Hi
I would like to have something like
str <- "df$JT == 12"
fun <- function(df) {
b <- eval(parse(str))
return(b)
}
but for performance "eval(parse(a))" should not be evaluated at each
function call, but should work as
fun <- function(df) {
b <- df$JT == 12
return(b)
}
Do you have
Is it possible to use "paste" to write out an expression and evaluate it?
Suppose I want to add two vectors X1 and X2, defined as follows:
X1 <- 1:6
X2 <- 6:1
If I write the following it looks like what I want but is a character:
noquote(paste(paste("X", 1, sep = ""), paste("X", 2, sep = ""), sep
Dear Berend,
Â
No need to devide what all you need a indentity matrix with same
dimesion.
 Following is one way to achieve this.
Â
 mat Hi,
>
> I have a matrix as below:
>
> mat=
>Â Â Â Â [,1]Â Â [,2]Â Â [,3]
> [1,]Â Â Â Â 1Â Â Â Â 4Â Â Â Â 7
> [2,]Â Â Â Â 2Â Â Â Â 5Â Â Â Â 8
> [3,]
I dont know how to tell to the function 'svm' in this two cases: tune.svm or
tune (svm...) the type (I want regression, but by default it works with
classification) and the specify kind of kernel (by default it work with
radial)...
thank you so much!!!
--
View this message in context:
htt
The test statistic above does not reproduce Stata results when there is more
than one endogenous variable. While the code to compute the test statistic
appears to be correct, the difference may lie in Stata computing the test
statistic based on canonical correlations rather than the formula in Stoc
Hi,
I guess this is what you are looking for:
dat1<-read.table(text="
f1 f2 f3 f4 f5 f6 f7 f9 f10 f11
t1 1 0 1 0 1 0 0 0 0 1
t2 1 0 0 0 0 1 1 1 1 1
t3 0 0 0 0 0 0 0 0 0 0
t4 1 0 0 0 1 0 0 0 0 0
t5 0 0 0 0 0 0 0 0 0 0
t6 0 0 0
It's not quite clear what you want to do. Is it just this?
a<-c(1,2,3,4)
dim(a)<-c(2,2)
a/a
This gives the element by element ratio.
HTH,
Daniel
shukor wrote
>
> Hi,
>
> I have a matrix as below:
>
> mat=
> [,1] [,2] [,3]
> [1,]147
> [2,]258
> [3,]36
On 17-09-2012, at 07:31, Jha, Ashutosh Kumar wrote:
> Dear Berend,
>
> No need to devide what all you need a indentity matrix with same dimesion.
> Following is one way to achieve this.
>
> mat<-matrix(1,ncol=ncol(mat),nrow=nrow(mat)).
That is not an identity matrix. It is a matrix with
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 th
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 dele
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
>
>
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 whe
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 I'
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 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 0
Pedro Mardones 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 = .054589
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
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 wrote:
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 matri
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-user
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", "
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. I've
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
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.
>
>
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:
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 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 m
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: $R_HOME/etc/M
Le mercredi 12 septembre 2012 à 07:08 -0700, Tim Hesterberg a écrit :
> One approach is to bootstrap the vector 1:n, where n is the number
> of individuals, with a function that does:
> f <- function(vectorOfIndices, theTable) {
> (1) create a new table with the same dimensions, but with the coun
On 16-09-2012, at 20:47, Eberle, Anthony wrote:
> I have a question about how one can modify or override the compilers
> that R uses for package installations? Or if perhaps this configuration
> is in some editable file somewhere.
>
> Initially I built the version of R 2.15.1 on Solaris SPARC (
I have a question about how one can modify or override the compilers
that R uses for package installations? Or if perhaps this configuration
is in some editable file somewhere.
Initially I built the version of R 2.15.1 on Solaris SPARC (virtual T4),
but found out the build was done as 32 bit. Af
On Sun, Sep 16, 2012 at 7:34 PM, Sam Steingold wrote:
> Is it possible to violate the identity sum(table(v)) == length(v) ??
Quite easily:
x <- c(1:5, NA)
sum(table(x)) # 5
length(x) # 6
Perhaps look at the exclude= argument.
Cheers,
Michael
> I see no way to do that and it holds in my smal
Is it possible to violate the identity sum(table(v)) == length(v) ??
I see no way to do that and it holds in my small examples,
but it is violated in the huge set I have:
system.time(z <- unique(data.frame(u=U,s=S)))
tab1 <- table(z$u)
tab1 <- tab1[tab1>0] # S is factor so some counts were 0
tab2
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 bu
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 la
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
To: r-help@r-projec
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 wrote:
> Hey all,
>
> Virgin post to this list - hope I've got it right ;o)
>
> I've been learning R in
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 tha
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,
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
> Hello,
>
> Here's another one.
>
> logic.result <- with(rep_data, know %in% c("very well", "fairly well") &
> getalong %in% c(4,5))
> rep_data$clo <- 1*logic.resu
Hello,
The obvious simplification is to call union() only once. With 10M rows
it should save time.
Then I've asked myself whether unique() wouldn't be faster.
f1 <- function(x){
x[[1]] <- factor(x[[1]], levels = union(x[[1]], x[[2]]))
x[[2]] <- factor(x[[2]], levels = union(x[[1]], x
Dear Rui Barradas and Michael Weylandt,
Many thanks for your replies.
My second question is solved now.
But I think I did not expressed my first wish in a clear way
Indeed,
in ex<-"cbind(data$response1,data$response2),
I want to extract the variable name between "$" and "," (corresponds to
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':
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: plotti
> -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
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
ex<-"c
On Sun, Sep 16, 2012 at 3:35 PM, Özgür Asar 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
"resp
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 t
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 "$" with "."
I am wondering that is it possible efficiently doing these in R?
Best
Ozgur
--
View this message i
HI,
Try this:
set.seed(1)
dat1<-data.frame(col1=sample(0:1,1000,replace=TRUE),col2=sample(0:10,1000,replace=TRUE))
count(dat1$col1==1 & dat1$col2>5)[2,2]
#[1] 209
A.K.
- Original Message -
From: SirRon
To: r-help@r-project.org
Cc:
Sent: Sunday, September 16, 2012 6:41 AM
Subject: [R]
On Sep 16, 2012, at 4:40 AM, peter dalgaard wrote:
>
> On Sep 16, 2012, at 07:48 , David Winsemius wrote:
>
>>
>> On Sep 15, 2012, at 7:15 PM, mcg wrote:
>>
>>> Hello R-users,
>>>
>>> I would like to use subscript in chemical formulas for the different
>>> treatments in a boxplot.
>>> Fot t
Thank you for the many replies on this issue. I turns out qplot is not
suited to multiple annotations, so the best suggestions were to use
ggplot. The following worked for making an annotated stacked bar plot:
ggplot(algaedata) +
geom_bar(aes(x = year, y = cellsperml, colour = DIV, group = DIV
Hello,
Here's another one.
logic.result <- with(rep_data, know %in% c("very well", "fairly well") &
getalong %in% c(4,5))
rep_data$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 wa
On Sun, Sep 16, 2012 at 9:06 AM, Dennis 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),
> mean.beta
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
-
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$start)
[1] 5291000 10988025 11767950 11840900 12267450 12276675
Levels: 5291000 10988025 11767950 11840
Hello,
Since logical values F/T are coded as integers 0/1, you can use this:
set.seed(5712) # make it reproducible
n <- 1e3
x <- data.frame(A = sample(0:1, n, TRUE), B = sample(0:10, n, TRUE))
count <- sum(x$A == 1 & x$B > 5) # 207
Hope this helps,
Rui Barradas
Em 16-09-2012 11:41, SirRon
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,
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
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 very
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)
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 cu
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 g
Thank you very much for very valuable comments.
They are very informative.
Bests,
Niklas
2012/9/16 Ted Harding
> [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 variable is
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