Re: [R] data frame manipulation with zero rows
Brian,
If I do understand correctly, I must use in my function something else than
ddply() if I want to avoid any error each time my df has zero rows?
Am I correct?
TY
> -Original Message-
> From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
> Sent: Tuesday, June 01, 2010 9:47 AM
> To: arnaud Gaboury
> Subject: Re: [R] data frame manipulation with zero rows
>
> On Tue, 1 Jun 2010, arnaud Gaboury wrote:
>
> > Dear group,
> >
> > Here is the kind of data.frame I obtain every day with my function :
> >
> > futures <-
> > structure(list(DESCRIPTION = c("CORN Jul/10", "CORN Jul/10",
> > "CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "LIVE CATTLE Aug/10",
> > "LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10",
> > "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10"
> > ), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406,
> > 18407, 18408, 18406, 18407, 18407, 18407, 18407), class = "Date"),
> >QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT =
> > c("373.2500",
> >"373.2500", "373.2500", "373.2500", "373.2500", "90.7750",
> >"90.7750", "14.9200", "14.9200", "14.9200", "14.9200", "14.9200"
> >)), .Names = c("DESCRIPTION", "CREATED.DATE", "QUANTITY",
> > "SETTLEMENT"), row.names = c(NA, 12L), class = "data.frame")
> >
> > I need then to apply to the df this following code line :
> >
> >> PosFut=ddply(futures, c("DESCRIPTION","SETTLEMENT"), summarise,
> POSITION=
> > sum(QUANTITY))[,c(1,3,2)]
> >
> > It works perfectly in most of case, BUT I have a new problem: it can
> > sometime occurs that my df "futures" is empty, with zero rows.
> >
> >
> > futures <-
> > structure(list(DESCRIPTION = character(0), CREATED.DATE =
> > structure(numeric(0), class = "Date"),
> >QUANTITY = numeric(0), SETTLEMENT = character(0)), .Names =
> > c("DESCRIPTION",
> > "CREATED.DATE", "QUANTITY", "SETTLEMENT"), row.names = integer(0),
> class =
> > "data.frame")
> >
> > It is not the usual case, but it can happen. With this df, when I
> pass the
> > above mentione line, I get an error :
> >
> >> PosFut=ddply(futures, c("DESCRIPTION","SETTLEMENT"), summarise,
> POSITION=
> > sum(QUANTITY))[,c(1,3,2)]
> > Error in tapply(1:nrow(data), splitv, list) :
> > arguments must have same length
> >
> >
> > How can I avoid this when my df is empty?
>
> Ask the author of the (missing) function ddply() to correct the error
> of using 1:nrow(data) by replacing it by seq_len(nrow(data)).
>
> It's helpful to give example code, but much more helpful if you test
> it: yours cannot work without the function ddply() -- this is what
> 'self-contained' means in the footer here.
>
>
> >
> > Any help is appreciated
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
> --
> Brian D. Ripley, rip...@stats.ox.ac.uk
> Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
> University of Oxford, Tel: +44 1865 272861 (self)
> 1 South Parks Road, +44 1865 272866 (PA)
> Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] MacOS X binary for lme4 not available on CRAN
Dear group, I have noticed that the MacOS X binary for lme4 is not available on CRAN at the moment. I am aware that it may be possible to install from source but I am not very familiar with that procedure and would rather avoid it. But I need the package for a statistics course next week. So can anybody give an update about the situation with the MacOS X binaries? Will it be resolved within the next days or should I rather install from source (in that case: is my assumption correct that it is possible or will it fail (I read the the check log and it seems that there is some error ?) Thank you for any advise or help -- Armin Goralczyk -- http://www.gwdg.de/~agoralc __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plot multiple columns
I'm running a long MCMC chain that is generating samples for 22 variables.
I have each run of the chain as a row in a matrix.
So: Chain[,1] is the column with all the samples for variable one.
Chain[,2] is the column with all the samples for variable 2, etc.
I'd like to fit all 22 on a single page to print a nice summary. It is
OK if the graphs are small, I just need to show the overall shape and
convergence.
Using par(mfrow=(11,2)) gives me the error: "figure margins too large"
when I try to draw a plot
I looked at the Lattice package, which seems very promising, but I can't
figure out how to have it plot each column in a separate box.
I need to do this same "one page" summary about 50 times, so it would be
a nightmare to make over 1,000 plots by hand.
Ideally, I'd create a loop for each of the 50 runs that would generate a
single page containing the 22 plots.
In pseudocode:
for( i in 1:50){
par(mfrow(11,2))
for(j in 1:22){
plot(Chain[,j], type="l")
}
}
BUT, this doesn't work.
Does anyone have any ideas about an easy way to do this?
Thanks!
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Re: [R] two questions about PLOT
thanks for your reply.
I have tried to use rseek.org.But still some problems.
When I add "axis(4)" and axis(1,at=1:6,labels=gradeinfo$gradenam),the old
tick or labels still
are there as shown in the figure,how could I delete them( the old tick
information in x-axis and left y axis )
My script is shown as below :
plot(avegrp,type='l',lty=1,col='black',lwd=4,xlab=xxlab,ylab=yylab,labels=TRUE)
axis(2)
axis(1,at=1:6,labels=gradeinfo$gradenam, tick=FALSE)
par(new=T)
plot(sdgrp,type='l',lty=3,col='red1',xlab=xxlab,ylab=yylab,lwd=4,labels=TRUE)
axis(1,at=1:6,labels=gradeinfo$gradenam, tick=FALSE)
#axis(2, col = "gold", lty = 2, lwd = 0.5)
axis(4)
legend("topright", legend, lty=llty, lwd=llwd,col =lgcol)
2010/6/1 Jannis
> I would wote this question one of the most often asked questions here on
> that list ;-). Try searching the help archiwe (www.rseek.org) and you will
> find solutions. I would guess that you need to use something like:
>
> axis(4)
>
> as the sides of the plot are always numbered from bottom,left,top,right
>
> HTH
> Jannis
> Jie TANG schrieb:
>
>> here ,I want to plot two lines in one figure.But I have two problems
>> 1) how to move one of the y-axis to be the right ? I tried to the
>> command"axis(2)",But I failed.
>> 2) how to add the axis information correctly.Since I have use the cmommand
>> "axis(1,at=1:6,labels=gradeinfo$gradenam)"
>> but it seems that the correct information that I want is superposition
>> with the old axis information.What can i do ?
>> the script and figure is shown as below .thanks .:)
>>
>> outflnm<-paste(Outdic,"meansd.jpg",sep="/")
>> jpeg(file=outflnm, bg="transparent")
>> legend<-c("average error","stand quare error")
>> lgcol<-c("black","red1")
>> par(las=1)
>> yylab<-c("forecast error")
>> xxlab<-c("typhoon class")
>> llty<-c(1,3)
>> llwd<-c(4,4)
>> #par(bg='yellow')
>>
>>
>> plot(avegrp,type='l',lty=1,col='black',lwd=4,xlab=xxlab,ylab=yylab)
>> par(new=T)
>> plot(sdgrp,type='l',lty=3,col='red1',xlab=xxlab,ylab=yylab,lwd=4)
>> #axis(2, col = "gold", lty = 2, lwd = 0.5)
>> legend("topright", legend, lty=llty, lwd=llwd,col =lgcol)
>> axis(1,at=1:6,labels=gradeinfo$gradenam)
>> dev.off()
>>
>>
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>
>
--
TANG Jie
Email: totang...@gmail.com
Tel: 0086-2154896104
Shanghai Typhoon Institute,China
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Re: [R] accessing a data frame with row names
On 31/05/2010, Gabor Grothendieck wrote: > Use read.csv or read.table(..., sep = ","). Also note that if you > delete the first comma of the header (as in the second example below) > you won't have to specify row.names since it can figure it out from > the fact that there is one fewer column name than data fields. > >> Lines <- ",column1,column2 > + row1,0.1,0.2 > + row2,0.3,0.4" >> >> read.csv(textConnection(Lines), row.names = 1) > column1 column2 > row1 0.1 0.2 > row2 0.3 0.4 Thank you. When I enter the command: max(dataframe[,2]) The response is: [1] 0.4 But I want to receive the row name, i.e.: [1] row2 0.4 Is this possible? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] accessing a data frame with row names
Hi r-help-boun...@r-project.org napsal dne 01.06.2010 10:20:35: > On 31/05/2010, Gabor Grothendieck wrote: > > Use read.csv or read.table(..., sep = ","). Also note that if you > > delete the first comma of the header (as in the second example below) > > you won't have to specify row.names since it can figure it out from > > the fact that there is one fewer column name than data fields. > > > >> Lines <- ",column1,column2 > > + row1,0.1,0.2 > > + row2,0.3,0.4" > >> > >> read.csv(textConnection(Lines), row.names = 1) > > column1 column2 > > row1 0.1 0.2 > > row2 0.3 0.4 > > Thank you. When I enter the command: > > max(dataframe[,2]) > > The response is: > > [1] 0.4 > > But I want to receive the row name, i.e.: > > [1] row2 0.4 It seems that you probably shall consult ?which with parameter arr.ind=T which(dataframe==max(dataframe), arr.ind=T) Regards Petr > > Is this possible? > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sweave & png
Thank you Ben Bolker a écrit : > Gildas Mazo curie.fr> writes: > > >> Is there a simple way to save my figures in png instead of pdf with >> Sweave ?? >> > > > See > > http://sites.google.com/site/thibautjombart/r-packages > > (scroll to the bottom) > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data frame manipulation ddply
Dear group,
Here is my data frame:
futures <-
structure(list(DESCRIPTION = c("CORN Jul/10", "CORN Jul/10",
"CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "LIVE CATTLE Aug/10",
"LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10",
"SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10"
), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406,
18407, 18408, 18406, 18407, 18407, 18407, 18407), class = "Date"),
QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT =
c("373.2500",
"373.2500", "373.2500", "373.2500", "373.2500", "90.7750",
"90.7750", "14.9200", "14.9200", "14.9200", "14.9200", "14.9200"
)), .Names = c("DESCRIPTION", "CREATED.DATE", "QUANTITY",
"SETTLEMENT"), row.names = c(NA, 12L), class = "data.frame")
Here is the line I pass :
>PosFut=ddply(futures, c("DESCRIPTION","SETTLEMENT"), summarise, POSITION=
sum(QUANTITY))[,c(1,3,2)]
And here the result :
PosFut <-
structure(list(DESCRIPTION = structure(1:3, .Label = c("CORN Jul/10",
"LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10"), class = "factor"),
POSITION = c(5, 4, 5), SETTLEMENT = structure(c(2L, 3L, 1L
), .Label = c("14.9200", "373.2500", "90.7750"), class = "factor")),
.Names = c("DESCRIPTION",
"POSITION", "SETTLEMENT"), class = "data.frame", row.names = c(NA,
-3L))
I can no more use ddply, as this above command line is in a function, and
this line should be able to work with a data frame with zero rows, and in
this case ddply doesn't work.
Any suggestion how to obtain the same result without ddply?
TY for any help.
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Re: [R] Faster matrix operation?
On 2010-06-01 0:16, bill.venab...@csiro.au wrote: xyzs<- matrix(rnorm(3*10,0,1),ncol=3) V<- c(2,3,4) system.time(vx<- apply(t(xyzs) * V, 2 ,sum)) user system elapsed 1.060.021.08 system.time(wx<- as.vector(xyzs %*% V)) user system elapsed 0 0 0 all.equal(vx, wx) [1] TRUE Or, for a very slight further reduction in time in the case of larger matrices/vectors: as.vector(tcrossprod(V, xyzs)) I mention this merely to remind new users of the excellent speed of [t]crossprod(). -Peter Ehlers ? -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Remko Duursma Sent: Tuesday, 1 June 2010 4:04 PM To: r-help@r-project.org Subject: [R] Faster matrix operation? Dear R-helpers, I have a three-column matrix with lots of rows: xyzs<- matrix(rnorm(3*10,0,1),ncol=3) # And I am multiplying it with some vector V, and summing the rows (columns after t()) in this way: V<- c(2,3,4) system.time(vx<- apply(t(xyzs) * V, 2 ,sum)) Ok, this does not take long (0.9 sec on my machine), but I have to do this lots of times, with frequently larger matrices. Is there a way to significantly speed this up, apart from writing it in Fortran or C and calling it from within R (which is what I am planning unless there is an alternative)? thanks, Remko __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Faster matrix operation?
On 1 June 2010 11:34, Peter Ehlers wrote: > Or, for a very slight further reduction in time in > the case of larger matrices/vectors: > > as.vector(tcrossprod(V, xyzs)) > > I mention this merely to remind new users of the > excellent speed of [t]crossprod(). > > -Peter Ehlers Thanks, I've been using t(x)%*%y for several years now, and had never found out there was a faster and more straight-forward function for this. Perhaps a link to [t]crossprod could be added on the help page for ?"%*%". Best, baptiste __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot multiple columns
Hi,
Try with:
par(mfrow=c(11,2))
It should work better!
Remember that mfrow is an argument of the function par(), not a function
itself.
One other tip: think about using pdf, ps, png, or SVG devices, I find it
easier and nicer.
HTH,
Ivan
Le 6/1/2010 10:02, Noah Silverman a écrit :
I'm running a long MCMC chain that is generating samples for 22 variables.
I have each run of the chain as a row in a matrix.
So: Chain[,1] is the column with all the samples for variable one.
Chain[,2] is the column with all the samples for variable 2, etc.
I'd like to fit all 22 on a single page to print a nice summary. It is
OK if the graphs are small, I just need to show the overall shape and
convergence.
Using par(mfrow=(11,2)) gives me the error: "figure margins too large"
when I try to draw a plot
I looked at the Lattice package, which seems very promising, but I can't
figure out how to have it plot each column in a separate box.
I need to do this same "one page" summary about 50 times, so it would be
a nightmare to make over 1,000 plots by hand.
Ideally, I'd create a loop for each of the 50 runs that would generate a
single page containing the 22 plots.
In pseudocode:
for( i in 1:50){
par(mfrow(11,2))
for(j in 1:22){
plot(Chain[,j], type="l")
}
}
BUT, this doesn't work.
Does anyone have any ideas about an easy way to do this?
Thanks!
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de
**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] New behavior of 'write.csv' append argument?
As noted in the help file under CSV files:
These wrappers are deliberately inflexible: they are designed to
ensure that the correct conventions are used to write a valid file.
Attempts to change append, col.names, sep, dec or qmethod are ignored,
with a warning.
On Mon, May 31, 2010 at 8:33 PM, Steven Worthington
wrote:
>
> Dear R users,
>
> I have recently begun to reuse some functions I made several months ago. The
> scripts write to a .csv file using the 'write.csv' function with the append
> option set to TRUE. This used to work fine, albeit with the warning
> "appending column names to file". I upgraded to R version 2.11.0 on OSX
> 10.5.8 and 'write.csv' will no longer append any files - I get the warning
> "attempt to set 'append' ignored". The exact same object will append when
> using 'write.table' without issue. Here's a minimal example. Has there been
> a change in 'write.csv'?
>
> #
> x <- rnorm(10, 5)
> y <- rnorm(10, 10)
> z <- rnorm(10, 15)
> dat <- data.frame(x, y, z)
> names(dat) <- c("first", "second", "third")
>
> write.csv(dat, file="dat.csv", append=TRUE)
> write.table(dat, file="dat2.csv", sep=",", row.names=FALSE, append=TRUE)
> #
>
> best,
>
> Steve
> --
> View this message in context:
> http://r.789695.n4.nabble.com/New-behavior-of-write-csv-append-argument-tp2237936p2237936.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
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Re: [R] two questions about PLOT
It may not be the nicest solution, but my suggestion should work.
Have you tried plot(type="n",...), plotting the axes with axis(), and
plotting the data with lines()?
Ivan
Le 6/1/2010 10:10, Jie TANG a écrit :
> thanks for your reply.
> I have tried to use rseek.org.But still some problems.
> When I add "axis(4)" and axis(1,at=1:6,labels=gradeinfo$gradenam),the old
> tick or labels still
> are there as shown in the figure,how could I delete them( the old tick
> information in x-axis and left y axis )
> My script is shown as below :
>
> plot(avegrp,type='l',lty=1,col='black',lwd=4,xlab=xxlab,ylab=yylab,labels=TRUE)
> axis(2)
> axis(1,at=1:6,labels=gradeinfo$gradenam, tick=FALSE)
> par(new=T)
>
> plot(sdgrp,type='l',lty=3,col='red1',xlab=xxlab,ylab=yylab,lwd=4,labels=TRUE)
> axis(1,at=1:6,labels=gradeinfo$gradenam, tick=FALSE)
> #axis(2, col = "gold", lty = 2, lwd = 0.5)
> axis(4)
> legend("topright", legend, lty=llty, lwd=llwd,col =lgcol)
>
>
>
> 2010/6/1 Jannis
>
>
>> I would wote this question one of the most often asked questions here on
>> that list ;-). Try searching the help archiwe (www.rseek.org) and you will
>> find solutions. I would guess that you need to use something like:
>>
>> axis(4)
>>
>> as the sides of the plot are always numbered from bottom,left,top,right
>>
>> HTH
>> Jannis
>> Jie TANG schrieb:
>>
>>
>>> here ,I want to plot two lines in one figure.But I have two problems
>>> 1) how to move one of the y-axis to be the right ? I tried to the
>>> command"axis(2)",But I failed.
>>> 2) how to add the axis information correctly.Since I have use the cmommand
>>> "axis(1,at=1:6,labels=gradeinfo$gradenam)"
>>>but it seems that the correct information that I want is superposition
>>> with the old axis information.What can i do ?
>>> the script and figure is shown as below .thanks .:)
>>>
>>> outflnm<-paste(Outdic,"meansd.jpg",sep="/")
>>> jpeg(file=outflnm, bg="transparent")
>>> legend<-c("average error","stand quare error")
>>> lgcol<-c("black","red1")
>>> par(las=1)
>>> yylab<-c("forecast error")
>>> xxlab<-c("typhoon class")
>>> llty<-c(1,3)
>>> llwd<-c(4,4)
>>> #par(bg='yellow')
>>>
>>>
>>> plot(avegrp,type='l',lty=1,col='black',lwd=4,xlab=xxlab,ylab=yylab)
>>> par(new=T)
>>> plot(sdgrp,type='l',lty=3,col='red1',xlab=xxlab,ylab=yylab,lwd=4)
>>> #axis(2, col = "gold", lty = 2, lwd = 0.5)
>>> legend("topright", legend, lty=llty, lwd=llwd,col =lgcol)
>>> axis(1,at=1:6,labels=gradeinfo$gradenam)
>>> dev.off()
>>>
>>>
>>>
>>> __
>>> R-help@r-project.org mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>>>
>>
>>
>
>
>
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de
**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation with zero rows
On 2010-06-01 1:53, arnaud Gaboury wrote:
Brian,
If I do understand correctly, I must use in my function something else than
ddply() if I want to avoid any error each time my df has zero rows?
Am I correct?
You could define a function to handle the zero-rows case:
f <- function(x){
if(nrow(x) < 1) out <- x[, c(1,3,2)] # or whatever
else
out <- ddply(x, c("DESCRIPTION","SETTLEMENT"), summarise,
POSITION=sum(QUANTITY))[,c(1,3,2)]
out
}
f(futures)
-Peter Ehlers
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: Tuesday, June 01, 2010 9:47 AM
To: arnaud Gaboury
Subject: Re: [R] data frame manipulation with zero rows
On Tue, 1 Jun 2010, arnaud Gaboury wrote:
Dear group,
Here is the kind of data.frame I obtain every day with my function :
futures<-
structure(list(DESCRIPTION = c("CORN Jul/10", "CORN Jul/10",
"CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "LIVE CATTLE Aug/10",
"LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10",
"SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10"
), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406,
18407, 18408, 18406, 18407, 18407, 18407, 18407), class = "Date"),
QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT =
c("373.2500",
"373.2500", "373.2500", "373.2500", "373.2500", "90.7750",
"90.7750", "14.9200", "14.9200", "14.9200", "14.9200", "14.9200"
)), .Names = c("DESCRIPTION", "CREATED.DATE", "QUANTITY",
"SETTLEMENT"), row.names = c(NA, 12L), class = "data.frame")
I need then to apply to the df this following code line :
PosFut=ddply(futures, c("DESCRIPTION","SETTLEMENT"), summarise,
POSITION=
sum(QUANTITY))[,c(1,3,2)]
It works perfectly in most of case, BUT I have a new problem: it can
sometime occurs that my df "futures" is empty, with zero rows.
futures<-
structure(list(DESCRIPTION = character(0), CREATED.DATE =
structure(numeric(0), class = "Date"),
QUANTITY = numeric(0), SETTLEMENT = character(0)), .Names =
c("DESCRIPTION",
"CREATED.DATE", "QUANTITY", "SETTLEMENT"), row.names = integer(0),
class =
"data.frame")
It is not the usual case, but it can happen. With this df, when I
pass the
above mentione line, I get an error :
PosFut=ddply(futures, c("DESCRIPTION","SETTLEMENT"), summarise,
POSITION=
sum(QUANTITY))[,c(1,3,2)]
Error in tapply(1:nrow(data), splitv, list) :
arguments must have same length
How can I avoid this when my df is empty?
Ask the author of the (missing) function ddply() to correct the error
of using 1:nrow(data) by replacing it by seq_len(nrow(data)).
It's helpful to give example code, but much more helpful if you test
it: yours cannot work without the function ddply() -- this is what
'self-contained' means in the footer here.
Any help is appreciated
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation ddply
Patrick,
When apply to this following df :
futures <-
structure(list(DESCRIPTION = character(0), CREATED.DATE =
structure(numeric(0), class = "Date"),
QUANTITY = numeric(0), SETTLEMENT = character(0)), .Names =
c("DESCRIPTION",
"CREATED.DATE", "QUANTITY", "SETTLEMENT"), row.names = integer(0), class =
"data.frame")
> PosFut <- aggregate(futures$QUANTITY, list(DESCRIPTION =
futures$DESCRIPTION,SETTLEMENT=futures$SETTLEMENT),sum)[,c(1,3,2)]
Error in aggregate.data.frame(as.data.frame(x), ...) :
no rows to aggregate
> -Original Message-
> From: Patrick Hausmann [mailto:patrick.hausm...@uni-bremen.de]
> Sent: Tuesday, June 01, 2010 11:38 AM
> To: arnaud Gaboury
> Subject: Re: [R] data frame manipulation ddply
>
> Hi Arnaud,
>
> maybe "aggregate" can help:
>
> PosFut <- aggregate(futures$QUANTITY, list(DESCRIPTION =
> futures$DESCRIPTION,
> SETTLEMENT = futures$SETTLEMENT),
> sum)[, c(1,3,2)]
>
> HTH,
> Patrick
>
> Am 01.06.2010 11:02, schrieb arnaud Gaboury:
> > Dear group,
> >
> > Here is my data frame:
> >
> >
> > futures<-
> > structure(list(DESCRIPTION = c("CORN Jul/10", "CORN Jul/10",
> > "CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "LIVE CATTLE Aug/10",
> > "LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10",
> > "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10"
> > ), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406,
> > 18407, 18408, 18406, 18407, 18407, 18407, 18407), class = "Date"),
> > QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT =
> > c("373.2500",
> > "373.2500", "373.2500", "373.2500", "373.2500", "90.7750",
> > "90.7750", "14.9200", "14.9200", "14.9200", "14.9200", "14.9200"
> > )), .Names = c("DESCRIPTION", "CREATED.DATE", "QUANTITY",
> > "SETTLEMENT"), row.names = c(NA, 12L), class = "data.frame")
> >
> > Here is the line I pass :
> >
> >> PosFut=ddply(futures, c("DESCRIPTION","SETTLEMENT"), summarise,
> POSITION=
> > sum(QUANTITY))[,c(1,3,2)]
> >
> > And here the result :
> >
> > PosFut<-
> > structure(list(DESCRIPTION = structure(1:3, .Label = c("CORN Jul/10",
> > "LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10"), class = "factor"),
> > POSITION = c(5, 4, 5), SETTLEMENT = structure(c(2L, 3L, 1L
> > ), .Label = c("14.9200", "373.2500", "90.7750"), class =
> "factor")),
> > .Names = c("DESCRIPTION",
> > "POSITION", "SETTLEMENT"), class = "data.frame", row.names = c(NA,
> > -3L))
> >
> > I can no more use ddply, as this above command line is in a function,
> and
> > this line should be able to work with a data frame with zero rows,
> and in
> > this case ddply doesn't work.
> > Any suggestion how to obtain the same result without ddply?
> >
> > TY for any help.
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> > and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] New behavior of 'write.csv' append argument?
Is there a reason why append=TRUE should not be used?
I actually use it a lot; that way I have less csv files created. I can
output everything to a single file. Since some statistical tests returns
lists with different number of elements, it is not always possible to
export the output all at once (or at least I don't know how).
All the packages that I know of that export to xls do not append either.
So from R2.11 is there no way to append to xls/csv anymore?
Ivan
Le 6/1/2010 11:56, jim holtman a écrit :
As noted in the help file under CSV files:
These wrappers are deliberately inflexible: they are designed to
ensure that the correct conventions are used to write a valid file.
Attempts to change append, col.names, sep, dec or qmethod are ignored,
with a warning.
On Mon, May 31, 2010 at 8:33 PM, Steven Worthington
wrote:
Dear R users,
I have recently begun to reuse some functions I made several months ago. The
scripts write to a .csv file using the 'write.csv' function with the append
option set to TRUE. This used to work fine, albeit with the warning
"appending column names to file". I upgraded to R version 2.11.0 on OSX
10.5.8 and 'write.csv' will no longer append any files - I get the warning
"attempt to set 'append' ignored". The exact same object will append when
using 'write.table' without issue. Here's a minimal example. Has there been
a change in 'write.csv'?
#
x<- rnorm(10, 5)
y<- rnorm(10, 10)
z<- rnorm(10, 15)
dat<- data.frame(x, y, z)
names(dat)<- c("first", "second", "third")
write.csv(dat, file="dat.csv", append=TRUE)
write.table(dat, file="dat2.csv", sep=",", row.names=FALSE, append=TRUE)
#
best,
Steve
--
View this message in context:
http://r.789695.n4.nabble.com/New-behavior-of-write-csv-append-argument-tp2237936p2237936.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de
**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot multiple columns
Hi,
You could use melt from the reshape package to create a long format
data.frame. This is more easy to plot with lattice or ggplot2, and you
can then use facetting to arrange several plots on the same page. The
dummy example below produces 10 pages of output with 10 graphs per
page.
library(ggplot2)
dl <- replicate(10, as.data.frame(matrix(rnorm(1e3), ncol=10)), simplify=FALSE)
names(dl) <- paste("column", seq_along(dl), sep="")
# dummy list of data.frames
str(dl)
# a function to plot one data.frame
plotone <- function(d, ...)
qplot(seq_along(value), value, data=melt(d)) +
facet_wrap(~variable, scales="free")
# call the function for each data.frame
pl <- llply(dl, plotone)
# print to a file
pdf("test.pdf")
l_ply(pl, print)
dev.off()
HTH,
baptiste
On 1 June 2010 10:02, Noah Silverman wrote:
> I'm running a long MCMC chain that is generating samples for 22 variables.
>
> I have each run of the chain as a row in a matrix.
>
> So: Chain[,1] is the column with all the samples for variable one.
> Chain[,2] is the column with all the samples for variable 2, etc.
>
> I'd like to fit all 22 on a single page to print a nice summary. It is
> OK if the graphs are small, I just need to show the overall shape and
> convergence.
>
> Using par(mfrow=(11,2)) gives me the error: "figure margins too large"
> when I try to draw a plot
> I looked at the Lattice package, which seems very promising, but I can't
> figure out how to have it plot each column in a separate box.
>
> I need to do this same "one page" summary about 50 times, so it would be
> a nightmare to make over 1,000 plots by hand.
>
> Ideally, I'd create a loop for each of the 50 runs that would generate a
> single page containing the 22 plots.
>
> In pseudocode:
>
> for( i in 1:50){
> par(mfrow(11,2))
> for(j in 1:22){
> plot(Chain[,j], type="l")
> }
> }
>
> BUT, this doesn't work.
>
> Does anyone have any ideas about an easy way to do this?
>
> Thanks!
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] data frame manipulation with zero rows
On Tue, 1 Jun 2010, Peter Ehlers wrote:
On 2010-06-01 1:53, arnaud Gaboury wrote:
Brian,
If I do understand correctly, I must use in my function something else than
ddply() if I want to avoid any error each time my df has zero rows?
Am I correct?
You could define a function to handle the zero-rows case:
f <- function(x){
if(nrow(x) < 1) out <- x[, c(1,3,2)] # or whatever
else
out <- ddply(x, c("DESCRIPTION","SETTLEMENT"), summarise,
POSITION=sum(QUANTITY))[,c(1,3,2)]
out
}
f(futures)
Or simply fix ddply. We don't know what that is or what it should do
for the case of zero rows: it may or may not be the one in package
plyr.
-Peter Ehlers
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: Tuesday, June 01, 2010 9:47 AM
To: arnaud Gaboury
Subject: Re: [R] data frame manipulation with zero rows
On Tue, 1 Jun 2010, arnaud Gaboury wrote:
Dear group,
Here is the kind of data.frame I obtain every day with my function :
futures<-
structure(list(DESCRIPTION = c("CORN Jul/10", "CORN Jul/10",
"CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "LIVE CATTLE Aug/10",
"LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10",
"SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10"
), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406,
18407, 18408, 18406, 18407, 18407, 18407, 18407), class = "Date"),
QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT =
c("373.2500",
"373.2500", "373.2500", "373.2500", "373.2500", "90.7750",
"90.7750", "14.9200", "14.9200", "14.9200", "14.9200", "14.9200"
)), .Names = c("DESCRIPTION", "CREATED.DATE", "QUANTITY",
"SETTLEMENT"), row.names = c(NA, 12L), class = "data.frame")
I need then to apply to the df this following code line :
PosFut=ddply(futures, c("DESCRIPTION","SETTLEMENT"), summarise,
POSITION=
sum(QUANTITY))[,c(1,3,2)]
It works perfectly in most of case, BUT I have a new problem: it can
sometime occurs that my df "futures" is empty, with zero rows.
futures<-
structure(list(DESCRIPTION = character(0), CREATED.DATE =
structure(numeric(0), class = "Date"),
QUANTITY = numeric(0), SETTLEMENT = character(0)), .Names =
c("DESCRIPTION",
"CREATED.DATE", "QUANTITY", "SETTLEMENT"), row.names = integer(0),
class =
"data.frame")
It is not the usual case, but it can happen. With this df, when I
pass the
above mentione line, I get an error :
PosFut=ddply(futures, c("DESCRIPTION","SETTLEMENT"), summarise,
POSITION=
sum(QUANTITY))[,c(1,3,2)]
Error in tapply(1:nrow(data), splitv, list) :
arguments must have same length
How can I avoid this when my df is empty?
Ask the author of the (missing) function ddply() to correct the error
of using 1:nrow(data) by replacing it by seq_len(nrow(data)).
It's helpful to give example code, but much more helpful if you test
it: yours cannot work without the function ddply() -- this is what
'self-contained' means in the footer here.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] New behavior of 'write.csv' append argument?
On 2010-05-31 18:33, Steven Worthington wrote:
Dear R users,
I have recently begun to reuse some functions I made several months ago. The
scripts write to a .csv file using the 'write.csv' function with the append
option set to TRUE. This used to work fine, albeit with the warning
"appending column names to file". I upgraded to R version 2.11.0 on OSX
10.5.8 and 'write.csv' will no longer append any files - I get the warning
"attempt to set 'append' ignored". The exact same object will append when
using 'write.table' without issue. Here's a minimal example. Has there been
a change in 'write.csv'?
#
x<- rnorm(10, 5)
y<- rnorm(10, 10)
z<- rnorm(10, 15)
dat<- data.frame(x, y, z)
names(dat)<- c("first", "second", "third")
write.csv(dat, file="dat.csv", append=TRUE)
write.table(dat, file="dat2.csv", sep=",", row.names=FALSE, append=TRUE)
#
It's always a good idea to check the NEWS file. Here is a quote:
"write.csv[2] no longer allow 'append' to be changed: as ever,
direct calls to write.table() give more flexibility as well as
more room for error."
So use write.table.
-Peter Ehlers
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] MacOS X binary for lme4 not available on CRAN
First, this is not the list to discuss MacOS X binaries: that is R-sig-mac. Second, only one person really knows about the situation with package binaries, the person who produces them (equally true for Windows and for Mac OS X). And that person does not read R-help, having asked for such issues to be discussed on R-sig-mac. I very much doubt that this will be solved this week: the failure of lme4 in its checks on Mac OS X has been going on for years. On Tue, 1 Jun 2010, Armin Goralczyk wrote: Dear group, I have noticed that the MacOS X binary for lme4 is not available on CRAN at the moment. I am aware that it may be possible to install from source but I am not very familiar with that procedure and would rather avoid it. But I need the package for a statistics course next week. So can anybody give an update about the situation with the MacOS X binaries? Will it be resolved within the next days or should I rather install from source (in that case: is my assumption correct that it is possible or will it fail (I read the the check log and it seems that there is some error ?) Installation will work but package checking will fail, at least on an i386 Mac. Thank you for any advise or help -- Armin Goralczyk -- http://www.gwdg.de/~agoralc __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Printing page nmbers in multi-page pdf
On May 31, 2010, at 11:40 PM, Nevil Amos wrote: Is it possible to print page numbers in pdf() with multiple pages? ?mtext thanks Nevil Amos __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] n-dimensional vector plot
On Jun 1, 2010, at 12:51 AM, suman dhara wrote: Sir, I have n p-dimensional regressor vectors corresponding to n responses. I want to plot n regressor vectors in R. Is it possible to plot them in R? If yes, please send me the code. ?matplot Thanks, Suman Dhara David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help barplots
Dear All, I am newbie to R, and I wanted to plot a barplots with R and in such a way that It will also show me position which I can plot on the bar line. Here is my code that I am using to plot, > chromosome <- c(40.2, 35.6, 36.1, 29.6, 31, 29.6, 31, 29.4, 28.2, 23, 23, 28.2) >barplot (chromosome, col="purple", xlab="Oryza sativa Chromosomes", border = NA, space = 5, ylim = c(0,45)) I wanted to mark the position say on chromosome 1 (40.2) I need to mark 10.2 and on other also. I also want to set the scale of y axis from 0,5,10,15,20,25,30,35,40,45 i.e gap of 5 instead of 10. please help me to solve my querygurus. Thank you Jeet [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Help barplots
Hi r-help-boun...@r-project.org napsal dne 01.06.2010 13:01:38: > Dear All, > > I am newbie to R, and I wanted to plot a barplots with R and in such a way > that It will also show me position which I can plot on the bar line. > > Here is my code that I am using to plot, > > > chromosome <- c(40.2, 35.6, 36.1, 29.6, 31, 29.6, 31, 29.4, 28.2, 23, 23, > 28.2) > >barplot (chromosome, col="purple", xlab="Oryza sativa Chromosomes", border > = NA, space = 5, ylim = c(0,45)) > > I wanted to mark the position say on chromosome 1 (40.2) I need to mark 10.2 > and on other also. I do not understand what you want? What is 10.2 maybe names.arg is what you want. barplot (chromosome, col="purple", xlab="Oryza sativa Chromosomes", border = NA, space = 5, ylim = c(0,45), names.arg=chromosome, axes=F) > I also want to set the scale of y axis from 0,5,10,15,20,25,30,35,40,45 i.e > gap of 5 instead of 10. use axes=F and axis(2, at=seq(0,45,5)) Regards Petr > > please help me to solve my querygurus. > > > Thank you > Jeet > >[[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] two questions about PLOT
On 06/01/2010 12:44 AM, Jie TANG wrote:
here ,I want to plot two lines in one figure.But I have two problems
1) how to move one of the y-axis to be the right ? I tried to the
command"axis(2)",But I failed.
2) how to add the axis information correctly.Since I have use the cmommand
"axis(1,at=1:6,labels=gradeinfo$gradenam)"
but it seems that the correct information that I want is superposition
with the old axis information.What can i do ?
the script and figure is shown as below .thanks .:)
outflnm<-paste(Outdic,"meansd.jpg",sep="/")
jpeg(file=outflnm, bg="transparent")
legend<-c("average error","stand quare error")
lgcol<-c("black","red1")
par(las=1)
yylab<-c("forecast error")
xxlab<-c("typhoon class")
llty<-c(1,3)
llwd<-c(4,4)
#par(bg='yellow')
plot(avegrp,type='l',lty=1,col='black',lwd=4,xlab=xxlab,ylab=yylab)
par(new=T)
plot(sdgrp,type='l',lty=3,col='red1',xlab=xxlab,ylab=yylab,lwd=4)
#axis(2, col = "gold", lty = 2, lwd = 0.5)
legend("topright", legend, lty=llty, lwd=llwd,col =lgcol)
axis(1,at=1:6,labels=gradeinfo$gradenam)
dev.off()
Hi Jie,
First problem:
# put this just after the jpeg device is opened
# to leave extra space on the right
par(mar=c(5,4,4,4)
# then just before "dev.off"
axis(4,...)
Second problem:
# don't display the default x axis
plot(avegrp,type='l',lty=1,col='black',lwd=4,
xlab=xxlab,ylab=yylab,xaxt="n")
# then display your custom axis
Jim
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Re: [R] Fancy Page layout
On 06/01/2010 04:16 AM, Noah Silverman wrote:
Hi,
Working on a report that is going to have a large number of graphs and
summaries. We have 80 "groups" with 20 variables each.
Ideally, I'd like to produce ONE page for each group. It would have two
columns of 10 graphs and then the 5 number summary of the variables at
the bottom.
So, perhaps the top 2/3 of the page has the graphs and the bottom third
has 20 rows of data summary(maybe a table of sorts.)
This COULD be done in Latex, but would have to be hand coded for each of
the 80 groups which would be painfully slow.
I can easily do the graphs with par(mfrow=c(5,2)) band then draw the
graphs in a loop.
But I am stuck from here:
1) How do I control the size of the plot window. (Ideally, it should
print to fill an 8.5 x 11 piece of paper)
2) Is there a way to "easily" insert a 5 number summary (summary
command) into the lower half of the page.
Does anybody have any ideas??
Hi Noah,
One easy way is to leave some space at the bottom, either by using:
par(mfrow=c(6,2))
or the more flexible "layout" function, and then use "text" or a fancier
function (textbox, boxed.labels, addtable2plot, etc.) to add your text
after:
par(xpd=NA)
allows you to display the text anywhere you please. If you use a bitmap
graphics device, make it big:
png("numberoneofeighty.png",850,1100)
so that it won't look lumpy on the printed page.
Jim
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Re: [R] Help barplots
Hi,
If you want to draw lines on your barchart then
aa = barplot(chromosome, col="purple", xlab="Oryza sativa Chromosomes",
border
= NA, space = 5, ylim = c(0,45))
returns the midpoints of each bar in the vector aa and then you can use
the lines() function to do the drawing.
Martyn
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of khush
Sent: 01 June 2010 12:02
To: r-help@r-project.org
Subject: [R] Help barplots
Dear All,
I am newbie to R, and I wanted to plot a barplots with R and in such a
way
that It will also show me position which I can plot on the bar line.
Here is my code that I am using to plot,
> chromosome <- c(40.2, 35.6, 36.1, 29.6, 31, 29.6, 31, 29.4, 28.2, 23,
23,
28.2)
>barplot (chromosome, col="purple", xlab="Oryza sativa Chromosomes",
border
= NA, space = 5, ylim = c(0,45))
I wanted to mark the position say on chromosome 1 (40.2) I need to mark
10.2
and on other also.
I also want to set the scale of y axis from 0,5,10,15,20,25,30,35,40,45
i.e
gap of 5 instead of 10.
please help me to solve my querygurus.
Thank you
Jeet
[[alternative HTML version deleted]]
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Re: [R] Help barplots
On 06/01/2010 09:01 PM, khush wrote: Dear All, I am newbie to R, and I wanted to plot a barplots with R and in such a way that It will also show me position which I can plot on the bar line. Here is my code that I am using to plot, chromosome<- c(40.2, 35.6, 36.1, 29.6, 31, 29.6, 31, 29.4, 28.2, 23, 23, 28.2) barplot (chromosome, col="purple", xlab="Oryza sativa Chromosomes", border = NA, space = 5, ylim = c(0,45)) I wanted to mark the position say on chromosome 1 (40.2) I need to mark 10.2 and on other also. I also want to set the scale of y axis from 0,5,10,15,20,25,30,35,40,45 i.e gap of 5 instead of 10. please help me to solve my querygurus. Hi Jeet, I think you want the x positions of the bars. Get them like this: xpos<-barplot (chromosome, col="purple", xlab="Oryza sativa Chromosomes", border = NA, space = 5, ylim = c(0,45)) Then you can place the extra labels using the values in xpos. For the custom y axis, add the argument yaxt="n" to your plot command and then add the axis later. I suspect you will have to use something like the staxlab function in the plotrix package to get all those labels to display. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help barplots
On 2010-06-01 5:01, khush wrote: Dear All, I am newbie to R, and I wanted to plot a barplots with R and in such a way that It will also show me position which I can plot on the bar line. Here is my code that I am using to plot, chromosome<- c(40.2, 35.6, 36.1, 29.6, 31, 29.6, 31, 29.4, 28.2, 23, 23, 28.2) barplot (chromosome, col="purple", xlab="Oryza sativa Chromosomes", border = NA, space = 5, ylim = c(0,45)) I wanted to mark the position say on chromosome 1 (40.2) I need to mark 10.2 and on other also. I also want to set the scale of y axis from 0,5,10,15,20,25,30,35,40,45 i.e gap of 5 instead of 10. please help me to solve my querygurus. If I understand correctly, maybe you want something like this: # save the x-locations of the bars in a vector bp; # use that to put marks on the bars with points(); # omit the default y-axis with yaxt = "n" or, as Petr # showed, with axes = FALSE; bp <- barplot(chromosome, border = NA, space = 5, ylim = c(0, 45), yaxt = "n") points(bp[1], 10.2, pch = 4, cex = 2) # add the y-axis: axis(2, at = seq(0, 45, 5), las = 1) #(you might want to add: box(bty = "l") Instead of points(), you could use segments() to place horizontal marks. -Peter Ehlers Thank you Jeet __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R on the iPhone/iPad? Not so much....a GPL violation
Hi all, Thanks to an offlist e-mail from Thomas (Lumley), I have spent the past few days getting a bit more edumacated on additional restrictions on the opportunity for R to appear on the iPhone/iPad. These in fact go well beyond the GPL issue that I raised in my initial post, which in and of itself, is not trivial by any means. I now know, or at least think I know, more about these issues than I probably wanted, but I also want to present a better picture of the situation. Note that I am not a lawyer and am not intending to represent my findings from a legal perspective. I am reporting them here using a common sense approach from my own reading of the relevant Apple iPhone SDK language, as well as being based upon specific examples and discussions that I located on the web. So, in summary, here are the key issues. I will follow each with some additional details below, but note that all such restrictions would need to be removed or otherwise overcome, before R could in fact appear on these two platforms, at least through Apple approved means in the App Store. 1. Distribution of GPL covered applications is not permissible via the App Store due to the Apple Terms of Service language, which infringes upon rights granted under the GPL. 'Nuff said. 2. The use of FORTRAN is precluded (explicitly from SDK version 4.x forward) Version 3.x of the iPhone SDK has the following language: 3.3.1 Applications may only use Documented APIs in the manner prescribed by Apple and must not use or call any private APIs. Apple of course does not offer a FORTRAN compiler, as those who build R from source on OSX, as I do, are keenly aware. Thanks to Simon, we have one to use for OSX, but one is not officially available for the iPhone/iPad in the SDK. Note that there is an important distinction here. The ability to build R versus the ability to have the resultant application pass Apple's review to be able to appear in the App Store. The above language has also been interpreted to apply to Java/Flash, precluding those environments from appearing on the iPhone/iPad. However, the beta release of the 4.x version of the iPhone SDK has the following language in the same section: 3.3.1 — Applications may only use Documented APIs in the manner prescribed by Apple and must not use or call any private APIs. Applications must be originally written in Objective-C, C, C++, or JavaScript as executed by the iPhone OS WebKit engine, and only code written in C, C++, and Objective-C may compile and directly link against the Documented APIs (e.g., Applications that link to Documented APIs through an intermediary translation or compatibility layer or tool are prohibited). Thus, in fact, one can only use Objective-C, C, C++ or JavaScript to develop applications for the iPhone/iPad. No FORTRAN, upon which of course, R is dependent. The above language has also been interpreted to further reinforce restrictions specifically on Java/Flash. 3. The implementation of programming language interpreters, of which R is one, is precluded. The following language appears in version 3.x of the iPhone SDK: 3.3.2 An Application may not itself install or launch other executable code by any means, including without limitation through the use of a plug-in architecture, calling other frameworks, other APIs or otherwise. No interpreted code may be downloaded or used in an Application except for code that is interpreted and run by Apple's Documented APIs and built-in interpreter(s). The above language has been (pardon the pun) interpreted to restrict the implementation of programming language interpreters on these devices. Some sites I found have narrowly focused on the "No interpreted code may be downloaded" part of the language as an "out" of sorts. However, upon review, they seem to ignore the "or used" wording, which would seem to be independent of the action of downloading. If the language was "and used", then one could envision the use of locally stored or entered code, but this is not the case. In either case, of course, R would not be one of Apple's "built-in" interpreters. I found two interesting examples of how Apple has either approved or rejected applications that can be considered interpreters. It is not clear where Apple drew the line between these two, such that it might enable one to better differentiate the reasoning and therefore design an app that would pass muster with them. Thus, one complication may be Apple's lack of internal consistency in making decisions on allowing or disallowing apps in the App Store. The first is an application which was intended to be a BASIC interpreter on the iPhone and which was apparently rejected by Apple under the name BasicMatrix: http://smartcalc.coollittlethings.com/?p=3 After the author substantially restricted the functionality of the application to being an "enhanced calculator" under the name "SmartCalc", Apple approved the ap
Re: [R] Plot multiple columns
Noah Silverman smartmediacorp.com> writes: > > I'm running a long MCMC chain that is generating samples for 22 variables. > > I have each run of the chain as a row in a matrix. > So: Chain[,1] is the column with all the samples for variable one. > Chain[,2] is the column with all the samples for variable 2, etc. The previous 2 paragraphs seemed contradictory until I realized that in the first paragraph you are using "run" to mean what I would usually call a "sample" ... > I'd like to fit all 22 on a single page to print a nice summary. It is > OK if the graphs are small, I just need to show the overall shape and > convergence. How about for example > x <- matrix(runif(22000),ncol=22) > library(coda) > m <- as.mcmc(x) > xyplot(m) > xyplot(m,layout=c(4,6)) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Mid-P value for a chi-squared test
Can anyone tell me how to calculate a mid-p value for a chi-squared test in R? Many thanks, Andrew Wilson __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] storing output data from a loop that has varying row numbers
Hi All,
I am trying to run a loop that will have varying numbers of rows with each
output.
Previously I have had the same number of rows so I would use (and I
appreciate that this will no doubt achieve some gasps as being thoroughly
inefficient!):
xdfrow<-(0)
xdfrow1<-(1:32)
xdfrow2<-(33:64)
xdfrow3<-(65:96)
xdfrow4<-(97:128)
xdfrow5<-(129:160)
xdfrow6<-(161:192)
xdfrow7<-(193:224)
and so on
xdf <- matrix(999, nrow=1024, ncol=7)
xdf <- as.data.frame(xdf)
NAM <- c("NAME","ID2","DAY","BEH", "B_FALSE", "B_TRUE","TOTAL")
colnames(xdf)<-NAM
I then use this matrix and then run the loop and assign the data to each of
the xdfrows just doing +1 on each loop. (If that makes sense? Not really
important, just trying to show that I do try and solve some of my own
problems, albeit perhaps not in the best manner!)
_
However, the data I'm working with now has a very varied number of rows
(0:2500) over a large data set and I can't work out how is best to do this.
So my loop would be:
for (i in 1:33){
SEL_DAY<-seal_dist[seal_dist[,10]==i,]
print(paste("DAY", i, "of 33"))
for (s in 1:11){
SEL_HR<-SEL_DAY[SEL_DAY[,5]==s,]
print(paste("HR", s, "of 11"))
indx <- subset(SEL_HR, SEL_HR$DIST == 0)
SEL_HR$TO_ID <- indx$ID[match(SEL_HR$TO, indx$TO)]}
}
where i is day and s is the hr within the day, the loop works fine because
it prints as i expect it too. I have not given any info on the data because
I assume this is more of a method question and will be very straight forward
to most people on here!? But I am happy to post data if it is needed.
I assume I need to set up a matrix before the loop,
e.g. DIST_LOOP<-matrix(NA,1000,ncol=11)
and then I should be able to put something before the first } that allows me
to add to the matrix, but everything I have tried doesn't work
e.g. DIST_LOOP[[i]]<-SEL_HR
Any help would be much appreciated,
Best wishes,
Ross
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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Re: [R] storing output data from a loop that has varying row numbers
Hi,
I might be a bit tired so that I don't understand everything but I'm a
bit confused.
How would you like your DIST_LOOP matrix to look like?
What do you intend to use xdfrow1...xdfrow7? As I said, I might not be
at the best of my shape!
A sample dataset would really help to see what you have and what you
want to do (maybe using dput).
But maybe just:
DIST_LOOP[i, ]<-SEL_HR
would be enough, if you want to fill rows, as I understood.
HTH,
Ivan
Le 6/1/2010 13:51, RCulloch a écrit :
Hi All,
I am trying to run a loop that will have varying numbers of rows with each
output.
Previously I have had the same number of rows so I would use (and I
appreciate that this will no doubt achieve some gasps as being thoroughly
inefficient!):
xdfrow<-(0)
xdfrow1<-(1:32)
xdfrow2<-(33:64)
xdfrow3<-(65:96)
xdfrow4<-(97:128)
xdfrow5<-(129:160)
xdfrow6<-(161:192)
xdfrow7<-(193:224)
and so on
xdf<- matrix(999, nrow=1024, ncol=7)
xdf<- as.data.frame(xdf)
NAM<- c("NAME","ID2","DAY","BEH", "B_FALSE", "B_TRUE","TOTAL")
colnames(xdf)<-NAM
I then use this matrix and then run the loop and assign the data to each of
the xdfrows just doing +1 on each loop. (If that makes sense? Not really
important, just trying to show that I do try and solve some of my own
problems, albeit perhaps not in the best manner!)
_
However, the data I'm working with now has a very varied number of rows
(0:2500) over a large data set and I can't work out how is best to do this.
So my loop would be:
for (i in 1:33){
SEL_DAY<-seal_dist[seal_dist[,10]==i,]
print(paste("DAY", i, "of 33"))
for (s in 1:11){
SEL_HR<-SEL_DAY[SEL_DAY[,5]==s,]
print(paste("HR", s, "of 11"))
indx<- subset(SEL_HR, SEL_HR$DIST == 0)
SEL_HR$TO_ID<- indx$ID[match(SEL_HR$TO, indx$TO)]}
}
where i is day and s is the hr within the day, the loop works fine because
it prints as i expect it too. I have not given any info on the data because
I assume this is more of a method question and will be very straight forward
to most people on here!? But I am happy to post data if it is needed.
I assume I need to set up a matrix before the loop,
e.g. DIST_LOOP<-matrix(NA,1000,ncol=11)
and then I should be able to put something before the first } that allows me
to add to the matrix, but everything I have tried doesn't work
e.g. DIST_LOOP[[i]]<-SEL_HR
Any help would be much appreciated,
Best wishes,
Ross
--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de
**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php
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Re: [R] What does LOESS stand for?
The closest "explanation" of the term "loess" I can find is in "Local regression models" by Cleveland, W. S.; Grosse, E. & Shyu, W. M. (Chapter 8 of Statistical models in S (1992) by Chambers, J. M. & Hastie, T. J.) ... local regression models is called "loess," which is short for "local regression," and was chosen as the name since a loess is a deposit of fine clay or silt ..., and is a surface of sorts. (p. 314). Tal Galili Sent by: r-help-boun...@r-project.org 05/31/2010 12:39 PM To Peter Neuhaus cc r-help@r-project.org Subject Re: [R] What does LOESS stand for? Hi Peter, If this article is correct: http://www.r-bloggers.com/abbreviations-of-r-commands-explained-250-r-abbreviations/ Loess stands for: [LO]cally [E]stimated [S]catterplot [S]moothing Best, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Mon, May 31, 2010 at 12:33 PM, Peter Neuhaus wrote: > Dear R-community, > > maybe someone can help me with this: > > I've been using the loess() smoother for quite a while now, and for > the matter of documentation I'd like to resolve the acronym LOESS. > Unfortunately there's no explanation in the help file, and I didn't > get anything convincing from google either. > > I know that the predecessor LOWESS stands for "Locally Weighted > Scatterplot Smoothing". But what does LOESS stand for, specifically? > "Locally Weighted Exponential Scatterplot Smoothing"? As far as > I understand LOESS is still a local polynomial regression, so that > would probably make no sense. > > Any help appreciated! > > Thanks in advance, > > Peter > > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R-help "spam" detection; please help the moderators
Dear readers of R-help
as most of you will *not* be aware, R-help has continued to work the
way it does, only thanks to a dozen of volunteers,
see https://stat.ethz.ch/mailman/listinfo/r-help .
The volunteers manually moderate e-mails that "look like spam" (and
sometimes are and sometimes are not).
While much more than 90% of the spam is filtered out long before
a human sees it, with the increasing sophistication of spammers,
manual intervention has deemed to be necessary and served the
community very well.
OTOH, in recent weeks, the amount of work for the volunteers has
increased, mainly because an increasingly number of non-spam postings are
erronously tagged as "possibly spam".
We have discussed about this and done some analysis and found
that most of these message that produce a considerable amount of
extra work share two properties :
1) they are posted via Nabble {which *always* attaches a small
pro-Nabble spam at the end of the message}
2) the e-mail address of the sender is from a freemail
provider, quite often 'at gmail dot com', and often the part
*before* the '@' (at-sign) ends with digits.
We hereby ask those among you who use a freemail account to
please no longer post via nabble.
Thank you for your support of R-help, *the* "community mailing
list" of the R project since even before that project existed
"formally", namely since 1997-04-01,
today 13 years and two months.
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[R] Issue with assigning text to matrix
My issue relates to adding text to a matrix and finding that the text is
converted to a number.
This is the section of code I'm having trouble with:
# First, I load in a list of names from a .csv file to 'names'
names <- read.csv(file("Names.csv"))
# Then I define a matrix which will be populated with various test
statistics, with several rows for each entry in names
testOutput <-matrix(nrow = 200, ncol = 5)
for (i in 1:nrow(names)){
testOutput[i,1] <- names[i,1]
testOutput[i,2] <- names[i,2]
# test statistics code here
}
If I look at names[,1], I get the following:
names[,1]
[1] EQ_Level_UK EQ_Level_EUR EQ_Level_US EQ_Level_Far East
[5] IR_PC 1_UKIR_PC 2_UKIR_PC 3_UKSwap_PC 1_UK
[9] Swap_PC 2_UK Swap_PC 3_UK FX_Level_EUR FX_Level_US
[13] FX_Level_Far East Infl_PC 1_UK Infl_PC 2_UK Infl_PC 3_UK
[17] Prop_Level_UK CreditAAA_PC 1_UK CreditAAA_PC 2_UK CreditAAA_PC 3_UK
[21] CreditAA_PC 1_UK CreditAA_PC 2_UK CreditAA_PC 3_UK CreditA_PC 1_UK
[25] CreditA_PC 2_UK CreditA_PC 3_UK CreditBBB_PC 1_UK CreditBBB_PC 2_UK
[29] CreditBBB_PC 3_UK
29 Levels: CreditA_PC 1_UK CreditA_PC 2_UK CreditA_PC 3_UK ... Swap_PC 3_UK
But if I look at testOutput[,1], I get:
testOutput[,1]
[1] "15" "13" "16" "14" "23" "24" "25" "27" "28" "29" "17" "19" "18" "20"
"21"
[16] "22" "26" "7" "8" "9" "4" "5" "6" "1" "2" "3" "10" "11" "12"
"17"
[31] NA NA "19" "18" NA NA NA "20" NA NA "21" NA NA "22"
NA
[46] NA "26" NA NA "7" NA NA "8" NA NA "9" NA NA "4"
NA
[61] NA "5" NA NA "6" NA NA "1" NA NA "2" NA NA "3"
NA
[76] NA "10" NA NA "11" NA NA "12" NA NA NA NA NA NA
NA
[91] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
NA
[106] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
NA
[121] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
NA
[136] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
NA
[151] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
NA
[166] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
NA
[181] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
NA
[196] NA NA NA NA NA
That is, the names are now converted to numbers. I think this might have
something to do with the way I've defined the testOutput matrix, but haven't
been able to find any information about how to fix it. Can anyone help?
Many thanks.
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Re: [R] storing output data from a loop that has varying row numbers
Hi Ivan,
Thanks for your help, your initial suggestion did not work, but that is no
doubt down to my lack of making sense!
Here is a short example of my dataset. Basically the loop is set up to match
the ID with the TO column based on DIST = 0. So A1 = 2, A1.1 =1, A2 = 4,
A2.1 = 3. That is fine for HR 9, but for HR 10 the numbers no longer match
those IDs so I need to loop the data and store each loop - if that makes
sense.
FROM TO DIST ID HR DD MM YY ANIMAL DAY
1 1 1 2.63981'A1' 9 30 9 7 1 1
2 1 2 0.0'A1' 9 30 9 7 1 1
3 1 3 6.95836'A1' 9 30 9 7 1 1
4 1 4 8.63809'A1' 9 30 9 7 1 1
5 1 1 0.0 'A1.1' 9 30 9 7 7 1
6 1 2 2.63981 'A1.1' 9 30 9 7 7 1
7 1 3 8.03071 'A1.1' 9 30 9 7 7 1
8 1 4 8.90896 'A1.1' 9 30 9 7 7 1
9 1 1 8.90896'A2' 9 30 9 7 1 1
101 2 8.63809'A2' 9 30 9 7 1 1
111 3 2.85602'A2' 9 30 9 7 1 1
121 4 0.0'A2' 9 30 9 7 1 1
131 1 8.03071 'A2.1' 9 30 9 7 7 1
141 2 6.95836 'A2.1' 9 30 9 7 7 1
151 3 0.0 'A2.1' 9 30 9 7 7 1
161 4 2.85602 A2.1' 9 30 9 7 7 1
171 1 3.53695'A1' 10 30 9 7 1 1
181 2 4.32457'A1' 10 30 9 7 1 1
191 3 0.0'A1' 10 30 9 7 1 1
201 4 8.85851'A1' 10 30 9 7 1 1
211 5 12.09194'A1' 10 30 9 7 1 1
221 1 7.44743 'A1.1' 10 30 9 7 7 1
231 2 0.0 'A1.1' 10 30 9 7 7 1
241 3 4.32457 'A1.1' 10 30 9 7 7 1
251 4 13.16728 'A1.1' 10 30 9 7 7 1
261 5 16.34761 'A1.1' 10 30 9 7 7 1
271 1 6.13176'A2' 10 30 9 7 1 1
281 2 13.16728'A2' 10 30 9 7 1 1
291 3 8.85851'A2' 10 30 9 7 1 1
301 4 0.0'A2' 10 30 9 7 1 1
311 5 3.40726'A2' 10 30 9 7 1 1
321 1 9.03345 'A2.1' 10 30 9 7 7 1
331 2 16.34761 'A2.1' 10 30 9 7 7 1
341 3 12.09194 'A2.1' 10 30 9 7 7 1
351 4 3.40726 'A2.1' 10 30 9 7 7 1
361 5 0.0 'A2.1' 10 30 9 7 7 1
371 1 0.0 'MALE1' 10 30 9 7 12 1
381 2 7.44743 'MALE1' 10 30 9 7 12 1
391 3 3.53695 'MALE1' 10 30 9 7 12 1
401 4 6.13176 'MALE1' 10 30 9 7 12 1
411 5 9.03345 'MALE1' 10 30 9 7 12 1
So the loop is:
DIST_LOOP<-matrix(NA,NA,ncol=11)
for (i in 1:33){
SEL_DAY<-seal_dist[seal_dist[,10]==i,]
SEL_DAY[i]=dist[i]
print(paste("DAY", i, "of 33"))
for (s in 1:11){
SEL_HR<-SEL_DAY[SEL_DAY[,5]==s,]
print(paste("HR", s, "of 11"))
indx <- subset(SEL_HR, SEL_HR$DIST == 0)
SEL_HR$TO_ID <- indx$ID[match(SEL_HR$TO, indx$TO)]
DIST_LOOP[i,]<-SEL_HR
}
}
But storing the data in the DIST_LOOP matrix doesn't work, I am just told in
another post that a list might be better than a matrix?
I hope this makes more sense!?
Many thanks,
Ross
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[R] discriminant analysis
Sir, Can you suggest some function for discriminant analysis for Binary response having continuous as well as categorical regressors. Thanks, Suman Dhara [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] loop
Can any one help it will be very kind, loop statements I have this table and some more records, I want to reshape it V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 TP53 Dis1 Dis2 Dis3 Dis4 Dis5 Dis6 DCI New1 New2 New3 New4 FDI Hi2 H3 H4 GHD I1 I3 I4 I5 I6 I7 I8 I want my new table or matrix to be some thing like this V1 V2 V3 Tp53 Dis1 Dis2 Tp53 Dis1 Dis3 Tp53 Dis1 Dis4 Tp53 Dis1 Dis5 Tp53 Dis1 Dis6 Tp53 Dis2 Dis3 Tp53 Dis2 Dis4 Tp53 Dis2 Dis5 Tp53 Dis2 Dis6 Tp53 Dis3 Dis4 Tp53 Dis3 Dis5 Tp53 Dis3 Dis6 Tp53 Dis4 Dis5 Tp53 Dis4 Dis6 Tp53 Dis5 Dis6 DCI New1 New2 DCI New1 New3 DCI New1 New4 DCI New2 New3 DCI New2 New4 DCI New3 New4 And so on for the rest of the table. -- View this message in context: http://r.789695.n4.nabble.com/loop-tp2238209p2238209.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Issue with assigning text to matrix
On Tue, Jun 1, 2010 at 5:04 AM, Jessica Queree
wrote:
> My issue relates to adding text to a matrix and finding that the text is
> converted to a number.
>
A matrix can only hold one type of data. Since you started with character,
that's what you get.
A dataframe can hold different types of data, both character and numeric.
If you convert your matrix to a dataframe with dataf.frame() it can hold
your output in the manner you expect.
testOutput <- data.frame(matrix(nrow = 200, ncol = 5))
Sarah
>
> This is the section of code I'm having trouble with:
>
>
>
> # First, I load in a list of names from a .csv file to 'names'
>
> names <- read.csv(file("Names.csv"))
>
>
>
> # Then I define a matrix which will be populated with various test
> statistics, with several rows for each entry in names
>
>
>
> testOutput <-matrix(nrow = 200, ncol = 5)
>
> for (i in 1:nrow(names)){
>
>
>
> testOutput[i,1] <- names[i,1]
>
> testOutput[i,2] <- names[i,2]
>
>
>
> # test statistics code here
>
>
>
> }
>
>
> That is, the names are now converted to numbers. I think this might have
> something to do with the way I've defined the testOutput matrix, but haven't
> been able to find any information about how to fix it. Can anyone help?
>
>
>
> Many thanks.
>
--
Sarah Goslee
http://www.functionaldiversity.org
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[R] save image to folder
Hello, is there an equivalent to write.csv for saving images to a folder? E.g. if I have a histogram or piechart I would need to write a command that saves the resulting image as png or jpg into a certain folder (working directory). Thanks -- View this message in context: http://r.789695.n4.nabble.com/save-image-to-folder-tp2238249p2238249.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] strsplit
Hello! I have the following problem: I have a file in R that has in the first row three informations in one row that I would like to in three different rows. The first row looks like this: GenusA_SpeciesC_Tree GenusA_SpeciesF_Tree GenusB_SpeciesA_Shrub ... I tried with strsplit and and substring but I don't get any solution. I know I can do this in Excel, but in R would be much nicer! Thanks for helping me. Joël Baumann __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] Yet Another Package for Time Data
Hi fellow R developers/users, I've recently revised a package called rtv, and now consider it reasonably stable. Description: A package for conveniently representing, manipulating and visualising time data. Here, time is regarded as a random variable, and objects are used to represent realisations of that random variable. This is particularly useful for change points, irregular timeseries and failure events. There's a strong emphasis on continuous representations of time, with user-specified origins and units. http://cran.r-project.org/web/packages/rtv/index.html The package contains classes notably similar to the POSIXt classes. However, the classes are based on a somewhat different philosophy. The major advantage of the package, is that crtv objects (similar to POSIXct objects) allow user-specified origins and units. It also allows time events to be represented as fractional months or fractional years. More info in the package vignette. Bug reports and feature requests welcome. kind regards -- Charlotte Maia http://sites.google.com/site/maiagx ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] storing output data from a loop that has varying row numbers
There's something very unlogic in your code. You have the whole time the
same datafra
On Tue, Jun 1, 2010 at 1:51 PM, RCulloch wrote:
>
> Hi All,
>
> I am trying to run a loop that will have varying numbers of rows with each
> output.
>
> Previously I have had the same number of rows so I would use (and I
> appreciate that this will no doubt achieve some gasps as being thoroughly
> inefficient!):
>
> xdfrow<-(0)
> xdfrow1<-(1:32)
> xdfrow2<-(33:64)
> xdfrow3<-(65:96)
> xdfrow4<-(97:128)
> xdfrow5<-(129:160)
> xdfrow6<-(161:192)
> xdfrow7<-(193:224)
>
> and so on
>
> xdf <- matrix(999, nrow=1024, ncol=7)
> xdf <- as.data.frame(xdf)
> NAM <- c("NAME","ID2","DAY","BEH", "B_FALSE", "B_TRUE","TOTAL")
> colnames(xdf)<-NAM
>
> I then use this matrix and then run the loop and assign the data to each of
> the xdfrows just doing +1 on each loop. (If that makes sense? Not really
> important, just trying to show that I do try and solve some of my own
> problems, albeit perhaps not in the best manner!)
>
> _
>
> However, the data I'm working with now has a very varied number of rows
> (0:2500) over a large data set and I can't work out how is best to do this.
>
> So my loop would be:
>
> for (i in 1:33){
>SEL_DAY<-seal_dist[seal_dist[,10]==i,]
>print(paste("DAY", i, "of 33"))
>for (s in 1:11){
>SEL_HR<-SEL_DAY[SEL_DAY[,5]==s,]
>print(paste("HR", s, "of 11"))
>indx <- subset(SEL_HR, SEL_HR$DIST == 0)
>SEL_HR$TO_ID <- indx$ID[match(SEL_HR$TO, indx$TO)]}
> }
>
> where i is day and s is the hr within the day, the loop works fine because
> it prints as i expect it too. I have not given any info on the data because
> I assume this is more of a method question and will be very straight
> forward
> to most people on here!? But I am happy to post data if it is needed.
>
> I assume I need to set up a matrix before the loop,
>
> e.g. DIST_LOOP<-matrix(NA,1000,ncol=11)
>
> and then I should be able to put something before the first } that allows
> me
> to add to the matrix, but everything I have tried doesn't work
>
> e.g. DIST_LOOP[[i]]<-SEL_HR
>
> Any help would be much appreciated,
>
> Best wishes,
>
> Ross
>
>
>
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/storing-output-data-from-a-loop-that-has-varying-row-numbers-tp2238396p2238396.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
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Re: [R] save image to folder
You might consider starting your search with ?png Sarah On Tue, Jun 1, 2010 at 5:34 AM, thoeb wrote: > > Hello, > > is there an equivalent to write.csv for saving images to a folder? E.g. if I > have a histogram or piechart I would need to write a command that saves the > resulting image as png or jpg into a certain folder (working directory). > -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] strsplit
How about this:
testdata <- data.frame(sp = c("GenusA_SpeciesC_Tree",
"GenusA_SpeciesF_Tree", "GenusB_SpeciesA_Shrub"),
stringsAsFactors=FALSE)
# for one
unlist(strsplit(testdata[1,1], split="_"))
# for all of them
do.call(rbind, sapply(testdata[,1], strsplit, split="_"))
Sarah
On Tue, Jun 1, 2010 at 5:45 AM, Joël Baumann wrote:
> Hello!
>
> I have the following problem:
>
> I have a file in R that has in the first row three informations in one row
> that I would like to in three different rows.
>
> The first row looks like this:
>
> GenusA_SpeciesC_Tree
> GenusA_SpeciesF_Tree
> GenusB_SpeciesA_Shrub
> ...
>
> I tried with strsplit and and substring but I don't get any solution. I know
> I can do this in Excel, but in R would be much nicer!
>
--
Sarah Goslee
http://www.functionaldiversity.org
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Re: [R] strsplit
Hi,
I don't know if it would help you since your goal is not really clear to
me, but here are some thoughts:
> x <- c("GenusA_SpeciesC_Tree", "GenusA_SpeciesF_Tree",
"GenusB_SpeciesA_Shrub")
> test <- strsplit(x, "_")
> test
[[1]]
[1] "GenusA" "SpeciesC" "Tree"
[[2]]
[1] "GenusA" "SpeciesF" "Tree"
[[3]]
[1] "GenusB" "SpeciesA" "Shrub"
> lapply(test, "[", 1)
[[1]]
[1] "GenusA"
[[2]]
[1] "GenusA"
[[3]]
[1] "GenusB"
> df <- data.frame(Genus=unlist(lapply(test,"[",1)),
Species=unlist(lapply(test,"[",2)), Veget=unlist(lapply(test,"[",3)))
> df
Genus Species Veget
1 GenusA SpeciesC Tree
2 GenusA SpeciesF Tree
3 GenusB SpeciesA Shrub
There are probably better and nicer ways to do each step.
HTH,
Ivan
Le 6/1/2010 11:45, Joël Baumann a écrit :
Hello!
I have the following problem:
I have a file in R that has in the first row three informations in one
row that I would like to in three different rows.
The first row looks like this:
GenusA_SpeciesC_Tree
GenusA_SpeciesF_Tree
GenusB_SpeciesA_Shrub
...
I tried with strsplit and and substring but I don't get any solution.
I know I can do this in Excel, but in R would be much nicer!
Thanks for helping me.
Joël Baumann
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Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
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Re: [R] strsplit
This might not be the most elegant way of doing it, but it should probably
work, given that data is always separated by a "_".
string.1<-strsplit(c("GenusA_SpeciesC_Tree","GenusA_SpeciesF_Tree",
"GenusB_SpeciesA_Shrub"),"_")
matrix(unlist(string.1),ncol=3,byrow=TRUE)
2010/6/1 Joël Baumann
> Hello!
>
> I have the following problem:
>
> I have a file in R that has in the first row three informations in one row
> that I would like to in three different rows.
>
> The first row looks like this:
>
> GenusA_SpeciesC_Tree
> GenusA_SpeciesF_Tree
> GenusB_SpeciesA_Shrub
> ...
>
> I tried with strsplit and and substring but I don't get any solution. I
> know I can do this in Excel, but in R would be much nicer!
>
> Thanks for helping me.
>
> Joël Baumann
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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[R] using R and for simulating a wireless network
Hi, I want to simulate a simple wireless network in R. There is a source and destination and source sends some data packets to the destination and some routing table and so on. Does anyone have any idea and knowledge about this or are there any packages for this. The thing that I am going to do is wiring a simple simulator using R that a node sends some packet and using some table to reach the destination (Like Ns-2 but with less details) Best Regards, Amir __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] any doc to understand arima state space model?
I am trying to understand R arima function. Any pointers would be appreciated. Thank you, Shakira. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] storing output data from a loop that has varying row numbers
could you just give us the output of dput() for the data you copied in the
mail?
eg dput(seal_dist[,1:100])
and an example of how you want your output. I guess I get what you want to
do, but it's not what your code is doing. And it will be difficult to put
that in a matrix, as you have different labels and different numbers of
TO-levels for different days and HR values.
Cheers
On Tue, Jun 1, 2010 at 3:32 PM, RCulloch wrote:
>
> Hi Ivan,
>
> Thanks for your help, your initial suggestion did not work, but that is no
> doubt down to my lack of making sense!
>
> Here is a short example of my dataset. Basically the loop is set up to
> match
> the ID with the TO column based on DIST = 0. So A1 = 2, A1.1 =1, A2 = 4,
> A2.1 = 3. That is fine for HR 9, but for HR 10 the numbers no longer match
> those IDs so I need to loop the data and store each loop - if that makes
> sense.
>
>
> FROM TO DIST ID HR DD MM YY ANIMAL DAY
> 1 1 1 2.63981'A1' 9 30 9 7 1 1
> 2 1 2 0.0'A1' 9 30 9 7 1 1
> 3 1 3 6.95836'A1' 9 30 9 7 1 1
> 4 1 4 8.63809'A1' 9 30 9 7 1 1
> 5 1 1 0.0 'A1.1' 9 30 9 7 7 1
> 6 1 2 2.63981 'A1.1' 9 30 9 7 7 1
> 7 1 3 8.03071 'A1.1' 9 30 9 7 7 1
> 8 1 4 8.90896 'A1.1' 9 30 9 7 7 1
> 9 1 1 8.90896'A2' 9 30 9 7 1 1
> 101 2 8.63809'A2' 9 30 9 7 1 1
> 111 3 2.85602'A2' 9 30 9 7 1 1
> 121 4 0.0'A2' 9 30 9 7 1 1
> 131 1 8.03071 'A2.1' 9 30 9 7 7 1
> 141 2 6.95836 'A2.1' 9 30 9 7 7 1
> 151 3 0.0 'A2.1' 9 30 9 7 7 1
> 161 4 2.85602 A2.1' 9 30 9 7 7 1
> 171 1 3.53695'A1' 10 30 9 7 1 1
> 181 2 4.32457'A1' 10 30 9 7 1 1
> 191 3 0.0'A1' 10 30 9 7 1 1
> 201 4 8.85851'A1' 10 30 9 7 1 1
> 211 5 12.09194'A1' 10 30 9 7 1 1
> 221 1 7.44743 'A1.1' 10 30 9 7 7 1
> 231 2 0.0 'A1.1' 10 30 9 7 7 1
> 241 3 4.32457 'A1.1' 10 30 9 7 7 1
> 251 4 13.16728 'A1.1' 10 30 9 7 7 1
> 261 5 16.34761 'A1.1' 10 30 9 7 7 1
> 271 1 6.13176'A2' 10 30 9 7 1 1
> 281 2 13.16728'A2' 10 30 9 7 1 1
> 291 3 8.85851'A2' 10 30 9 7 1 1
> 301 4 0.0'A2' 10 30 9 7 1 1
> 311 5 3.40726'A2' 10 30 9 7 1 1
> 321 1 9.03345 'A2.1' 10 30 9 7 7 1
> 331 2 16.34761 'A2.1' 10 30 9 7 7 1
> 341 3 12.09194 'A2.1' 10 30 9 7 7 1
> 351 4 3.40726 'A2.1' 10 30 9 7 7 1
> 361 5 0.0 'A2.1' 10 30 9 7 7 1
> 371 1 0.0 'MALE1' 10 30 9 7 12 1
> 381 2 7.44743 'MALE1' 10 30 9 7 12 1
> 391 3 3.53695 'MALE1' 10 30 9 7 12 1
> 401 4 6.13176 'MALE1' 10 30 9 7 12 1
> 411 5 9.03345 'MALE1' 10 30 9 7 12 1
>
>
> So the loop is:
>
> DIST_LOOP<-matrix(NA,NA,ncol=11)
>
> for (i in 1:33){
>SEL_DAY<-seal_dist[seal_dist[,10]==i,]
> SEL_DAY[i]=dist[i]
> print(paste("DAY", i, "of 33"))
>for (s in 1:11){
>SEL_HR<-SEL_DAY[SEL_DAY[,5]==s,]
>print(paste("HR", s, "of 11"))
>indx <- subset(SEL_HR, SEL_HR$DIST == 0)
>SEL_HR$TO_ID <- indx$ID[match(SEL_HR$TO, indx$TO)]
> DIST_LOOP[i,]<-SEL_HR
>}
>}
>
> But storing the data in the DIST_LOOP matrix doesn't work, I am just told
> in
> another post that a list might be better than a matrix?
>
> I hope this makes more sense!?
>
> Many thanks,
>
> Ross
> --
> View this message in context:
> http://r.789695.n4.nabble.com/storing-output-data-from-a-loop-that-has-varying-row-numbers-tp2238396p2238483.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
---
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Re: [R] strsplit
Another option is to use the colsplit() function (in the reshape package):
testdata <- data.frame(sp = c("GenusA_SpeciesC_Tree",
"GenusA_SpeciesF_Tree", "GenusB_SpeciesA_Shrub"),
stringsAsFactors=FALSE)
library(reshape)
testdata <- cbind(testdata, colsplit(testdata$sp, split="_", names=c("Genus",
"Species", "Type")))
-Ista
On Tuesday 01 June 2010 10:10:22 am Ivan Calandra wrote:
> Hi,
>
> I don't know if it would help you since your goal is not really clear to
>
> me, but here are some thoughts:
> > x <- c("GenusA_SpeciesC_Tree", "GenusA_SpeciesF_Tree",
>
> "GenusB_SpeciesA_Shrub")
>
> > test <- strsplit(x, "_")
> > test
>
> [[1]]
> [1] "GenusA" "SpeciesC" "Tree"
> [[2]]
> [1] "GenusA" "SpeciesF" "Tree"
> [[3]]
> [1] "GenusB" "SpeciesA" "Shrub"
>
> > lapply(test, "[", 1)
>
> [[1]]
> [1] "GenusA"
> [[2]]
> [1] "GenusA"
> [[3]]
> [1] "GenusB"
>
> > df <- data.frame(Genus=unlist(lapply(test,"[",1)),
>
> Species=unlist(lapply(test,"[",2)), Veget=unlist(lapply(test,"[",3)))
>
> > df
>
> Genus Species Veget
> 1 GenusA SpeciesC Tree
> 2 GenusA SpeciesF Tree
> 3 GenusB SpeciesA Shrub
>
> There are probably better and nicer ways to do each step.
>
> HTH,
> Ivan
>
> Le 6/1/2010 11:45, Joël Baumann a écrit :
> > Hello!
> >
> > I have the following problem:
> >
> > I have a file in R that has in the first row three informations in one
> > row that I would like to in three different rows.
> >
> > The first row looks like this:
> >
> > GenusA_SpeciesC_Tree
> > GenusA_SpeciesF_Tree
> > GenusB_SpeciesA_Shrub
> > ...
> >
> > I tried with strsplit and and substring but I don't get any solution.
> > I know I can do this in Excel, but in R would be much nicer!
> >
> > Thanks for helping me.
> >
> > Joël Baumann
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
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Re: [R] storing output data from a loop that has varying row numbers
Hi Ross,
I'm still really confused about your question.
Let me ask you some specific questions (maybe someone more experienced
would understand at once, but I'm no expert; I hope I can still help
you! In any case, I would like to understand for myself ;) )
- is "seal_dist" the name of your data.frame?
- what do you want to do with
SEL_DAY[i]=dist[i]
? What is "dist"? If I understand well, you want to replace the values
in FROM (then TO, then DIST...) with the values from the same column
number in dist?
- Since I still haven't understood your goal completely, I still don't
understand why you add the column TO_ID to SEL_HR.
- In any case, a matrix cannot work because you want to store data of
different classes in DIST_LOOP (ID is character and the others are
numeric). You can either use a data.frame (if you really want to have
the table-like structure, which is a list) or a list.
- Moreover, the output from dput(your data) would really help to see
what you have!
HTH,
Ivan
Le 6/1/2010 15:32, RCulloch a écrit :
Hi Ivan,
Thanks for your help, your initial suggestion did not work, but that is no
doubt down to my lack of making sense!
Here is a short example of my dataset. Basically the loop is set up to match
the ID with the TO column based on DIST = 0. So A1 = 2, A1.1 =1, A2 = 4,
A2.1 = 3. That is fine for HR 9, but for HR 10 the numbers no longer match
those IDs so I need to loop the data and store each loop - if that makes
sense.
FROM TO DIST ID HR DD MM YY ANIMAL DAY
1 1 1 2.63981'A1' 9 30 9 7 1 1
2 1 2 0.0'A1' 9 30 9 7 1 1
3 1 3 6.95836'A1' 9 30 9 7 1 1
4 1 4 8.63809'A1' 9 30 9 7 1 1
5 1 1 0.0 'A1.1' 9 30 9 7 7 1
6 1 2 2.63981 'A1.1' 9 30 9 7 7 1
7 1 3 8.03071 'A1.1' 9 30 9 7 7 1
8 1 4 8.90896 'A1.1' 9 30 9 7 7 1
9 1 1 8.90896'A2' 9 30 9 7 1 1
101 2 8.63809'A2' 9 30 9 7 1 1
111 3 2.85602'A2' 9 30 9 7 1 1
121 4 0.0'A2' 9 30 9 7 1 1
131 1 8.03071 'A2.1' 9 30 9 7 7 1
141 2 6.95836 'A2.1' 9 30 9 7 7 1
151 3 0.0 'A2.1' 9 30 9 7 7 1
161 4 2.85602 A2.1' 9 30 9 7 7 1
171 1 3.53695'A1' 10 30 9 7 1 1
181 2 4.32457'A1' 10 30 9 7 1 1
191 3 0.0'A1' 10 30 9 7 1 1
201 4 8.85851'A1' 10 30 9 7 1 1
211 5 12.09194'A1' 10 30 9 7 1 1
221 1 7.44743 'A1.1' 10 30 9 7 7 1
231 2 0.0 'A1.1' 10 30 9 7 7 1
241 3 4.32457 'A1.1' 10 30 9 7 7 1
251 4 13.16728 'A1.1' 10 30 9 7 7 1
261 5 16.34761 'A1.1' 10 30 9 7 7 1
271 1 6.13176'A2' 10 30 9 7 1 1
281 2 13.16728'A2' 10 30 9 7 1 1
291 3 8.85851'A2' 10 30 9 7 1 1
301 4 0.0'A2' 10 30 9 7 1 1
311 5 3.40726'A2' 10 30 9 7 1 1
321 1 9.03345 'A2.1' 10 30 9 7 7 1
331 2 16.34761 'A2.1' 10 30 9 7 7 1
341 3 12.09194 'A2.1' 10 30 9 7 7 1
351 4 3.40726 'A2.1' 10 30 9 7 7 1
361 5 0.0 'A2.1' 10 30 9 7 7 1
371 1 0.0 'MALE1' 10 30 9 7 12 1
381 2 7.44743 'MALE1' 10 30 9 7 12 1
391 3 3.53695 'MALE1' 10 30 9 7 12 1
401 4 6.13176 'MALE1' 10 30 9 7 12 1
411 5 9.03345 'MALE1' 10 30 9 7 12 1
So the loop is:
DIST_LOOP<-matrix(NA,NA,ncol=11)
for (i in 1:33){
SEL_DAY<-seal_dist[seal_dist[,10]==i,]
SEL_DAY[i]=dist[i]
print(paste("DAY", i, "of 33"))
for (s in 1:11){
SEL_HR<-SEL_DAY[SEL_DAY[,5]==s,]
print(paste("HR", s, "of 11"))
indx<- subset(SEL_HR, SEL_HR$DIST == 0)
SEL_HR$TO_ID<- indx$ID[match(SEL_HR$TO, indx$TO)]
DIST_LOOP[i,]<-SEL_HR
}
}
But storing the data in the DIST_LOOP matrix doesn't work, I am just told in
another post that a list might be better than a matrix?
I hope this makes more sense!?
Many thanks,
Ross
--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de
**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php
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[R] Help on aggregate method
Dear R experts,
I would really appreciate if you had an idea on how to use more
efficiently the aggregate method:
More specifically, I would like to calculate the mean of certain
values on a data frame, grouped by various attributes, and then
create a new column in the data frame that will have the corresponding
mean for every row. I attach part of my code:
matchMean <- function(ind,dataTable,aggrTable)
{
index <- which((aggrTable[,1]==dataTable[["Attr1"]][ind]) &
(aggrTable[,2]==dataTable[["Attr2"]][ind]))
as.numeric(aggrTable[index,3])
}
avgDur <- aggregate(ap.dat[["Dur"]], by = list(ap.dat[["Attr1"]],
ap.dat[["Attr2"]]), FUN="mean")
meanDur <- sapply((1:length(ap.dat[,1])), FUN=matchMean, ap.dat, avgDur)
ap.dat <- cbind (ap.dat, meanDur)
As I deal with very large dataset, it takes long time to run my
matching function, so if you had an idea on how to automate more this
matching process I would be really grateful.
Thank you very much in advance!
Kind regards,
Stella
--
Stella Pachidi
Master in Business Informatics student
Utrecht University
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Re: [R] storing output data from a loop that has varying row numbers
Is this what you're looking for?
seal_list <- split(seal_dist,sel)
out <- lapply(seal_list,function(x){
indx <- subset(x, x$DIST == 0)
x$TO_ID <- indx$ID[match(x$TO, indx$TO)]
return(x)
})
output <- unsplit(out,sel)
Cheers
Joris
On Tue, Jun 1, 2010 at 3:32 PM, RCulloch wrote:
>
> Hi Ivan,
>
> Thanks for your help, your initial suggestion did not work, but that is no
> doubt down to my lack of making sense!
>
> Here is a short example of my dataset. Basically the loop is set up to
> match
> the ID with the TO column based on DIST = 0. So A1 = 2, A1.1 =1, A2 = 4,
> A2.1 = 3. That is fine for HR 9, but for HR 10 the numbers no longer match
> those IDs so I need to loop the data and store each loop - if that makes
> sense.
>
>
> FROM TO DIST ID HR DD MM YY ANIMAL DAY
> 1 1 1 2.63981'A1' 9 30 9 7 1 1
> 2 1 2 0.0'A1' 9 30 9 7 1 1
> 3 1 3 6.95836'A1' 9 30 9 7 1 1
> 4 1 4 8.63809'A1' 9 30 9 7 1 1
> 5 1 1 0.0 'A1.1' 9 30 9 7 7 1
> 6 1 2 2.63981 'A1.1' 9 30 9 7 7 1
> 7 1 3 8.03071 'A1.1' 9 30 9 7 7 1
> 8 1 4 8.90896 'A1.1' 9 30 9 7 7 1
> 9 1 1 8.90896'A2' 9 30 9 7 1 1
> 101 2 8.63809'A2' 9 30 9 7 1 1
> 111 3 2.85602'A2' 9 30 9 7 1 1
> 121 4 0.0'A2' 9 30 9 7 1 1
> 131 1 8.03071 'A2.1' 9 30 9 7 7 1
> 141 2 6.95836 'A2.1' 9 30 9 7 7 1
> 151 3 0.0 'A2.1' 9 30 9 7 7 1
> 161 4 2.85602 A2.1' 9 30 9 7 7 1
> 171 1 3.53695'A1' 10 30 9 7 1 1
> 181 2 4.32457'A1' 10 30 9 7 1 1
> 191 3 0.0'A1' 10 30 9 7 1 1
> 201 4 8.85851'A1' 10 30 9 7 1 1
> 211 5 12.09194'A1' 10 30 9 7 1 1
> 221 1 7.44743 'A1.1' 10 30 9 7 7 1
> 231 2 0.0 'A1.1' 10 30 9 7 7 1
> 241 3 4.32457 'A1.1' 10 30 9 7 7 1
> 251 4 13.16728 'A1.1' 10 30 9 7 7 1
> 261 5 16.34761 'A1.1' 10 30 9 7 7 1
> 271 1 6.13176'A2' 10 30 9 7 1 1
> 281 2 13.16728'A2' 10 30 9 7 1 1
> 291 3 8.85851'A2' 10 30 9 7 1 1
> 301 4 0.0'A2' 10 30 9 7 1 1
> 311 5 3.40726'A2' 10 30 9 7 1 1
> 321 1 9.03345 'A2.1' 10 30 9 7 7 1
> 331 2 16.34761 'A2.1' 10 30 9 7 7 1
> 341 3 12.09194 'A2.1' 10 30 9 7 7 1
> 351 4 3.40726 'A2.1' 10 30 9 7 7 1
> 361 5 0.0 'A2.1' 10 30 9 7 7 1
> 371 1 0.0 'MALE1' 10 30 9 7 12 1
> 381 2 7.44743 'MALE1' 10 30 9 7 12 1
> 391 3 3.53695 'MALE1' 10 30 9 7 12 1
> 401 4 6.13176 'MALE1' 10 30 9 7 12 1
> 411 5 9.03345 'MALE1' 10 30 9 7 12 1
>
>
> So the loop is:
>
> DIST_LOOP<-matrix(NA,NA,ncol=11)
>
> for (i in 1:33){
>SEL_DAY<-seal_dist[seal_dist[,10]==i,]
> SEL_DAY[i]=dist[i]
> print(paste("DAY", i, "of 33"))
>for (s in 1:11){
>SEL_HR<-SEL_DAY[SEL_DAY[,5]==s,]
>print(paste("HR", s, "of 11"))
>indx <- subset(SEL_HR, SEL_HR$DIST == 0)
>SEL_HR$TO_ID <- indx$ID[match(SEL_HR$TO, indx$TO)]
> DIST_LOOP[i,]<-SEL_HR
>}
>}
>
> But storing the data in the DIST_LOOP matrix doesn't work, I am just told
> in
> another post that a list might be better than a matrix?
>
> I hope this makes more sense!?
>
> Many thanks,
>
> Ross
> --
> View this message in context:
> http://r.789695.n4.nabble.com/storing-output-data-from-a-loop-that-has-varying-row-numbers-tp2238396p2238483.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
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[[alternative HTML version deleted]]
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Re: [R] storing output data from a loop that has varying row numbers
Sorry, forgot to add the sel. This is the first line, then just run the
rest.
sel <- as.factor(paste(seal_dist[,10],"-",seal_dist[,5],sep=""))
cheers
Joris
On Tue, Jun 1, 2010 at 4:56 PM, Joris Meys wrote:
> Is this what you're looking for?
>
> seal_list <- split(seal_dist,sel)
>
> out <- lapply(seal_list,function(x){
> indx <- subset(x, x$DIST == 0)
> x$TO_ID <- indx$ID[match(x$TO, indx$TO)]
> return(x)
> })
>
> output <- unsplit(out,sel)
>
> Cheers
> Joris
>
> On Tue, Jun 1, 2010 at 3:32 PM, RCulloch wrote:
>
>>
>> Hi Ivan,
>>
>> Thanks for your help, your initial suggestion did not work, but that is no
>> doubt down to my lack of making sense!
>>
>> Here is a short example of my dataset. Basically the loop is set up to
>> match
>> the ID with the TO column based on DIST = 0. So A1 = 2, A1.1 =1, A2 = 4,
>> A2.1 = 3. That is fine for HR 9, but for HR 10 the numbers no longer match
>> those IDs so I need to loop the data and store each loop - if that makes
>> sense.
>>
>>
>> FROM TO DIST ID HR DD MM YY ANIMAL DAY
>> 1 1 1 2.63981'A1' 9 30 9 7 1 1
>> 2 1 2 0.0'A1' 9 30 9 7 1 1
>> 3 1 3 6.95836'A1' 9 30 9 7 1 1
>> 4 1 4 8.63809'A1' 9 30 9 7 1 1
>> 5 1 1 0.0 'A1.1' 9 30 9 7 7 1
>> 6 1 2 2.63981 'A1.1' 9 30 9 7 7 1
>> 7 1 3 8.03071 'A1.1' 9 30 9 7 7 1
>> 8 1 4 8.90896 'A1.1' 9 30 9 7 7 1
>> 9 1 1 8.90896'A2' 9 30 9 7 1 1
>> 101 2 8.63809'A2' 9 30 9 7 1 1
>> 111 3 2.85602'A2' 9 30 9 7 1 1
>> 121 4 0.0'A2' 9 30 9 7 1 1
>> 131 1 8.03071 'A2.1' 9 30 9 7 7 1
>> 141 2 6.95836 'A2.1' 9 30 9 7 7 1
>> 151 3 0.0 'A2.1' 9 30 9 7 7 1
>> 161 4 2.85602 A2.1' 9 30 9 7 7 1
>> 171 1 3.53695'A1' 10 30 9 7 1 1
>> 181 2 4.32457'A1' 10 30 9 7 1 1
>> 191 3 0.0'A1' 10 30 9 7 1 1
>> 201 4 8.85851'A1' 10 30 9 7 1 1
>> 211 5 12.09194'A1' 10 30 9 7 1 1
>> 221 1 7.44743 'A1.1' 10 30 9 7 7 1
>> 231 2 0.0 'A1.1' 10 30 9 7 7 1
>> 241 3 4.32457 'A1.1' 10 30 9 7 7 1
>> 251 4 13.16728 'A1.1' 10 30 9 7 7 1
>> 261 5 16.34761 'A1.1' 10 30 9 7 7 1
>> 271 1 6.13176'A2' 10 30 9 7 1 1
>> 281 2 13.16728'A2' 10 30 9 7 1 1
>> 291 3 8.85851'A2' 10 30 9 7 1 1
>> 301 4 0.0'A2' 10 30 9 7 1 1
>> 311 5 3.40726'A2' 10 30 9 7 1 1
>> 321 1 9.03345 'A2.1' 10 30 9 7 7 1
>> 331 2 16.34761 'A2.1' 10 30 9 7 7 1
>> 341 3 12.09194 'A2.1' 10 30 9 7 7 1
>> 351 4 3.40726 'A2.1' 10 30 9 7 7 1
>> 361 5 0.0 'A2.1' 10 30 9 7 7 1
>> 371 1 0.0 'MALE1' 10 30 9 7 12 1
>> 381 2 7.44743 'MALE1' 10 30 9 7 12 1
>> 391 3 3.53695 'MALE1' 10 30 9 7 12 1
>> 401 4 6.13176 'MALE1' 10 30 9 7 12 1
>> 411 5 9.03345 'MALE1' 10 30 9 7 12 1
>>
>>
>> So the loop is:
>>
>> DIST_LOOP<-matrix(NA,NA,ncol=11)
>>
>> for (i in 1:33){
>>SEL_DAY<-seal_dist[seal_dist[,10]==i,]
>> SEL_DAY[i]=dist[i]
>> print(paste("DAY", i, "of 33"))
>>for (s in 1:11){
>>SEL_HR<-SEL_DAY[SEL_DAY[,5]==s,]
>>print(paste("HR", s, "of 11"))
>>indx <- subset(SEL_HR, SEL_HR$DIST == 0)
>>SEL_HR$TO_ID <- indx$ID[match(SEL_HR$TO, indx$TO)]
>> DIST_LOOP[i,]<-SEL_HR
>>}
>>}
>>
>> But storing the data in the DIST_LOOP matrix doesn't work, I am just told
>> in
>> another post that a list might be better than a matrix?
>>
>> I hope this makes more sense!?
>>
>> Many thanks,
>>
>> Ross
>> --
>> View this message in context:
>> http://r.789695.n4.nabble.com/storing-output-data-from-a-loop-that-has-varying-row-numbers-tp2238396p2238483.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Joris Meys
> Statistical Consultant
>
> Ghent University
> Faculty of Bioscience Engineering
> Department of Applied mathematics, biometrics and process control
>
> Coupure Links 653
> B-9000 Gent
>
> tel : +32 9 264 59 87
> joris.m...@ugent.be
> ---
> Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
>
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathemati
Re: [R] Help on aggregate method
It's easiest for us to help if you give us a reproducible example. We
don't have your datasets (ap.dat), so we can't run your code below.
It's easy to create sample data with the random number generators in R,
or use ?dput to give us a sample of your actual data.frame.
I would guess your problem is solved by ?ave though.
Stella Pachidi wrote:
Dear R experts,
I would really appreciate if you had an idea on how to use more
efficiently the aggregate method:
More specifically, I would like to calculate the mean of certain
values on a data frame, grouped by various attributes, and then
create a new column in the data frame that will have the corresponding
mean for every row. I attach part of my code:
matchMean <- function(ind,dataTable,aggrTable)
{
index <- which((aggrTable[,1]==dataTable[["Attr1"]][ind]) &
(aggrTable[,2]==dataTable[["Attr2"]][ind]))
as.numeric(aggrTable[index,3])
}
avgDur <- aggregate(ap.dat[["Dur"]], by = list(ap.dat[["Attr1"]],
ap.dat[["Attr2"]]), FUN="mean")
meanDur <- sapply((1:length(ap.dat[,1])), FUN=matchMean, ap.dat, avgDur)
ap.dat <- cbind (ap.dat, meanDur)
As I deal with very large dataset, it takes long time to run my
matching function, so if you had an idea on how to automate more this
matching process I would be really grateful.
Thank you very much in advance!
Kind regards,
Stella
--
Stella Pachidi
Master in Business Informatics student
Utrecht University
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Re: [R] data frame manipulation with zero rows
It is indeed ddply() from package plyr.
> -Original Message-
> From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
> Sent: Tuesday, June 01, 2010 12:24 PM
> To: Peter Ehlers
> Cc: arnaud Gaboury; r-help@r-project.org
> Subject: Re: [R] data frame manipulation with zero rows
>
> On Tue, 1 Jun 2010, Peter Ehlers wrote:
>
> > On 2010-06-01 1:53, arnaud Gaboury wrote:
> >> Brian,
> >>
> >> If I do understand correctly, I must use in my function something
> else than
> >> ddply() if I want to avoid any error each time my df has zero rows?
> >> Am I correct?
> >>
> >
> > You could define a function to handle the zero-rows case:
> >
> > f <- function(x){
> > if(nrow(x) < 1) out <- x[, c(1,3,2)] # or whatever
> > else
> > out <- ddply(x, c("DESCRIPTION","SETTLEMENT"), summarise,
> >POSITION=sum(QUANTITY))[,c(1,3,2)]
> > out
> > }
> > f(futures)
>
> Or simply fix ddply. We don't know what that is or what it should do
> for the case of zero rows: it may or may not be the one in package
> plyr.
>
> >
> > -Peter Ehlers
> >
> >>
> >>
> >>> -Original Message-
> >>> From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
> >>> Sent: Tuesday, June 01, 2010 9:47 AM
> >>> To: arnaud Gaboury
> >>> Subject: Re: [R] data frame manipulation with zero rows
> >>>
> >>> On Tue, 1 Jun 2010, arnaud Gaboury wrote:
> >>>
> Dear group,
>
> Here is the kind of data.frame I obtain every day with my function
> :
>
> futures<-
> structure(list(DESCRIPTION = c("CORN Jul/10", "CORN Jul/10",
> "CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "LIVE CATTLE Aug/10",
> "LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10",
> "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10"
> ), CREATED.DATE = structure(c(18403, 18406, 18406, 18406, 18406,
> 18407, 18408, 18406, 18407, 18407, 18407, 18407), class = "Date"),
> QUANTITY = c(1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1), SETTLEMENT =
> c("373.2500",
> "373.2500", "373.2500", "373.2500", "373.2500", "90.7750",
> "90.7750", "14.9200", "14.9200", "14.9200", "14.9200",
> "14.9200"
> )), .Names = c("DESCRIPTION", "CREATED.DATE", "QUANTITY",
> "SETTLEMENT"), row.names = c(NA, 12L), class = "data.frame")
>
> I need then to apply to the df this following code line :
>
> > PosFut=ddply(futures, c("DESCRIPTION","SETTLEMENT"), summarise,
> >>> POSITION=
> sum(QUANTITY))[,c(1,3,2)]
>
> It works perfectly in most of case, BUT I have a new problem: it
> can
> sometime occurs that my df "futures" is empty, with zero rows.
>
>
> futures<-
> structure(list(DESCRIPTION = character(0), CREATED.DATE =
> structure(numeric(0), class = "Date"),
> QUANTITY = numeric(0), SETTLEMENT = character(0)), .Names =
> c("DESCRIPTION",
> "CREATED.DATE", "QUANTITY", "SETTLEMENT"), row.names = integer(0),
> >>> class =
> "data.frame")
>
> It is not the usual case, but it can happen. With this df, when I
> >>> pass the
> above mentione line, I get an error :
>
> > PosFut=ddply(futures, c("DESCRIPTION","SETTLEMENT"), summarise,
> >>> POSITION=
> sum(QUANTITY))[,c(1,3,2)]
> Error in tapply(1:nrow(data), splitv, list) :
> arguments must have same length
>
>
> How can I avoid this when my df is empty?
> >>>
> >>> Ask the author of the (missing) function ddply() to correct the
> error
> >>> of using 1:nrow(data) by replacing it by seq_len(nrow(data)).
> >>>
> >>> It's helpful to give example code, but much more helpful if you
> test
> >>> it: yours cannot work without the function ddply() -- this is what
> >>> 'self-contained' means in the footer here.
>
> --
> Brian D. Ripley, rip...@stats.ox.ac.uk
> Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
> University of Oxford, Tel: +44 1865 272861 (self)
> 1 South Parks Road, +44 1865 272866 (PA)
> Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] Question about the license of an R package
Dear R-users, I'm developing a package that heavily depends on another package released under the GPL-2 license. In addition, some few functions depends on other packages released under the following licences (as described in the corresponding pages of http://cran.r-project.org/web/packages/): GPL GPL >=2. I want to release my package under a GPL >= 2 license, and after reading the section 1.1.1 of the R-exts manual and looking at the compatibility matrix found on http://gplv3.fsf.org/dd3-faq, I'm still in doubt if I can do that. I would appreciate if you could tell me if can I release my package under a GPL >= 2 license considering the aforementioned licenses ? Finally, a general question: When a package is released under GPL >=2, does it mean that the terms of the GPL-3 license apply or not ? Thanks in advance for any help. Mauricio __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] BreastCancer Dataset for Classification in kknn
Dear All,
I'm getting a error while trying to apply the BreastCancer dataset
(package=mlbench) to kknn (package=kknn) that I don't understand as I'm new
to R.
The codes are as follow:
rm = (list = ls())
library(mlbench)
data(BreastCancer)
library(kknn)
BCancer = na.omit(BreastCancer)
d = dim(BCancer)[1]
i1 = seq(1, d, 2)
i2 = seq(2, d, 2)
t1 = BCancer[i1, ]
t2 = BCancer[i2, ]
y2 = BCancer[i2, 11]
x = 10
k = array(1:x, dim = c(x,1))
ker = array(c( "rectangular", "triangular", "epanechnikov", "biweight",
"triweight", "cos", "inv", "gaussian"), dim = c(8,1))
f = function(x, ker){
BreastCancer.kknn <- kknn(Class~., train = t1, test = t2, k = x,
kernel = ker, distance = 1)
fit = fitted(BreastCancer.kknn)
z <- (fit==y2)
z.e <- (100 - (length(y2)-length(z[!z]))/length(y2)*100 )
}
err.k = function(ker){
error.BreastCancer = apply(k,1,function(y) f(y, ker))
}
err.ker = apply(ker, 1, err.k)
colnames(err.ker) = c("rectangular", "triangular", "epanechnikov",
"biweight",
"triweight", "cos", "inv", "gaussian")
print(err.ker)
It throws a error: Error in as.matrix(learn[, ind == i]) :
(subscript) logical subscript too long
In addition: Warning messages:
1: In model.matrix.default(mt, mf) : variable 'Id' converted to a factor
2: In model.matrix.default(mt, test) : variable 'Id' converted to a factor
I tried the codes with other datasets in mlbench package and most of them
working. That is the mistake here for this particular dataset and how can I
solve it?
Thanks
Nitin
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Re: [R] Issue with assigning text to matrix
Jessica,
Two further comments:
1. When you read your data, your character vectors are
stored as factors; is that what you want?
2. You may be better off using a list.
-Peter Ehlers
On 2010-06-01 7:47, Sarah Goslee wrote:
On Tue, Jun 1, 2010 at 5:04 AM, Jessica Queree
wrote:
My issue relates to adding text to a matrix and finding that the text is
converted to a number.
A matrix can only hold one type of data. Since you started with character,
that's what you get.
A dataframe can hold different types of data, both character and numeric.
If you convert your matrix to a dataframe with dataf.frame() it can hold
your output in the manner you expect.
testOutput<- data.frame(matrix(nrow = 200, ncol = 5))
Sarah
This is the section of code I'm having trouble with:
# First, I load in a list of names from a .csv file to 'names'
names<- read.csv(file("Names.csv"))
# Then I define a matrix which will be populated with various test
statistics, with several rows for each entry in names
testOutput<-matrix(nrow = 200, ncol = 5)
for (i in 1:nrow(names)){
testOutput[i,1]<- names[i,1]
testOutput[i,2]<- names[i,2]
# test statistics code here
}
That is, the names are now converted to numbers. I think this might have
something to do with the way I've defined the testOutput matrix, but haven't
been able to find any information about how to fix it. Can anyone help?
Many thanks.
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Re: [R] storing output data from a loop that has varying row numbers
Hi Joris,
Thanks for your help!
The data as requested:
structure(list(FROM = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L),
TO = c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L,
2L, 3L, 4L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L,
3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L), DIST = c(2.63981,
0, 6.95836, 8.63809, 0, 2.63981, 8.03071, 8.90896, 8.90896,
8.63809, 2.85602, 0, 8.03071, 6.95836, 0, 2.85602, 3.53695,
4.32457, 0, 8.85851, 12.09194, 7.44743, 0, 4.32457, 13.16728,
16.34761, 6.13176, 13.16728, 8.85851, 0, 3.40726, 9.03345,
16.34761, 12.09194, 3.40726, 0, 0, 7.44743, 3.53695, 6.13176,
9.03345), ID = structure(c(12L, 12L, 12L, 12L, 11L, 11L,
11L, 11L, 14L, 14L, 14L, 14L, 13L, 13L, 13L, 143L, 12L, 12L,
12L, 12L, 12L, 11L, 11L, 11L, 11L, 11L, 14L, 14L, 14L, 14L,
14L, 13L, 13L, 13L, 13L, 13L, 94L, 94L, 94L, 94L, 94L), .Label =
c("'11.1'",
"'15.1'", "'15.5'", "'18.1'", "'24.2'", "'26.1'", "'26.2'",
"'28.3'", "'4.2'", "'7.1'", "'A1.1'", "'A1'", "'A2.1'", "'A2'",
"'B1'", "'C1'", "'D1.1'", "'D1'", "'D2.1'", "'D2'", "'D3.1'",
"'D3'", "'D4.1'", "'D4'", "'D5.1'", "'D5'", "'D6.1'", "'D6'",
"'E1.1'", "'E1'", "'E2.1'", "'E2'", "'E4'", "'E5'", "'F1.1'",
"'F1'", "'F10.1'", "'F10'", "'F11'", "'F2'", "'F3'", "'F4.1'",
"'F4'", "'F5.1'", "'F5'", "'F7'", "'F8.1'", "'F8'", "'G2.1'",
"'G2'", "'G3.1'", "'G3'", "'G4.1'", "'G4'", "'G5.1'", "'G5'",
"'H1.1'", "'H1'", "'H2'", "'H3.1'", "'H3'", "'H8'", "'I1.1'",
"'I1'", "'I2'", "'I4.1'", "'I4'", "'J1.1'", "'J1'", "'J2.1'",
"'J2'", "'J3'", "'J6'", "'J7'", "'JUV'", "'K1.1'", "'K1'",
"'K2'", "'K3'", "'K4.1'", "'K4'", "'L1.1'", "'L1'", "'L2.1'",
"'L2'", "'L4'", "'M1'", "'M2.1'", "'M2'", "'M3.1'", "'M3'",
"'M4.1'", "'M4'", "'MALE1'", "'N1.1'", "'N1'", "'N2'", "'N3'",
"'N4.1'", "'N4'", "'O1'", "'O2'", "'O3.1'", "'O3'", "'O4.1'",
"'O4'", "'O5'", "'P1.1'", "'P1'", "'Q1'", "'Q2'", "'Q3'",
"'R1.1'", "'R1'", "'R2'", "'R3.1'", "'R3'", "'R4.1'", "'R4'",
"'R5.1'", "'R5'", "'S1.1'", "'S1'", "'S2.1'", "'S2'", "'S3.1'",
"'S3'", "'S4.1'", "'S4'", "'T1'", "'U1.1'", "'U1'", "'U2'",
"'U3'", "'UKFEM'", "'UKMAL'", "'UKPUP'", "'V1.1'", "'V1'",
"'W1.1'", "'W1'", "'WR'", "A2.1'"), class = "factor"), HR = c(9L,
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L,
10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L,
10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L,
10L), DD = c(30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L,
30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L,
30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L,
30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L), MM = c(9L, 9L, 9L,
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L,
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L,
9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L), YY = c(7L, 7L, 7L, 7L, 7L,
7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L,
7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L,
7L, 7L, 7L, 7L, 7L, 7L), ANIMAL = c(1L, 1L, 1L, 1L, 7L, 7L,
7L, 7L, 1L, 1L, 1L, 1L, 7L, 7L, 7L, 7L, 1L, 1L, 1L, 1L, 1L,
7L, 7L, 7L, 7L, 7L, 1L, 1L, 1L, 1L, 1L, 7L, 7L, 7L, 7L, 7L,
12L, 12L, 12L, 12L, 12L), DAY = c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L)), .Names = c("FROM", "TO", "DIST", "ID",
"HR", "DD", "MM", "YY", "ANIMAL", "DAY"), row.names = c(NA, 41L
), class = "data.frame")
The output should be as the original file is, but it should have an
additional column for 'TO_ID'
I hope that makes sense?
Cheers,
Ross
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Re: [R] storing output data from a loop that has varying row numbers
Hi Ivan, Thanks again for your help! I'll just go through your questions... I'm still really confused about your question. -Sorry!!! Let me ask you some specific questions (maybe someone more experienced would understand at once, but I'm no expert; I hope I can still help you! In any case, I would like to understand for myself ;) ) - is "seal_dist" the name of your data.frame? yes so.. head(seal_dist) FROM TODIST ID HR DD MM YY ANIMAL DAY 11 1 2.63981 'A1' 9 30 9 7 1 1 21 2 0.0 'A1' 9 30 9 7 1 1 31 3 6.95836 'A1' 9 30 9 7 1 1 41 4 8.63809 'A1' 9 30 9 7 1 1 51 1 0.0 'A1.1' 9 30 9 7 7 1 61 2 2.63981 'A1.1' 9 30 9 7 7 1 - what do you want to do with SEL_DAY[i]=dist[i] That was a (desperate) attempt to do 'something', but didn't work - so shouldn't have been in the script I posted, sorry! ? What is "dist"? It is a measure of distance from one point (ID) to another i.e. the distance between A1 and A1.1 If I understand well, you want to replace the values in FROM (then TO, then DIST...) with the values from the same column number in dist? The problem is that Arc doesn't output the data as I'd like, so I want to create a new column to add to the data. What Arc has done is taken a distance between each ID for each hour, but because the number of IDs in each hour don't match it means that the TO number is not unique to the ID throughout the entire dataset, only on that given hour. So when distance = 0 in the TO column then that TO number -s equal to the ID i.e. the distance to A1 to A1 is 0, so I then want to use that information to create a new column that will tell me the actual ID. If that is any clearer? - Since I still haven't understood your goal completely, I still don't understand why you add the column TO_ID to SEL_HR. see above - In any case, a matrix cannot work because you want to store data of different classes in DIST_LOOP (ID is character and the others are numeric). You can either use a data.frame (if you really want to have the table-like structure, which is a list) or a list. I see, can you advise on how to set up a list to write to? - Moreover, the output from dput(your data) would really help to see what you have! I have not long posted it, I hope it helps!! Thanks again for your help Ivan, much appreciated, Ross -- View this message in context: http://r.789695.n4.nabble.com/storing-output-data-from-a-loop-that-has-varying-row-numbers-tp2238396p2238630.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to make R automatic?
Hello, I have a question about how R can run automatically. Here is the story: A file called "data.csv" will be generated every couple of minutes in a folder (overwrite itself). What I want R to do is: 1) scan the folder to find the file. 2) if the file is a newly generated file, process the file 3) output some file identifier on the screen, such as the time when the file was generated. 4) pause for 20 seconds (doesnt need to be precise) and repeat 1) Idealy, I would like R to do this everyday from 9:00am to 6:00pm. What I got so far is: I can use list.files() and file.info to get 1) and 2). For 4), I can write some trivial loop to kill some time, but I would like to be more accurate tham that, considering different PC running the loop. I have no idea about 3). Any suggestion/better solution? Thank you very much!! Best regards, Ted -- View this message in context: http://r.789695.n4.nabble.com/How-to-make-R-automatic-tp2238541p2238541.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] using the design matrix to correctly configure contrasts
Esteemed R-forum subscribers,
I'm having a tough time configuring contrasts for my 3-way ANOVA. In short:
I don't know how to configure (all) my contrasts correctly in order to
specify (all) my comparisons of interest.
I succeeded getting my contrasts of interest set up for a simpler 2-way
ANOVA based on the fairly intuitive logic of the design col.names.
But i'm not able to extend this logic to a more complex three way ANOVA.
Obviously there *is* a logic to the "0's" and "1's" of R's design
'treatment contrast' matrix- i see that the "1's" correspond to each of
the factors that an observation is associated with. But exactly how this
information can be used to ascribe contrasts of interest remains beyond
my understanding, especially where interaction is concerned. Googling
for days didn't yield the help i was looking for.
Can some one point me to a detailed explanation (suitable for R
dilettantes) how to specify any contrast of interest given a design matrix?
For anyone really needing some diversion, i'd *really* appreciate a
personal explanation using my design as an example. Warning- its not
small...
#load limma package to analyse affy gene expression data
library(limma)
targets <- read.delim("targets.txt")
#factors & levels
Time <- factor(targets$Time, levels = c("t1", "t2", "t3", "t4",
"t5", "t6", "t7", "t8",
"t9", "t10", "t11", "t12",
"t13", "t14", "t15", "t16"))
Tissue <- factor(targets$Tissue, levels = c("R", "C"))
Pperiod <- factor(targets$Pperiod, levels = c("E", "L", "S"))
#model excluding the 3-way interaction (written for convenience to
# build on previous 2-way ANOVA
design <- model.matrix(~Time + Tissue + Pperiod + Time:Tissue +
Time:Pperiod + Tissue:Pperiod)
fit <- lmFit(rma.pp, design)
#how does this design look:
> colnames(design)
[1] "(Intercept)" "Timet2"
[3] "Timet3" "Timet4"
[5] "Timet5" "Timet6"
[7] "Timet7" "Timet8"
[9] "Timet9" "Timet10"
[11] "Timet11" "Timet12"
[13] "Timet13" "Timet14"
[15] "Timet15" "Timet16"
[17] "TissueC" "PperiodL"
[19] "PperiodS" "Timet2:TissueC"
[21] "Timet3:TissueC" "Timet4:TissueC"
[23] "Timet5:TissueC" "Timet6:TissueC"
[25] "Timet7:TissueC" "Timet8:TissueC"
[27] "Timet9:TissueC" "Timet10:TissueC"
[29] "Timet11:TissueC" "Timet12:TissueC"
[31] "Timet13:TissueC" "Timet14:TissueC"
[33] "Timet15:TissueC" "Timet16:TissueC"
[35] "Timet2:PperiodL" "Timet3:PperiodL"
[37] "Timet4:PperiodL" "Timet5:PperiodL"
[39] "Timet6:PperiodL" "Timet7:PperiodL"
[41] "Timet8:PperiodL" "Timet9:PperiodL"
[43] "Timet10:PperiodL" "Timet11:PperiodL"
[45] "Timet12:PperiodL" "Timet13:PperiodL"
[47] "Timet14:PperiodL" "Timet15:PperiodL"
[49] "Timet16:PperiodL" "Timet2:PperiodS"
[51] "Timet3:PperiodS" "Timet4:PperiodS"
[53] "Timet5:PperiodS" "Timet6:PperiodS"
[55] "Timet7:PperiodS" "Timet8:PperiodS"
[57] "Timet9:PperiodS" "Timet10:PperiodS"
[59] "Timet11:PperiodS" "Timet12:PperiodS"
[61] "Timet13:PperiodS" "Timet14:PperiodS"
[63] "Timet15:PperiodS" "Timet16:PperiodS"
[65] "TissueC:PperiodL" "TissueC:PperiodS"
#the contrasts that i believe i *can* work out correctly:
cont.matrix <- cbind(
t1.S.RvsAll.S.R.Times =
c( 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
0, 0),
t1.E.RvsAll.E.R.Times =
c( 0, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0),
t1.L.RvsAll.L.R.Times =
c( 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0)
)
fit2 <- contrasts.fit(fit, cont.matrix)
fit3 <- eBayes(fit2)
#sensical out put follows suggesting i'm on the right track
BUT! Most basically, the contrasts i can NOT work out are:
t1.S.CvsAll.Times & t1.L.CvsAll.Times
ie., similar to t1.S.RvsAll.Times & t1.S.RvsAll.Times above but for the
"C" tissue, not the "S".
--
Karl Brand
Department of Genetics
Erasmus MC
Dr Molewaterplein 50
3015 GE Rotterdam
P +31 (0)10 704 3409 | F +31 (0)10 704 4743 | M +31 (0)642 777 268
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Re: [R] Help on aggregate method
Take a look at
?split (and unsplit)
eg:
Dur <- rnorm(100)
Attr1=rep(c("A","B"),each=50)
Attr2=rep(c("A","B"),times=50)
ap.dat <-data.frame(Attr1,Attr2,Dur)
split.fact <- paste(ap.dat$Attr1,ap.dat$Attr2)
ap.list <-split(ap.dat,split.fact)
ap.mean <-lapply(ap.list,function(x){
x$meanDur=rep(mean(x$Dur),dim(x)[1])
return(x)
})
ap.dat.fast <- unsplit(ap.mean,split.fact)
system.time on 1000 replicates gives :
> system.time(replicate(1000,{
+ split.fact <- paste(ap.dat$Attr1,ap.dat$Attr2)
+ ap.list <-split(ap.dat,split.fact)
+ ap.mean <-lapply(ap.list,functi [TRUNCATED]
user system elapsed
4.880.004.88
> source(.trPaths[5], echo=TRUE, max.deparse.length=150)
> system.time(replicate(1000,{
+ avgDur <- aggregate(ap.dat[["Dur"]], by = list(ap.dat[["Attr1"]],
+ ap.dat[["Attr2"]]), FUN="mean")
+ meanDur <- sapp [TRUNCATED]
user system elapsed
58.000.11 58.13
>
It should be a tenfold faster.
Cheers
Joris
On Tue, Jun 1, 2010 at 4:48 PM, Stella Pachidi wrote:
> Dear R experts,
>
> I would really appreciate if you had an idea on how to use more
> efficiently the aggregate method:
>
> More specifically, I would like to calculate the mean of certain
> values on a data frame, grouped by various attributes, and then
> create a new column in the data frame that will have the corresponding
> mean for every row. I attach part of my code:
>
> matchMean <- function(ind,dataTable,aggrTable)
> {
>index <- which((aggrTable[,1]==dataTable[["Attr1"]][ind]) &
> (aggrTable[,2]==dataTable[["Attr2"]][ind]))
>as.numeric(aggrTable[index,3])
> }
>
> avgDur <- aggregate(ap.dat[["Dur"]], by = list(ap.dat[["Attr1"]],
> ap.dat[["Attr2"]]), FUN="mean")
> meanDur <- sapply((1:length(ap.dat[,1])), FUN=matchMean, ap.dat, avgDur)
> ap.dat <- cbind (ap.dat, meanDur)
>
> As I deal with very large dataset, it takes long time to run my
> matching function, so if you had an idea on how to automate more this
> matching process I would be really grateful.
>
> Thank you very much in advance!
>
> Kind regards,
> Stella
>
>
>
> --
> Stella Pachidi
> Master in Business Informatics student
> Utrecht University
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
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Re: [R] How to make R automatic?
?Sys.sleep On Tue, 1 Jun 2010, zhangted001 wrote: Hello, I have a question about how R can run automatically. Here is the story: A file called "data.csv" will be generated every couple of minutes in a folder (overwrite itself). What I want R to do is: 1) scan the folder to find the file. 2) if the file is a newly generated file, process the file 3) output some file identifier on the screen, such as the time when the file was generated. 4) pause for 20 seconds (doesnt need to be precise) and repeat 1) Idealy, I would like R to do this everyday from 9:00am to 6:00pm. What I got so far is: I can use list.files() and file.info to get 1) and 2). For 4), I can write some trivial loop to kill some time, but I would like to be more accurate tham that, considering different PC running the loop. I have no idea about 3). Any suggestion/better solution? Thank you very much!! Best regards, Ted -- View this message in context: http://r.789695.n4.nabble.com/How-to-make-R-automatic-tp2238541p2238541.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:cbe...@tajo.ucsd.edu UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to make R automatic?
On 2010-06-01 8:07, zhangted001 wrote:
Hello, I have a question about how R can run automatically. Here is the
story:
A file called "data.csv" will be generated every couple of minutes in a
folder (overwrite itself). What I want R to do is:
1) scan the folder to find the file.
2) if the file is a newly generated file, process the file
3) output some file identifier on the screen, such as the time when the file
was generated.
4) pause for 20 seconds (doesnt need to be precise) and repeat 1)
Idealy, I would like R to do this everyday from 9:00am to 6:00pm.
What I got so far is: I can use list.files() and file.info to get 1) and 2).
For 4), I can write some trivial loop to kill some time, but I would like to
be more accurate tham that, considering different PC running the loop. I
have no idea about 3).
For (3), assign the value of file.info:
fi <- file.info("data.csv")
print(fi[, "mtime"])
For (4) see help(Sys.sleep)
-Peter Ehlers
Any suggestion/better solution?
Thank you very much!!
Best regards,
Ted
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Re: [R] storing output data from a loop that has varying row numbers
OK, then I was right. It's exactly what my code does.
Enjoy.
Cheers
On Tue, Jun 1, 2010 at 4:25 PM, RCulloch wrote:
>
> Hi Joris,
>
> Thanks for your help!
>
> The data as requested:
>
> structure(list(FROM = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L),
>TO = c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L,
>2L, 3L, 4L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L,
>3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L), DIST = c(2.63981,
>0, 6.95836, 8.63809, 0, 2.63981, 8.03071, 8.90896, 8.90896,
>8.63809, 2.85602, 0, 8.03071, 6.95836, 0, 2.85602, 3.53695,
>4.32457, 0, 8.85851, 12.09194, 7.44743, 0, 4.32457, 13.16728,
>16.34761, 6.13176, 13.16728, 8.85851, 0, 3.40726, 9.03345,
>16.34761, 12.09194, 3.40726, 0, 0, 7.44743, 3.53695, 6.13176,
>9.03345), ID = structure(c(12L, 12L, 12L, 12L, 11L, 11L,
>11L, 11L, 14L, 14L, 14L, 14L, 13L, 13L, 13L, 143L, 12L, 12L,
>12L, 12L, 12L, 11L, 11L, 11L, 11L, 11L, 14L, 14L, 14L, 14L,
>14L, 13L, 13L, 13L, 13L, 13L, 94L, 94L, 94L, 94L, 94L), .Label =
> c("'11.1'",
>"'15.1'", "'15.5'", "'18.1'", "'24.2'", "'26.1'", "'26.2'",
>"'28.3'", "'4.2'", "'7.1'", "'A1.1'", "'A1'", "'A2.1'", "'A2'",
>"'B1'", "'C1'", "'D1.1'", "'D1'", "'D2.1'", "'D2'", "'D3.1'",
>"'D3'", "'D4.1'", "'D4'", "'D5.1'", "'D5'", "'D6.1'", "'D6'",
>"'E1.1'", "'E1'", "'E2.1'", "'E2'", "'E4'", "'E5'", "'F1.1'",
>"'F1'", "'F10.1'", "'F10'", "'F11'", "'F2'", "'F3'", "'F4.1'",
>"'F4'", "'F5.1'", "'F5'", "'F7'", "'F8.1'", "'F8'", "'G2.1'",
>"'G2'", "'G3.1'", "'G3'", "'G4.1'", "'G4'", "'G5.1'", "'G5'",
>"'H1.1'", "'H1'", "'H2'", "'H3.1'", "'H3'", "'H8'", "'I1.1'",
>"'I1'", "'I2'", "'I4.1'", "'I4'", "'J1.1'", "'J1'", "'J2.1'",
>"'J2'", "'J3'", "'J6'", "'J7'", "'JUV'", "'K1.1'", "'K1'",
>"'K2'", "'K3'", "'K4.1'", "'K4'", "'L1.1'", "'L1'", "'L2.1'",
>"'L2'", "'L4'", "'M1'", "'M2.1'", "'M2'", "'M3.1'", "'M3'",
>"'M4.1'", "'M4'", "'MALE1'", "'N1.1'", "'N1'", "'N2'", "'N3'",
>"'N4.1'", "'N4'", "'O1'", "'O2'", "'O3.1'", "'O3'", "'O4.1'",
>"'O4'", "'O5'", "'P1.1'", "'P1'", "'Q1'", "'Q2'", "'Q3'",
>"'R1.1'", "'R1'", "'R2'", "'R3.1'", "'R3'", "'R4.1'", "'R4'",
>"'R5.1'", "'R5'", "'S1.1'", "'S1'", "'S2.1'", "'S2'", "'S3.1'",
>"'S3'", "'S4.1'", "'S4'", "'T1'", "'U1.1'", "'U1'", "'U2'",
>"'U3'", "'UKFEM'", "'UKMAL'", "'UKPUP'", "'V1.1'", "'V1'",
>"'W1.1'", "'W1'", "'WR'", "A2.1'"), class = "factor"), HR = c(9L,
>9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L,
>10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L,
>10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L,
>10L), DD = c(30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L,
>30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L,
>30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L,
>30L, 30L, 30L, 30L, 30L, 30L, 30L, 30L), MM = c(9L, 9L, 9L,
>9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L,
>9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L,
>9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L), YY = c(7L, 7L, 7L, 7L, 7L,
>7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L,
>7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L,
>7L, 7L, 7L, 7L, 7L, 7L), ANIMAL = c(1L, 1L, 1L, 1L, 7L, 7L,
>7L, 7L, 1L, 1L, 1L, 1L, 7L, 7L, 7L, 7L, 1L, 1L, 1L, 1L, 1L,
>7L, 7L, 7L, 7L, 7L, 1L, 1L, 1L, 1L, 1L, 7L, 7L, 7L, 7L, 7L,
>12L, 12L, 12L, 12L, 12L), DAY = c(1L, 1L, 1L, 1L, 1L, 1L,
>1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
>1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
>1L, 1L, 1L, 1L, 1L)), .Names = c("FROM", "TO", "DIST", "ID",
> "HR", "DD", "MM", "YY", "ANIMAL", "DAY"), row.names = c(NA, 41L
> ), class = "data.frame")
>
>
>
> The output should be as the original file is, but it should have an
> additional column for 'TO_ID'
>
> I hope that makes sense?
>
> Cheers,
>
> Ross
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/storing-output-data-from-a-loop-that-has-varying-row-numbers-tp2238396p2238576.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
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R-he
Re: [R] How to make R automatic?
?difftime ?file.info file.info(filename)$mtime Sys.sleep(20) Nikhil Kaza Asst. Professor, City and Regional Planning University of North Carolina nikhil.l...@gmail.com On Jun 1, 2010, at 10:07 AM, zhangted001 wrote: Hello, I have a question about how R can run automatically. Here is the story: A file called "data.csv" will be generated every couple of minutes in a folder (overwrite itself). What I want R to do is: 1) scan the folder to find the file. 2) if the file is a newly generated file, process the file 3) output some file identifier on the screen, such as the time when the file was generated. 4) pause for 20 seconds (doesnt need to be precise) and repeat 1) Idealy, I would like R to do this everyday from 9:00am to 6:00pm. What I got so far is: I can use list.files() and file.info to get 1) and 2). For 4), I can write some trivial loop to kill some time, but I would like to be more accurate tham that, considering different PC running the loop. I have no idea about 3). Any suggestion/better solution? Thank you very much!! Best regards, Ted -- View this message in context: http://r.789695.n4.nabble.com/How-to-make-R-automatic-tp2238541p2238541.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on aggregate method
Dear Erik and R experts,
Thank you for the fast response!
I include an example with the ChickWeight dataset:
ap.dat <- ChickWeight
matchMeanEx <- function(ind,dataTable,aggrTable)
{
index <- which((aggrTable[,1]==dataTable[["Diet"]][ind]) &
(aggrTable[,2]==dataTable[["Chick"]][ind]))
as.numeric(aggrTable[index,3])
}
avgW <- aggregate(ap.dat[["weight"]], by = list(ap.dat[["Diet"]],
ap.dat[["Chick"]]), FUN="mean")
meanW <- sapply((1:length(ap.dat[,1])), FUN=matchMeanEx, ap.dat, avgW)
ap.dat <- cbind (ap.dat, meanW)
Best regards,
Stella
On Tue, Jun 1, 2010 at 4:58 PM, Erik Iverson wrote:
> It's easiest for us to help if you give us a reproducible example. We
> don't have your datasets (ap.dat), so we can't run your code below. It's
> easy to create sample data with the random number generators in R, or use
> ?dput to give us a sample of your actual data.frame.
>
> I would guess your problem is solved by ?ave though.
>
> Stella Pachidi wrote:
>
>> Dear R experts,
>>
>> I would really appreciate if you had an idea on how to use more
>> efficiently the aggregate method:
>>
>> More specifically, I would like to calculate the mean of certain
>> values on a data frame, grouped by various attributes, and then
>> create a new column in the data frame that will have the corresponding
>> mean for every row. I attach part of my code:
>>
>> matchMean <- function(ind,dataTable,aggrTable)
>> {
>>index <- which((aggrTable[,1]==dataTable[["Attr1"]][ind]) &
>> (aggrTable[,2]==dataTable[["Attr2"]][ind]))
>>as.numeric(aggrTable[index,3])
>> }
>>
>> avgDur <- aggregate(ap.dat[["Dur"]], by = list(ap.dat[["Attr1"]],
>> ap.dat[["Attr2"]]), FUN="mean")
>> meanDur <- sapply((1:length(ap.dat[,1])), FUN=matchMean, ap.dat, avgDur)
>> ap.dat <- cbind (ap.dat, meanDur)
>>
>> As I deal with very large dataset, it takes long time to run my
>> matching function, so if you had an idea on how to automate more this
>> matching process I would be really grateful.
>>
>> Thank you very much in advance!
>>
>> Kind regards,
>> Stella
>>
>>
>>
>> --
>> Stella Pachidi
>> Master in Business Informatics student
>> Utrecht University
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
--
Stella Pachidi
Master in Business Informatics student
Utrecht University
email: s.pach...@students.uu.nl
tel: +31644478898
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] storing output data from a loop that has varying row numbers
Hi Ross, I think Joris answered your question, but to keep with your own code: To set up a list, you can use: DIST_LOOP <- list() Or, if you know the length: DIST_LOOP <- vector(mode="list", length=11) You would then fill up with (in your for loop): DIST_LOOP[[i]] <- SEL_HR ##Actually, what you had at the beginning, but DIST_LOOP is now a list so it should work. Regarding what you're trying to do, wouldn't aggregate() work? Or with unique()? I still haven't completely understood what you looking for, so I might be completely wrong! But as I said, I guess (and hope for you) Joris fixed the problem already Ivan Le 6/1/2010 16:56, RCulloch a écrit : Hi Ivan, Thanks again for your help! I'll just go through your questions... I'm still really confused about your question. -Sorry!!! Let me ask you some specific questions (maybe someone more experienced would understand at once, but I'm no expert; I hope I can still help you! In any case, I would like to understand for myself ;) ) - is "seal_dist" the name of your data.frame? yes so.. head(seal_dist) FROM TODIST ID HR DD MM YY ANIMAL DAY 11 1 2.63981 'A1' 9 30 9 7 1 1 21 2 0.0 'A1' 9 30 9 7 1 1 31 3 6.95836 'A1' 9 30 9 7 1 1 41 4 8.63809 'A1' 9 30 9 7 1 1 51 1 0.0 'A1.1' 9 30 9 7 7 1 61 2 2.63981 'A1.1' 9 30 9 7 7 1 - what do you want to do with SEL_DAY[i]=dist[i] That was a (desperate) attempt to do 'something', but didn't work - so shouldn't have been in the script I posted, sorry! ? What is "dist"? It is a measure of distance from one point (ID) to another i.e. the distance between A1 and A1.1 If I understand well, you want to replace the values in FROM (then TO, then DIST...) with the values from the same column number in dist? The problem is that Arc doesn't output the data as I'd like, so I want to create a new column to add to the data. What Arc has done is taken a distance between each ID for each hour, but because the number of IDs in each hour don't match it means that the TO number is not unique to the ID throughout the entire dataset, only on that given hour. So when distance = 0 in the TO column then that TO number -s equal to the ID i.e. the distance to A1 to A1 is 0, so I then want to use that information to create a new column that will tell me the actual ID. If that is any clearer? - Since I still haven't understood your goal completely, I still don't understand why you add the column TO_ID to SEL_HR. see above - In any case, a matrix cannot work because you want to store data of different classes in DIST_LOOP (ID is character and the others are numeric). You can either use a data.frame (if you really want to have the table-like structure, which is a list) or a list. I see, can you advise on how to set up a list to write to? - Moreover, the output from dput(your data) would really help to see what you have! I have not long posted it, I hope it helps!! Thanks again for your help Ivan, much appreciated, Ross -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question about the license of an R package
Mauricio Zambrano wrote: Dear R-users, I'm developing a package that heavily depends on another package released under the GPL-2 license. Are you including code from that package in yours, or just making use of it? The former requires that you follow all the GPL rules about your own. If you just use their package, then license your package any way you like. In addition, some few functions depends on other packages released under the following licences (as described in the corresponding pages of http://cran.r-project.org/web/packages/): GPL GPL >=2. I want to release my package under a GPL >= 2 license, and after reading the section 1.1.1 of the R-exts manual and looking at the compatibility matrix found on http://gplv3.fsf.org/dd3-faq, I'm still in doubt if I can do that. I would appreciate if you could tell me if can I release my package under a GPL >= 2 license considering the aforementioned licenses ? Finally, a general question: When a package is released under GPL >=2, does it mean that the terms of the GPL-3 license apply or not ? I believe that's an offer to license under whatever version of the GPL (from 2 up) that the person using your code chooses. If GPL-3 offers something that GPL-2 doesn't (e.g. compatibility with other GPL-3 code), they can use that. If GPL-3 places restrictions they don't like (e.g. incompatibility with GPL-2 code), they can use GPL-2. Duncan Murdoch Thanks in advance for any help. Mauricio __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R-help "spam" detection; please help the moderators
Hi all,
I also couldn't help but notice that some of my messages are bounced for
following reason:
The message headers matched a filter rule
I included the header of one of the messages below, but neither of these
messages is sent trough Nabble, nor does any mail address has digits in it.
I also never had that before. Did you change some of the rules somehow?
Cheers
Joris
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References:
Date: Fri, 28 May 2010 14:32:32 +0200
Delivered-To: jorism...@gmail.com
Message-ID:
Subject: Re: [R] How to get values out of a string using regular expressions?
From: Joris Meys
To: Gabor Grothendieck
Cc: R mailing list
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On Tue, Jun 1, 2010 at 3:25 PM, Martin Maechler
wrote:
> Dear readers of R-help
>
> as most of you will *not* be aware, R-help has continued to work the
> way it does, only thanks to a dozen of volunteers,
> see https://stat.ethz.ch/mailman/listinfo/r-help .
>
> The volunteers manually moderate e-mails that "look like spam" (and
> sometimes are and sometimes are not).
> While much more than 90% of the spam is filtered out long before
> a human sees it, with the increasing sophistication of spammers,
> manual intervention has deemed to be necessary and served the
> community very well.
>
> OTOH, in recent weeks, the amount of work for the volunteers has
> increased, mainly because an increasingly number of non-spam postings are
> erronously tagged as "possibly spam".
> We have discussed about this and done some analysis and found
> that most of these message that produce a considerable amount of
> extra work share two properties :
> 1) they are posted via Nabble {which *always* attaches a small
> pro-Nabble spam at the end of the message}
> 2) the e-mail address of the sender is from a freemail
>provider, quite often 'at gmail dot com', and often the part
>*before* the '@' (at-sign) ends with digits.
>
> We hereby ask those among you who use a freemail account to
> please no longer post via nabble.
>
> Thank you for your support of R-help, *the* "community mailing
> list" of the R project since even before that project existed
> "formally", namely since 1997-04-01,
> today 13 years and two months.
>
> Martin Maechler, ETH Zurich
> (and R-help creator and principal manager)
>
> __
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
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[R] as.date
Dear group,
Here is my df (obtained with a read.csv2()):
df <-
structure(list(DESCRIPTION = c("COTTON NO.2 Jul/10", "COTTON NO.2 Jul/10",
"PALLADIUM Jun/10", "PALLADIUM Jun/10", "SUGAR NO.11 Jul/10",
"SUGAR NO.11 Jul/10"), CREATED.DATE = c("13/05/2010", "13/05/2010",
"14/05/2010", "14/05/2010", "10/05/2010", "10/05/2010"), QUANITY = c(1,
1, -1, -1, 1, 1), CLOSING.PRICE = c("81.2000", "81.2000", "503.6000",
"503.6000", "13.8900", "13.8900")), .Names = c("DESCRIPTION",
"CREATED.DATE", "QUANITY", "CLOSING.PRICE"), row.names = c(NA,
6L), class = "data.frame")
> str(df)
'data.frame': 6 obs. of 4 variables:
$ DESCRIPTION : chr "COTTON NO.2 Jul/10" "COTTON NO.2 Jul/10" "PALLADIUM
Jun/10" "PALLADIUM Jun/10" ...
$ CREATED.DATE : chr "13/05/2010" "13/05/2010" "14/05/2010" "14/05/2010"
...
$ QUANITY : num 1 1 -1 -1 1 1
$ CLOSING.PRICE: chr "81.2000" "81.2000" "503.6000" "503.6000" ...
I want to change the class of df$CREATED.DATE from Chr to Date:
>pose$CREATED.DATE=as.Date(pose$CREATED.DATE,"%d/%m/%y")
Here is what I get :
df <-
structure(list(DESCRIPTION = c("COTTON NO.2 Jul/10", "COTTON NO.2 Jul/10",
"PALLADIUM Jun/10", "PALLADIUM Jun/10", "SUGAR NO.11 Jul/10",
"SUGAR NO.11 Jul/10"), CREATED.DATE = structure(c(18395, 18395,
18396, 18396, 18392, 18392), class = "Date"), QUANITY = c(1,
1, -1, -1, 1, 1), CLOSING.PRICE = c("81.2000", "81.2000", "503.6000",
"503.6000", "13.8900", "13.8900")), .Names = c("DESCRIPTION",
"CREATED.DATE", "QUANITY", "CLOSING.PRICE"), row.names = c(NA,
6L), class = "data.frame")
Where does the problem comes from?? Maybe from my sytem date ??
TY for any help
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] using the design matrix to correctly configure contrasts
Please look at the maiz example, the last example in ?MMC in the HH package.
Follow the example all the way to the end. It illustrates a problem and
then the resolution
of the problem.
install.packages("HH") ## if you don't have it yet.
library(HH)
?MMC
Rich
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Re: [R] as.date
Where does the problem comes from?? Maybe from my sytem date ?? Simply from not reading the options carefully enough, from ?strptime, ‘%y’ Year without century (00-99). If you use this on input, which century you get is system-specific. So don't! Most often values 00 to 68 are prefixed by 20 and 69 to 99 by 19 - that is the behaviour specified by the 2001 POSIX standard, but it does also say ‘it is expected that in a future version of IEEE Std 1003.1-2001 the default century inferred from a 2-digit year will change’. ‘%Y’ Year with century. You want %Y, not %y. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] as.date
Hello,
The lowercase 'y' is year without century. This should work for you:
as.Date(x=df$CREATED.DATE, format="%d/%m/%Y")
HTH,
Josh
On Tue, Jun 1, 2010 at 8:57 AM, arnaud Gaboury wrote:
> Dear group,
>
> Here is my df (obtained with a read.csv2()):
>
>
> df <-
> structure(list(DESCRIPTION = c("COTTON NO.2 Jul/10", "COTTON NO.2 Jul/10",
> "PALLADIUM Jun/10", "PALLADIUM Jun/10", "SUGAR NO.11 Jul/10",
> "SUGAR NO.11 Jul/10"), CREATED.DATE = c("13/05/2010", "13/05/2010",
> "14/05/2010", "14/05/2010", "10/05/2010", "10/05/2010"), QUANITY = c(1,
> 1, -1, -1, 1, 1), CLOSING.PRICE = c("81.2000", "81.2000", "503.6000",
> "503.6000", "13.8900", "13.8900")), .Names = c("DESCRIPTION",
> "CREATED.DATE", "QUANITY", "CLOSING.PRICE"), row.names = c(NA,
> 6L), class = "data.frame")
>
>> str(df)
> 'data.frame': 6 obs. of 4 variables:
> $ DESCRIPTION : chr "COTTON NO.2 Jul/10" "COTTON NO.2 Jul/10" "PALLADIUM
> Jun/10" "PALLADIUM Jun/10" ...
> $ CREATED.DATE : chr "13/05/2010" "13/05/2010" "14/05/2010" "14/05/2010"
> ...
> $ QUANITY : num 1 1 -1 -1 1 1
> $ CLOSING.PRICE: chr "81.2000" "81.2000" "503.6000" "503.6000" ...
>
> I want to change the class of df$CREATED.DATE from Chr to Date:
>
>
>>pose$CREATED.DATE=as.Date(pose$CREATED.DATE,"%d/%m/%y")
>
> Here is what I get :
>
> df <-
> structure(list(DESCRIPTION = c("COTTON NO.2 Jul/10", "COTTON NO.2 Jul/10",
> "PALLADIUM Jun/10", "PALLADIUM Jun/10", "SUGAR NO.11 Jul/10",
> "SUGAR NO.11 Jul/10"), CREATED.DATE = structure(c(18395, 18395,
> 18396, 18396, 18392, 18392), class = "Date"), QUANITY = c(1,
> 1, -1, -1, 1, 1), CLOSING.PRICE = c("81.2000", "81.2000", "503.6000",
> "503.6000", "13.8900", "13.8900")), .Names = c("DESCRIPTION",
> "CREATED.DATE", "QUANITY", "CLOSING.PRICE"), row.names = c(NA,
> 6L), class = "data.frame")
>
> Where does the problem comes from?? Maybe from my sytem date ??
>
> TY for any help
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joshua Wiley
Senior in Psychology
University of California, Riverside
http://www.joshuawiley.com/
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Re: [R] as.date
Change this line to :
>pose$CREATED.DATE=as.Date(pose$CREATED.DATE,"%d/%m/%Y") # mind the capital
Y
>pose
DESCRIPTION CREATED.DATE QUANITY CLOSING.PRICE
1 COTTON NO.2 Jul/10 2010-05-13 1 81.2000
2 COTTON NO.2 Jul/10 2010-05-13 1 81.2000
3 PALLADIUM Jun/10 2010-05-14 -1 503.6000
4 PALLADIUM Jun/10 2010-05-14 -1 503.6000
5 SUGAR NO.11 Jul/10 2010-05-10 1 13.8900
6 SUGAR NO.11 Jul/10 2010-05-10 1 13.8900
Cheers
Joris
On Tue, Jun 1, 2010 at 5:57 PM, arnaud Gaboury wrote:
> Dear group,
>
> Here is my df (obtained with a read.csv2()):
>
>
> df <-
> structure(list(DESCRIPTION = c("COTTON NO.2 Jul/10", "COTTON NO.2 Jul/10",
> "PALLADIUM Jun/10", "PALLADIUM Jun/10", "SUGAR NO.11 Jul/10",
> "SUGAR NO.11 Jul/10"), CREATED.DATE = c("13/05/2010", "13/05/2010",
> "14/05/2010", "14/05/2010", "10/05/2010", "10/05/2010"), QUANITY = c(1,
> 1, -1, -1, 1, 1), CLOSING.PRICE = c("81.2000", "81.2000", "503.6000",
> "503.6000", "13.8900", "13.8900")), .Names = c("DESCRIPTION",
> "CREATED.DATE", "QUANITY", "CLOSING.PRICE"), row.names = c(NA,
> 6L), class = "data.frame")
>
> > str(df)
> 'data.frame': 6 obs. of 4 variables:
> $ DESCRIPTION : chr "COTTON NO.2 Jul/10" "COTTON NO.2 Jul/10" "PALLADIUM
> Jun/10" "PALLADIUM Jun/10" ...
> $ CREATED.DATE : chr "13/05/2010" "13/05/2010" "14/05/2010" "14/05/2010"
> ...
> $ QUANITY : num 1 1 -1 -1 1 1
> $ CLOSING.PRICE: chr "81.2000" "81.2000" "503.6000" "503.6000" ...
>
> I want to change the class of df$CREATED.DATE from Chr to Date:
>
>
> >pose$CREATED.DATE=as.Date(pose$CREATED.DATE,"%d/%m/%y")
>
> Here is what I get :
>
> df <-
> structure(list(DESCRIPTION = c("COTTON NO.2 Jul/10", "COTTON NO.2 Jul/10",
> "PALLADIUM Jun/10", "PALLADIUM Jun/10", "SUGAR NO.11 Jul/10",
> "SUGAR NO.11 Jul/10"), CREATED.DATE = structure(c(18395, 18395,
> 18396, 18396, 18392, 18392), class = "Date"), QUANITY = c(1,
> 1, -1, -1, 1, 1), CLOSING.PRICE = c("81.2000", "81.2000", "503.6000",
> "503.6000", "13.8900", "13.8900")), .Names = c("DESCRIPTION",
> "CREATED.DATE", "QUANITY", "CLOSING.PRICE"), row.names = c(NA,
> 6L), class = "data.frame")
>
> Where does the problem comes from?? Maybe from my sytem date ??
>
> TY for any help
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
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Re: [R] Issue with assigning text to matrix
Hi Jessica,
this tells me that your text is saved as a factor.
Try :
names <- read.csv(file="Names.csv",stringsAsFactors=F)
Cheers
Joris
On Tue, Jun 1, 2010 at 11:04 AM, Jessica Queree
wrote:
> My issue relates to adding text to a matrix and finding that the text is
> converted to a number.
>
>
>
> This is the section of code I'm having trouble with:
>
>
>
> # First, I load in a list of names from a .csv file to 'names'
>
> names <- read.csv(file("Names.csv"))
>
>
>
> # Then I define a matrix which will be populated with various test
> statistics, with several rows for each entry in names
>
>
>
> testOutput <-matrix(nrow = 200, ncol = 5)
>
> for (i in 1:nrow(names)){
>
>
>
>testOutput[i,1] <- names[i,1]
>
>testOutput[i,2] <- names[i,2]
>
>
>
># test statistics code here
>
>
>
> }
>
>
>
>
>
> If I look at names[,1], I get the following:
>
>
>
> names[,1]
>
> [1] EQ_Level_UK EQ_Level_EUR EQ_Level_US EQ_Level_Far
> East
>
> [5] IR_PC 1_UKIR_PC 2_UKIR_PC 3_UKSwap_PC 1_UK
>
> [9] Swap_PC 2_UK Swap_PC 3_UK FX_Level_EUR FX_Level_US
>
> [13] FX_Level_Far East Infl_PC 1_UK Infl_PC 2_UK Infl_PC 3_UK
>
> [17] Prop_Level_UK CreditAAA_PC 1_UK CreditAAA_PC 2_UK CreditAAA_PC
> 3_UK
>
> [21] CreditAA_PC 1_UK CreditAA_PC 2_UK CreditAA_PC 3_UK CreditA_PC 1_UK
>
> [25] CreditA_PC 2_UK CreditA_PC 3_UK CreditBBB_PC 1_UK CreditBBB_PC
> 2_UK
>
> [29] CreditBBB_PC 3_UK
>
> 29 Levels: CreditA_PC 1_UK CreditA_PC 2_UK CreditA_PC 3_UK ... Swap_PC 3_UK
>
>
>
> But if I look at testOutput[,1], I get:
>
>
>
> testOutput[,1]
>
> [1] "15" "13" "16" "14" "23" "24" "25" "27" "28" "29" "17" "19" "18" "20"
> "21"
>
> [16] "22" "26" "7" "8" "9" "4" "5" "6" "1" "2" "3" "10" "11" "12"
> "17"
>
> [31] NA NA "19" "18" NA NA NA "20" NA NA "21" NA NA "22"
> NA
>
> [46] NA "26" NA NA "7" NA NA "8" NA NA "9" NA NA "4"
> NA
>
> [61] NA "5" NA NA "6" NA NA "1" NA NA "2" NA NA "3"
> NA
>
> [76] NA "10" NA NA "11" NA NA "12" NA NA NA NA NA NA
> NA
>
> [91] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> NA
>
> [106] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> NA
>
> [121] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> NA
>
> [136] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> NA
>
> [151] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> NA
>
> [166] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> NA
>
> [181] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> NA
>
> [196] NA NA NA NA NA
>
>
>
> That is, the names are now converted to numbers. I think this might have
> something to do with the way I've defined the testOutput matrix, but
> haven't
> been able to find any information about how to fix it. Can anyone help?
>
>
>
> Many thanks.
>
>[[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
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Re: [R] as.date
TY for the tip. The lower case is in fact the culprit. > -Original Message- > From: Erik Iverson [mailto:er...@ccbr.umn.edu] > Sent: Tuesday, June 01, 2010 6:05 PM > To: arnaud Gaboury > Cc: r-help@r-project.org > Subject: Re: [R] as.date > > > > Where does the problem comes from?? Maybe from my sytem date ?? > > Simply from not reading the options carefully enough, from ?strptime, > > '%y' Year without century (00-99). If you use this on input, > which >century you get is system-specific. So don't! Most often >values 00 to 68 are prefixed by 20 and 69 to 99 by 19 - that >is the behaviour specified by the 2001 POSIX standard, but > it >does also say 'it is expected that in a future version of >IEEE Std 1003.1-2001 the default century inferred from a >2-digit year will change'. > > '%Y' Year with century. > > > You want %Y, not %y. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on aggregate method
Stella Pachidi wrote:
Dear Erik and R experts,
Thank you for the fast response!
I include an example with the ChickWeight dataset:
ap.dat <- ChickWeight
matchMeanEx <- function(ind,dataTable,aggrTable)
{
index <- which((aggrTable[,1]==dataTable[["Diet"]][ind]) &
(aggrTable[,2]==dataTable[["Chick"]][ind]))
as.numeric(aggrTable[index,3])
}
avgW <- aggregate(ap.dat[["weight"]], by = list(ap.dat[["Diet"]],
ap.dat[["Chick"]]), FUN="mean")
meanW <- sapply((1:length(ap.dat[,1])), FUN=matchMeanEx, ap.dat, avgW)
ap.dat <- cbind (ap.dat, meanW)
How about simply using ave.
ap.dat$meanW <- ave(ap.dat$weight, list(ap.dat$Diet, ap.dat$Chick))
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] any doc to understand arima state space model?
Type in Google Arima R Read the first hit, the third, the fifth, and any other that says "tutorial" Cheers Joris On Tue, Jun 1, 2010 at 4:14 PM, shakira M wrote: > I am trying to understand R arima function. Any pointers would be > appreciated. > > Thank you, > Shakira. > >[[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Joris Meys Statistical Consultant Ghent University Faculty of Bioscience Engineering Department of Applied mathematics, biometrics and process control Coupure Links 653 B-9000 Gent tel : +32 9 264 59 87 joris.m...@ugent.be --- Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on aggregate method
Dear Erik,
Thank you very much. Indeed ave did the same job amazingly fast! I did not
know the function before.
Many thanks to all R experts who answer to this mailing list, it's amazing
how much help you offer to the newbies :)
Kind regards,
Stella
On Tue, Jun 1, 2010 at 6:11 PM, Erik Iverson wrote:
>
>
> Stella Pachidi wrote:
>
>> Dear Erik and R experts,
>>
>> Thank you for the fast response!
>>
>> I include an example with the ChickWeight dataset:
>>
>> ap.dat <- ChickWeight
>>
>> matchMeanEx <- function(ind,dataTable,aggrTable)
>> {
>> index <- which((aggrTable[,1]==dataTable[["Diet"]][ind]) &
>> (aggrTable[,2]==dataTable[["Chick"]][ind]))
>> as.numeric(aggrTable[index,3])
>> }
>>
>> avgW <- aggregate(ap.dat[["weight"]], by = list(ap.dat[["Diet"]],
>> ap.dat[["Chick"]]), FUN="mean")
>> meanW <- sapply((1:length(ap.dat[,1])), FUN=matchMeanEx, ap.dat, avgW)
>> ap.dat <- cbind (ap.dat, meanW)
>>
>>
>
> How about simply using ave.
>
> ap.dat$meanW <- ave(ap.dat$weight, list(ap.dat$Diet, ap.dat$Chick))
>
--
Stella Pachidi
Master in Business Informatics student
Utrecht University
email: s.pach...@students.uu.nl
tel: +31644478898
[[alternative HTML version deleted]]
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Re: [R] Help barplots
Dear All, Thank you very much for your kind help and support I got it. Jeet On Tue, Jun 1, 2010 at 5:32 PM, Jim Lemon wrote: > On 06/01/2010 09:01 PM, khush wrote: > >> Dear All, >> >> I am newbie to R, and I wanted to plot a barplots with R and in such a way >> that It will also show me position which I can plot on the bar line. >> >> Here is my code that I am using to plot, >> >> chromosome<- c(40.2, 35.6, 36.1, 29.6, 31, 29.6, 31, 29.4, 28.2, 23, 23, >>> >> 28.2) >> >>> barplot (chromosome, col="purple", xlab="Oryza sativa Chromosomes", >>> border >>> >> = NA, space = 5, ylim = c(0,45)) >> >> I wanted to mark the position say on chromosome 1 (40.2) I need to mark >> 10.2 >> and on other also. >> I also want to set the scale of y axis from 0,5,10,15,20,25,30,35,40,45 >> i.e >> gap of 5 instead of 10. >> >> please help me to solve my querygurus. >> >> Hi Jeet, > I think you want the x positions of the bars. Get them like this: > > xpos<-barplot (chromosome, col="purple", > > xlab="Oryza sativa Chromosomes", > border = NA, space = 5, ylim = c(0,45)) > > Then you can place the extra labels using the values in xpos. > For the custom y axis, add the argument yaxt="n" to your plot command and > then add the axis later. I suspect you will have to use something like the > staxlab function in the plotrix package to get all those labels to display. > > Jim > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] discriminant analysis
Which kind of discriminant analysis? If you mean LDA, use lda() in package MASS and read the correpsonding book "Modern Applied Statistics with S" by Venables and Ripley published by Springer. Uwe Ligges On 01.06.2010 09:24, suman dhara wrote: Sir, Can you suggest some function for discriminant analysis for Binary response having continuous as well as categorical regressors. Thanks, Suman Dhara [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Ignoring initial rows in a text file import
On 01.06.2010 02:19, David Winsemius wrote:
On May 31, 2010, at 8:14 PM, jim holtman wrote:
try this:
input <- readLines("yourfile.txt")
# determine start
start <- grep("\tBegin Main\t", input)[1] # first line if many
Puzzled. I thought backslashes in grepping patterns needed to be
doubled? I guess not.
Well, yes, for a backslash, but here a tab is specified, but no
backslash at all.
Additionally, see ?regex:
"The current implementation interprets \a as BEL, \e as ESC, \f as FF,
\n as LF, \r as CR and \t as TAB."
Uwe Ligges
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Re: [R] Plot multiple columns
Hi, I used the term "run", as each iteration of the Gibbs sampler produces 22 variables (coefficients for Beta in a regression model) The example wont work On 6/1/10 5:54 AM, Ben Bolker wrote: > Noah Silverman smartmediacorp.com> writes: > > >> I'm running a long MCMC chain that is generating samples for 22 variables. >> >> I have each run of the chain as a row in a matrix. >> > >> So: Chain[,1] is the column with all the samples for variable one. >> Chain[,2] is the column with all the samples for variable 2, etc. >> > The previous 2 paragraphs seemed contradictory until I realized > that in the first paragraph you are using "run" to mean what I would > usually call a "sample" ... > > >> I'd like to fit all 22 on a single page to print a nice summary. It is >> OK if the graphs are small, I just need to show the overall shape and >> convergence. >> > How about for example > > >> x <- matrix(runif(22000),ncol=22) >> library(coda) >> m <- as.mcmc(x) >> xyplot(m) >> xyplot(m,layout=c(4,6)) >> > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Set resolution of embedded plots in pdf() or CairoPDF()
Both pdf() and CairoPDF() produce vector graphics, hence there is no things such as resolution required here. Uwe Ligges On 27.05.2010 11:09, Will Eagle wrote: Dear all, how can I set the resolution of embedded plots in PDF using pdf() or CairoPDF() to a value of e.g. 600 dpi to meet journal requirements? Thanks in advance, Will __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] library installation problem, invalid regular expression in help indices
I guess something strange is loaded or changed in your personal startup scripts (i.e. through your Renviron or Rprofile settings. Uwe Ligges On 27.05.2010 15:34, Rainer Machne wrote: Hi, I have a strange package installation problem after update to R 2.11.0 on Fedora Core 12. A colleague of mine with the very same Fedora and R versions doesn't have this problem, while logging on to his computer but installing from my settings and into my local library path also shows the problem, so it seems to be related to some of my environment settings. Installation of several packages fails during installation of help indices. The error is: "Error: invalid regular expression '^[[:blank:]]*" The packages for which installation failes, with the respective help files after which the error occurs: robustbase: NOxEmissions tuneR: lilyinput kernlab: kha rrcov: CovSest seewave: wasp Any ideas? btw, i just copied my colleague's installation folders to mine, and it seems to work; for me the problem is solved for now. But it would still be good to know what the problem is. Thanks, Rainer __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] BreastCancer Dataset for Classification in kknn
Hi Nitin,
It can be solved by splitting your data a bit different. You need more
training data than you have evaluation data, eg :
i1 = 1:400
i2=401:d
Then it works on my computer. No clue as to where the error originates from
though.
Cheers
Joris
On Tue, Jun 1, 2010 at 4:27 PM, Nitin wrote:
> Dear All,
>
> I'm getting a error while trying to apply the BreastCancer dataset
> (package=mlbench) to kknn (package=kknn) that I don't understand as I'm new
> to R.
> The codes are as follow:
>
> rm = (list = ls())
> library(mlbench)
> data(BreastCancer)
> library(kknn)
>
> BCancer = na.omit(BreastCancer)
> d = dim(BCancer)[1]
> i1 = seq(1, d, 2)
> i2 = seq(2, d, 2)
>
> t1 = BCancer[i1, ]
> t2 = BCancer[i2, ]
> y2 = BCancer[i2, 11]
>
> x = 10
> k = array(1:x, dim = c(x,1))
> ker = array(c( "rectangular", "triangular", "epanechnikov", "biweight",
>"triweight", "cos", "inv", "gaussian"), dim = c(8,1))
>
> f = function(x, ker){
>
>BreastCancer.kknn <- kknn(Class~., train = t1, test = t2, k = x,
>kernel = ker, distance = 1)
>fit = fitted(BreastCancer.kknn)
>
>z <- (fit==y2)
>z.e <- (100 - (length(y2)-length(z[!z]))/length(y2)*100 )
> }
>
> err.k = function(ker){
>error.BreastCancer = apply(k,1,function(y) f(y, ker))
> }
>
> err.ker = apply(ker, 1, err.k)
> colnames(err.ker) = c("rectangular", "triangular", "epanechnikov",
> "biweight",
>"triweight", "cos", "inv", "gaussian")
> print(err.ker)
>
> It throws a error: Error in as.matrix(learn[, ind == i]) :
> (subscript) logical subscript too long
> In addition: Warning messages:
> 1: In model.matrix.default(mt, mf) : variable 'Id' converted to a factor
> 2: In model.matrix.default(mt, test) : variable 'Id' converted to a factor
>
> I tried the codes with other datasets in mlbench package and most of them
> working. That is the mistake here for this particular dataset and how can I
> solve it?
>
> Thanks
> Nitin
>
>[[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joris Meys
Statistical Consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
Coupure Links 653
B-9000 Gent
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fancy Page layout
Thanks Jim,
That helps.
Ben Bolker had a nice suggestion on how to get the lattice package to
easily plot all 22 variables in one window.
Ultimately, I'd like to generate a PDF that will print on a standard
(8.5 x 11) page.
A few things I'm still stuck are:
1) How to use the lattice command AND leave room at the bottom for a
text block
2) How to tell lattice the size of the window
3) How to integrate all this together - draw a big window, plot
trellis in the top half and then text box in the bottom.
Any thoughts?
-N
On 6/1/10 4:53 AM, Jim Lemon wrote:
> On 06/01/2010 04:16 AM, Noah Silverman wrote:
>> Hi,
>>
>> Working on a report that is going to have a large number of graphs and
>> summaries. We have 80 "groups" with 20 variables each.
>>
>> Ideally, I'd like to produce ONE page for each group. It would have two
>> columns of 10 graphs and then the 5 number summary of the variables at
>> the bottom.
>> So, perhaps the top 2/3 of the page has the graphs and the bottom third
>> has 20 rows of data summary(maybe a table of sorts.)
>> This COULD be done in Latex, but would have to be hand coded for each of
>> the 80 groups which would be painfully slow.
>>
>> I can easily do the graphs with par(mfrow=c(5,2)) band then draw the
>> graphs in a loop.
>>
>> But I am stuck from here:
>>
>> 1) How do I control the size of the plot window. (Ideally, it should
>> print to fill an 8.5 x 11 piece of paper)
>> 2) Is there a way to "easily" insert a 5 number summary (summary
>> command) into the lower half of the page.
>>
>> Does anybody have any ideas??
>>
> Hi Noah,
> One easy way is to leave some space at the bottom, either by using:
>
> par(mfrow=c(6,2))
>
> or the more flexible "layout" function, and then use "text" or a
> fancier function (textbox, boxed.labels, addtable2plot, etc.) to add
> your text after:
>
> par(xpd=NA)
>
> allows you to display the text anywhere you please. If you use a
> bitmap graphics device, make it big:
>
> png("numberoneofeighty.png",850,1100)
>
> so that it won't look lumpy on the printed page.
>
> Jim
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question about the license of an R package
2010/6/1 Duncan Murdoch : > Mauricio Zambrano wrote: >> >> Dear R-users, >> >> I'm developing a package that heavily depends on another package >> released under the GPL-2 license. >> > > Are you including code from that package in yours, or just making use of it? > The former requires that you follow all the GPL rules about your own. If > you just use their package, then license your package any way you like. Thank you very much for the information!. I'm just using the packages, without including a single line of code. >> >> In addition, some few functions depends on other packages released >> under the following licences (as described in the corresponding pages >> of http://cran.r-project.org/web/packages/): >> >> GPL >> GPL >=2. >> >> I want to release my package under a GPL >= 2 license, and after >> reading the section 1.1.1 of the R-exts manual and looking at the >> compatibility matrix found on http://gplv3.fsf.org/dd3-faq, I'm still >> in doubt if I can do that. >> >> I would appreciate if you could tell me if can I release my package >> under a GPL >= 2 license considering the aforementioned licenses ? >> >> >> Finally, a general question: >> >> When a package is released under GPL >=2, does it mean that the terms >> of the GPL-3 license apply or not ? >> > > I believe that's an offer to license under whatever version of the GPL (from > 2 up) that the person using your code chooses. If GPL-3 offers something > that GPL-2 doesn't (e.g. compatibility with other GPL-3 code), they can use > that. If GPL-3 places restrictions they don't like (e.g. incompatibility > with GPL-2 code), they can use GPL-2. > > Duncan Murdoch Thanks Duncan. Knds, Mauricio Zambrano B. -- == Dept. of Civil and Env. Engineering University of Trento, Italy > > Thanks in advance for any help. >> >> Mauricio >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R-help "spam" detection; please help the moderators
Hi Joris,
The "matched a filter rule" is the principal reason for holding
messages for moderation. Please don't become anxious about the
situation -- one of the reasons we have become concerned about
the situation is that people whose messages get held up do tend
to become worried about it. This is unnecessary -- they, and you,
are not really doing anything wrong!
As I understand it, the filter rules are set by the ethz.ch admins,
and even Martin does not seem to know, in detail, what they are.
Also, it is likely that the filters "learn", and they may well
be "learning" from a lot of other emails received at eth.ch which
have nothing to do with R-help but which are true spam -- then the
headers of such messages could be folded into "Bayes spam scores"
which can trigger "matched a filter rule".
As far as R-help is concerned, the situation seems to be that
gmail.com and nabble.com are important triggers (though there
are plenty of others). Lots of email addresses which do not have
the username ending in digits are trapped in this way.
In round figures, proportions I have logged myself amongst the
messages held because they "matched a filter rule" are:
non-gmail, non-nabble: 30%
non-gmail, nabble: 25%
gmail, non-nabble: 32%
gmail, nabble: 13%
gmail : 45%
nabble: 52%
The true nature of this situation is still unclear!
Ted.
On 01-Jun-10 16:00:40, Joris Meys wrote:
> Hi all,
>
> I also couldn't help but notice that some of my messages are bounced
> for following reason:
>
>The message headers matched a filter rule
>
> I included the header of one of the messages below, but neither of
> these messages is sent trough Nabble, nor does any mail address has
> digits in it.
> I also never had that before. Did you change some of the rules somehow?
>
> Cheers
> Joris
>
> ---
>
> MIME-Version: 1.0
> Received: by 10.140.173.9 with HTTP; Fri, 28 May 2010 05:32:32 -0700
> (PDT)
> In-Reply-To:
>
> References:
>
>
> Date: Fri, 28 May 2010 14:32:32 +0200
> Delivered-To: jorism...@gmail.com
> Message-ID:
>
> Subject: Re: [R] How to get values out of a string using regular
> expressions?
> From: Joris Meys
> To: Gabor Grothendieck
> Cc: R mailing list
> Content-Type: multipart/alternative;
> boundary=000e0cd2295481515c0487a6b3be
>
> --000e0cd2295481515c0487a6b3be
> Content-Type: text/plain; charset=ISO-8859-1
>
>
>
> On Tue, Jun 1, 2010 at 3:25 PM, Martin Maechler
> wrote:
>
>> Dear readers of R-help
>>
>> as most of you will *not* be aware, R-help has continued to work the
>> way it does, only thanks to a dozen of volunteers,
>> see https://stat.ethz.ch/mailman/listinfo/r-help .
>>
>> The volunteers manually moderate e-mails that "look like spam" (and
>> sometimes are and sometimes are not).
>> While much more than 90% of the spam is filtered out long before
>> a human sees it, with the increasing sophistication of spammers,
>> manual intervention has deemed to be necessary and served the
>> community very well.
>>
>> OTOH, in recent weeks, the amount of work for the volunteers has
>> increased, mainly because an increasingly number of non-spam postings
>> are
>> erronously tagged as "possibly spam".
>> We have discussed about this and done some analysis and found
>> that most of these message that produce a considerable amount of
>> extra work share two properties :
>> 1) they are posted via Nabble {which *always* attaches a small
>> pro-Nabble spam at the end of the
>> message}
>> 2) the e-mail address of the sender is from a freemail
>>provider, quite often 'at gmail dot com', and often the part
>>*before* the '@' (at-sign) ends with digits.
>>
>> We hereby ask those among you who use a freemail account to
>> please no longer post via nabble.
>>
>> Thank you for your support of R-help, *the* "community mailing
>> list" of the R project since even before that project existed
>> "formally", namely since 1997-04-01,
>> today 13 years and two months.
>>
>> Martin Maechler, ETH Zurich
>> (and R-help creator and principal manager)
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Joris Meys
> Statistical Consultant
>
> Ghent University
> Faculty of Bioscience Engineering
> Department of Applied mathematics, biometrics and process control
>
> Coupure Links 653
> B-9000 Gent
>
> tel : +32 9 264 59 87
> joris.m...@ugent.be
> ---
> Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
[R] Problem using apply
Hello ,
I can not get apply function to do what I want when doing stuff on rows.
Let's say I want to divide the rows of matrix by their mean, the below show
you get the same result weather you use 1 or 2, i.e. same result for columns
than for means..:(
Thanks a lot for the help,
m = matrix( c(1,4,7,4,5,8,7,8,9), nrow = 3 )
divideByMean = function( v ){
return( v/mean(v))
}
rowMean = c( mean( m[1,]),mean( m[2,]),mean( m[3,]) )
rowMean
colMean = c( mean( m[,1]),mean( m[,2]),mean( m[,3]) )
colMean
m
print("ByRow")
apply( m, 1, divideByMean)
print("ByCol")
apply( m, 2, divideByMean)
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem using apply
The mean by col and by row are the same:
colMeans(m) == rowMeans(m)
So:
m / rowMeans(m) # You don't need apply here
m / colMeans(m)
On Tue, Jun 1, 2010 at 2:26 PM, Joachim de Lezardiere <
joachim.lez...@gmail.com> wrote:
> Hello ,
>
>
>
> I can not get apply function to do what I want when doing stuff on rows.
> Let's say I want to divide the rows of matrix by their mean, the below show
> you get the same result weather you use 1 or 2, i.e. same result for
> columns
> than for means..:(
>
>
>
> Thanks a lot for the help,
>
>
>
>
>
>
>
> m = matrix( c(1,4,7,4,5,8,7,8,9), nrow = 3 )
>
>
>
> divideByMean = function( v ){
>
> return( v/mean(v))
>
> }
>
>
>
> rowMean = c( mean( m[1,]),mean( m[2,]),mean( m[3,]) )
>
> rowMean
>
>
>
> colMean = c( mean( m[,1]),mean( m[,2]),mean( m[,3]) )
>
> colMean
>
>
>
> m
>
> print("ByRow")
>
> apply( m, 1, divideByMean)
>
>
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> print("ByCol")
>
> apply( m, 2, divideByMean)
>
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> __
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> PLEASE do read the posting guide
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> and provide commented, minimal, self-contained, reproducible code.
>
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O
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[R] textbox in lattice
Hi, I want to add a box at the bottom of a lattice window (device/page?). Lattice has drawn a nice group of panels with all the plots I need. How do I add my own summary text at the bottom (several lines worth?) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem using apply
Well, your example matrix is symmetric, so row and column operations
naturally return the same values.
You may want to note though that if you apply your function to a matrix
along rows, the results will be stored in the __columns__ of the resulting
matrix. Thus, if you want to simply divide the rows of a matrix by their
mean, you need to transpose the result.
Example:
m = matrix(c(0,1,2,5), 2,2)
m
apply( m, 1, divideByMean)
t(apply( m, 1, divideByMean)) # This is what you probably want.
HTH,
Peter
On Tue, Jun 1, 2010 at 10:26 AM, Joachim de Lezardiere <
joachim.lez...@gmail.com> wrote:
> Hello ,
>
>
>
> I can not get apply function to do what I want when doing stuff on rows.
> Let's say I want to divide the rows of matrix by their mean, the below show
> you get the same result weather you use 1 or 2, i.e. same result for
> columns
> than for means..:(
>
>
>
> Thanks a lot for the help,
>
>
>
>
>
>
>
> m = matrix( c(1,4,7,4,5,8,7,8,9), nrow = 3 )
>
>
>
> divideByMean = function( v ){
>
> return( v/mean(v))
>
> }
>
>
>
> rowMean = c( mean( m[1,]),mean( m[2,]),mean( m[3,]) )
>
> rowMean
>
>
>
> colMean = c( mean( m[,1]),mean( m[,2]),mean( m[,3]) )
>
> colMean
>
>
>
> m
>
> print("ByRow")
>
> apply( m, 1, divideByMean)
>
>
>
> print("ByCol")
>
> apply( m, 2, divideByMean)
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>[[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot multiple columns
Hi: Ben's example worked for me as follows: x <- matrix(runif(22000),ncol=22) library(lattice) library(coda) m <- as.mcmc(x) xyplot(m, layout = c(2, 11)) layout = uses columns first, then rows...I always forget that :) It also has an optional third argument for pages, which might come in handy here. If you want the order of the panels to be reversed, use the additional option as.table = TRUE. HTH, Dennis On Tue, Jun 1, 2010 at 9:37 AM, Noah Silverman wrote: > Hi, > > I used the term "run", as each iteration of the Gibbs sampler produces > 22 variables (coefficients for Beta in a regression model) > > > The example wont work > > > On 6/1/10 5:54 AM, Ben Bolker wrote: > > Noah Silverman smartmediacorp.com> writes: > > > > > >> I'm running a long MCMC chain that is generating samples for 22 > variables. > >> > >> I have each run of the chain as a row in a matrix. > >> > > > >> So: Chain[,1] is the column with all the samples for variable one. > >> Chain[,2] is the column with all the samples for variable 2, etc. > >> > > The previous 2 paragraphs seemed contradictory until I realized > > that in the first paragraph you are using "run" to mean what I would > > usually call a "sample" ... > > > > > >> I'd like to fit all 22 on a single page to print a nice summary. It is > >> OK if the graphs are small, I just need to show the overall shape and > >> convergence. > >> > > How about for example > > > > > >> x <- matrix(runif(22000),ncol=22) > >> library(coda) > >> m <- as.mcmc(x) > >> xyplot(m) > >> xyplot(m,layout=c(4,6)) > >> > > __ > > R-help@r-project.org mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
