-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of darnold
Sent: Friday, August 03, 2012 9:18 PM
To: r-help@r-project.org
Subject: Re: [R] Head or Tails game
Wow! Some great responses!
I am getting some great responses.
Hello,
Works with me:
d1 - data.frame(V1 = 1:3,
V2 = c(some text = 9, some tèxt = 9, some other text = 9))
regexpr(some text = 9, d1$V2)
[1] 1 -1 -1
attr(,match.length)
[1] 13 -1 -1
regexpr(some tèxt = 9, d1$V2)
[1] -1 1 -1
attr(,match.length)
[1] -1 13 -1
d1$V1[regexpr(some text =
HI,
It works with me. I am using R 2.15 on Ubuntu 12.04.
d1 - data.frame(V1 = 1:5, V2=c(some text = 9, some téxt=9,sóme tèxt=9,
söme text=9, some têxt=9))
d1
# V1 V2
#1 1 some text = 9
#2 2 some téxt=9
#3 3 sóme tèxt=9
#4 4 söme text=9
#5 5 some têxt=9
Thank you. It works great.
Sent from my iPhone
On Aug 5, 2012, at 9:08 PM, Jorge I Velez jorgeivanve...@gmail.com wrote:
Hi Faz,
Here is one way of doing it where x is your data frame:
x[, colMeans(is.na(x)) = .15]
HTH,
Jorge.-
On Sun, Aug 5, 2012 at 9:04 PM, Faz Jones wrote:
Thank you.. It was very informative and helpful. It works
Sent from my iPhone
On Aug 5, 2012, at 10:21 PM, arun smartpink...@yahoo.com wrote:
HI,
Try this:
dat1-data.frame(x=c(NA,NA,rnorm(6,15),NA),y=c(NA,rnorm(8,15)),z=c(rnorm(7,15),NA,NA))
dat1[which(colMeans(is.na(dat1))=.15)]
On Aug 5, 2012, at 3:52 PM, alan.x.simp...@nab.com.au wrote:
Dear all
I have a Windows Server 2008 R2 Enterprise machine, with 64bit R
installed
running on 2 x Quad-core Intel Xeon 5500 processor with 24GB DDR3
1066 Mhz
RAM. I am seeking to analyse very large data sets (perhaps as much
Thank you very much John, can you read it now?
Hello,
I'd like to do next, see if you could help me please:
I have a csv called datuak with a id called calee_id and a colunm
called poids.
I have another csv called datuak2 with the same id called calee_id,
(although there are calee_id that are in
Thanks arun and Rui; 3 fantastic suggestions.
The Season interval is not always a month so arun's suggestion works better
for this dataset. I couldn't get the as.between function to work on arun's
second suggestion, it only returned NAs.
However, arun's first suggestion worked a treat!
Many
On 06.08.2012 09:34, David Winsemius wrote:
On Aug 5, 2012, at 3:52 PM, alan.x.simp...@nab.com.au wrote:
Dear all
I have a Windows Server 2008 R2 Enterprise machine, with 64bit R
installed
running on 2 x Quad-core Intel Xeon 5500 processor with 24GB DDR3 1066
Mhz
RAM. I am seeking to
On 06/08/2012 09:42, Uwe Ligges wrote:
On 06.08.2012 09:34, David Winsemius wrote:
On Aug 5, 2012, at 3:52 PM, alan.x.simp...@nab.com.au wrote:
Dear all
I have a Windows Server 2008 R2 Enterprise machine, with 64bit R
installed
running on 2 x Quad-core Intel Xeon 5500 processor with 24GB
On 2/08/2012 8:06 a.m., Thomas Steiner wrote:
Hi Ray,
2012/7/31 Ray Brownrigg ray.brownr...@ecs.vuw.ac.nz:
On 07/28/12 23:46, Thomas Steiner wrote:
Hi,
I'd like to have a low resolution word map in R.
The maps package has this option, but if I use the argument, the map
looses sense: Russia
Hi
Hi,
I appreciate your help with the segmented function. I am relatively new
to
R. I followed the introduction of the 'segmented'-package by Vito
Muggeo,
but still it does not work.
Here are the lines I wrote:
data_test-data.frame(x=c(1:10),y=c(1,1,1,1,1,2,3,4,5,6))
Hi
It is better to use dput for presenting data for others. You probably want
?merge.
Something like
merge(datuak, datuak2, by = calee_id, all.x=TRUE)
However calee_id seems to be a floating point number and it may be rounded
so you shall beware of it.
Regards
Petr
Thank you very much
I hope the original poster fixed this a long time ago, but I had the same
problem and here is how I fixed it:
- go to the application Fontbook
- check if the Arial font has duplicates, and delete them, even if they are
set to Off
- restart the computer.
emmats wrote
I was re-running some
http://www.google.com/url?sa=trct=jq=lty+in+r+line+typessource=webcd=1ved=0CFQQFjAAurl=http%3A%2F%2Fstudents.washington.edu%2Fmclarkso%2Fdocuments%2Fline%2520styles%2520Ver2.pdfei=HYgfUMPgGYLJrQfWjIGYBwusg=AFQjCNGL8xBzLN2je0RQFc5e8Hk5eRnS9Q
--
View this message in context:
Hi, All,
I am using the supervised lda function (slda) from 'lda' package in R for
topic modeling (http://cran.r-project.org/web/packages/lda/index.html), my
data is a collection of documents, and within which each doc has a label.
There are about 97 different categories and 18K documents in
On Sat, Aug 4, 2012 at 11:27 PM, Roberto rmosce...@unitus.it wrote:
Hi,
I need to remove collinear variables to my Near-Infrared table of spectra.
What package can I use?
Something simple, because I am a novice about statistic.
There many methods of assessing multicollinearlity but to
Sorry but my previous email did not go through properly. Instead of the ? you
should really read an egrave or #232 according to
http://www.lookuptables.com/.
So there are extended ASCII characters I need to deal with.
I have tried
d1$V1[regexpr(some tegravext = 9,d1$V2)0] - 9
and
Thanks Arun,
It works all right, I just found out that my problem was not with accents but
with the correct spelling of some text.
Kind regards,
Luca
Il giorno 06/ago/2012, alle ore 15.01, arun ha scritto:
Hi,
Here, the string with in the quotes are read exactly like that. So,
Alan,
More RAM will definitely help. But if you have an object needing more than
2^31-1 ~ 2 billion elements, you'll hit a wall regardless. This could be
particularly limiting for matrices. It is less limiting for data.frame
objects (where each column could be 2 billion elements). But many R
Dear all
I'm a bit surprised by the results output from nzchar(). The help page
says: nzchar is a fast way to find out if elements of a character
vector are *non-empty strings*. (my emphasis. However, if you do
x - c(letters, NA, '')
nzchar(x)
[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
On Mon, Aug 6, 2012 at 4:48 PM, Liviu Andronic landronim...@gmail.com wrote:
string, something that I find strange. At best NA is the equivalent of
an empty string. In this sense, if you Hmisc::describe() the vector
you get, as I would expect, that in the context of character vectors
NA and ''
I have a complex 2D polygon with thousands of vertices, and I'd like to be
able to identify points from a large set contained within the polygon, and
was wondering if there might be an efficient way of doing this? Any advice
would be useful! Here is a small example of what I mean:
# make
Thanks Uwe but, actually, I did so.
Since #8220;filetorun.exe#8221; looks in the current folder for
#8220;input.txt#8221;, I tried moving all needed files to a newly created
temporary folder #8220;tmp.id#8221; (say, tmp.1) and running the executable.
This works fine by doing it directly from
Hello,
With your example run and click 10 black points inside the area.
ploc - locator(n=10)
points(ploc$x, ploc$y, pch = 19, col = green, cex = 1)
Hope this helps,
Rui Barradas
Em 06-08-2012 16:05, Ally escreveu:
I have a complex 2D polygon with thousands of vertices, and I'd like to be
On Mon, Aug 6, 2012 at 9:53 AM, Liviu Andronic landronim...@gmail.com wrote:
On Mon, Aug 6, 2012 at 4:48 PM, Liviu Andronic landronim...@gmail.com wrote:
string, something that I find strange. At best NA is the equivalent of
an empty string.
Certainly not to my mind, unless you think that zero
I have two versions of a bibtex database which have gotten badly out of
sync. I need to find find all the entries in
bib2 which are not contained in bib1, according to their bibtex keys.
But I can't figure out how to extract a list of the bibentry keys in
these databases.
A minor question: Is
This is off-topic (not about R), and a quick Web search of test within
polygon yields many results, and adding R to the search when using Google
provides hints about applying the algorithms in R.
---
Jeff Newmiller
Liviu:
Well, as usual, to a certain extent this is arbitrary and the only
issue is whether it is documented correctly.
To me, NA (of whatever mode) means indeterminate or unknown, so
since is known and of length 0, I would have expected NA as a
return. But the point is, not what our particular
Thanks for the suggestion, got exactly what I needed from
library(splancs)
?pip
Alastair
Jeff Newmiller wrote
This is off-topic (not about R), and a quick Web search of test within
polygon yields many results, and adding R to the search when using
Google provides hints about applying
On Mon, 6 Aug 2012, Michael Friendly wrote:
I have two versions of a bibtex database which have gotten badly out of
sync. I need to find find all the entries in bib2 which are not
contained in bib1, according to their bibtex keys. But I can't figure
out how to extract a list of the bibentry
Dear all
I'm pretty sure that I'm approaching the problem in a wrong way.
Suppose the following character vector:
(x[1:10] - paste(x[1:10], sample(1:10, 10), sep=''))
[1] a10 b7 c2 d3 e6 f1 g5 h8 i9 j4
x
[1] a10 b7 c2 d3 e6 f1 g5 h8 i9 j4 k
l m n
[15] o p q r s t
Sorry, forgot to Cc the list.
Em 06-08-2012 17:29, Rui Barradas escreveu:
Hello,
I'm glad it helped.
The result of function cut() is a factor variable so you can coerce it
to integer, giving more normal names, or, if you want to keep track
of the intervals the adjusted r2 belong to, got
After searching online, I found that clusterCall or foreach might be the
solution.
Best wishes,
Jie
On Sun, Aug 5, 2012 at 10:23 PM, Jie jimmycl...@gmail.com wrote:
Dear All,
Suppose I have a program as below: Outside is a loop for simulation (with
random generated data), inside there are
nzchar(x) !is.na(x)
No?
-- Bert
On Mon, Aug 6, 2012 at 9:25 AM, Liviu Andronic landronim...@gmail.com wrote:
Dear all
I'm pretty sure that I'm approaching the problem in a wrong way.
Suppose the following character vector:
(x[1:10] - paste(x[1:10], sample(1:10, 10), sep=''))
[1] a10 b7
Not that I've had a chance to really look at the problem, but I've removed
outer loops using parLapply from the parallel package. Works great.
On Mon, Aug 6, 2012 at 11:41 AM, Jie jimmycl...@gmail.com wrote:
After searching online, I found that clusterCall or foreach might be the
solution.
On 08/06/2012 09:41 AM, Jie wrote:
After searching online, I found that clusterCall or foreach might be the
solution.
Re-write your outer loop as an lapply, then on non-Windows use
parallel::mclapply. Or on windows use makePSOCKcluster and parLapply. I
ended with
library(parallel)
Hello,
Fun as an exercise in vectorization. 30 times faster. Don't look, guess.
Gave it up? Ok, here it is.
is_letter - function(x, pattern=c(letters, LETTERS)){
sapply(x, function(y){
any(sapply(pattern, function(z) grepl(z, y, fixed=T)))
})
}
# test ascii codes, just one
On 08/06/2012 09:51 AM, Rui Barradas wrote:
Hello,
Fun as an exercise in vectorization. 30 times faster. Don't look, guess.
system.time(res0 - grepl([[:alpha:]], x))
user system elapsed
0.060 0.000 0.061
system.time(res1 - has_letter(x))
user system elapsed
3.728 0.008
Perhaps I am missing something, but why use sapply() when grepl() is already
vectorized?
is.letter - function(x) grepl([:alpha:], x)
is.number - function(x) grepl([:digit:], x)
x - c(letters, 1:26)
x[1:10] - paste(x[1:10], sample(1:10, 10), sep='')
x - rep(x, 1e3)
str(x)
chr [1:52000] a2
It would be nice to be able to trigger NA returning NA with an argument to
the function, but you can easily get that result:
ifelse(is.na(x), NA, nzchar(x))
[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[13] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
Dear all,
I would like to test the differences in dependent variable X depending on 2
grouping variables of each 10 levels.
I do this with a 2-way ANOVA, followed by a Tukey HSD test (TukeyHSD(x)).
However, since a lot of combinations are possible with 2 grouping variables,
each of 10 levels,
Greetings,
I am using S4 classes and the code is organized by putting each class into a
separate source file
in a separate folder:
In Folder base/base.R:defines class base and does
setGeneric(showSelf,
function(this) standardGeneric(showSelf)
)
setMethod(showSelf,
Hi,
I run the second list of codes (is.between()) again from the sent mail. It
works fine for me. I am using R 2.15 on Ubuntu 12.04.
sessionInfo()
R version 2.15.0 (2012-03-30)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3]
Hi,
Here, the string with in the quotes are read exactly like that. So, you may
have to use the symbol instead of friendly or numeric from the link. Or
you have to convert those.
d1 - data.frame(V1 = 1:4,
V2 = c(some text = 9, some tegravext = 9, some tèxt = 9, some
t#232xt = 9))
Hi
can ANY body help me to programme this formula:
c[lj] and c[l'j] are matrix
A[j]^-1 is an invertible diagonal matrix
g[ll']=i[ll'] - sum *#from j=1 to k#* c[lj]c[l'j]A[j]^-1
WHERE
i[ll']= 1/n sum from i=1 to n z[il] z[il']
n,k,m are given. j=1...k,l,l'=1...m,
it s
Hi,
I am trying to run the command forward.sel.par, however I receive
the error message: Error: could not find function 'simpleRDA2'. I
have the vegan library loaded. The documentation on varpart has not
helped me to understand why I cannot call this function. Maybe I am
missing something obvious
Hi,
Not sure whether this is you wanted.
x-letters
(x[1:10] - paste(x[1:10], sample(1:10, 10), sep=''))
x1-c(x,1:26)
x1
[1] a4 b3 c5 d2 e9 f6 g1 h8 i10 j7 k l
[13] m n o p q r s t u v w x
[25] y z 1 2 3 4 5 6 7 8 9 10
[37] 11 12
Dear R users
I read two csv data files into R and called them Tem1 and Tem5.
For the first column, data in Tem1 has 13 digits where in Tem5 there are 14
digits for each observation.
Originally there are 'numerical' as can be seen in my code below. But how
can I display/convert them using
On Aug 6, 2012, at 12:06 PM, Marc Schwartz marc_schwa...@me.com wrote:
Perhaps I am missing something, but why use sapply() when grepl() is already
vectorized?
is.letter - function(x) grepl([:alpha:], x)
is.number - function(x) grepl([:digit:], x)
Sorry, typos in the above from my CP.
Dear R Community,
I'm trying to write a loop to split my data into different series. I
need to make a
new matrix (or series) according to the series code.
For instance, every time the code column assumes the value 433 I need to
save date, value, and code into the dados433 matrix.
Please take a
- Forwarded Message -
From: arun smartpink...@yahoo.com
To: Liviu Andronic landronim...@gmail.com
Cc: R help r-help@r-project.org
Sent: Monday, August 6, 2012 12:56 PM
Subject: Re: [R] test if elements of a character vector contain letters
Hi,
Not sure whether this is you wanted.
On Aug 6, 2012, at 2:01 AM, tibr wrote:
I hope the original poster fixed this a long time ago, but I had the
same
problem and here is how I fixed it:
- go to the application Fontbook
- check if the Arial font has duplicates, and delete them, even if
they are
set to Off
- restart the
You would make it much easier for R-help readers to solve your problem if
you provided a small example data set with your code, so that we could
reproduce your results and troubleshoot the issues.
Jean
Naidraug white@wright.edu wrote on 08/05/2012 09:08:25 AM:
I've looked everywhere
Only an extra set of brackets:
is.letter - function(x) grepl([[:alpha:]], x)
is.number - function(x) grepl([[:digit:]], x)
Without them, the functions are fast, but wrong.
x
[1] a8 b5 c10 d1 e6 f2 g4 h3 i7 j9 k l
[13] m n o p q r s t u v w x
[25] y z 1
On 8/6/2012 11:54 AM, Achim Zeileis wrote:
On Mon, 6 Aug 2012, Michael Friendly wrote:
I have two versions of a bibtex database which have gotten badly out
of sync. I need to find find all the entries in bib2 which are not
contained in bib1, according to their bibtex keys. But I can't figure
Dominic,
It's great that you provided some example data, but a much smaller data
frame would have sufficed. For example, 10 randomly selected rows from
your data ...
LF - structure(list(Serra.da.Foladoira = c(27.335652173913,
25.4632608695652,
24.464652173913, 22.550652173913,
On Aug 6, 2012, at 7:34 AM, Dominic Roye wrote:
Hello everyone,
I have a dataset with 5 colums (4 colums with thresholds of weather
stations and one with month - data of 5 years). Now I would like to
calculate the average for each month.
I tried this unsuccessfully:
lf.med -
HJ,
You don't provide any reproducible code, so I had to make up my own.
dat - data.frame(a=letters[1:5], x=c(20110911001084, 20110911001084,
20110911001084, 20110911001084, 20110911001084),
y=c(2.10004e+12, 2.10004e+12, 2.10004e+12, 2.10004e+12,
2.10004e+12))
In my example,
On Aug 6, 2012, at 8:04 AM, hafida wrote:
Hi
can ANY body help me to programme this formula:
c[lj] and c[l'j] are matrix
A[j]^-1 is an invertible diagonal matrix
g[ll']=i[ll'] - sum *#from j=1 to k#* c[lj]c[l'j]A[j]^-1
WHERE
i[ll']= 1/n sum from i=1 to n z[il] z[il']
n,k,m are
Hello,
Try the following.
split(data.frame(dados), dados[, code])
Also, it's better to have data like 'dados' in a data.frame, like this
you would have dates of class Date, and numbers of classes numeric or
integer:
dados2 - data.frame(dados)
dados2$date - as.Date(dados2$date)
Well, I just posted the fourth copy to the list which I apologize
for. (I meant to delete the response I wrote.)
Re-re-posting an unclear message seems unwise on your part, 'hafida'.
You are not following the advice in the footer to all messages and you
are not following the advice in the
Would just saving the results onto an object saving it work?
From the help page example
summary(fm1 - aov(breaks ~ wool + tension, data = warpbreaks))
myresults- TukeyHSD(fm1, tension, ordered = TRUE)
write.table (myresults, file = ksksk)
John Kane
Kingston ON Canada
On Mon, Aug 6, 2012 at 6:42 PM, Bert Gunter gunter.ber...@gene.com wrote:
nzchar(x) !is.na(x)
No?
It doesn't work for what I need:
x
[1] a10 b8 c9 d2 e3 f4 g1 h7 i6 j5 k
l m n
[15] o p q r s t u v w x y
z 1 2
[29] 3 4 5 6 7 8 9 10 11 12
Dear all,
For two sets of random variables, say, x - rnorm(1000, 10, 10) and y
- rnorm(1000. 3, 20).
Is there any way to overlay the histograms (and density curves) of x and y
on the plot of y vs. x?
The histogram of x is on the x axis and that of y is on the y axis.
The density curve here
See
example(layout)
for one idea. I think you might also want to look into rug plots.
Best,
Michael
On Mon, Aug 6, 2012 at 2:40 PM, li li hannah@gmail.com wrote:
Dear all,
For two sets of random variables, say, x - rnorm(1000, 10, 10) and y
- rnorm(1000. 3, 20).
Is there any way to
You probably mean grepl('[a-zA-Z]', x)
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 Mon, Aug 6, 2012 at 3:29 PM, Liviu Andronic landronim...@gmail.com wrote:
On Mon, Aug
A OK I MAKE A MISTAKE
OK MR DAVID I WILL DO IT
THANK YOU
--
View this message in context:
http://r.789695.n4.nabble.com/program-of-matrix-tp4639288p4639334.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org
I CANT FIND ANY ANSWER MR DAVID
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
__
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Dear Rui and Arun,
Thanks a lot for your help. I will test all the proposed solutions ;-)
Best regards,
Henrique Andrade
2012/8/6 Rui Barradas ruipbarra...@sapo.pt:
Hello,
Try the following.
split(data.frame(dados), dados[, code])
Also, it's better to have data like 'dados' in a
HI,
You can subset the data
dados433-subset(dados,dados[,3]==433)
is.matrix(dados433)
#[1] TRUE
dados433
date value code
1 2012-01-01 0.56 433
2 2012-02-01 0.45 433
3 2012-03-01 0.21 433
4 2012-04-01 0.64 433
5 2012-05-01 0.36 433
6 2012-06-01 0.08 433
HI,
#These links
http://stackoverflow.com/questions/8545035/scatterplot-with-marginal-histograms-in-ggplot2
http://stackoverflow.com/questions/11022675/rotate-histogram-in-r-or-overlay-a-density-in-a-barplot
# might be helpful for you.
A.K.
- Original Message -
From: li li
Please find some reference online or textbook. This must be contained in
the model assessment part.
AIC, BIC, rolling prediction/forecasting error might be what you want.
Best wishes,
Jie
On Fri, Aug 3, 2012 at 4:07 AM, Soham soham.tommarvolorid...@gmail.comwrote:
Hi I am trying to fit a time
On Aug 6, 2012, at 11:16 AM, hafida wrote:
I CANT FIND ANY ANSWER MR DAVID
When I suggested that you learn to use the shift key, I was hoping for
a sparing use of that key, such as at the beginning of sentences. The
caps-lock key is different than the shift key.
You are also posting to
Hi,
simpleRDA2 is still in the vegan package, but it is not exported.
I.e., the author only intends it for internal use and he doesn't make
it available to end users directly. If you need to get at it, you can
use
getAnywhere(simpleRDA2)
which will show it.
If you need to make it available to
I am porting a program in matlab to R,
The problem is that Matlab has a feature where symbols that aren't arguments
are evaluated immediately.
That is:
Y=3
F=@(x) x*Y
Will yield a function such that F(2)=6.
If later say. Y=4 then F(2) will still equal 6.
R on the other hand has lazy
You could use local(), as in
F - local({
+Y - 3
+function(x) x * Y
+})
F(7)
[1] 21
Y - 19
F(5)
[1] 15
Look into 'environments' for more.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From:
On Sun, Aug 5, 2012 at 4:49 PM, Douglas Karabasz
doug...@sigmamonster.com wrote:
I have a xts object made of daily closing prices I have acquired using
quantmod.
Here is my code:
library(xts)
library(quantmod)
library(lubridate)
# Gets SPY data
getSymbols(SPY)
# Subset Prices
On Mon, Aug 6, 2012 at 4:30 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
On Sun, Aug 5, 2012 at 4:49 PM, Douglas Karabasz
doug...@sigmamonster.com wrote:
I have a xts object made of daily closing prices I have acquired using
quantmod.
Here is my code:
library(xts)
On Mon, Aug 6, 2012 at 7:35 PM, Marc Schwartz marc_schwa...@me.com wrote:
is.letter - function(x) grepl([[:alpha:]], x)
is.number - function(x) grepl([[:digit:]], x)
This does exactly what I wanted:
x
[1] a10 b8 c9 d2 e3 f4 g1 h7 i6 j5 k
l m n
[15] o p q r s t u v
Thanks to both: Cute question, clever, informative answer.
However, Bill, I don't think you **quite** answered him, although the
modification needed is completely trivial. Of course, I could never
have figured it out without your response.
Anyway, I interpret the question as asking for the
Both of those approaches require the function to be created
at the same time that the environment containing some of its
bindings is created. You can also take an existing function and
assign a new environment to it. E.g.,
f - function(x) y * x
ys - c(2,3,5,7,11)
fs - lapply(ys,
Thank you both, this was very helpful. I need to study environments more. Do
either of you know a good source?
-Original Message-
From: Bert Gunter [mailto:gunter.ber...@gene.com]
Sent: Monday, August 06, 2012 6:03 PM
To: William Dunlap
Cc: Schoenfeld, David Alan,Ph.D.,Biostatistics;
I would like to find out how to apply commands found in the bayesm package, to
analyze data gathered via a choice-based conjoint study. Is there a web
resource where I can seek an R-Project consultant experienced in this, who I
could hire to walk me through the appropriate bayesm commands to
On Mon, Aug 6, 2012 at 9:03 PM, Schoenfeld, David
Alan,Ph.D.,Biostatistics dschoenf...@partners.org wrote:
Thank you both, this was very helpful. I need to study environments more. Do
either of you know a good source?
Disclaimer: I really have no idea what I'm talking about.
They are a
Hello fellow R users,
I would need your help on GAM/GAMM models and interpolation on a marked
spatial point process (cases and controls).
I use the mgcv package to fit a GAMM model with a binary outcome, a
parametric part (var1+..+varn), a spline used for the spatial variation, and
a random
Hi,
Try this:
F-function(x,type=local){Y=3
x*Y}
F(3)
#[1] 9
Y-4
F(3)
#[1] 9
Y-5
F(3)
#[1] 9
A.K.
- Original Message -
From: Schoenfeld, David Alan,Ph.D.,Biostatistics dschoenf...@partners.org
To: 'r-help@r-project.org' r-help@r-project.org
Cc:
Sent: Monday, August 6, 2012 5:07
Hi,
How do I do a Duncan Multiple Range Test, with Nested ANOVA.
I am not sure how to write the nested variable within dmrt.
Any suggestions?
Archana
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R-help@r-project.org mailing list
Hi, all
I am trying to use the label_wrap_gen function in this website.
https://github.com/hadley/ggplot2/wiki/labeller
I tried to make a long name like this
Light and heavy good vehicles (diesel) -\nGVX
f2 = facet_grid(vehicle ~ ., labeller=label_wrap_gen(width=15))
eventually, I got
lty =1 denotes the single continuous line
lty = 2 denotes the broken line
lty = 3 dotted line
--
View this message in context:
http://r.789695.n4.nabble.com/line-type-lty-tp3466345p4639365.html
Sent from the R help mailing list archive at Nabble.com.
Dear Jean
Thanks a lot for your help.
The reason I did not provide producible code is that my work started with
reading in some large csv files, e.g. the data is not created by myself.
But the data is from the same data provider so I would expect to receive
data in exactly same data format.
I
Well, i believe writing correct date format would have served the purpose.
Suppose tfr contains Date as column and is a factor by class.
tft$Date - as.Date(as.character(tfr$Date),%d/%m%Y) should give you the
desired output.
--
View this message in context:
Dan, google refine http://goo.gl/AeKml can actually transform zip codes
into longitude/latitude - http://goo.gl/1HDWb will show you how to do this
from street adresses, but it should also work from city names -- i think it
will allocate a default long/lat for a city, but not sure of the exact
Hi everyone,
I have a time series data set and I want to fill my line plot of this time
series with different colors
e.g
I want to fill portion related to 1995-1996 with blue , portion related to
1996-1997 with orange and then portion related to 1997-1998 with red
can anyone please help me.
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