Hello,
Or maybe to avoid the typo (Market is column 5), use variables names, in
something like
myData <- read.table(text="
Date Stock1 Stock2 Stock3Market
01/01/2000 1 2 3 4
01/02/2000 5 6 7 8
0
Hi,I
want to be able to print in colour from the R console i.e. commands (in navy)
with
results (in red) as on the console.
From
the console if I click on File -> Print, the commands with results print
on my printer but only in monochrome, not in colour. Similarly, if I Edit ->
Select All, Edit
Hello,
It seems that there is a problem with ":".
If you only need the date, you can use as.Date(Sys.time()) instead of the
complete form.
If you need the time, you can try the following commands which change the
":" into "-" :
newsystime<-<-format(Sys.time(),"%Y-%m-%d-%H-%M-%S")
write.csv(data.f
Hi,
Try this:
Year<-c("01/2000","02/2000","03/2000")
#If you want to convert it directly to month/year
library(zoo)
as.yearmon(Year,format="%m/%Y")
[1] "Jan 2000" "Feb 2000" "Mar 2000"
#As your intention is to have DD/MM/ format,
Year1<-paste("01/",Year,sep="")
Year1
[1] "01/01/2000" "01/02
Dear All,
I am a research student in environment. I have only little
programming knowledge. I am currently doing the last project about rainfall
impact on ground water quality in an area. It happens that I have to use R to
read rainfall data (3 dimension) from ASC file (*.asc), and t
Hi,
Try this:
myData = data.frame(Name = c('a', 'a', 'b', 'b'), length = c(1,2,3,4), type=
c('x','x','y','z'))
z<-aggregate(length~Name,myData,mean)
z1<-aggregate(length~type,myData,mean)
merge(z,merge(z,z1),all=TRUE)
Name length type
1 a 1.5 x
2 b 3.5
A.K.
- Original
Dear all,
I can't figure out a way to have more than one plot using filled.contour() in a
single plate. I tried to use layout() or par(), but the way filled.contour() is
written seems to override those commands.
Any suggestions would be really appreciated.
Jonathan
Hello,
It is not what happens.
Function "convexhull" exists in both "siar" and "spatstat" packages. As
you already load "spatstat", when you are loading "siar", the
"convexhull" in "spatstat" is masked by the one in "siar".
Thus, when you will run "convexhull" function, it will be the one fr
Hi,
I have a problem while loading the "siar" program in R.
When I am loading siar, system does not load convexhull. On the screen I
have seen such writings.
The following object(s) are masked from package:spatstat:
convexhull
How can I load the convexhull, how can I unmask from t
Dear Mr Newmiller and Mr Oettli,
Thanks a lot for your valuable guidance. Task is done. Thanks again.
Regards
Vincy
--- On Wed, 7/4/12, Jeff Newmiller wrote:
From: Jeff Newmiller
Subject: Re: [R] How to use Sys.time() while writing a csv file name
To: "Vincy Pyne" , r-help@r-project.org
Re
Hi,
A few comments. First a for loop is probably not optimally efficient.
Consider instead (using a bulit in example dataset):
lm(cbind(mpg, hp) ~ cyl + vs, data = mtcars)
which gives:
Call:
lm(formula = cbind(mpg, hp) ~ cyl + vs, data = mtcars)
Coefficients:
mpg hp
(Inter
You forgot to follow the posting guide and tell us what operating system you
are using (sessionInfo), but I am going to guess that you are on Windows where
the colon (":") is an illegal symbol in filenames. Try formatting the time
explicitly in the conversion to character using the format string
Hello,
Try something like that:
> lgd <- format(Sys.time(), "%Y_%m_%d_%H_%M_%S")
Regards,
Pascal
Le 04/07/2012 14:21, Vincy Pyne a écrit :
Dear R helpers,
I am using Beta distribution to generate the random no.s (recovery rates in my
example). However, each time I need to save these random
On 3 July 2012 22:03, Akhil dua wrote:
> and I need to run a seperate regression of every stock on market
> so I want to write a "for loop" so that I wont have to write codes again
> and again to run the regression...
>
1. Do give a subject line -- a blank one is commonly used by a virus.
2. I
?lm
and note in particular the section beginning "If response is a matrix..."
-- Bert
On Tue, Jul 3, 2012 at 10:08 PM, Akhil dua wrote:
> -- Forwarded message --
> From: Akhil dua
> Date: Wed, Jul 4, 2012 at 10:33 AM
> Subject:
> To: r-help@r-project.org
>
>
> Hi everyone I
> h
Dear R helpers,
I am using Beta distribution to generate the random no.s (recovery rates in my
example). However, each time I need to save these random no.s in a csv format.
To distinguish different csv files, one way I thought was use of Sys.time in
the file name. My code is as follows -
# My
Homework? (We don't do homework here).
-- Bert
On Tue, Jul 3, 2012 at 10:08 PM, Akhil dua wrote:
> -- Forwarded message --
> From: Akhil dua
> Date: Wed, Jul 4, 2012 at 10:33 AM
> Subject:
> To: r-help@r-project.org
>
>
> Hi everyone I
> have data on stock prices and market ind
-- Forwarded message --
From: Akhil dua
Date: Wed, Jul 4, 2012 at 10:33 AM
Subject:
To: r-help@r-project.org
Hi everyone I
have data on stock prices and market indices
and I need to run a seperate regression of every stock on market
so I want to write a "for loop" so that I wo
Hi everyone I
have data on stock prices and market indices
and I need to run a seperate regression of every stock on market
so I want to write a "for loop" so that I wont have to write codes again
and again to run the regression...
my data is in the format given below
Date Stock
?paste
Please (re-) read the "Introduction to R" document supplied with the software
for faster answers.
Also, please read the Posting Guide mentioned at the bottom of every message on
this list. In particular, providing data in raw tabular form is often
ambiguous, and the use of the dput func
Thanks for your validation. Yes Peter's solution is the fastest, faster than
the previous one by saving 25% time. It was missed out in my previous testing.
Jin
-Original Message-
From: Pascal Oettli [mailto:kri...@ymail.com]
Sent: Wednesday, 4 July 2012 2:07 PM
To: Li Jin
Cc: r-help@r-p
Hi
I have monthly data and the dates are in MM/YY Format
I need to convert them into DD/MM/YY format by pasting 01 in place of DD to
all the observations in my Year Column
ex:
YearStock Prices
01/2000 1
02/2000 2
03/2000 3
I need to convert them to
Yea
Le 04/07/2012 12:43, Peter Ehlers a écrit :
On 2012-07-03 17:23, jin...@ga.gov.au wrote:
Thank you all for providing various alternatives. They are all pretty
fast. Great help! Based on a test of a dataset with 800,000 rows, the
time used varies from 0.04 to 11.56 s. The champion is:
a1$h2 <-
On 2012-07-03 17:23, jin...@ga.gov.au wrote:
Thank you all for providing various alternatives. They are all pretty fast.
Great help! Based on a test of a dataset with 800,000 rows, the time used
varies from 0.04 to 11.56 s. The champion is:
a1$h2 <- 0
a1$h2[a1$h1=="H"] <- 1
Interesting. My t
Thank you all for providing various alternatives. They are all pretty fast.
Great help! Based on a test of a dataset with 800,000 rows, the time used
varies from 0.04 to 11.56 s. The champion is:
> a1$h2 <- 0
> a1$h2[a1$h1=="H"] <- 1
Regards,
Jin
Geoscience Australia Disclaimer: This e-mail (and
I found this [1] book interesting. About "big data" It really depends from
a number of things... if can help, I know hdf5 work pretty Well with huge
dataset .
[1] http://www.ggobi.org/book/index.html
On Jul 3, 2012 7:14 PM, "Michael" wrote:
> Hi all,
>
> Could you please help me?
>
> I am look
On Tue, 3 Jul 2012, Michael wrote:
I am looking for books/pointers/resources/tutorials on visualizing
complex/big data and on understanding multivariate relations in
complicated data.
Michael,
You need to become familiar with the works of Edward Tufte, the dean of
complex data visualization
Hi all,
Could you please help me?
I am looking for books/pointers/resources/tutorials on visualizing
complex/big data and on understanding multivariate relations in complicated
data.
More specifically, we have categorical variables and are interested in how
to visualize the categorical data and
On Tue, Jul 3, 2012 at 12:41 PM, jimmycloud wrote:
> I have a general question about coefficients estimation of the mixed model.
>
I have 2 ideas for you.
1. Fit with lme4 package, using the lmer function. That's what it is for.
2. If you really want to write your own EM algorithm, I don't feel
try this:
> myData = data.frame(Name = c('a', 'a', 'b', 'b'), length = c(1,2,3,4), type
+ = c('x','x','y','z'))
>
> result <- do.call(rbind, lapply(split(myData, myData$Name), function(.name){
+ data.frame(Name = .name$Name[1L]
+ , length = mean(.name$length)
+ , type = if (all(.name$type[1L] == .
If David's response is what you were seeking, I would then ask: homework??
In general, explicit model matrices are not needed for linear modeling in R.
Still not clear what you meant by "reparametrization," but if David's
interpretation is correct, my "impossible" comment is clearly wrong.
-- Be
Marck,
A little late, but perhaps this will help someone in the future. I am
guessing that some of your "integer" fields contain scientific notation, and
for some reason read.table is not interpreting those as integers. Consider
changing the affected column classes from "integer" to "numeric" an
Hi,
Try this:
data1<-data.frame(date,time,count)
dat1<-data1[with(data1,rev(order(count))),]
data2<-subset(dat1,rle(dat1$count)$lengths==1)
dat3<-aggregate(data2$count,list(data2$date),max)
colnames(dat3)<-c("date","count")
data4<-merge(dat3,data2)
data4<-data4[,c(1,3,2)]
data4
date
On 03/07/2012 4:10 PM, mms...@comcast.net wrote:
Hello,
I want to create a design matrix using R. Can you explain the code which
creates the following please? I understand the first part.
b=g1(?) does what?
That is gl, not g1 (i.e. gee ell not gee one).See ?gl for a
description.
Duncan
On Jul 3, 2012, at 4:10 PM, mms...@comcast.net wrote:
Hello,
I want to create a design matrix using R. Can you explain the code
which creates the following please? I understand the first part.
What first part if the next part is ... this ?
b=g1(?) does what?
That's just generating a f
1. You need to learn to use R Help. It is there for a purpose.
help (help)
## or
?help
is where to start.
2. Before posting further, please read "An Introduction to R." Ships with
every distro.
3. ?gl
4. This is not the best way to do this anyway.
dd <- expand.grid(b=1:4,a=1:3) ## is preferab
Hello,
I want to create a design matrix using R. Can you explain the code which
creates the following please? I understand the first part.
b=g1(?) does what?
dd <- data.frame(a = gl(3,4), b = gl(4,1,12)) # balanced 2-way
dd
a b
1 1 1
2 1 2
3 1 3
4 1 4
5 2 1
6 2 2
7 2 3
8 2 4
9 3 1
Of course it _should_ be:
ifelse(x[-length(x)] < x[-1], ...,...)
Sorry...
-- Bert
On Tue, Jul 3, 2012 at 1:00 PM, Bert Gunter wrote:
> Vectorize vectorize vectorize!
>
> if(x[-length(x)] < x[-1]) {...}
>
> (where x is the whole vector of entries)
>
> Bill Dunlap has posted some elegant code
Vectorize vectorize vectorize!
if(x[-length(x)] < x[-1]) {...}
(where x is the whole vector of entries)
Bill Dunlap has posted some elegant code within the last month or 2 aimed
at this sort of thing, so search on his posts in the archive.
-- Bert
On Tue, Jul 3, 2012 at 12:10 PM, jcrosbie wro
I've taken another look, and I actually think you're right. I'm going with
Gabor's. Changing d to a list is such a simple solution that I can't
believe I didn't try it earlier.
On Tue, Jul 3, 2012 at 1:39 PM, Peter Ehlers wrote:
> On 2012-07-03 09:47, Spencer Maynes wrote:
>
>> Thanks guys for t
Hello,
I've changed the way you create data.frame 'tests', like it was the
conditions were factors, which are coded as integers. We need them of
class character to be parsed. Note that the same could be done with the
data.frame 'info' but it's not absolutely needed.
tests <- data.frame(rule
I would like to remove a loop to speed up my code.
I want to remove a loop which references the last row.
In general I want to a remove a loop which looks something like this:
for 2 to number of rows in a matrix do{
if indextrow-1 is < currentIndexRow then do something.
}
My R code:
fo
Got it! Thank you Rui!
cp
On Tue, Jul 3, 2012 at 10:14 AM, Rui Barradas wrote:
> Hello,
>
> I'm glad it helped. See answer inline.
>
> Em 03-07-2012 17:09, Claudia Penaloza escreveu:
>
> Thank you Rui and Jim, both 'i1' and 'i1new' worked perfectly
>> because there are no instances of 'Dd' or '
On 2012-07-03 09:47, Spencer Maynes wrote:
Thanks guys for the help, I'm going to go with Patrick Burns answer because
it seems to work the best for my situation, but these all seem like they
should work.
Patrick's solution is similar to Gabor's, but, personally,
I favour Gabor's. Seems neatest
Dear all,
I have an excel file that contains 6 sheets
1,2,3,4,5,6
The analysis is repeated every 3 sheets
Sheets 1, 2, 3:
I want to add (horizontally) the data contained in the matrix : sheet2
(5:end,3:end )
of *Sheet2 * to the sheet3 such that the first element of the matrix
*sheet2 (5:end,3
On 02.07.2012 22:48, Mathias Worni wrote:
Sorry for the misunderstanding. What I meant to say is that I cannot get
the code for the specific graphic that I am running, so that I can keep
my code reproducible.
Which code for the graphics?
To be reproducible, you need the R vesion, the package
Dear All,
have a general question about coefficients estimation of the mixed model.
I simulated a very basic model: Y|b=X*\beta+Z*b +\sigma^2* diag(ni);
b follows
N(0,\psi) #i.e. bivariate normal
where b is the latent variable, Z and
I have a general question about coefficients estimation of the mixed model.
I simulated a very basic model: Y|b=X*\beta+Z*b +\sigma^2* diag(ni);
b follows
N(0,\psi) #i.e. bivariate normal
where b is the latent variable, Z and X are ni*2
Hi,
Glad all of them worked. In my reply to you, my first solution was:
list2<-lapply(1:10,function(x) vec1)
The more generic form should be:
list2<-lapply(1:length(list1),function(x) vec1)
A.K.
- Original Message -
From: Spencer Maynes
To: r-help@r-project.org
Cc:
Sent: Tuesday, Ju
Al Ehan wrote
>
> Hi guys,
>
> I'm trying to use the the integral function to estimate the area under a
> PDF and a crossing curve. first I stated the function with several vectors
> in it:
>
> fn=function(a,b,F,mu,alpha,xi)
> {
> x<-vector()
> fs<-function(x)
> {
> c <- (mu+(alpha*(1-(1-F)^xi
On 03/07/2012 06:56, Erin Hodgess wrote:
Dear R People:
I'm back to installing R from source, this time on a 64 bit machine.
What OS?
I'm using the R-Patched.tar.gz as my source.
When I have the openmp option set to -fopenmp, I get the error that
libgomp.spec is not found. Ok, so I comment
Hi guys,
I'm trying to use the the integral function to estimate the area under a
PDF and a crossing curve. first I stated the function with several vectors
in it:
fn=function(a,b,F,mu,alpha,xi)
{
x<-vector()
fs<-function(x)
{
c <- (mu+(alpha*(1-(1-F)^xi)/xi))
tmp <- (1 + (xi * (x - mu))/alpha)
I have water chemistry data with censored values (i.e., those less than
reporting levels) in a data frame with a narrow (i.e., database table)
format. The structure is:
$ site: Factor w/ 64 levels "D-1","D-2","D-3",..: 1 1 1 1 1 1 1 1 ...
$ sampdate: Date, format: "2007-12-12" "2007-12-12
#I have a dataframe called "tests" that contain "character expressions". These
characters are rules that use data from within another dataframe. Is there
any way within R I can access the rules in the dataframe called tests, and
then "evaluate" these "rules"?
#An example may better explain wha
Hi everyone.
I have these data :
myData = data.frame(Name = c('a', 'a', 'b', 'b'), length = c(1,2,3,4), type
= c('x','x','y','z'))
which gives me:
Name length type
1a 1x
2a 2x
3b 3y
4b 4 z
I would group (mean) this DF using 'Name' as grouping
Hi,
Hope this helps.
date1<- c("thursday November 20, 2012", "friday November 21, 2012", "saturday
November 22, 2012")
date2<- as.Date(date1, format= "%A %B %d, %Y")
date2
[1] "2012-11-20" "2012-11-21" "2012-11-22"
A.K.
- Original Message -
From: arunkumar
To: r-help@r-projec
Dear all,
I produced the following graph with ggplot which is almost fine, yet I don't
like that the legend for "Means" and "Observations" includes a line, though no
line is used in the plot for those two (the line for "Overall Mean" on the
other hand is wanted):
library(ggplot2)
ddf <- data.f
Dear David,
In fact just updating R seems to have fixed the problem. There's a lesson.
Thanks a lot,
George
On 3 July 2012 17:02, David Winsemius [via R] <
ml-node+s789695n4635292...@n4.nabble.com> wrote:
>
> On Jul 3, 2012, at 4:01 AM, georgeshirreff wrote:
>
> > Dear all,
> >
> > I am trying
Thanks guys for the help, I'm going to go with Patrick Burns answer because
it seems to work the best for my situation, but these all seem like they
should work.
On Tue, Jul 3, 2012 at 2:51 AM, Patrick Burns wrote:
> b <- rep(list(d), length(b))
>
>
> On 02/07/2012 23:16, Spencer Maynes wrote:
>
I did not suggest that you don't HAVE an operating system... simply that you
don't know how to use yours.
However you accomplish starting programs automatically, you WILL need to
specify both the R interpreter (which you do seem to be accomplishing) and the
name of the script you wish to run (w
On Jul 3, 2012, at 4:43 AM, cindy.dol wrote:
Do you know how can I run a script on R from Excel without rExcel
but with
VBA and batch?
It would seem that this question should be directed to a VBA mailing
list.
--
David Winsemius, MD
West Hartford, CT
_
Hello,
I'm glad it helped. See answer inline.
Em 03-07-2012 17:09, Claudia Penaloza escreveu:
Thank you Rui and Jim, both 'i1' and 'i1new' worked perfectly
because there are no instances of 'Dd' or 'dD' in the data set (that I
would/not want to include/exclude)... but I understand that 'i1new'
Thank you Rui and Jim, both 'i1' and 'i1new' worked perfectly because there
are no instances of 'Dd' or 'dD' in the data set (that I would/not want to
include/exclude)... but I understand that 'i1new' targets precisely what I
want.
Why isn't a leader of zero's required for either 'i1' or 'i1new',
Hello,
The trick is to use seq() for date classes.
First of all, when creating data.frames use stringsAsFactors=FALSE, in
order not to convert them to factors. I've added this option to your
data.frame() instructions. And then,
mydata1$dates <- as.POSIXct(mydata1$dates)
mydata1b$dates <- as
Many thanks for these ideas ... I'll try them, and report back
Cheers Bob Kinley
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: 03 July 2012 15:54
To: Robert Douglas Kinley
Cc: r-help@r-project.org
Subject: Re: [R] saving contour() plot info
O
Hello,
Inline.
Em 03-07-2012 09:22, Sajeeka Nanayakkara escreveu:
I have already fitted several models
using R code; arima(rates,c(p,d,q))
As I heard, best model produce the
smallest AIC value, but maximum likelihood estimation procedure optimizer
should converge.
How to check whether maximum
Hello,
Try, with 'x' your date(s),
as.Date(x, format="%A, %B %d,%Y")
strptime(x, format="%A, %B %d,%Y")
Note: this is in the help page for ?strptime
Hope this helps,
Rui Barradas
Em 03-07-2012 07:04, arunkumar escreveu:
hi
I have a data like thursday, November 20,2012. I'm not able
On 03/07/2012 10:36 AM, Robert Douglas Kinley wrote:
{ I think this message got rejected at the 1st attempt - trying again}
R 2.15.1 , windows XP
I have a very non-stationary bivariate time-series - say {xt,yt} t=1 ...
lots.
I want to do a bivariate density contour-plot of the whole seri
On Jul 3, 2012, at 4:01 AM, georgeshirreff wrote:
Dear all,
I am trying to write figures directly to a file using the jpeg()
function.
this_ylab=expression(paste("P",sep="")~lambda)
this_xlab=expression(rho)
jpeg("file_name.jpeg",width=100,height=100,units="mm",res=300)
Try instead some
Dear,
I would like to use the siar program, when I lodging package the message
appear, so I couldn't use the related calculations.
The following object(s) are masked from package:spatstat:convexhull
Could you please write me what shoud I do?
Sukran Yalcin Ozdilek
[[alternative HTML ve
Thank you for your help
So i have tried many ways on different computers, but i believe i have an
operating system, as I open RScript up in my local directory it just
flashes at me and them disappears
I tried to task schedule, using the following code
"C:\Program Files\R\R-2.12.1\bin\i386\Rscri
Try using polygon() or grid.polygon() from the grid package.
You can find code example on the R project title page.
Go to http://r-project.org and click on the plot (the example plot showing
PCA, clustering and factors, lower left are two graphs, similar to what you
want).
This will bring you to
hi list,
used versions: 2.12.1 and 2.14.0 under ubuntu and macosx.
I recently stumbled over a problem with `nls', which occurs if the model
is not specified explicitly but via an evaluation of a 'call' object.
simple example:
8<--
That I correct. I have only 9.
Thanks for the explanation :)
--
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__
R Graph Gallery contains lots of plot examples with source code:
http://addictedtor.free.fr/graphiques/
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Sent from the R help mailing list archive at Nabble.com.
In R you should slashes instead of backslashes:
C:\PROGRA~1\R\R-2.11.1\bin\RScript.exe
C:/Users/Vincent/Documents/temp/test.r
Bart
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Sent from the R help mailing list archive at
Hi,
If you want to assign a vector to every element of a list,
vec1<-11:20
list1<-split(LETTERS[1:10],1:length(LETTERS[1:10]))
list2<-lapply(1:10,function(x) vec1)
or,
list3<-lapply(list1,function(x) list1=vec1)
or
list4<-list()
vec2<-1:5
list4[1:length(list1)]<-list(vec2)
# if you want to assi
Hello
I have dataframes.
mydata1 <-data.frame(value=c(15,20,25,30,45,50),dates=c("2005-05-25 07:00:00
","2005-05-25 19:00:00","2005-06-25 07:00:00","2005-06-25 19:00:00
","2005-07-25 07:00:00","2005-8-25 19:00:00"))
or
mydata2 <-data.frame(value=c(15,20,25,30,45,50),dates=c("2005-05-25 00:00:00
","
I have already fitted several models
using R code; arima(rates,c(p,d,q))
As I heard, best model produce the
smallest AIC value, but maximum likelihood estimation procedure optimizer
should converge.
How to check whether maximum likelihood estimation procedure optimizer has
converged or not?
hi
I have a data like thursday, November 20,2012. I'm not able to convert
into data format in R
Can anyone please help
-
Thanks in Advance
Arun
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Sent from the R help mailing list ar
And one more alternative:
a1$h2 <- apply(a1,1, function(x) if (x["h1"]=="H") 1 else 0 )
--
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___
Dear all,
I am trying to write figures directly to a file using the jpeg() function.
this_ylab=expression(paste("P",sep="")~lambda)
this_xlab=expression(rho)
jpeg("file_name.jpeg",width=100,height=100,units="mm",res=300)
plot(input_values,output_values,pch=16,cex=cex_pt,xlab="",frame.plot=T,an
{ I think this message got rejected at the 1st attempt - trying again}
R 2.15.1 , windows XP
I have a very non-stationary bivariate time-series - say {xt,yt} t=1 ...
lots.
I want to do a bivariate density contour-plot of the whole series and then step
through the series 1 second at a time
Anything here that might help
http://learnr.wordpress.com/2010/01/26/ggplot2-quick-heatmap-plotting/
John Kane
Kingston ON Canada
> -Original Message-
> From: joeclar...@hotmail.com
> Sent: Mon, 2 Jul 2012 13:21:25 -0700
> To: mueller.eisb...@googlemail.com, r-help@r-project.org
> Subjec
Any sample data for us to work with? See ?dput for a good method of supplying
sample data.
John Kane
Kingston ON Canada
> -Original Message-
> From: denissearchun...@yahoo.com.mx
> Sent: Mon, 2 Jul 2012 14:04:53 -0700 (PDT)
> To: r-help@r-project.org
> Subject: [R] how to do a graph wi
Your data is effectively unreadable. Please use dput() to supply sample data.
Here is one way to do what you want for data frames using the ggplot2 package.
library(ggplot2)
mydata <- data.frame(x = 1:10, y = rnorm(10), z = c(rep(1,4), rep(2, 6)))
ggplot(mydata, aes(x, y, colour= as.factor( z)
On Jul 3, 2012, at 5:08 AM, Jim Lemon wrote:
On 07/03/2012 05:18 PM, jin...@ga.gov.au wrote:
Hi all,
I would like create a new column in a data.frame (a1) to store 0, 1
data converted from a factor as below.
a1$h2<-NULL
for (i in 1:dim(a1)[1]) {
if (a1$h1[i]=="H") a1$h2[i]<-1 else a
On Jul 2, 2012, at 10:09 PM, Tjun Kiat Teo wrote:
I am trying to to write a wrapper function for the ode solver (under
the package desolve) to enable it to take multivariate arrays. I know
how to do it for 1 dimension arrays but my code breaks down when I try
to do it for 2 dimensional arrays.
On Mon, Mar 5, 2012 at 2:47 PM, vincent.deluard
wrote:
> Hi Jim,
>
> Please disregard my earlier post -- I have done some research and realized
> the space between after "program " was the the issue. I can now open
> RScript.exe from the command prompt using the abbreviated form:
>
> C:\PROGRA~1\R
Homework?? We don't do homework here. If not, ?optim or look at the CRAN
"Optimize" task view for optimizers. There is even a maxLik package that
might be useful.
-- Bert
On Mon, Jul 2, 2012 at 8:58 PM, Ali Tamaddoni wrote:
> Hi All
>
>
>
> I have a data frame called "nbd" with two columns (x a
Thanks Joshua for the clear explanation.
I've previously posted in r sig mixed, but I got no response... :(
Cheers,
- Camila
On Mon, Jul 2, 2012 at 9:31 PM, Joshua Wiley wrote:
> Hi Camila,
>
> In mixed equation form instead of multilevel, it would be:
>
> Y_it = gamma_00 + gamma_10*X_it + ga
On 07/03/2012 05:18 PM, jin...@ga.gov.au wrote:
Hi all,
I would like create a new column in a data.frame (a1) to store 0, 1 data
converted from a factor as below.
a1$h2<-NULL
for (i in 1:dim(a1)[1]) {
if (a1$h1[i]=="H") a1$h2[i]<-1 else a1$h2[i]<-0
}
My question: is it possible
Do you know how can I run a script on R from Excel without rExcel but with
VBA and batch?
--
View this message in context:
http://r.789695.n4.nabble.com/R-with-VBA-Excel-tp4635265.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@
Hello,
In order to avoid messing up the data, use dput. See below, in the end.
As for your question, try this:
set.seed(1234)
xyz <- data.frame(x=sample(20, 10), y=sample(20, 10), z=sample(0:1, 10,
TRUE))
# pch=16 --> solid circle; cex=4 --> 4 fold expansion
with(xyz, plot(x, y, col=z+1, pch
Hi
> Hi all,
>
> I would like create a new column in a data.frame (a1) to store 0, 1 data
> converted from a factor as below.
>
> a1$h2<-NULL
> for (i in 1:dim(a1)[1]) {
> if (a1$h1[i]=="H") a1$h2[i]<-1 else a1$h2[i]<-0
> }
>
> My question: is it possible to remove the loop from a
Hello,
This is here for some days now, and I've decided to give it a try.
I've rewritten your fitfunction(), making it simpler. And include the
gamma distribution in the list.
require(MASS)
fitfunction <- function(Type, x) list(Type=Type, Fit=fitdistr(x, Type))
fun <- function(x, data){
Hello,
Inline.
Em 03-07-2012 01:15, jim holtman escreveu:
You will have to change the 'i1' expression as follows:
i1 <- grepl("^([0D]|[0d])*$", dd$ch)
i1 # matches strings with d & D in them
[1] TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# second string had 'd' & 'D' in
b <- rep(list(d), length(b))
On 02/07/2012 23:16, Spencer Maynes wrote:
I have a vector d of unknown length, and a list b of unknown length. I
would like to replace every element of b with d. Simply writing b<-d does
not work as R tries to fit every element of d to a different element of d,
and
Hello,
Sorry, but it was you that misread some of the suggestions. I have
written raw=TRUE not raw=raw. Just see
m <- matrix(1:6, ncol=2) # your example
p2 <- poly(m, degree=2, raw=TRUE) # it's raw=TRUE, not raw=raw !!!
deg2 <- attr(p2, 'degree') == 2
p2[, deg2]
p6 <- poly(m, degree=6, raw
On 07/03/2012 06:21 AM, Joseph Clark wrote:
These "carpet plots" are also called "heat maps" and there's a current thread with the
subject line "Heat Maps" in which I've given a couple of examples of code for them. The R function
image() is very easy to use:
image( x=(x values), y=(y value
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