Note:
libmpi.dylib: cannot open shared object file: No such file or directory
Linux does not have .dylib's: Darwin (part of Mac OS X) does.
This is not an R question (it seems to be about your MPI installation)
and you need to discuss it with the package maintainer (see the posting
quide).
Hi Everyone! :D
Just need a little help on data frames. I was wondering if anyone would know
a way to rename the row numbers on a data frame.
e.g.
head(df)
Hats Dogs Horses
52 Glen9.53 0.00 0.00
18 Jane 1.48
Hi,
I am a developer from Oracle. I wanted to create a png file out of png device
with a bit depth of 8. I am using R on linux. But I noticed that R
automatically switches between 8 and 24 bit depth. i tried a lot of things from
your grDevices package but unable to control the bitdepth.
which package from CRAN used for Big-Data analysis ?
is there any separate package for Big-Data analysis?
or for making reports Business intelligence
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Hi,
You can try this and see. I'm assuming that the initial text file
named test.txt.
x-read.table(test.txt,header=T) # if headers are present in test.txt
or
x-read.table(test.txt)
# Actually, read.table() command skips the blank lines.
n-256
for (i in 1:100){
Hello,
Try the following.
rownames(df) - seq_len(nrow(df))
Hope this helps,
Rui Barradas
Em 10-10-2012 08:41, CrimMagic escreveu:
Hi Everyone! :D
Just need a little help on data frames. I was wondering if anyone would know
a way to rename the row numbers on a data frame.
e.g.
head(df)
estimate- optimx(init.par,Linn,gr=NULL,method= L-BFGS-B, hessian=TRUE,
control =
list(trace=1),lower=c(0,0,0,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf),upper=c(1,1,1,Inf,Inf,Inf,Inf,Inf,Inf,Inf))
fn is Linn
Function has 10 arguments
par[ 1 ]: 0 ? 0 ? 1 In Bounds
par[ 2 ]: 0 ? 0
This is operator error. Do not attempt to optimize over an infinite range.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
I'm completely new at R... I have sinkhole survey data (lat, long, and
elevation) and have been trying to plot a rotatable 3d plot
for several hours... cannot get it right.
You can use loess and predict.loess to generate a fitted 3d surface that can be
plotted using contour() and similar.
On 10/10/2012 07:23, Anshul Gupta wrote:
Hi,
I am a developer from Oracle. I wanted to create a png file out of
png
device with a bit depth of 8. I am using R on linux. But I noticed that
R automatically switches between 8 and 24 bit depth. i tried a lot of
things from your grDevices package
I keep getting this Error in x^2 : non-numeric argument to
binary operator
using multiple different codes, ones which have been verified
to work by my professor and other students.
I confirm the previous poster's lack of error in running the code supplied. I
also agree that something
learn how to debug your program. I think the call is:
options(error = recover)
don't have R on my iPad so cannot check. When you do this, you drop into the
'browser' at which point you can examine the value and see that you have a
non-numeric object.
Sent from my iPad
On Oct 10, 2012, at
With the following code :
dat1 - matrix(nrow=4, ncol=2)
dat1[1,] - c(-2, 1)
dat1[2,] - c(-1.7, 0.9)
dat1[3,] - c(0.1, 0.6)
dat1[4,] - c(0.5, 0.5)
theplot - xyplot(V2 ~ V1, as.data.frame(dat1), pch=c(4,1,5,4))
plot(theplot, prefix=theplot) # for a predictable name
On 10-10-2012, at 00:21, Adam Zeilinger wrote:
Dear R help,
Thanks again for the responses. I increased the lower constraint to:
lower = list(p1 = 0.0001, p2 = 0.0001, mu1 = 0.0001, mu2 = 0.0001).
I also included an upper box constraint of:
upper = list(p1 = Inf, p2 = Inf, mu1 =
Hi,
I am trying to fit the HoltWinters exponential smoothing on a monthly time
series data in R. My questions are:
1. I know that the level, trend and seasonality are updated over time. So
are the output coefficients a, b and s1-s12 for a specific time t (a,b and
s12 for last observataion for
Hello,
i have been doing browns exponential smooting for myself and have a little
trouble with plotting values:
par(xpd=TRUE)
plot(vector,xlab=Period,ylab=Values)
legend(max(vector), legend = c(Original values, Estimated values),
col=c(blue,red),lwd=0.5, cex=1, xjust=0.1, yjust=-0.3)
Sorry but I don't understand what your opinion.
Also try this initial values : ( 0.5, 0.5, 0.5, 1 ,1,1,1,1,1,1)
Then I got same error.
Regards,
Serdar
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Hi all,
I wrote a function that actually does what I want it to do, but it tends to be
very slow for large amount of data. On my computer it takes 5.37 seconds for
16000 data points and 21.95 seconds for 32000 data points. As my real data
consists of 1800 data points it would take ages to
Recently I have been using R in UBUNTU 11.10.
Can anyone please tell me how to connect R and Lyx in UBUNTU? I have been
successful in doing that under Windows but I got stuck for Ubuntu.
Thanks in advance
-Atanu
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View this message in context:
On 12-10-10 5:31 AM, piranha piranha wrote:
Hello,
i have been doing browns exponential smooting for myself and have a little
trouble with plotting values:
par(xpd=TRUE)
The line above says allow plotting outside the frame.
plot(vector,xlab=Period,ylab=Values)
You could set xlim and slim when using plot()
plot(vector,xlab=Period,ylab=Values,xlim=range(0,length(vector)+1),ylim=range(vector,est_vector,forecast))
i think - you forgot to provide data for the vectors :)
On 10.10.2012, at 11:31, piranha piranha wrote:
Hello,
i have been doing browns
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 10/10/12 11:57, ATANU wrote:
Recently I have been using R in UBUNTU 11.10. Can anyone please tell me how
to connect R and
Lyx in UBUNTU? I have been successful in doing that under Windows but I got
stuck for Ubuntu.
This is a LyX question,
From: 발송실패알림 naver-mai...@naver.com
Subject:[발송실패 안내]
envy721c@naver.cobSDsnLzroZwg66mU7J287J20IOyghOyGoeuQmOyngCDrqrvtlojsig==teuLiOuLpC4=
The only plain english in the message is that the mail was denied by the
receiver
Anyone else getting this?
Hello,
Yes, since yesterday. I thought it was a virus in my system but after
running the anti-virus twice and after your mail I guess it's something
else.
Rui Barradas
Em 10-10-2012 12:50, Jessica Streicher escreveu:
From: 발송실패알림 naver-mai...@naver.com
Subject:[발송실패 안내]
Hi,
From yesterday onwards, I also got the same message. I contacted
r-help-ow...@r-project.org. David replied:
I just got one too. It is an automated response from the mail server that
handles bobo:
bobo bleza...@gmail.com
Just put him on your PLONK list and you will not be bothered
Hello,
'outer' is a bad name for a function, it's already an R one. See ?outer.
As for your algorithm, it runs quadratically in the length of x and y so
you should expect a quadratic time behavior. What are you trying to do?
Your code gets max(x), max(y) and some other points near those. Can
hi
My string contain
string = The sales is good when my num1 between 1 to 5 . else the sales is
poor.
i want to find the patten between 1 to 5 . between and to will be constant
but the numbers would be changing. How to search for that pattern.
Please help
-
Thanks in Advance
With the following code :
I would like to plot 4 points, and have the circle and diamond shapes filled
with grey. What am I missing ?
Thanks by advance for your help,
Pierrick Bruneau
Research Fellow
CRP Gabriel Lippmann
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View this message in context:
(sorry for repetition: the previous mail resulted from a weird manipulation
in the forum)
With the following code :
dat1 - matrix(nrow=4, ncol=2)
dat1[1,] - c(-2, 1)
dat1[2,] - c(-1.7, 0.9)
dat1[3,] - c(0.1, 0.6)
dat1[4,] - c(0.5, 0.5)
theplot - xyplot(V2 ~ V1, as.data.frame(dat1),
I use by() to generate a summary statistics like so:
Lbys - by(dat[Nidx], dat$LipTest, summary)
where Nidx is an index vector with names picking out the columns in the
data frame dat.
This returns a list of character arrays (see below for str() output) where
the columns are named correctly but
On 10/9/2012 12:27 PM, Virgilio Gómez-Rubio wrote:
Dear all,
Perhaps this is not the place but... we are going to print some stickers
to take them to a conference for a sticker exchange with other free
software organisations. Last year I printed very simple stickers and
this year I would like
On Oct 10, 2012, at 14:53 , Michael Friendly wrote:
How about something along the lines of
http://www.ilovegenerator.com/i-love-r-1842509
I think something similar was produced for a previous useR! conference
-Michael
...as a promo for Revolution R. Been there, done that, got the
Dear all
I wonder if someone can explain what is the main difference between omega
and alpha reliabilities?
I understand an omega reliability is based on hierarchical factor model as
shown in this
graphhttp://rgm2.lab.nig.ac.jp/RGM_results/psych:omega.graph/omega.graph_002_big.png,
and
alpha
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Alex van der Spek
Sent: Wednesday, October 10, 2012 2:48 PM
To: r-help@r-project.org
Subject: [R] Summary using by() returns character arrays in a list
I use by() to
This looks like a conceptual question rather than a R question. PLEASE do read
the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Arun Prasad Gurubaramurugeshan,
Senior Research Analyst
-Original Message-
Advice:
Do not post here. This does not appear to be an R question. Post to a
statistics list like stats.stackexchange.com instead.
-- Bert
On Wed, Oct 10, 2012 at 6:42 AM, codec cat v.code...@gmail.com wrote:
Dear all
I wonder if someone can explain what is the main difference between omega
Thank you Petr,
Try this
str(by(iris, iris$Species, summary))
and you will see what is actually returned is a list of 3, each element
containing a character table, not a numeric table. The rownames of these
tables are empty but should contain the names of the summary stats.
I have a workaround
Hello,
Try the following.
pattern - between [[:digit:]]+ to [[:digit:]]+
re - regexpr(pattern, string)
regmatches(string, re)
Hope this helps,
Rui Barradas
Em 10-10-2012 12:45, arunkumar escreveu:
hi
My string contain
string = The sales is good when my num1 between 1 to 5 . else the
Dear all,
I have two coordinates vectors, say X and Y of length n.
I want to generate for each couple of coordinates X1,Y1 X2,Y2
X3,Y3Xn,Yn a random coordinate which is located in a square define
as X +/- dx and Y +/- dy.
I saw the runif function which can generate for just one value at a
I don't understand why my questions never get posted on the mailing list...
The one above is still unresolved.
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Sent from the R help mailing list archive at
I’m entirely stumped on this particular issue and really hoping someone has
some advice for me.
I am running a covariant model in lm I would like to give the standard
errors or the confidence intervals for the fitted lines. I’ve been using the
dataset OrangeSprays where I want lines for each
Hi,
I have a question about using lm on matrix, have to admit it is very
trivial but I just couldn't find the answer after searched the mailing
list and other online tutorial. It would be great if you could help.
I have a matrix trainx of 492(rows) by 220(columns) that is my x,
and trainy is 492
?predict
Have you read An Inrtroduction to R?
-- Bert
On Wed, Oct 10, 2012 at 6:49 AM, Sigrid s.stene...@gmail.com wrote:
I’m entirely stumped on this particular issue and really hoping someone has
some advice for me.
I am running a covariant model in lm I would like to give the standard
On Wed, Oct 10, 2012 at 4:56 AM, Pierrick Bruneau pbrun...@gmail.comwrote:
With the following code :
dat1 - matrix(nrow=4, ncol=2)
dat1[1,] - c(-2, 1)
dat1[2,] - c(-1.7, 0.9)
dat1[3,] - c(0.1, 0.6)
dat1[4,] - c(0.5, 0.5)
theplot - xyplot(V2 ~ V1, as.data.frame(dat1), pch=c(4,1,5,4))
Is there a way to make any of the write functions (.table or .csv) append
to the same file?
I get this warning message and do not know how to enable the appending.
Warning messages:
1: In write.csv(names(Lbys)[c], fo, append = TRUE) :
attempt to set 'append' ignored
2: In write.csv(Lbys[[c]],
On Wed, Oct 10, 2012 at 4:56 AM, Pierrick Bruneau pbrun...@gmail.comwrote:
With the following code :
dat1 - matrix(nrow=4, ncol=2)
dat1[1,] - c(-2, 1)
dat1[2,] - c(-1.7, 0.9)
dat1[3,] - c(0.1, 0.6)
dat1[4,] - c(0.5, 0.5)
theplot - xyplot(V2 ~ V1, as.data.frame(dat1), pch=c(4,1,5,4))
Your subject line also suggests you might consult with your local priest. ;-)
Cheers,
RMW
On Wed, Oct 10, 2012 at 2:51 PM, Bert Gunter gunter.ber...@gene.com wrote:
Advice:
Do not post here. This does not appear to be an R question. Post to a
statistics list like stats.stackexchange.com
On Wed, Oct 10, 2012 at 2:07 PM, pbruneau pbrun...@gmail.com wrote:
I don't understand why my questions never get posted on the mailing list...
The one above is still unresolved.
Probably because you post through Nabble which requires manual
intervention to allow posts rather than subscribing
write.table() should append just fine. write.csv() doesn't append I'm
told because there's no guarantee it matches the column headings of
the original file. That said, write.table(..., append = TRUE, sep =
,) is a pretty good approximation.
Cheers,
Michael
On Wed, Oct 10, 2012 at 3:44 PM, Alex
Change 1 to some other number to get more points from runif()
More generally, take a look at An Introduction to R and read most
everything you can find on the topic of vectorization. If you don't
know how to get An Introduction to R, try typing help.start() at
your prompt and it should happen
On 10-10-2012, at 16:15, Poizot Emmanuel emmanuel.poi...@cnam.fr wrote:
Dear all,
I have two coordinates vectors, say X and Y of length n.
I want to generate for each couple of coordinates X1,Y1 X2,Y2 X3,Y3Xn,Yn
a random coordinate which is located in a square define as X +/- dx and Y
On Wed, Oct 10, 2012 at 4:09 PM, Poizot Emmanuel
emmanuel.poi...@cnam.fr wrote:
Le 10/10/2012 17:02, R. Michael Weylandt a écrit :
Change 1 to some other number to get more points from runif()
More generally, take a look at An Introduction to R and read most
everything you can find on the
Hi Bert,
I just looked at An Introduction to R - and I do apologize if my questions
are trivial. I see that they use predict as a function in lm, but I'm not
sure how to incorporate it into a command.
Thank you,
S
--
View this message in context:
On 10/10/2012 10:56 AM, R. Michael Weylandt wrote:
write.table() should append just fine. write.csv() doesn't append I'm
told because there's no guarantee it matches the column headings of
the original file. That said, write.table(..., append = TRUE, sep =
,) is a pretty good approximation.
Hello,
If 'by' is giving you trouble, why not 'aggregate'?
agg.df - aggregate(iris, list(iris$Species), FUN = summary)
str(agg.df)
Hope this helps,
Rui Barradas
Em 10-10-2012 15:02, Alex van der Spek escreveu:
Thank you Petr,
Try this
str(by(iris, iris$Species, summary))
and you will see
Hello,
Your model is equivalent to the model below. As for standard errors, try
predict.lm with the appropriate argument.
?predict.lm
model - lm(decrease ~ rowpos + colpos*treatment, data = OrchardSprays)
predict(model, se.fit = TRUE, interval = confidence)
Hope this helps,
Rui Barradas
Em
i want to show histogram in innovative way (good if interactive) in html
report..is threre any tutorials/hint
also there is any package to show correalation plot
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Hi,
This answer may be a bit overdue, but I had a similar problem and it took
me quite some time to find a good solution. When searching with google,
your post kept showing up, so I post this to help others.
The command that worked for me was interaction.plot.
Example:
x -
Hi,
history()
gives Error in savehistory(file) : no history available to save
although I can scroll throu history with C^uparrow an C^downarrow.
How can I make history() work and/or show the current history in a file,
so that I can choose from previous commands?
The web did not throw up
Hi,
On Wed, Oct 10, 2012 at 12:03 PM, Christian Hoffmann
c-w.hoffm...@sunrise.ch wrote:
Hi,
history()
gives Error in savehistory(file) : no history available to save
although I can scroll throu history with C^uparrow an C^downarrow.
How can I make history() work and/or show the current
In addition to Bert's answer. If the 0 and/or 100 are hard boundaries
(you know that values cannot be outside those values) and you have
data points near one or both of the bounds, then the functions in the
logspline package may be of use.
On Tue, Oct 9, 2012 at 2:38 PM, Bert Gunter
What do you mean by connect? And how did you succeed under Windows?
i.e. what is your expectation?
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Wed, Oct 10, 2012 at 4:57
* Prof Brian Ripley evc...@fgngf.bk.np.hx [2012-10-08 06:37:07 +0100]:
On 08/10/2012 02:57, Peter Ehlers wrote:
On 2012-10-07 14:44, Sam Steingold wrote:
* Peter Ehlers ruy...@hpnytnel.pn [2012-10-07 10:03:42 -0700]:
On 2012-10-07 08:34, Sam Steingold wrote:
I know it does not look very
Are the points you are looking for (those data points with no other data
points above or to the right of them) a subset of the convex hull of the
data points? If so, chull(x,y) can quickly give you the points on the convex
hull (typically a fairly small number) and you can look through them for
No, the desired points are not a subset of the convex hull.
E.g., x=c(0,1:5), y=c(0,1/(1:5)).
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: William Dunlap
Sent: Wednesday, October 10, 2012 9:46 AM
To: 'tonja.krue...@web.de'; r-help@r-project.org
What does the sudden appearance of Contacting Delphi ..the oracle
is unavailable.
We apologize for any inconvenience. mean? A bug? It appears at plotting.
Thanks
Christian
--
Christian W. Hoffmann,
CH - 8915 Hausen am Albis, Switzerland
Rigiblickstrasse 15 b, Tel.+41-44-7640853
Hi all,
I'm running into some computer issues when trying to run a binomial model
for spatially correlated data using glmmPQL and was wondering if anyone
could help me out.
My whole dataset consists of about 300,000 points for which I have a suite
of environmental variables (I'm trying to come up
On Wed, Oct 10, 2012 at 12:52 PM, Christian Hoffmann
c-w.hoffm...@sunrise.ch wrote:
Am 10.10.12 18:17, schrieb Steve Lianoglou:
Hi,
On Wed, Oct 10, 2012 at 12:03 PM, Christian Hoffmann
c-w.hoffm...@sunrise.ch wrote:
Hi,
history()
gives Error in savehistory(file) : no history available
Hi,
history()
gives Error in savehistory(file) : no history available to save
although I can scroll throu history with C^uparrow an C^downarrow.
How can I make history() work
and/or show the current history in a *file*, so that I can choose from
previous commands?
The notion of having
Dear all,
I am trying to export my fixed effect results to Latex. I am using the plm
package with the summary function. However, it does not look like apsrtable,
stargazer, or any other package can accompany using the plm package.
I am interested in a classic table with the coefficient in
HI,
May be this helps you:
Using the dataset iris:
by.list-by(iris, iris$Species, summary)
dat1-do.call(rbind,lapply(by.list,function(x) gsub(.*\\:,,x)))
row.names(dat1)-paste(rep(unlist(dimnames(by.list),use.names=F),each=6),unlist(lapply(lapply(by.list,`[`,1:6),function(x)
On Oct 10, 2012, at 7:58 AM, namrata mohapatra wrote:
Hello
I am interested to plot a contour plot using the colour : rainbow , however I
want to reverse the order of the colour ( such that red represents max value
and blue min value) and also remove the lines in white in the plot .
On Oct 10, 2012, at 9:59 AM, Christian Hoffmann wrote:
What does the sudden appearance of Contacting Delphi ..the oracle is
unavailable.
We apologize for any inconvenience. mean? A bug? It appears at plotting.
In this instance an unsuccessful attempt at humor:
Apologies - I feel this is a very simple thing to do yet I am failing
massively. I keep finding information about how to do much more complicated
things (usually on this mailing list!), which then fail when I try to apply
it to my simple task.
Anyway, all I want to do is read in a series of
Hello,
It's in the source code for `?`, file src/library/utils/R/question.R,
lines 32 to 35.
32cat(Contacting Delphi...)
33 flush.console()
34Sys.sleep(2+rpois(1,2))
35cat(the oracle is unavailable.\nWe apologize for any
inconvenience.\n)
36
Hi,
May be this:
set.seed(1)
dat1-data.frame(keys=paste0(key,1:5),value=sample(1:15,5,replace=TRUE))
list1-lapply(split(dat1,dat1$keys),`[`,2)
list1$key2
# value
#2 6
A.K.
- Original Message -
From: VA Smith v...@st-andrews.ac.uk
To: r-help@r-project.org
Cc:
Sent: Wednesday,
On Tue, Oct 9, 2012 at 10:35 AM, Jessica Streicher
j.streic...@micromata.de wrote:
Can i somehow append objects to an .Rdata file?
I didn't see an option for it in the save() method.
dump() won't work since i have s4 objects in there.
I'm not sure I completely understand the issues you're
HI,
By modifying the earlier solution using sapply()
set.seed(1)
dat1-data.frame(keys=paste0(key,1:5),value=sample(1:15,5,replace=TRUE))
list2-sapply(split(dat1,dat1$keys),`[`,2)
names(list2)-dat1[,1]
list2$key2
#[1] 6
A.K.
- Original Message -
From: VA Smith v...@st-andrews.ac.uk
To:
Its suddenness is only as sudden as your input that triggered it. Its oblique
nature is a reflection of the nature of your input.
http://lmgtfy.com/?q=R+contacting+Delphi+oracle+is+unavailable
---
Jeff Newmiller
I finally was able to compile/load it under windows 7. I had similar
problems to what you show below.
I set the MYSQL_HOME environmental variable through windows (start
button control panel System and Security system Advanced
System Settings Environmental variables). I had to set it to the
Hello!
I'd like to know if it is correct to
test with anova two models specified like this:
m1=y~x1+s(x2,by=x3),family=poisson
m0=y~x1+s(x2),family=poisson
anova(m1,m0)
Cheers
Anna
Anna Freni Sterrantino
Department of Statistics
University of Bologna, Italy
via Belle Arti 41, 40124 BO.
Unsuccessfull? Why unsuccessfull? Have you noticed the call to rpois? It
beats runif by potential infinity.
Rui Barradas
Em 10-10-2012 18:30, David Winsemius escreveu:
On Oct 10, 2012, at 9:59 AM, Christian Hoffmann wrote:
What does the sudden appearance of Contacting Delphi ..the oracle
It might help if you would bore us with at least one or two of the things
you have tried. It seems logical to read the file into a data frame using
read.table(). Then you can change it into any format you want:
Listname - read.table(text=key value
Key1 1
Key2 2
Key3 3, header=TRUE)
Listname
Sorry, I'm sure I'm not using the appropriate vocab here, which is
undoubtedly why I can't seem to find a fix to this (hopefully very
easy) problem.
Suppose you have a factor
abc - factor(c(2,2,3,4,7,7))
And you want to know what the number in the nth spot in that would be
abc[1]
[1] 2
Levels:
Hi Brigid,
as.numeric() extracts the index of the factor level, which is the way
R handles the likelihood that a factor is not actually numeric. Try:
as.numeric(as.character(abc[1]))
[1] 2
and see also ?factor particularly the section on the interpretation of a factor.
Sarah
On Wed, Oct 10,
Hello,
Try instead
?levels
abc - factor(c(2,2,3,4,7,7))
as.numeric(levels(abc)[1])
Hope this helps,
Rui Barradas
Em 10-10-2012 19:39, Brigid Mooney escreveu:
Sorry, I'm sure I'm not using the appropriate vocab here, which is
undoubtedly why I can't seem to find a fix to this (hopefully very
Rui, that doesn't answer the question as I understood it:
Your suggestion returns the numeric value of the second value of the levels:
as.numeric(levels(abc)[2])
[1] 3
But I read the question as wanting the numeric value of the second
element of abc:
as.numeric(as.character(abc[2]))
[1] 2
On
HI,
May be you can use library(texreg):
library(plm)
#generating some data
x - rnorm(270)
y - rnorm(270)
t - rep(1:3,30)
i - rep(1:90, each=3)
data - data.frame(i,t,x,y)
fe - plm(y~x,data=data,model=within)
summary(fe)
library(texreg)
fe1-extract.plm(fe) #extract the plm object
I think David was saying the humor was unsuccessful (i.e., Christian
didn't get the joke), not that the code was.
Cheers,
RMW
On Wed, Oct 10, 2012 at 7:16 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Unsuccessfull? Why unsuccessfull? Have you noticed the call to rpois? It
beats runif by
Sorry, not one of my days. Forgot to Cc the list.
Rui barradas
Em 10-10-2012 20:28, Sarah Goslee escreveu:
Sent just to me?
On Wed, Oct 10, 2012 at 3:26 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
You're right, apologies to the op and the list. I was thinking of the more
complicated
Did not see a simple way to make it faster. However, this is a piece of
code which can be made to run much faster in C. See below.
I don't know if you are familiar with running c-code from R. If not, the
official documentation is in the R Extensions manual. However, this is
not the most
There are packages for big data analysis, which is best depends on
what you want to do. The High Performance Computing task view on CRAN
has a section on packages that deal with big data which gives some
more detail and may help you choose which package(s) to use.
On Wed, Oct 10, 2012 at 12:36
Your original method would be the following function
f - function (x, y)
{
xy - cbind(x, y)
outside - function(z) {
!any(x z[1] y z[2])
}
j - apply(xy, 1, outside)
which(j)
}
and the following one quickly computes the same thing as the above
as long as there are no
On Wed, Oct 10, 2012 at 4:47 PM, sagarnikam123 sagarnikam...@gmail.com wrote:
i want to show histogram in innovative way (good if interactive) in html
report..is threre any tutorials/hint
also there is any package to show correalation plot
The histogram is well defined, so I'm not really sure
Can someone more capable than I help Martin out with this? I'm feeling
out of my league (that or I've missed something obvious)
Shot in the dark: you aren't running this in some sort of debug mode, are you?
RMW
On Sun, Oct 7, 2012 at 5:10 PM, Martin Ivanov tra...@abv.bg wrote:
Thank You very
Hello,
I am learning to use the metafor package to conduct meta-regression analyses
for a systematic review on multidisciplinary care interventions in chronic
kidney disease. For the forest plots, I can't figure out how to plot
unadjusted and adjusted models on the same plot. From top to
Hello,
I am trying to re-code all my programs from SAS into R.
In SAS I use the following code:
proc sort data=upper;
by tdate stock_symbol expire strike;
run;
data upper1;
set upper;
by tdate stock_symbol expire strike;
if first.expire then output;
rename strike=astrike;
run;
on the
New to R and having issues with loops. I am aware that I should use
vectorization whenever possible and use the apply functions, however,
sometimes a loop seems necessary.
I have a data set of 2 million rows and have tried run a couple of loops of
varying complexity to test efficiency. If I do a
I need a Hessian matrix in nlmnib package to discuss whether parameters
are significant or not.
Please let me know how to obtain hessian matrix and how to evaluate the
significancy of parameters.
Regards
Serdar
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View this message in context:
#optim package
estimate-optim(init.par,Linn,hessian=TRUE, method=c(L-BFGS-B),control =
list(trace=1,abstol=0.001),lower=c(0,0,0,0,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf),upper=c(1,1,1,1,Inf,Inf,Inf,Inf,Inf,Inf,Inf,Inf,Inf))
#nlminb package
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