuration when
| running R v2.15.1 on UNIX/LINUX? Through some benchmarking across
| multiple hardware (UNIX, LINUX, SPARC, x86, Windows, physical, virtual)
| it seems that the smaller
Thank you John,
you are giving me two precious tips (in addition, well explained!):
1. to use the package plyr (I didn't know it before, but it seems to
make the deal!)
2. a smart and promising way to use it
I can finally plot the partial results, to have a first glance and
compare to them
On 2012-09-16 05:04, Rui Barradas wrote:
Hello,
Since logical values F/T are coded as integers 0/1, you can use this:
set.seed(5712) # make it reproducible
n - 1e3
x - data.frame(A = sample(0:1, n, TRUE), B = sample(0:10, n, TRUE))
count - sum(x$A == 1 x$B 5) # 207
Another way:
On Sep 16, 2012, at 3:41 AM, SirRon wrote:
Hello,
I'm working with a dataset that has 2 columns and 1000 entries. Column 1 has
either value 0 or 1, column 2 has values between 0 and 10. I would like to
count how often Column 1 has the value 1, while Column 2 has a value greater
5.
This
On 2012-09-16 08:32, David Winsemius wrote:
On Sep 16, 2012, at 4:40 AM, peter dalgaard wrote:
On Sep 16, 2012, at 07:48 , David Winsemius wrote:
On Sep 15, 2012, at 7:15 PM, mcg wrote:
Hello R-users,
I would like to use subscript in chemical formulas for the different treatments
in a
Dear R-users;
I'm working with a a dataset that was previously used to fit a
nonlinear model of the form:
Y ~ a * (1 + b * log(1 - c * X^d))
The parameters published elsewhere are:
a = 1.758863, b = .217217, c = .99031, and d = .054589
However, there is no way I can replicate this result.
Dear all,
In the following code, I was trying to compute each row of the param
iteratively based
on the first row.
This likely is not the best way. Can anyone suggest a simpler way to
improve the code.
Thanks a lot!
Hannah
param - matrix(0, 11, 5)
colnames(param) - c(p, q, r,
On Sep 16, 2012, at 2:58 PM, Peter Ehlers wrote:
On 2012-09-16 08:32, David Winsemius wrote:
On Sep 16, 2012, at 4:40 AM, peter dalgaard wrote:
On Sep 16, 2012, at 07:48 , David Winsemius wrote:
On Sep 15, 2012, at 7:15 PM, mcg wrote:
Hello R-users,
I would like to use
i m using a package NbClust for cluster analysis. in the following algorithm
-NbClust(m, diss=NULL, distance = euclidean, min.nc=2, max.nc=15, method =
ward, index = all, alphaBeale = 0.1)
i want to define my own dissimilarity matrix of dimension 38*38. my original
data m is a matrix of
My first criteria is to make sure my application never swaps/pages due
to memory issues -- have enough physical memory so it never happens
and control what else is running on the machine. Once you start
paging, performance takes a real hit.
On Sun, Sep 16, 2012 at 2:30 PM, Eberle, Anthony
Hi there,
I used the command
sudo apt-get install r-base
to install R on an EC2 server as shown below:
http://www.r-bloggers.com/ec2-micro-instance-of-rstudio/
It works but the version of R installed is:
R.version.string
[1] R version 2.12.1 (2010-12-16)
I want to the latest version with
Pedro Mardones mardones.p at gmail.com writes:
Dear R-users;
I'm working with a a dataset that was previously used to fit a
nonlinear model of the form:
Y ~ a * (1 + b * log(1 - c * X^d))
The parameters published elsewhere are:
a = 1.758863, b = .217217, c = .99031, and d =
Hi,
I have big .csv file. I would like to filter that file into a new table.
For example, I have .csv file as below:
f1 f2 f3 f4 f5 f6 f7 f9 f10 f11
t1 1 0 1 0 1 0 0 0 01
t2 1 0 0 0 0 1 1 1 11
t3 0 0 0 0 0 0 0 0 00
t4 1 0 0
Hi,
I have big .csv file. I would like to filter that file into a new table.
For example, I have .csv file as below:
f1 f2 f3 f4 f5 f6 f7 f9 f10 f11
t1 1 0 1 0 1 0 0 0 01
t2 1 0 0 0 0 1 1 1 11
t3 0 0 0 0 0 0 0 0 00
t4 1 0 0
Hi,
I'm using the XML package to scrape data and I'm trying to figure out
how to eliminate the memory leak I'm currently experiencing. In the
searches I've done, it sounds like the existence of the leak is fairly
well known. What isn't as clear is exactly how to solve it. The
general process
Hi All,
I am analyzing a set of data collected by two-stage cluster sampling. My
model is
y_ij = mu + T_i + e_ij
where T_i is the ith treatment and e_ij is random error for the ijth
individual. I have MSE_within and MSE_between, which lead to MSE_T for the
model.
Suppose I have balanced data
On 17-09-2012, at 00:51, li li wrote:
Dear all,
In the following code, I was trying to compute each row of the param
iteratively based
on the first row.
This likely is not the best way. Can anyone suggest a simpler way to
improve the code.
Thanks a lot!
Hannah
param -
Hi,
I have a matrix as below:
mat=
[,1] [,2] [,3]
[1,]147
[2,]258
[3,]369
What I want to do is, I would like to divide each column with its own
value, in order to get value 1.
Is there any simple script for that?
[[alternative HTML version
On 17-09-2012, at 06:50, s.s.m. fauzi wrote:
Hi,
I have a matrix as below:
mat=
[,1] [,2] [,3]
[1,]147
[2,]258
[3,]369
What I want to do is, I would like to divide each column with its own
value, in order to get value 1.
Is there any
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