hello,
I have a question regarding the function WebCorpus..When I am using this
function its showing error: couldnt find the function...I have downloaded
the tm package...Can you help me in this regard? Also I want to know how we
can data/information about a product or topic from google,facebook e
Hello,
Did you upgrade your R version from a previous one to the current one? If yes,
probably you should use:
> update.packages(checkBuilt=TRUE).
Best Regards,
Pascal
- Mail original -
De : RobMusk
À : r-help@r-project.org
Cc :
Envoyé le : Samedi 9 juin 2012 9h52
Objet : [R] Matri
Hello, I am new to R appreciate all the help from the R experts in the
forum
I am doing a click tracking project , the click tracking tracks the click,
mouse over events . I am tracking what are the follow up events before some
one clicks on the main zone 4 from which we generate the revenue
I wanted to round rather than truncate timestamps
# create a data set to work with
s <- 30 * rnorm(10, 1, 0.1)
gps.timestamp <- Sys.time() + s*1:10
gps.timestamp <- c(round(gps.timestamp[1], "min"), gps.timestamp,
as.POSIXct("2012-06-08 17:32:59"), as.POSIXct("2012-06-08 17:32:30"))
f <- funct
There are almost surely a brazillion ways. Here's one:
Call your data frame "X". Do:
XT <- xtabs(AvailableMW~Bands+FuelTypeNum,data=X)
PXT <- 100*XT/apply(XT,1,sum)
print(round(PXT,2))
This gives percent of fuel type 1 in band PB0 as
0.54 53.87 3.93 41.65
to two decimal places
Hello,
I'm using R code that includes a residual permutation that was written as a
supplement to the paper:
Turner et al. 2010. A general hypothesis-testing framework for stable isotopes
ratios in ecological studies. Ecology 91:2227-2233.
The supplemental code is available at:
http://www.esa
Hi R users all ,
I have a clean install of R-2.15.0 in a win32 machine and a problem loading
Matrix 1.0-6 that looks like this:
> library(Matrix)
Loading required package: lattice
Error : object ‘kronecker’ is not exported by 'namespace:methods'
Error: package/namespace load failed for ‘Matrix’
>
On Jun 8, 2012, at 6:15 PM, jcrosbie wrote:
How would I find the Percent of FuelTypeNum within the Band given
AvailableMW?
example:
type 1 is 1% of PB0
type 2 is 54% of PB0
type 4 is 4% of PB0
type 5 is 42% of PB0
format( 100*prop.table(table(dfrm$FuelTypeNum)), 0 )
... should do
On Jun 8, 2012, at 4:22 PM, Angelr wrote:
Hi, the global row indexing for my cluster assignment is as :
buf <- tapply(test1, test2, FUN = function(x){x;})
$`1`
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
$`2`
[1] 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
On 2012-06-08 14:33, Steve E. wrote:
Dear R Community - I hope you might be able to provide some guidance
regarding the use of the rle function. I have a set of time-series data
where a measured value is recorded every 30 seconds after the start of an
experiment. Many of the measured values repea
How would I find the Percent of FuelTypeNum within the Band given
AvailableMW?
example:
type 1 is 1% of PB0
type 2 is 54% of PB0
type 4 is 4% of PB0
type 5 is 42% of PB0
Note: the Bands and fuel types are not always constant.
Data:
FuelTypeNum Bands AvailableMW Avai
Dear R Community - I hope you might be able to provide some guidance
regarding the use of the rle function. I have a set of time-series data
where a measured value is recorded every 30 seconds after the start of an
experiment. Many of the measured values repeat and I am interested only in
the value
Hi, the global row indexing for my cluster assignment is as :
buf <- tapply(test1, test2, FUN = function(x){x;})
$`1`
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
$`2`
[1] 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
I used "tapply" function to find minimum as:
On Fri, Jun 8, 2012 at 5:31 PM, R. Michael Weylandt
wrote:
> For the matter and hand, you can probably get to it by simply trying
> base:::diag. In this case, it's not too hard because what you're
> seeing is the S4 generic that the Matrix package is defining _over_
> the regular base function gen
For the matter and hand, you can probably get to it by simply trying
base:::diag. In this case, it's not too hard because what you're
seeing is the S4 generic that the Matrix package is defining _over_
the regular base function generic.
More generally, going down the rabbit hole of S4:
As it sugg
How can one get the source code for diag? I tried the following:
> diag
standardGeneric for "diag" defined from package "base"
function (x = 1, nrow, ncol)
standardGeneric("diag")
Methods may be defined for arguments: x, nrow, ncol
Use showMethods("diag") for currently available ones.
I quickly looked at it, and the difference comes from:
n <- 5e3
system.time(x <- array(0, c(n, n))) # from diag()
system.time(x <- matrix(0, n, n)) # from Rdiag()
Replaced in R-devel.
Best,
Uwe Ligges
On 07.06.2012 12:11, Spencer Graves wrote:
On 6/7/2012 2:27 AM, Rui Barradas wrote:
Hel
Yes, I've been messing with that. I've also been using the hexView package.
Reading as characters first is just helping me figure out the structure of
this binary file. In this situation it really helped. For example:
È \001 \002
20012
This probably isn't how I'll do it in my final dra
When reading binary files, it is usually best to use readBin's
what=, size=, signed=, and endian= arguments to get what you want.
Reading as characters and then converting them as you are doing
is a very hard way to do things (and this particular conversion doesn't
make much sense).
Bill Dunlap
Sp
Thank you, that was very helpful.
Would it be possible to rename the legend values as well. Example 1 as
Biomass, 2 as coal, 4 as gas, 5 as hydro?
ggplot(data=tempTable, aes(x=Bands8, y=AvailableMWNewFormat,
fill=as.factor(FuelTypeNum))) +
geom_bar(position="stack", stat="identity")+
guides(
Okay, Bill smelt something wrong, so I must revise.
This works for large numbers:
prds = sapply(sapply(cnt_str,charToRaw),as.integer)
PS - this also solves an issue I've been having elsewhere...
PPS- Bill - I'm reading binary files...and learning.
thanks!
ben
On Fri, Jun 8, 2012 at 12:16 PM,
Of course, if you have data points at hand.
Thanks Bert, for pointing out.
Ozgur
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Can you tell us why you are interested in this mapping?
I.e., how did the "\001" and "\102" arise and why do you
want to convert them to the integers 1 and 102?
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help
Thanks for all your help. I did it this way:
> x = sapply(cnt_str,deparse)
> x
\002\001\002
"\"\\002\"" "\"\\001\"" "\"\\102\""
> as.numeric(substr(x,3,5))
[1] 2 1 102
...which is a bit of a hack, but gets me where I want to go.
Thanks,
Ben
On Fri, Jun 8, 2012 at 11:
On 08/06/2012 1:50 PM, Peter Langfelder wrote:
On Fri, Jun 8, 2012 at 10:25 AM, David Winsemius wrote:
>
> On Jun 8, 2012, at 1:11 PM, Ben quant wrote:
>
>> Hello,
>>
>> How do I change this:
>>>
>>> cnt_str
>>
>> [1] "\002" "\001" "\102"
>>
>> ...to this:
>>>
>>> cnt_str
>>
>> [1] "2" "
On Jun 8, 2012, at 19:25 , David Winsemius wrote:
>
> On Jun 8, 2012, at 1:11 PM, Ben quant wrote:
>
>> Hello,
>>
>> How do I change this:
>>> cnt_str
>> [1] "\002" "\001" "\102"
>>
>> ...to this:
>>> cnt_str
>> [1] "2" "1" "102"
>>
>> Having trouble because of this:
>>> nchar(cnt_str[1])
>>
On Fri, Jun 8, 2012 at 10:25 AM, David Winsemius wrote:
>
> On Jun 8, 2012, at 1:11 PM, Ben quant wrote:
>
>> Hello,
>>
>> How do I change this:
>>>
>>> cnt_str
>>
>> [1] "\002" "\001" "\102"
>>
>> ...to this:
>>>
>>> cnt_str
>>
>> [1] "2" "1" "102"
>>
>> Having trouble because of this:
>>>
>>> nc
On Jun 8, 2012, at 1:11 PM, Ben quant wrote:
Hello,
How do I change this:
cnt_str
[1] "\002" "\001" "\102"
...to this:
cnt_str
[1] "2" "1" "102"
Having trouble because of this:
nchar(cnt_str[1])
[1] 1
"\001" is ASCII cntrl-A, a single character.
?Quotes # not the first, second or
Hello,
How do I change this:
> cnt_str
[1] "\002" "\001" "\102"
...to this:
> cnt_str
[1] "2" "1" "102"
Having trouble because of this:
> nchar(cnt_str[1])
[1] 1
Thanks!
Ben
[[alternative HTML version deleted]]
__
R-help@r-project.org maili
I doubt that there is anythig like certfication in R
The same applies for interview questions.
For tutorials just google "r statistics tutorial"
John Kane
Kingston ON Canada
> -Original Message-
> From: sagarnikam...@gmail.com
> Sent: Fri, 8 Jun 2012 02:48:09 -0700 (PDT)
> To: r-help@r
yes you are right it was my inattention it is friday and my head needs
to start the week end sorry;-)
2012/6/8 David Winsemius
>
> On Jun 8, 2012, at 12:33 PM, Guido Leoni wrote:
>
> Dear list
>> Is there a way to extract a random sample without duplicated row from a
>> dataframe ?.
>> a
On Jun 8, 2012, at 12:33 PM, Guido Leoni wrote:
Dear list
Is there a way to extract a random sample without duplicated row
from a
dataframe ?.
a=c(1,2,3,1,1,1,2,1)
b=c(1,2,3,1,2,1,2,1)
c=c(1,1,1,1,1,1,1,1)
d=c(1,2,3,1,1,1,2,1)
prov<-data.frame(a,b,c,d)
prov2<-prov[sample(1:nrow(prov),5,repl
Dear Guido,
Try
prov2<-prov[sample(1:nrow(prov),5,replace=F),]
which corresponds to without replacement sampling.
Best
Ozgur
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Dear list
Is there a way to extract a random sample without duplicated row from a
dataframe ?.
a=c(1,2,3,1,1,1,2,1)
b=c(1,2,3,1,2,1,2,1)
c=c(1,1,1,1,1,1,1,1)
d=c(1,2,3,1,1,1,2,1)
prov<-data.frame(a,b,c,d)
prov2<-prov[sample(1:nrow(prov),5,replace=T),]
prov2
a b c d
3 3 3 1 3
6 1 1 1 1
On 06/08/2012 07:39 AM, R. Michael Weylandt wrote:
Well, I know that ROCR and gplots are on CRAN so you should probably
be installing them with install.packages() on your magical and
unspecified OS and version of R.
I don't really know the details, but you might check the version /
reinstall g
Dear Andras,
A possible option might be checking the fit of your data to a specific
distribution and generating the data from the distribution to which your
data fit, with the parameters such as the MLE of them obtained by using the
avaiable data.
For instance, you can check whether your data is
On Jun 8, 2012, at 11:33 AM, Peter Milne wrote:
I am interested in knowing whether and how I can test the
significance of
the relationship between my continuous predictor variable (a
covariate) and
my binary response variable according to two different groups, my
categorical predictor varia
You need to set the frequency attribute of your time series.
E.g.,
HoltWinters(ts(sample(100)))
HoltWinters(ts(sample(100), frequency = 4)) # Quarterly data
See ?ts for what frequency is and to figure out what makes sense for
your problem.
Best,
Michael
On Fri, Jun 8, 2012 at 9:49 AM, anindya
Dear netters,
Sorry for cross-posting this question. I am sure R-Help is not a
research methods discussion list, but we have many statisticians in
the list and I would like to hear from them. Any function/package in R
would be able to deal with the problem from this researcher?
-- Forward
Hello,
Try something along the lines of
res <- lapply(split(dat, list(dat$id, dat$trip)), function(x) x[x$time
== min(x$time), ])
do.call(rbind, res)
Note that 'data' is an R function, so I've called it 'dat'.
And that the column names are not exactly the same.
Also, you should follow the
I am interested in knowing whether and how I can test the significance of
the relationship between my continuous predictor variable (a covariate) and
my binary response variable according to two different groups, my
categorical predictor variable, in a logistic regression model (glm).
Specifically,
On Jun 7, 2012, at 12:55 , Gabor Grothendieck wrote:
> On Wed, Jun 6, 2012 at 4:40 PM, May Katharina
> wrote:
>> On Jun 6, 2012, at 10:19 , Gabor Grothendieck wrote:
>>
>>> On Wed, Jun 6, 2012 at 3:55 PM, May Katharina
>>> wrote:
Hello,
I'm trying to use na.spline (package zoo)
I am interested in knowing whether and how I can test the significance of
the relationship between my continuous predictor variable (a covariate) and
my binary response variable according to two different groups, my
categorical predictor variable, in a logistic regression model (glm).
Specifically,
On 06.06.2012 22:44, Santosh wrote:
Hello experts,
Sorry for posting the SPlus related question here.. I have not found any
solution yet after some attempts and hence, sending it to a wider spectrum
of users! I was successful in processing files uing R's XML librariy.
Not that we know, we ju
>
> Dear Duncan
>
> Many thanks again for your help!
>
> Now I've mended my code in the following simplified form:
>
>
> pdf("filename.pdf",paper="a4",width=8,height=12,encoding="default")
> par(mfrow=c(2,2))
>
> plot(1:length(SortedDataInList1[[3]][,1]),SortedDataInList1[[3]][,4],xlim=c(1,length(S
Dear all,
I am reading C++ code that is called in an R package. It is not clear to me
what SET_STRING_ELT, getAttrib, etc exactly do.
I understand that these are macros that have been declared in Rinternals.h and
who are used for the correct "translation" of the R data structures, but I have
n
I'm fairly new to R and still learning how to use it. I could really use some
help with the following problem.
I have a huge .csv file containing thousands of measurements on 34 different
birds. Measurements include longitude, latitude, altitude, speed, time, etc.
All birds have a different number
Hi,
I have the following data in a csv file
49893878, 54350306, 68914033, 46888379, 75506404,
54164263, 62846960, 78304638, 63721932, 70269568, 60440103, 79784327,
65918962, 76581629, 72016677, 47225594, 93944513, 65793666, 82709931,
87852261, 75876270, 88715213, 65496028, 80160380, 66089429, 1
On 08/06/2012 10:42 AM, HJ YAN wrote:
>
> Dear Duncan
>
> Many thanks again for your help!
>
> Now I've mended my code in the following simplified form:
>
>
> pdf("filename.pdf",paper="a4",width=8,height=12,encoding="default")
> par(mfrow=c(2,2))
>
>
plot(1:length(SortedDataInList1[[3]][,1
?chol2inv
kjetil
On Fri, Jun 8, 2012 at 1:43 AM, wrote:
> Dear R list members,
>
> I have a vector of Cholesky parameterization of a matrix let say A. I would
> like to compute the determinant and inverse of the original matrix A from the
> vector of cholesky parameters , would you suggest an
On Fri, Jun 8, 2012 at 10:08 AM, Özgür Asar wrote:
> Hi,
>
> Isn't the Cholesky decomposition of A=L (L)^T where T stands for "transpose"
> and L is the Cholesky factor of A.
>
> You say you have the Cholesky decomposition, isn't it L (above)?
>
> A<-L%*%t(L)
> det(A)
> solve(A)
>
> would be your
On Jun 8, 2012, at 16:08 , Özgür Asar wrote:
> Hi,
>
> Isn't the Cholesky decomposition of A=L (L)^T where T stands for "transpose"
> and L is the Cholesky factor of A.
>
> You say you have the Cholesky decomposition, isn't it L (above)?
>
> A<-L%*%t(L)
> det(A)
> solve(A)
>
> would be your
Well, I know that ROCR and gplots are on CRAN so you should probably
be installing them with install.packages() on your magical and
unspecified OS and version of R.
Michael
On Fri, Jun 8, 2012 at 2:07 AM, guoshicheng2005
wrote:
> I meet lots of problem when installing the package ROCR, do you ha
I'll summarize the results in terms of total run time for the suggestions
that have been made as well as post the code for those that come across this
post in the future. First the results (the code for which is provided
second):
What I tried to do using suggestions from Bert and Dan:
t1
# user
I wouldn't go quite so far as to say there's absolutely nothing else
-- one could, e.g., also fit lognormal, gamma, beta or most any other
two parameters distributions from the supplied data [assuming the
support matches].
What I did say is that you need domain specific knowledge to pick a
distrib
Hi,
Isn't the Cholesky decomposition of A=L (L)^T where T stands for "transpose"
and L is the Cholesky factor of A.
You say you have the Cholesky decomposition, isn't it L (above)?
A<-L%*%t(L)
det(A)
solve(A)
would be your answer.
Hope this helps
Ozgur
--
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ht
On 06.06.2012 23:53, Gurubaramurugeshan, Arun wrote:
Hi,
I am trying to download data off of this web site
http://www.eia.gov/oil_gas/petroleum/data_publications/wrgp/mogas_history.html
I used to set the proxy following code to set the proxy
Sys.setenv(wget="http://username:password@"proxy se
Hello,
Your cbind() are wrong, they are NOT binding columns, just column.
The problem would be solved with
beta <- cbind(beta, summary(Res)$coef[, 1])
and the same for 't.value' and 'sd'.
By the way, 'sd' is a bad name for a variable, it's already an R function.
A better way of solving the pro
On Jun 8, 2012, at 3:52 AM, maxbre wrote:
thanks david,
yes, you are right PART of the confusion is because of what you
mentioned
(sorry for that) but going back to my own data this is JUST PART of
the
problem…
...see my reproducible example
teq<-structure(list(site = structure(c(4L, 2L,
Thanks Gabor,
The trick to solve my problem is the %j, many thanks. I've seen now
it's documented in the help page of strptime as you indicate.
But please note that using format="ymd" I do get a chron object
and note that the value is the date without the 2 first digits
> require(chron)
Loading
Hi all,
I was wondering why I get errors trying to solve this:
*simeq <- function(x) {
f <- numeric(length(x))
f[1] <- x[1] * dnorm((log(x[1]/D) + (r + x[2]^2/2) * T)/(x[2] * sqrt(T)))
- D * exp(-r * T) * dnorm((log(x[1]/D) + (r + x[2]^2/2) * T)/(x[2] *
sqrt(T)) - x[2] * sqrt(T))
f[2] <- dn
Hey,
thanks a lot.
The polar.plot function from plotrix did exactly what I wanted to have!
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_
Dear Duncan
Thanks a lot for your hints.
As you can see from my code (just one line above the command using dev.copy) I
have tried using pdf but got same problem, so I hushed it out.
Any ideas??
Many thanks!
HJ
Sent using BlackBerry® from Orange
-Original Message-
From: Duncan Murd
Hi,
Try this,
df.sort<-apply(df,2,sort)
> df.sort
x
A 1
D 2
I 2
B 3
J 8
G 12
H 33
E 34
F 44
C 51
df.sort<-as.data.frame(df.sort)
A.K.
- Original Message -
From: Johannes Radinger
To: R-help@r-project.org
Cc:
Sent: Friday, June 8, 2012 3:22 AM
Subject: [R] Sort 1-column
Hello Martin,
Did you try with 'polar.plot' from the 'plotrix' package?
Best Regards,
Pascal
- Mail original -
De : MartinD
À : r-help@r-project.org
Cc :
Envoyé le : Vendredi 8 juin 2012 18h01
Objet : [R] Rose plot (like a windrose)
Dear R Gurus,
I spent some time in looking for hel
oracle has their SCJP certification for java language
like that i want R certification for my job interview
Also
1) i want interview questions on R(any website/pdb/any source)
2)any video about R learning(tutorials), any CD/DVD learning software
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http://
Hi all,
Sorry for the rather uninformative subject, but the error I get is not very
informative either.
When using the XML and RCurl package to retrieve the content of an html page,
htmlTreeParse fails, printing out the beginning of the HTML:
Error in htmlTreeParse(getURL(url)) :
File htt
Hi R-listers,
Savings regression results after a loop is straightforward. But what about when
you have nested loops?
I am running a regression of the form
lm(y~1+x+M+ D[,i] + D[,j] + D[,k])
where x is the variable of interest. M and D are vectors with other covariates.
Vectors "M" and "x" ar
Dear Vito,
Thank you very much for the advice - I will try it and see how the regression
looks.
Best,
Lucas
On Jun 6, 2012, at 11:45 AM, Vito Muggeo (UniPa) wrote:
> dear Lucas,
> If you are interested in selecting the number of breakpoints here a possible
> remedy:
>
> 1. Fit a segmented
On Fri, Jun 8, 2012 at 7:16 AM, Jannis wrote:
> The following appears to be pretty "standard" (though not included in chron)
> to me:
>
> b = ISOdate(2001,12,12)
> format(b, '%j')
>
Using POSIXct here is error prone because it needlessly introduces
time zones into a problem where they are not r
The following appears to be pretty "standard" (though not included in
chron) to me:
b = ISOdate(2001,12,12)
format(b, '%j')
or is that what you refer to with
"I know I can make my own using julian" ?
No idea about the other questions though ...
Jannis
On 08.06.2012 10:01, Agustin Lobo w
On 12-06-08 7:06 AM, Jannis wrote:
Hi R users,
i use do.call() to run some functions on large datasets. For debugging
purposes I occasionally use browser. In the case of large datasets
supplied to do.call, this however results in R printing huge amounts of
numbers on the screen. Is there any wa
Hi R users,
i use do.call() to run some functions on large datasets. For debugging
purposes I occasionally use browser. In the case of large datasets
supplied to do.call, this however results in R printing huge amounts of
numbers on the screen. Is there any way to reduce this output?
I foun
On Fri, Jun 8, 2012 at 4:01 AM, Agustin Lobo wrote:
> Hi!
> Is not there an standard R function to retrieve the day of the year
> (since 1st Jan of the same year)?
> I know I can make my own using julian, but find it weird that having
> days(), months() etc doy() does not exist as an standard func
On 12-06-08 6:46 AM, yhj...@googlemail.com wrote:
Dear Duncan
Thanks a lot for your hints.
As you can see from my code (just one line above the command using dev.copy) I
have tried using pdf but got same problem, so I hushed it out.
That's not the right place to put the pdf() call, it shou
On Wed, Jun 6, 2012 at 12:54 PM, emorway wrote:
> useRs-
>
> I'm attempting to scan a more than 1Gb text file and read and store the
> values that follow a specific key-phrase that is repeated multiple time
> throughout the file. A snippet of the text file I'm trying to read is
> attached. The t
On Thu, Jun 7, 2012 at 1:40 PM, emorway wrote:
> Thanks for your suggestions. Bert, in your response you raised my awareness
> to "regular expressions". Are regular expressions the same across various
> languages? Consider the following line of text:
>
> txt_line<-" PERCENT DISCREPANCY =
Ola, outra vez.
Desculpa, está errado!
É
a[[2]]
a[["Deviance"]] # claro!
com a[[1]] ou a[["Df"]] graus de liberdade. Os p-values estão correctos.
Rui Barradas
Em 08-06-2012 11:14, Rui Barradas escreveu:
Ola,
Eu sou português, não sei escrever castellano/espanhol mas compreendo.
Aqui é m
Ola,
Eu sou português, não sei escrever castellano/espanhol mas compreendo.
Aqui é melhor em inglês.
Para responder à questão,
modelo <- glm(...etc...)
a <- anova(modelo, test="Chisq")
a
length(a) # 'a' tem 5 elementos
sapply(a, print) # ver o objecto 'a'
a[[4]] # valores da estatistica, co
Hi,
Please check function radial.plot() in package plotrix.
Regards,
Carlos Ortega
www.qualityexcellence.es
2012/6/8 MartinD
> Dear R Gurus,
> I spent some time in looking for help but didn't find a way to do what I
> want.
> I do have a vector (in Degrees) containing of 360 elements, one elem
I found the mac console and tried the first solution as well. It works fine.
R -e "Sweave('test.Rnw')" --args PatientId=2
<>=
argString<-commandArgs(TRUE)
print(argString)
argList<-strsplit(argString,"=")[[1]]
print(argList)
assign(argList[1],as.numeric(argList[2]))
print(PatientId)
@
results in
Hi
I have a problem on how to proceed with further steps in my analysis. I did
a linear OLS regression (ri,t=alpha*beta*rm,t+et) with my daily data of
stock and index returns. There is now the problem of arch in my error terms.
Thus I used the following r command:
/garch(resid_desn, order=c(0,2)
Dear R Gurus,
I spent some time in looking for help but didn't find a way to do what I
want.
I do have a vector (in Degrees) containing of 360 elements, one element per
degree on a circle.
The data is dimensionless and in the range of -0.2 to 0.5.
An Example:
Wind Dir [degrees], Value
1, 0.1
On 12-06-07 10:08 PM, HJ YAN wrote:
Dear R users
I am trying to exporting plots from R to an external folder, or to the
working directory, but the resolution of plots (pdf file) largely reduced.
Any way I can get same quality as my original plots?? e.g. I tested the
plotting part using one examp
On 06/08/2012 12:43 AM, David Studer wrote:
Could anyone please tell me what is the most elegant
way to divide an ordinal variable in equal groups? (as
cut() does with continous variables)
for example I'd like to have the factor "educational level"
in three groups "low" "medium" and "high"
Hi
Well, i have my System not set up to run from command line, so i can't help you
there, but the scipt solution
id<-1
Sweave('test.Rnw') #printing out the id
works fine for me. There were several other suggestions in the answers given in
the link.
Am 07.06.2012 um 09:30 schrieb Manish Gupta:
>
Hi Agustin,
You can check and adapt this solution to your needs:
http://stackoverflow.com/questions/9465817/count-days-per-year
Regards,
Carlos Ortega
www.qualityexcellence.es
2012/6/8 Agustin Lobo
> Hi!
> Is not there an standard R function to retrieve the day of the year
> (since 1st Jan of
On Fri, 8 Jun 2012, Kazu Nada wrote:
Hi, I need help to do "mlogit" including "l( ) function.
When I put following command,
"Error in parse(text = x) : :2:0:" is shown.
(1) Your example is not reproducible for us, see the footer of this mail
and follow the posting guide.
(2) You manually
Hello,
Just put the entire regexp between parenthesis.
extracted <-
strsplit(gsub("([+-]?(?:\\d+(?:\\.\\d*)|\\.\\d+)(?:[eE][+-]?\\d+)?)","\\1%&",txt_line),"%&")
extracted
sapply(strsplit(unlist(extracted), "="), "[", 2)
As for speed, I believe that this might take longer. It will have to
Hi.
I need to finde some sections in a vector by a spesific pattern. I
need to recognise monotonically increasing sections in a one dimension
array. But there are some noises, which should not be taken into
account. Is there some way to do pattern matching similar to
Continuous Query Language (CQL)
Dear R list members,
I have a vector of Cholesky parameterization of a matrix let say A. I would
like to compute the determinant and inverse of the original matrix A from the
vector of cholesky parameters , would you suggest an R function to do the task.
I have tried hard but unable to find any
Dear R users
I am trying to exporting plots from R to an external folder, or to the
working directory, but the resolution of plots (pdf file) largely reduced.
Any way I can get same quality as my original plots?? e.g. I tested the
plotting part using one example and obtained pretty good (/readable
HI,
Have you tried italic()?
legend(expression(italic("Study area")).
It is untested as I am not in front of my linux system.
A.K.
legend(expression(italic("Expression species")
- Original Message -
From: Vikram Chhatre
To: r-help@r-project.org
Cc:
Sent: Thursday, June 7, 2012 8:27 P
Estimados amigos,
Estoy familiarizándome con los modelos lineales generalizados en R. Estoy
interesado en realizar un análisis lig linear y me gustaría saber cuáles
son o como extraer los valores correspondientes al chi cuadrado en el
análisis para cada grupo y para las interacciones. Desde ya mu
I meet lots of problem when installing the package ROCR, do you have meet such
problems?
1, biocLite("ROCR")
2, biocLite("gplots")
3, biocLite("Rgraphviz")
4, sudo apt-get install graphviz
oh, no, unlimited question, what's wrong with R in ROCR or gplots or et al
Error : object ânobsâ is no
Hi, I need help to do "mlogit" including "l( ) function.
When I put following command,
"Error in parse(text = x) : :2:0:" is shown.
res1<-(mlogit(choice~train+bus+plane+taxi+year+cost
+I(gen*train)+I(gen*bus)+I(gen*plane)+I(gen*taxi)+I(gen*year)+I(gen*cost)
+I(age*train)+I(age*bus)+I(age
Thanks, Don. the Cairo package was just what I needed.
It was a challenge to get it all set up on the serve since I don't have root
privileges. Had to install Pixman and Cairo, figure out to set TMPDIR so R
could build the package.
This page was critical is getting Cairo to compile from sou
Hello R-Experts,
I want to use "BBoptim" function to get mle. Is this possible? Practically I
want to check whether moment estimate of mle is really mle for a given data
set.
Thanks in advance.
Regards,
rehena
[[alternative HTML version deleted]]
Hi!
Is not there an standard R function to retrieve the day of the year
(since 1st Jan of the same year)?
I know I can make my own using julian, but find it weird that having
days(), months() etc doy() does not exist as an standard function.
Also, is the following not a bit inconsistent?
> a <- c
thanks david,
yes, you are right PART of the confusion is because of what you mentioned
(sorry for that) but going back to my own data this is JUST PART of the
problem…
...see my reproducible example
teq<-structure(list(site = structure(c(4L, 2L, 2L, 4L, 2L, 4L, 4L,
3L, 1L, 3L, 1L, 1L, 3L, 4L, 5
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