Re: [R] Loading same libraries (old version newest version) in r
On Mon, 12-Oct-2009 at 10:54AM +1100, Steven Kang wrote:
| Hi all,
|
| I require 2 RMySQL libraries in order to query from a database.
You mean you require 2 RMySQL packages. We must be pedantic to answer
your question.
|
| 'RMySQL_0.7-4' (newest version) results in an error when more than 1 field
| is queried.
|
| ''RMySQL_0.5-11' (old version) resolves this issue where more than 1 field
| can be queried without any errors. However, other queries results in an
| error messeges (but no error messeges with 'RMySQL_0.7-4' package)
|
| Thus, I require two scripts each with different 'RMySQL' packages loaded to
| query from a database.
If you put your *packages* in separate *libraries*, you can load the
one you require using the lib.loc parameter, e.g.
require(RMySQL, lib.loc = /path/to/oldPackage)
When you wish to use the one in /path/to/newPackage, unload the old
one and change the lib.loc value.
HTH
|
| Are there any commands that will do the job (i.e. loading 2 different
| libraries from different folder) without having to click 'Pakages' tool bar
| in th RGui console?
|
| Your help in resolving this issue would be appreciated.
|
|
|
|
|
|
| Steven
|
| [[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.
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
(_)-(_) . Eleanor Roosevelt
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Re: [R] user input in R
And try also search statsrus The first hit shall be Paul Johnsons's howto's which helped me several years ago especially with basic issues. Regards Petr r-help-boun...@r-project.org napsal dne 10.10.2009 18:00:19: I'm just learning R (I don't know any other programming languages), and I have a question. I am trying to figure out how to ask for user input (say, a set of 3 numbers) then put those numbers into an array. I've looked around, but I haven't been able to find any answers that I understand. Thanks! Astoundingly, finding out how to load my own datasets was one of the hardest things for me to find information on when I was new to R and I had considerable programming experience. Part of the problem was that I didn't yet know exactly what R documentation was available or how to easily access it. Further, veteran users tend to assume loading datasets is easy, and much documentation and sample code uses 'generated data'. Thus, sample code demonstrating 'loading datasets' is a little hard to find when you are starting out! The official recommendation would go something like, You probably want to read the Import/Export documentation.You can find this on Windows GUI under help | manuals | Data Import/Export. Of course you proably also need to read chapter 7 of An Introduction to R which is also linked under manuals in the help menu. That said, I offer some beginner cheats you might find useful below. I particularly recommend the Verzzani Appendix. Things you could type at the command line to learn more about different ways to enter data are: ?scan ?read.table ?read.delim ?read.csv Beginner cheats Suggestions If you want GUI data input (of dataframes), I'd suggest installing and running the Rcmdr package. It lets you load many types of datasets, including Excel files. These can be used within Rcmdr or from the command line. If you are really new to R, I suggest looking at some of the contributed documentation to get a feel for how things are done as well as how to load user data http://cran.us.r-project.org/other-docs.html In particular, look at the appendix in John Verzzani's document to see a pragmatic description of using scan to input user data. (see Appendix: Entering Data in R on p. 103 or so). http://cran.us.r-project.org/doc/contrib/Verzani-SimpleR.pdf __ 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] field index given name.
Hi,
How do I access the index number of a field given I only know the field
name?
eg - I want to set the probability of the field 'species' higher than the
other fields to use in sampling.
colprob - array(dim=NCOL(iris))
for(i in 1:NCOL(iris)){colprob[i]=0.5}
colprob[iris$species] = 1 #this doesn't work
colprob
[1] 0.5 0.5 0.5 0.5 0.5
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Re: [R] Loading same libraries (old version newest version) in r
On Mon, 12 Oct 2009, Patrick Connolly wrote: On Mon, 12-Oct-2009 at 10:54AM +1100, Steven Kang wrote: | Hi all, | | I require 2 RMySQL libraries in order to query from a database. You mean you require 2 RMySQL packages. We must be pedantic to answer your question. | | 'RMySQL_0.7-4' (newest version) results in an error when more than 1 field | is queried. | | ''RMySQL_0.5-11' (old version) resolves this issue where more than 1 field | can be queried without any errors. However, other queries results in an | error messeges (but no error messeges with 'RMySQL_0.7-4' package) | | Thus, I require two scripts each with different 'RMySQL' packages loaded to | query from a database. If you put your *packages* in separate *libraries*, you can load the one you require using the lib.loc parameter, e.g. require(RMySQL, lib.loc = /path/to/oldPackage) Or library(): require() is just a wrapper. When you wish to use the one in /path/to/newPackage, unload the old one and change the lib.loc value. Unfortunately that will not work (but starting a new session and loading from a different lib.loc will). First, RMySQL has a namespace, so you need to unload the namespace, not just the package. But when you do that, there is AFAICS no .onUnload hook so the DLL remains loaded and will not be reloaded when the other version is loaded. You could try playing with library.dynam.unload, but my hazy recollection is that did not work either (something to do with libmysqlclient only getting initialized once in a process). I don't see an OS mentioned here, but basically do not believe that two versions are really needed: my guess is that this is Windows and you need to install RMySQL *from the sources* against your installation of MySQL. There is a long history of binary incompatibility between versions of MySQL, even those which differ only by one in the 'patchlevel'. | | Are there any commands that will do the job (i.e. loading 2 different | libraries from different folder) without having to click 'Pakages' tool bar | in th RGui console? | | Your help in resolving this issue would be appreciated. | | | | | | | Steven -- 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] field index given name.
Hi Phil
Try the following
which(names(iris)=='Species')
[1] 5
HTH
Schalk Heunis
On Mon, Oct 12, 2009 at 8:53 AM, tdm ph...@philbrierley.com wrote:
Hi,
How do I access the index number of a field given I only know the field
name?
eg - I want to set the probability of the field 'species' higher than the
other fields to use in sampling.
colprob - array(dim=NCOL(iris))
for(i in 1:NCOL(iris)){colprob[i]=0.5}
colprob[iris$species] = 1 #this doesn't work
colprob
[1] 0.5 0.5 0.5 0.5 0.5
--
View this message in context:
http://www.nabble.com/field-index-given-name.-tp25851216p25851216.html
Sent from the R help mailing list archive at Nabble.com.
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[[alternative HTML version deleted]]
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Re: [R] field index given name.
Thanks - would never have guessed that. I eventually got the following to do
what I want...
colprob - array(dim=NCOL(iris))
for(i in 1:NCOL(iris)){
+ colprob[i]=
+ ifelse(names(iris)[i] == 'Species',1,0.5)
+ }
colprob
[1] 0.5 0.5 0.5 0.5 1.0
Schalk Heunis-2 wrote:
Hi Phil
Try the following
which(names(iris)=='Species')
[1] 5
HTH
Schalk Heunis
On Mon, Oct 12, 2009 at 8:53 AM, tdm ph...@philbrierley.com wrote:
Hi,
How do I access the index number of a field given I only know the field
name?
eg - I want to set the probability of the field 'species' higher than the
other fields to use in sampling.
colprob - array(dim=NCOL(iris))
for(i in 1:NCOL(iris)){colprob[i]=0.5}
colprob[iris$species] = 1 #this doesn't work
colprob
[1] 0.5 0.5 0.5 0.5 0.5
--
View this message in context:
http://www.nabble.com/field-index-given-name.-tp25851216p25851216.html
Sent from the R help mailing list archive at Nabble.com.
__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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--
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[R] Help Error
Hi R-users,
I would like to ask question related to error output.
If an error comments come out, then the program will automatically stop.
I want to ask , how I can still continue the program even though there is an
error comment?
var=VAR(Canada,p=3,type=const)
for (j in 1:nrow(com))
{
mat=ma
{
for (i in 1:ncol(com))
{
y=which(mat==com[j,i])
mat[y]=NA
}
z=which(mat!=0)
mat[z]=0
hitt=SVAR(var , estmethod = scoring, Amat = mat, Bmat = NULL,max.iter =
100, maxls = 1000, conv.crit = 1.0e-8)
hit[j]=hitt$LR$statistic
}
}
There will be an error comment,e.g. : Error in solve.default(BinvA) :
Lapack routine dgesv: system is exactly singular
But I still want to continue to compute the SVAR.. How can I do it?
Thanks for answering it..
Regards,
Arif
_
Facebook.
k-basics.aspx?ocid=PID23461::T:WLMTAGL:ON:WL:en-id:SI_SB_2:092009
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Re: [R] Help Error
This may be of some help:?try
On Mon, Oct 12, 2009 at 9:42 AM, Arif Chandra arif.chan...@hotmail.comwrote:
Hi R-users,
I would like to ask question related to error output.
If an error comments come out, then the program will automatically stop.
I want to ask , how I can still continue the program even though there is
an error comment?
var=VAR(Canada,p=3,type=const)
for (j in 1:nrow(com))
{
mat=ma
{
for (i in 1:ncol(com))
{
y=which(mat==com[j,i])
mat[y]=NA
}
z=which(mat!=0)
mat[z]=0
hitt=SVAR(var , estmethod = scoring, Amat = mat, Bmat = NULL,max.iter
= 100, maxls = 1000, conv.crit = 1.0e-8)
hit[j]=hitt$LR$statistic
}
}
There will be an error comment,e.g. : Error in solve.default(BinvA) :
Lapack routine dgesv: system is exactly singular
But I still want to continue to compute the SVAR.. How can I do it?
Thanks for answering it..
Regards,
Arif
_
Facebook.
k-basics.aspx?ocid=PID23461::T:WLMTAGL:ON:WL:en-id:SI_SB_2:092009
[[alternative HTML version deleted]]
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[[alternative HTML version deleted]]
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[R] Position of legend box
Good morning to you. I have about 4 different lines in one plot. I have used legend to indicate the colour of each plot. But the box contain the legend covers part of the lines thereby blurring the legend. There are some spaces in the plot that are empty and large enough to accommodate the legend box. If there a command I could use to set the position of the legend myself. I added the legend using: legend(oo1$daymon[100],3600,c(2005,2006,2004,2003),col=c(4,2,3,1),lty=1) Thank you. Ogbos [[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] package nlme
Hi R Users, When I use package nlme for linear model with random effects, there exists errors and I don't know the data structure of lme. Here is my data: Rice-data.frame(Yield=c(8,7,4,9,7,6,9,8,8,8,7,5,9,9,5,7,7,8,8,8,4,8,6,4,8,8,9), Variety=rep(rep(c(A1,A2,A3),each=3),3), Stand=rep(c(B1,B2,B3),9), Block=rep(1:3,each=9)) Rice.lme-lme(Yield ~ Variety + Stand + Block,data=Rice) And error will be like below: Wrong at getGroups.data.frame(dataMix, groups) : Invalid formula for groups can you give me some advice and you will be appreciated. Thanks. Wenjun. [[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] package nlme
You did not specify the random effects. Try something like Rice.lme-lme(Yield ~ Variety + Stand,data=Rice, random = ~1|Block) HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens wenjun zheng Verzonden: maandag 12 oktober 2009 10:57 Aan: R-help@r-project.org Onderwerp: [R] package nlme Hi R Users, When I use package nlme for linear model with random effects, there exists errors and I don't know the data structure of lme. Here is my data: Rice-data.frame(Yield=c(8,7,4,9,7,6,9,8,8,8,7,5,9,9,5,7,7,8,8,8,4,8,6,4 ,8,8,9), Variety=rep(rep(c(A1,A2,A3),each=3),3), Stand=rep(c(B1,B2,B3),9), Block=rep(1:3,each=9)) Rice.lme-lme(Yield ~ Variety + Stand + Block,data=Rice) And error will be like below: Wrong at getGroups.data.frame(dataMix, groups) : Invalid formula for groups can you give me some advice and you will be appreciated. Thanks. Wenjun. [[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. Druk dit bericht a.u.b. niet onnodig af. Please do not print this message unnecessarily. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ 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] Problem Fixed
I write to thank all those who have helped me to fix this problem. Best regards Ogbos 2009/10/7 ogbos okike ogbos.ok...@gmail.com Good morning. I wish to plot two data on the same axis. I tried plot(x,y, type = l) for the first and tried to use lines or points(x,y, lty = 2, col = 4) to add or plot the second data on alongside the first. However, what I got was not encouraging. I have attached the two data and would be pleased if anybody could be of help. Thank you Best regards Ogbos [[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] Error: cannot allocate vector of size 1.2 Gb
Dear List, today I turn to you with a next problem. I'm trying to compare species richness between various datasets (locations) using species accumulation curves (Chapter 4, page 54 in Tree diversity analysishttp://www.worldagroforestry.org/treesandmarkets/tree_diversity_analysis.aspby Kindt Coe). To accomplish this I'm using package BiodiversityR. My data is comprised of species community (PoCom) (10 locations with 83 species) and environmental factors (PoEnv) (10 locations with 17 factors). In attempt to calculate the function (accumcomp) I receive the following error. I can not imagine how a 10x83+10x17 matrix can grow to a GB or more. Unless I'm missing something? How can I combat this? poacc2 - accumcomp(PoCom, y=PoEnv, factor1=HM_sprem, method=exact) Error: cannot allocate vector of size 1.2 Gb In addition: Warning messages: 1: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) 2: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) 3: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) 4: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) Cheers, Roman [[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] Creating a Clustered-Stacked Column Chart
On 10/12/2009 01:53 AM, zhijie zhang wrote:
Thanks. I think there may be no easy method to achive it.
library(lattice)
barchart(Titanic, scales = list(x = free),auto.key = list(title
=Survived),layout=c(4,1),horizontal = FALSE)
The above method generates four graphs, two graphs in the left are for
children's male and female,respectively and the right two graphs are for
adult's male and female,respectively .
Actually, i hope to generate two graphs finally. Say the right two graphs
for adult are overlaid with the left two graphs for children,respectively.
Take the 1st of x variable as an example, in the place of 1st, the
stacked bar for both children and adult should be displayed. Maybe the data
for children and adult should be first shifted certain values to different
directions and then applying the overlay function to get it.
My above ideas to display a data may be bad. Anyway, thanks a lot.
Hi Zhijie,
This looked like an interesting challenge, so I bent the barp function a
bit to do it. The barp3 function attached will accept a 3D array in
which the rows represent groups, the columns subgroups and the files
sub-subgroups. The arrangement of bars is that of the example you gave.
In order to translate the 2D matrix of the example to the 3D array, I
have written a little conversion function df2array. Try this:
clustack-structure(list(Country = structure(c(3L, 2L, 1L),
.Label = c(Asia,Europe, N Amer), class = factor),
Q1.pencils = c(16L, 14L,18L), Q1.pens = c(12L, 9L, 14L),
Q2.pencils = c(18L, 15L, 18L), Q2.pens = c(14L, 11L, 15L),
Q3.pencils = c(17L, 11L, 20L), Q3.pens = c(11L, 8L, 15L),
Q4.pencils = c(20L, 14L, 21L), Q4.pens = c(14L, 12L, 16L)),
.Names = c(Country, Q1.pencils,Q1.pens, Q2.pencils, Q2.pens,
Q3.pencils, Q3.pens,Q4.pencils, Q4.pens),
class = data.frame, row.names = c(NA,-3L))
# x is the original data frame with the group names in the
# first column and the column names as in the example
# I read it in as a CSV file.
# depth is the length of the files (3rd dimension)
df2array-function(x,depth) {
dimx-dim(x)
if(dimx[2]%%depth) stop(depth must divide number of columns without
remainder)
column.order-NULL
for(d in 1:depth)
column.order-c(column.order,seq(d,dimx[2]-(depth-d),by=depth))
return(array(unlist(x[,column.order]),c(dimx[1],dimx[2]/depth,depth)))
}
# this converts the data frame to the array
pp.array-df2array(clustack[,2:9],2)
# this function is pretty much the same as barp
barp3-function(height,width=0.4,names.arg=NULL,
legend.lab=NULL,legend.pos=e,col=NULL,border=par(fg),
main=NULL,xlab=,ylab=,xlim=NULL,ylim=NULL,
staxx=FALSE,staxy=FALSE,height.at=NULL,
height.lab=NULL,cex.axis=par(cex.axis),
do.first=NULL) {
if(is.data.frame(height)) its_ok-is.numeric(unlist(height))
else its_ok-is.numeric(height)
if(!its_ok) stop(barp3 can only display bars with numeric heights)
hdim-dim(height)
if(is.null(hdim) || length(hdim) != 3)
stop(barp3 can only plot 3 dimensional arrays)
ngroups-hdim[1]
if(length(col)==hdim[3])
barcol-array(rep(col,each=hdim[1]*hdim[2]),hdim)
else barcol-col
if(is.null(xlim)) xlim-c(0.4,ngroups+0.6)
if(any(height0,na.rm=TRUE))
stop(Can't have negative bar heights in barp3)
if(is.null(ylim)) {
maxstack-0
for(group in 1:hdim[1]) {
for(subgroup in 1:hdim[2]) {
thistack-sum(height[group,subgroup,],na.rm=TRUE)
if(thistack maxstack) maxstack-thistack
}
}
ylim-c(0,maxstack*1.05)
}
plot(0,type=n,main=main,xlab=xlab,ylab=ylab,
axes=FALSE,xlim=xlim,ylim=ylim,xaxs=i,yaxs=i)
if(!is.null(do.first)) eval(do.first)
if(is.null(names.arg)) names.arg-1:ngroups
if(staxx) {
axis(1,at=1:ngroups,labels=rep(,ngroups),
cex.axis=cex.axis)
staxlab(1,at=1:ngroups,labels=names.arg,cex=cex.axis)
}
else axis(1,at=1:ngroups,labels=names.arg,cex.axis=cex.axis)
if(is.null(height.at)) height.at-pretty(ylim)
if(is.null(height.lab)) height.lab-pretty(ylim)
if(staxy) {
axis(2,at=height.at,labels=rep(,length(height.lab)),
cex.axis=cex.axis)
staxlab(2,at=height.at,labels=height.lab,cex=cex.axis)
}
else axis(2,at=height.at,labels=height.lab,cex.axis=cex.axis)
barwidth-2*width/hdim[2]
for(group in 1:hdim[1]) {
left-group-hdim[2]*barwidth/2
for(subgroup in 1:hdim[2]) {
bottom-0
for(subsub in 1:hdim[3]) {
rect(left,bottom,left+barwidth,
bottom+height[group,subgroup,subsub],
col=barcol[group,subgroup,subsub],border=border)
bottom-bottom+height[group,subgroup,subsub]
}
left-left+barwidth
}
}
if(!is.null(legend.lab)) {
xjust-yjust-0.5
if(is.null(legend.pos)) {
cat(Click at the lower left corner of the legend\n)
legend.pos-locator(1)
xjust-yjust-0
}
if(legend.pos[1] == e)
legend.pos-emptyspace(barpinfo,bars=TRUE)
legend(legend.pos,legend=legend.lab,fill=col,
xjust=xjust,yjust=yjust)
}
box()
}
# the colors should be an array of the same form as the data
barp3(pp.array,col=array(rep(1:8,each=3),c(3,4,2)),
names.arg=clustack[,1])
Have
Re: [R] plot the same types of graphics on the same R graphic device
one example is to plot two vectors against each other and then, to plot the CI (confidence interval) of each point on the same graphics. The first could be done by plot and the second by plotCI in gplots package but how to plot on the same R graphic device? Look forward to your reply, --- On Fri, 10/9/09, Sarah Goslee sarah.gos...@gmail.com wrote: From: Sarah Goslee sarah.gos...@gmail.com Subject: Re: [R] plot the same types of graphics on the same R graphic device To: carol white wht_...@yahoo.com, r-help r-help@r-project.org Date: Friday, October 9, 2009, 3:04 PM Hi Carol, It isn't at all clear exactly what you are trying to do, but you might want to read the help for either points() and lines() [to put more than one data pair on a single plot], or for par, specifically mfrow and mfcol, or for layout [to put more than one plot on a single device window]. If none of those do it, then try again with a more detailed description of what you are looking for. Sarah On Fri, Oct 9, 2009 at 5:10 PM, carol white wht_...@yahoo.com wrote: Hi, How to plot the same types of graphics on the same R graphic device? Suppose that we want to plot a vector y against x (using plot for instance). How is it possible to plot y against x for different values of these two vectors on the same device so that the plots could be compared? Cheers, Carol -- Sarah Goslee http://www.functionaldiversity.org [[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] plot the same types of graphics on the same R graphic device
On 10/12/2009 08:47 PM, carol white wrote: one example is to plot two vectors against each other and then, to plot the CI (confidence interval) of each point on the same graphics. The first could be done by plot and the second by plotCI in gplots package but how to plot on the same R graphic device? Hi Carol, I would: plot(the first vector) dispersion(for the first vector) points(the second vector) dispersion(for the second vector) There are a number of other functions to display CIs. 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] Error: cannot allocate vector of size 1.2 Gb
Dear Roman, could you give us the trace given by traceback() ? I suspect the error is resulting from the permutations and/or jackknife procedure in the underlying functions specaccum and specpool. You can take a look at the package R.huge, but that one is deprecated already. There are other packages around too, but I have no experience with them. You find some more tips here : http://www.matthewckeller.com/html/memory.html This should give you a place to start looking. Kind regards Joris On Mon, Oct 12, 2009 at 11:39 AM, romunov romu...@gmail.com wrote: Dear List, today I turn to you with a next problem. I'm trying to compare species richness between various datasets (locations) using species accumulation curves (Chapter 4, page 54 in Tree diversity analysishttp://www.worldagroforestry.org/treesandmarkets/tree_diversity_analysis.aspby Kindt Coe). To accomplish this I'm using package BiodiversityR. My data is comprised of species community (PoCom) (10 locations with 83 species) and environmental factors (PoEnv) (10 locations with 17 factors). In attempt to calculate the function (accumcomp) I receive the following error. I can not imagine how a 10x83+10x17 matrix can grow to a GB or more. Unless I'm missing something? How can I combat this? poacc2 - accumcomp(PoCom, y=PoEnv, factor1=HM_sprem, method=exact) Error: cannot allocate vector of size 1.2 Gb In addition: Warning messages: 1: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) 2: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) 3: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) 4: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) Cheers, Roman [[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.
[R] Non linear fitting of 2 distributions with shared parameters
Hi, I need to compare non-linear fittings of 2 different experimental distributions. I use nls() to the fit the 2 distribution and it works pretty well. The statistical comparison of 2 non linear fits is well described in the book Fitting Models to Biological Data using Linear and Nonlinear Regression from Harvey Motulsky and Arthur Christopoulos. This test is based on the calculation of an F-value that compares the residuals of 1°) the global fit of both data sets (with one value for each parameter) on one hand, 2°) with the 2 fits of each separate data sets (with thus 2 values for each parameter) on the other hand. This methods allows to decide wether the 2 data sets belong to the same curve. No problem this far. But the next step is about comparing each parameter to decide wether or not they are different between each data set. The comparison still lies on the calculation of an F-value, but here I compare 1°) the residuals of the 2 fits of each separate data sets, 2°) with the residuals the 2 fits of each data sets which share a same value of the parameter that is compared (i.e. the 2 curve fits find the same value for the shared parameter). This point is where I need some help : is there any function to make non-linear fitting of (at least) 2 data sets with shared parameters ? Hope someone will have some good news for me Etienne Toffin [[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] Error: cannot allocate vector of size 1.2 Gb
Hello joris, this is the traceback() output. Hopefully you can make some sense out of it. Thank you for the tips as well (R.huge looks promising)! traceback() 7: vector(integer, length) 6: integer(nbins) 5: tabulate(bin, pd) 4: as.vector(data) 3: array(tabulate(bin, pd), dims, dimnames = dn) 2: table(y[, factor]) 1: accumcomp(PoCom, y = PoEnv, factor1 = HM_sprem, method = exact) Cheers, Roman On Mon, Oct 12, 2009 at 12:52 PM, joris meys jorism...@gmail.com wrote: Dear Roman, could you give us the trace given by traceback() ? I suspect the error is resulting from the permutations and/or jackknife procedure in the underlying functions specaccum and specpool. You can take a look at the package R.huge, but that one is deprecated already. There are other packages around too, but I have no experience with them. You find some more tips here : http://www.matthewckeller.com/html/memory.html This should give you a place to start looking. Kind regards Joris On Mon, Oct 12, 2009 at 11:39 AM, romunov romu...@gmail.com wrote: Dear List, today I turn to you with a next problem. I'm trying to compare species richness between various datasets (locations) using species accumulation curves (Chapter 4, page 54 in Tree diversity analysis http://www.worldagroforestry.org/treesandmarkets/tree_diversity_analysis.asp by Kindt Coe). To accomplish this I'm using package BiodiversityR. My data is comprised of species community (PoCom) (10 locations with 83 species) and environmental factors (PoEnv) (10 locations with 17 factors). In attempt to calculate the function (accumcomp) I receive the following error. I can not imagine how a 10x83+10x17 matrix can grow to a GB or more. Unless I'm missing something? How can I combat this? poacc2 - accumcomp(PoCom, y=PoEnv, factor1=HM_sprem, method=exact) Error: cannot allocate vector of size 1.2 Gb In addition: Warning messages: 1: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) 2: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) 3: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) 4: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) Cheers, Roman [[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.
Re: [R] Error: cannot allocate vector of size 1.2 Gb
Hi Roman, that throws a different light on the problem. It goes wrong from the start, so it has little to do with the bootstrap or jackknife procedures. R.huge won't help you either. Likely your error comes from the fact that factor1 is not an argument of the function accumcomp. the argument is factor. As R doesn't find this, it'll try to tabulate the complete environmental dataset, and this gives the memory overflow. Try : poacc2 - accumcomp(PoCom, y=PoEnv, factor=HM_sprem, method=exact) That should work. I can't try it out without dataset off course. If it doesn't work, post the traceback again, I'll take another look. Kind regards Joris On Mon, Oct 12, 2009 at 1:07 PM, romunov romu...@gmail.com wrote: Hello joris, this is the traceback() output. Hopefully you can make some sense out of it. Thank you for the tips as well (R.huge looks promising)! traceback() 7: vector(integer, length) 6: integer(nbins) 5: tabulate(bin, pd) 4: as.vector(data) 3: array(tabulate(bin, pd), dims, dimnames = dn) 2: table(y[, factor]) 1: accumcomp(PoCom, y = PoEnv, factor1 = HM_sprem, method = exact) Cheers, Roman On Mon, Oct 12, 2009 at 12:52 PM, joris meys jorism...@gmail.com wrote: Dear Roman, could you give us the trace given by traceback() ? I suspect the error is resulting from the permutations and/or jackknife procedure in the underlying functions specaccum and specpool. You can take a look at the package R.huge, but that one is deprecated already. There are other packages around too, but I have no experience with them. You find some more tips here : http://www.matthewckeller.com/html/memory.html This should give you a place to start looking. Kind regards Joris On Mon, Oct 12, 2009 at 11:39 AM, romunov romu...@gmail.com wrote: Dear List, today I turn to you with a next problem. I'm trying to compare species richness between various datasets (locations) using species accumulation curves (Chapter 4, page 54 in Tree diversity analysishttp://www.worldagroforestry.org/treesandmarkets/tree_diversity_analysis.aspby Kindt Coe). To accomplish this I'm using package BiodiversityR. My data is comprised of species community (PoCom) (10 locations with 83 species) and environmental factors (PoEnv) (10 locations with 17 factors). In attempt to calculate the function (accumcomp) I receive the following error. I can not imagine how a 10x83+10x17 matrix can grow to a GB or more. Unless I'm missing something? How can I combat this? poacc2 - accumcomp(PoCom, y=PoEnv, factor1=HM_sprem, method=exact) Error: cannot allocate vector of size 1.2 Gb In addition: Warning messages: 1: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) 2: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) 3: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) 4: In vector(integer, length) : Reached total allocation of 1023Mb: see help(memory.size) Cheers, Roman [[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] Error: cannot allocate vector of size 1.2 Gb
Hi Joris,
thanks for spotting that one. This little mistake has gotten in when I was
trying desperate things with the analysis (factor1 is used in diversitycomp).
Nevertheless, here is the result:
poacc2 - accumcomp(PoCom, y=PoEnv, factor=HM_sprem, method=exact)
Error in if (p == 1) { : argument is of length zero
And the traceback():
traceback()
3: specaccum(x, method = method, permutations = permutations, conditioned =
conditioned,
gamma = gamma, ...)
2: accumresult(x, y, factor, level = levels[i], scale = scale, method =
method,
permutations = permutations, conditioned = conditioned, gamma =
gamma)
1: accumcomp(PoCom, y = PoEnv, factor = HM_sprem, method = exact)
Best wishes,
Roman
On Mon, Oct 12, 2009 at 1:56 PM, joris meys jorism...@gmail.com wrote:
Hi Roman,
that throws a different light on the problem. It goes wrong from the
start, so it has little to do with the bootstrap or jackknife
procedures. R.huge won't help you either.
Likely your error comes from the fact that factor1 is not an
argument of the function accumcomp. the argument is factor. As R
doesn't find this, it'll try to tabulate the complete environmental
dataset, and this gives the memory overflow.
Try :
poacc2 - accumcomp(PoCom, y=PoEnv, factor=HM_sprem, method=exact)
That should work. I can't try it out without dataset off course. If it
doesn't work, post the traceback again, I'll take another look.
Kind regards
Joris
On Mon, Oct 12, 2009 at 1:07 PM, romunov romu...@gmail.com wrote:
Hello joris,
this is the traceback() output. Hopefully you can make some sense out of
it.
Thank you for the tips as well (R.huge looks promising)!
traceback()
7: vector(integer, length)
6: integer(nbins)
5: tabulate(bin, pd)
4: as.vector(data)
3: array(tabulate(bin, pd), dims, dimnames = dn)
2: table(y[, factor])
1: accumcomp(PoCom, y = PoEnv, factor1 = HM_sprem, method = exact)
Cheers,
Roman
On Mon, Oct 12, 2009 at 12:52 PM, joris meys jorism...@gmail.com
wrote:
Dear Roman,
could you give us the trace given by traceback() ? I suspect the error
is resulting from the permutations and/or jackknife procedure in the
underlying functions specaccum and specpool.
You can take a look at the package R.huge, but that one is deprecated
already. There are other packages around too, but I have no experience
with them. You find some more tips here :
http://www.matthewckeller.com/html/memory.html
This should give you a place to start looking.
Kind regards
Joris
On Mon, Oct 12, 2009 at 11:39 AM, romunov romu...@gmail.com wrote:
Dear List,
today I turn to you with a next problem. I'm trying to compare species
richness between various datasets (locations) using species
accumulation
curves (Chapter 4, page 54 in Tree diversity
analysis
http://www.worldagroforestry.org/treesandmarkets/tree_diversity_analysis.asp
by
Kindt Coe). To accomplish this I'm using package
BiodiversityR. My data is comprised of species community (PoCom) (10
locations with 83 species) and environmental factors (PoEnv) (10
locations
with 17 factors).
In attempt to calculate the function (accumcomp) I receive the
following
error. I can not imagine how a 10x83+10x17 matrix can grow to a GB or
more.
Unless I'm missing something? How can I combat this?
poacc2 - accumcomp(PoCom, y=PoEnv, factor1=HM_sprem,
method=exact)
Error: cannot allocate vector of size 1.2 Gb
In addition: Warning messages:
1: In vector(integer, length) :
Reached total allocation of 1023Mb: see help(memory.size)
2: In vector(integer, length) :
Reached total allocation of 1023Mb: see help(memory.size)
3: In vector(integer, length) :
Reached total allocation of 1023Mb: see help(memory.size)
4: In vector(integer, length) :
Reached total allocation of 1023Mb: see help(memory.size)
Cheers,
Roman
[[alternative HTML version deleted]]
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Re: [R] field index given name.
On Oct 12, 2009, at 3:22 AM, tdm wrote:
Thanks - would never have guessed that. I eventually got the
following to do
what I want...
colprob - array(dim=NCOL(iris))
for(i in 1:NCOL(iris)){
+ colprob[i]=
+ ifelse(names(iris)[i] == 'Species',1,0.5)
+ }
colprob
[1] 0.5 0.5 0.5 0.5 1.0
This would be more in keeping with the approah in R of avoiding loops
when possble:
colprob - array(dim=NCOL(iris))
colprob[] - 0.5
colprob
[1] 0.5 0.5 0.5 0.5 0.5
colprob[names(iris) == Species] - 1
colprob
[1] 0.5 0.5 0.5 0.5 1.0
--
David
Schalk Heunis-2 wrote:
Hi Phil
Try the following
which(names(iris)=='Species')
[1] 5
HTH
Schalk Heunis
On Mon, Oct 12, 2009 at 8:53 AM, tdm ph...@philbrierley.com wrote:
Hi,
How do I access the index number of a field given I only know the
field
name?
eg - I want to set the probability of the field 'species' higher
than the
other fields to use in sampling.
colprob - array(dim=NCOL(iris))
for(i in 1:NCOL(iris)){colprob[i]=0.5}
colprob[iris$species] = 1 #this doesn't work
colprob
[1] 0.5 0.5 0.5 0.5 0.5
--
View this message in context:
http://www.nabble.com/field-index-given-name.-tp25851216p25851216.html
Sent from the R help mailing list archive at Nabble.com.
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and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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--
View this message in context:
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and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] Position of legend box
On Oct 12, 2009, at 4:07 AM, ogbos okike wrote: Good morning to you. I have about 4 different lines in one plot. I have used legend to indicate the colour of each plot. But the box contain the legend covers part of the lines thereby blurring the legend. There are some spaces in the plot that are empty and large enough to accommodate the legend box. If there a command I could use to set the position of the legend myself. I added the legend using: legend(oo1$daymon[100],3600,c(2005,2006, 2004,2003),col=c(4,2,3,1),lty=1) This may be of help, courtesy of plotmeister Jim Lemon: http://finzi.psych.upenn.edu/Rhelp08/2009-June/202740.html There are also facilities in Harrell's Hmisc that do something similar: library(Hmisc) ?labcurve __ David Winsemius, MD Heritage Laboratories 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] is that possible to graph 4 dimention plot
I'm basically put off by the question itself. Plotting a 4-dimensional
graph is rather complicated if the world has only 3 dimensions. A
4-dimensional representation is typically a movie (with time as the
4th dimension). You could try to project a heatmap on a 3D surface
graph, but I doubt this will make things much more clear.
So the standard (and correct) way of solving this problem, is to think
about a clever way to represent the needed information in less
dimensions. Ryan gives some nice examples and tips of how to do that,
but those are dismissed as not helpful.
People on the list answer voluntarily. They do not like to be told
that you don't think it will really help. Did you actually try it
out? You certainly don't give that impression. Maybe you should have
another look at it, and question your own approach to the problem as
well.
Keep it in mind for next time, you're not making yourself popular this way.
Kind regards
Joris
On Sat, Oct 10, 2009 at 10:01 PM, Duncan Murdoch murd...@stats.uwo.ca wrote:
On 07/10/2009 5:50 PM, gcheer3 wrote:
Thanks for your reply.
But I don't think it will really help. My problem is as follows:
I have 20 observations
y - rnorm(N,mean= rep(th[1:2],N/2),sd=th[3])
I have a loglikelihood function for 3 variables mu-(mu1,mu2) and sig
loglike - function(mu,sig){
temp-rep(0,length(y))
for (i in 1:(length(y)))
{
temp[i]-log((1/2)*dnorm(y[i],mu[1],sig)+(1/2)*dnorm(y[i],mu[2],sig))}
return(sum(temp))
}
for example
mu-c(1,1.5)
sig-2
loglike(mu,sig)
[1] -34.1811
I am interested how mu[1], mu[2], and sig changes, will effect the
loglikelihood surface. At what values of mu and sig will make
loglikelihood
the maximum and at what values of mu and sig will make loglikelihood has
local max (smaller hills) and at what values of mu and sig the
loglikelihood
is flat , etc.
I tried contour3d also, seems doesn't work
I haven't seen any replies to this. One explanation would be that everyone
was turned off (as I was) by the rude remark above.
On this list, before saying that something doesn't work, it's polite to
give a simple, nicely formatted, self-contained reproducible example of what
went wrong, and to ask whether it is your error or an error in the package.
Taking that approach will usually result in someone pointing out your error
(and fixing your code); sometimes it will result in a package author
agreeing it's a bug, and fixing it.
Duncan Murdoch
Thanks for any advice
Ryan-50 wrote:
Suppose there are 4 variables
d is a function of a , b and c
I want to know how a, b and c change will make d change
It will be straightforward to see it if we can graph the d surface
if d is only a function of a and b, I can use 'persp' to see the surface
of
d. I can easily see at what values of a and b, d will get the maxium or
minium or multiple modes, etc
But for 4 dimention graph, is there a way to show the surface of d
Will use color help
Thanks a lot
Not sure what your data looks like, but you might also consider looking
at a 2 dimensional version. See ?coplot
for example:
coplot(lat ~ long | depth * mag, data = quakes)
Or you can make 2 or 3-dimensional plots using the lattice package
conditioning on some of the variables - e.g. d ~ a | b * c,
etc.
If a, b, and c are continuous, you can use equal.count. Here is
an uninteresting example, considering a, b, and c as points along
a grid:
a - b - c - seq(1:10)
dat - data.frame(expand.grid(a, b, c))
names(dat) - letters[1:3]
dat$d - with(dat, -(a-5)^2 - (b-5)^2 - (c-5)^2)
library(lattice)
# 2-d:
xyplot(d ~ a | equal.count(b)*equal.count(c), data=dat, type=l)
# etc.
# 3-d:
contourplot(d ~ a * b | equal.count(c), data=dat)
wireframe(d ~ a * b | equal.count(c), data=dat)
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[R] List mappings and variable creation
Hi All,
I have a questions about associative list mappings in R, and if they are
possible?
I have data in the form show below, and want to make a new 'bucket' variable
called combined. Which is the sum of the control and the exposed metric
values
This combined variable is a many to many matching as values only appear in
the file if they have a value 0.
conversion.type filteredIDbucketID Metric Value
countertrue control a 1
countertrue control b 1
countertrue control c 2
countertrue control d 3
countertrue exposed a 4
countertrue exposed e 1
ASIDE:
At the minute i read the data into my file and and then create all the
'missing' row values
(in this case,
countertrue control e 0
countertrue exposed b 0
countertrue exposed c 0
countertrue exposed d 0)
and then run a sort on the data, and count the number of times control
appears, and then use this as an index matcher.
saw.aggr.data - [order(saw.aggr.data$bucketID, saw.aggr.data$metric), ]
no.of.metrics - length(saw.aggr.data$bucketID[grep(control,
saw.aggr.data$bucketID)])
for (i in (1:no.of.metrics)) {
assign(paste(combined, as.character(saw.aggr.data$metric[i])),
(saw.aggr.data$value[i] + saw.aggr.data$value[i + no.of.metrics]))
}
This does what i want it to but is very very weak and could be open to large
errors, ( error handling currently via grepping the names of the metric[i]
== name of metric [i + no.of.metrics])
Is there a more powerful way of doing this using some kind of list mapping?
I've looked at the older threads in this area and it looks like something
that should be possible but i can't figure out how to do this?
Ideally i'd like a final dataset / list that is of the following form.
conversion.type filteredIDbucketID Metric Value
countertrue control a 1
countertrue control b 1
countertrue control c 2
countertrue control d 3
countertrue exposed a 4
countertrue exposed e 1
countertrue combineda 5
countertrue combinedb 1
countertrue combinedc 2
countertrue combinedd 3
countertrue combinede 1
So i dont have to create the dummy variables.
does this make sense?
Many thanks in advance
Mike
--
Michael Pearmain
I abhor averages. I like the individual case. A man may have six meals
one day and none the next, making an average of three meals per day, but
that is not a good way to live. ~Louis D. Brandeis
f you received this communication by mistake, please don't forward it to
anyone else (it may contain confidential or privileged information), please
erase all copies of it, including all attachments, and please let the sender
know it went to the wrong person. Thanks.
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[R] R and computer size
Dear R-mailing list Hope you can help me. I am using R for windows to analyze my 107 HGU133Plus2.0 chips, however, R chrash when I try to use ReadAffy(). I want to buy a computer that can handle all these arrays, do you know how big a computer I need to buy? Best, Skov, Denmark [[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] svy / weighted regression
Dear Peter, Thanks for the input. The zero rates in some strata occurs because sampling depended on case status: In Finland only 50% of the non-cases were sampled, while all others were sampled with 100% probability. Best Laust On Sat, Oct 10, 2009 at 11:02 AM, Peter Dalgaard p.dalga...@biostat.ku.dk wrote: Sorry, forgot to reply all... Laust wrote: Dear list, I am trying to set up a propensity-weighted regression using the survey package. Most of my population is sampled with a sampling probability of one (that is, I have the full population). However, for a subset of the data I have only a 50% sample of the full population. In previous work on the data, I analyzed these data using SAS and STATA. In those packages I used a propensity weight of 1/[sampling probability] in various generalized linear regression-procedures, but I am having trouble setting this up. I bet the solution is simple, but I’m a R newbie. Code to illustrate my problem below. Hi Laust, You probably need the package author to explain fully, but as far as I can see, the crux is that a dispersion parameter is being used, based on Pearson residuals, even in the Poisson case (i.e. you effectively get the same result as with quasipoisson()). I don't know what the rationale is for this, but it is clear that with your data, an estimated dispersion parameter is going to be large. E.g. the data has both 0 cases in 75 person-years and 1000 cases in 5000 person-years for Denmark, and in your model they are supposed to have the same Poisson rate. summary.svyglm starts off with est.disp - TRUE and AFAICS there is no way it can get set to FALSE. Knowing Thomas, there is probably a perfectly good reason not to just set the dispersion to 1, but I don't get it either... Thanks Laust # loading survey library(survey) # creating data listc - c(Denmark,Finland,Norway,Sweden,Denmark,Finland,Norway,Sweden) listw - c(1,2,1,1,1,1,1,1) listd - c(0,0,0,0,1000,1000,1000,2000) listt - c(75,50,90,190,5000,5000,5000,1) list.cwdt - c(listc, listw, listd, listt) country - data.frame(country=listc,weight=listw,deaths=listd,yrs_at_risk=listt) # running a frequency weighted regression to get the correct point estimates for comparison glm - glm(deaths ~ country + offset(log(yrs_at_risk)), weights=weight, data=country, family=poisson()) summary(glm) regTermTest(glm, ~ country) # running survey weighted regression svy - svydesign(~0,,data=country, weight=~weight) svyglm - svyglm(deaths ~ country + offset(log(yrs_at_risk)), design=svy, data=country, family=poisson()) summary(svyglm) # point estimates are correct, but standard error is way too large regTermTest(svyglm, ~ country) # test indicates no country differences __ 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. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907 __ 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] Nelder-Mead with output of simplex vertices
-- begin included Greetings! I want to follow the evolution of a Nelder-Mead function minimisation (a function of 2 variables). Hence each simplex will have 3 vertices. Therefore I would like to have a function which can output the coordinates of the 3 vertices after each new simplex is generated. However, there seems to be no way (which I can detect) of extracting this information from optim() (the 'trace' argument to 'control' does not seem to have provision for this, according to '?optim', and I have tried it out without success). --- end include - Why not put a cat() statement into fn, the function that you supply which optim is calling? That will give the vertices that it tries one by one. Terry T. __ 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] svy / weighted regression
I think you are missing the point. You have 4 zero death counts associated with much higher person years of exposure followed by 4 death counts in the thousands associated with lower degrees of exposures. It seems unlikely that these are real data as there are not cohorts that would exhibit such lower death-rates. So it appears that in setting up your test case, you have created an impossibly unrealistic test problem. -- David On Oct 12, 2009, at 9:12 AM, Laust wrote: Dear Peter, Thanks for the input. The zero rates in some strata occurs because sampling depended on case status: In Finland only 50% of the non-cases were sampled, while all others were sampled with 100% probability. Best Laust On Sat, Oct 10, 2009 at 11:02 AM, Peter Dalgaard p.dalga...@biostat.ku.dk wrote: Sorry, forgot to reply all... Laust wrote: Dear list, I am trying to set up a propensity-weighted regression using the survey package. Most of my population is sampled with a sampling probability of one (that is, I have the full population). However, for a subset of the data I have only a 50% sample of the full population. In previous work on the data, I analyzed these data using SAS and STATA. In those packages I used a propensity weight of 1/[sampling probability] in various generalized linear regression-procedures, but I am having trouble setting this up. I bet the solution is simple, but I’m a R newbie. Code to illustrate my problem below. Hi Laust, You probably need the package author to explain fully, but as far as I can see, the crux is that a dispersion parameter is being used, based on Pearson residuals, even in the Poisson case (i.e. you effectively get the same result as with quasipoisson()). I don't know what the rationale is for this, but it is clear that with your data, an estimated dispersion parameter is going to be large. E.g. the data has both 0 cases in 75 person-years and 1000 cases in 5000 person-years for Denmark, and in your model they are supposed to have the same Poisson rate. summary.svyglm starts off with est.disp - TRUE and AFAICS there is no way it can get set to FALSE. Knowing Thomas, there is probably a perfectly good reason not to just set the dispersion to 1, but I don't get it either... Thanks Laust # loading survey library(survey) # creating data listc - c (Denmark ,Finland,Norway,Sweden,Denmark,Finland,Norway,Sweden) listw - c(1,2,1,1,1,1,1,1) listd - c(0,0,0,0,1000,1000,1000,2000) listt - c(75,50,90,190,5000,5000,5000,1) list.cwdt - c(listc, listw, listd, listt) country - data .frame(country=listc,weight=listw,deaths=listd,yrs_at_risk=listt) # running a frequency weighted regression to get the correct point estimates for comparison glm - glm(deaths ~ country + offset(log(yrs_at_risk)), weights=weight, data=country, family=poisson()) summary(glm) regTermTest(glm, ~ country) # running survey weighted regression svy - svydesign(~0,,data=country, weight=~weight) svyglm - svyglm(deaths ~ country + offset(log(yrs_at_risk)), design=svy, data=country, family=poisson()) summary(svyglm) # point estimates are correct, but standard error is way too large regTermTest(svyglm, ~ country) # test indicates no country differences __ 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. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907 __ 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 Heritage Laboratories 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] Nelder-Mead with output of simplex vertices
On 12-Oct-09 13:24:01, Terry Therneau wrote: -- begin included Greetings! I want to follow the evolution of a Nelder-Mead function minimisation (a function of 2 variables). Hence each simplex will have 3 vertices. Therefore I would like to have a function which can output the coordinates of the 3 vertices after each new simplex is generated. However, there seems to be no way (which I can detect) of extracting this information from optim() (the 'trace' argument to 'control' does not seem to have provision for this, according to '?optim', and I have tried it out without success). --- end include - Why not put a cat() statement into fn, the function that you supply which optim is calling? That will give the vertices that it tries one by one. Terry T. That's neat and simple! It hadn't occurred to me. Thanks! Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 12-Oct-09 Time: 14:32:08 -- XFMail -- __ 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 and computer size
Vibe Henriette Skov wrote: Dear R-mailing list Hope you can help me. I am using R for windows to analyze my 107 HGU133Plus2.0 chips, however, R chrash when I try to use ReadAffy(). I want to buy a computer that can handle all these arrays, do you know how big a computer I need to buy? Hi Skov -- this is a Bioconductor question, so ask on the Bioconductor mailing list. http://bioconductor.org/docs/mailList.html There are alternatives to affy::ReadAffy that might work on your current machine, including affy::justRMA, the aroma.affymetrix package, and the xps package (which requires a system dependency that some find difficult to install). Others on the bioc list might have more guidance on memory requirements on 64 bit linux systems, which is what you'll want for large data analysis. Martin Best, Skov, Denmark [[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. -- Martin Morgan Computational Biology / Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: Arnold Building M1 B861 Phone: (206) 667-2793 __ 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] Error: cannot allocate vector of size 1.2 Gb
That's the file I needed, and the problem I expected. The factor you
specify is not a factor, but a numerical variable. Even more, if you
tabulate it, you have 2 times the values 1, 2 and 5, and all
other values only once. That's what gives you the error.
p is an internal variable of the function specaccum. It's the number
of columns in a dataframe containing only locations with a certain
value for the specified factor. If there is only one location with
that factor level, you get only one row and thus a vector. The
function ncol() gives result NULL if you try it on a vector, so p is
NULL, and has zero length.
You can for example convert HM_sprem to a factor, indicating low and
high values.
PoEnv$test
-factor(ifelse(PoEnv$HM_sprem2.5,low,high),levels=c(low,high),ordered=T)
poacc2 - accumcomp(PoCom, y=PoEnv, factor=test, method=exact)
works like a charm for me.
In the future, check carefully which type of arguments are asked for,
and use the function str() to check if they really are what you think
they are.
Kind regards
Joris
-- Forwarded message --
From: romunov romu...@gmail.com
Date: Mon, Oct 12, 2009 at 3:14 PM
Subject: Re: [R] Error: cannot allocate vector of size 1.2 Gb
To: joris meys jorism...@gmail.com
I hope this workspace file attached is what you were looking for.
Cheers,
Roman
On Mon, Oct 12, 2009 at 2:07 PM, joris meys jorism...@gmail.com wrote:
It's easier if you just load the data in R, save the workspace and
send that one. I really don't have time to read in csv data and find
my way trough it, sorry. It also makes it possible for me to find
errors due to the wrong data format (factors that are not defined as
factor and the likes).
Kind regards
Joris
On Mon, Oct 12, 2009 at 2:04 PM, romunov romu...@gmail.com wrote:
Hi again,
thank you for your time, I really appreciate you taking time dealing with my
problem. I am sending you my dataset (community (zdruzbe_analiza.csv) and
environmental (okoljski_analiza.csv) factors) in case you want to try it out
for yourself.
Cheers,
Roman
On Mon, Oct 12, 2009 at 2:02 PM, romunov romu...@gmail.com wrote:
Hi Joris,
thanks for spotting that one. This little mistake has gotten in when I was
trying desperate things with the analysis (factor1 is used in
diversitycomp). Nevertheless, here is the result:
poacc2 - accumcomp(PoCom, y=PoEnv, factor=HM_sprem, method=exact)
Error in if (p == 1) { : argument is of length zero
And the traceback():
traceback()
3: specaccum(x, method = method, permutations = permutations, conditioned
= conditioned,
gamma = gamma, ...)
2: accumresult(x, y, factor, level = levels[i], scale = scale, method =
method,
permutations = permutations, conditioned = conditioned, gamma =
gamma)
1: accumcomp(PoCom, y = PoEnv, factor = HM_sprem, method = exact)
Best wishes,
Roman
On Mon, Oct 12, 2009 at 1:56 PM, joris meys jorism...@gmail.com wrote:
Hi Roman,
that throws a different light on the problem. It goes wrong from the
start, so it has little to do with the bootstrap or jackknife
procedures. R.huge won't help you either.
Likely your error comes from the fact that factor1 is not an
argument of the function accumcomp. the argument is factor. As R
doesn't find this, it'll try to tabulate the complete environmental
dataset, and this gives the memory overflow.
Try :
poacc2 - accumcomp(PoCom, y=PoEnv, factor=HM_sprem, method=exact)
That should work. I can't try it out without dataset off course. If it
doesn't work, post the traceback again, I'll take another look.
Kind regards
Joris
On Mon, Oct 12, 2009 at 1:07 PM, romunov romu...@gmail.com wrote:
Hello joris,
this is the traceback() output. Hopefully you can make some sense out
of it.
Thank you for the tips as well (R.huge looks promising)!
traceback()
7: vector(integer, length)
6: integer(nbins)
5: tabulate(bin, pd)
4: as.vector(data)
3: array(tabulate(bin, pd), dims, dimnames = dn)
2: table(y[, factor])
1: accumcomp(PoCom, y = PoEnv, factor1 = HM_sprem, method = exact)
Cheers,
Roman
On Mon, Oct 12, 2009 at 12:52 PM, joris meys jorism...@gmail.com
wrote:
Dear Roman,
could you give us the trace given by traceback() ? I suspect the error
is resulting from the permutations and/or jackknife procedure in the
underlying functions specaccum and specpool.
You can take a look at the package R.huge, but that one is deprecated
already. There are other packages around too, but I have no experience
with them. You find some more tips here :
http://www.matthewckeller.com/html/memory.html
This should give you a place to start looking.
Kind regards
Joris
On Mon, Oct 12, 2009 at 11:39 AM, romunov romu...@gmail.com wrote:
Dear List,
today I turn to you with a next problem. I'm trying to compare
species
[R] RPostgreSQL and needed .dlls
Dear List, I am trying to connect from R 2.9.2 on Win XP SP3 to a remotely installed PostgreSQL DB (8.3.7 on Ubuntu Server 9.04). Everything seems to be properly installed, as I can connect to the DB from within Excel and RKWard (running on another machine). But regarding R on the Win XP, I cannot load RPostgreSQL. libpq.dll is missing. What can I do now, without having to install PostgreSQL on this Windows, where I do not need it as I have my remote DB? Thanks in advance __ 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] need help
Sample Size λ=5     α = 4      β = 3 Min. 1St Qu. Median Mean 3rd Qu. Max. 100 0.00 1.740638 4.040032 4.433828 5.607589 22.450405 500 0.00 2.212375 3.915889 4.750014 6.356894 22.860806 1000 0.00 2.295969 4.207002 5.054784 6.763213 27.419783 5000 0.00 2.288180 4.060344 4.997187 6.747790 53.890014 1 0.00 2.323913 4.165010 5.055568 6.787189 45.338561 Sample Size λ =10     α = 4      β = 3 Min. 1St Qu. Median Mean 3rd Qu. Max. 100 1.796326 5.634923 8.586495 9.432652 12.250539 26.535398 500 0.1435474 6.2096963 9.352632 9.9507770 12.5914827 39.0010363 1000 0.1491454 5.9161795 .6888265 9.8011127 2.7219316 35.0477795 5000 0.1116321 6.1443172 9.072187 10.0330122 12.6700472 65.7772109 1 0.00 6.116205 9.044711 10.004620 12.767728 54.560908 Sample Size λ =20     α = 4      β = 3 Min. 1St Qu. Median Mean 3rd Qu. Max. 100 4.825701 14.720441 17.41218 19.743268 23.279030 63.946636 500 6.305691 14.132502 18.90554 19.977387 24.180544 54.408182 1000 4.350899 14.581057 18.96924 20.261373 24.740963 52.379454 5000 2.07920 14.73828 19.20401 20.19534 24.52991 72.49832 1 2.070761 14.458651 19.12765 20.035136 24.329874 138.388379 Sample Size λ =30     α = 4      β = 3 Min. 1St Qu. Median Mean 3rd Qu. Max. 100 9.783008 23.858371 29.60942 30.458481 36.540193 50.774556 500 9.686187 24.006261 29.14230 30.520085 35.921312 69.158839 1000 8.78857 23.50711 28.47166 29.69044 34.83602 72.65655 5000 8.248564 23.094087 28.75213 29.722024 35.276848 90.079316 1 4.508184 23.422685 28.96963 30.044882 35.502004 107.360714   My friends, group R . I hope you help me to comment on these results? And how can compute the variance of  Fs.  model.freq=expression(data=rpois(30)) model.sev=expression(data=rpareto(30,30)) Fs=aggregateDist(simulation,nb.simul=1000,model.freq,model.sev)  Thanks  Mohd PhD Student  Malaysia [[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] function: ploting an igraph object within lattice
Hi, I would like to be able to develop a function to plot an igraph object
with lattice (trellis type displays will be usefull for grouping etc).
Anyway, I mostly feeble
Igraph requires that you convert two columns of data two an igraph object
and to be able to plot the graph...I have tried a very, very simplistic (if
not naive) approach and surprise, surprise, it didn't work.
First of all, does anyone know any good tutorial packages/books that will
help someone like me unused to programming develop their function writing
skills, and secondly, can anyone suggest a way to generate an igraph object
in lattice.
Thanks
Andrew
##generates the igraph plot
library(igraph)
LE-read.csv(LE.csv)
LG-graph.data.frame(LE, directed=F)
LG$layout-layout.kamada.kawai(LG)
plot(LG, vertex.size=4, vertex.label=NA, vertex.color=red,
vertex.shape=square, edge.color=blue4)
##
##
library(lattice)
##
##
xyplot.igraph - function(plot.igraph, ...) {
dat - graph.data.frame(x = object, directed=FALSE)
prepanel=prepanel.igraph
}
prepanel.igraph - function(x, axes=FALSE, xlab=, ylab=, add=FALSE,
xlim=c(-1,1), ylim=c(-1,1), main=, sub=)
http://www.nabble.com/file/p25855220/LE.csv LE.csv
--
View this message in context:
http://www.nabble.com/function%3A-ploting-an-igraph-object-within-lattice-tp25855220p25855220.html
Sent from the R help mailing list archive at Nabble.com.
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[R] R commander.
i have two RData files,,i want to print them to check the format of the
tables in these files,,,i can load both the files and can read it as well
load('ann.RData')
str(ann)
List of 4
$ Name : chr [1:561466] rs3094315 rs12562034 rs3934834
rs9442372 ...
$ Position : int [1:561466] 742429 758311 995669 1008567 1011278 1011521
1020428 1021403 1038818 1039813 ...
$ Chromosome: chr(0)
$ Chr.num : num [1:561466] 1 1 1 1 1 1 1 1 1 1 ...
but when i try to display all the table by using the R commander.i have got
the display of 'pheno.RData' file,,but the other file 'ann.RData' show me an
error i.e.
ann - as.data.frame(ann)
Error in data.frame(Name = c(rs3094315, rs12562034, rs3934834,
rs9442372, :
arguments imply differing number of rows: 561466, 0
i am sending you both files,,,hope u will help me solve this problm
Ilyas
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to output profile plots for groups using lattice package
Hi Peter (and anyone else willing to help me out), Many thanks for your help. Having used your code plus a few other modifications, I only get the points plotted but without the two lines. I just cannot figure out what the problem is. My code is as follows: library(lattice) datos2 - subset(datos, samplesize != 10 parm != Theta0) unq - sort(unique(datos2$samplesize)) datos2$fsamplesize - factor(datos2$samplesize, labels = paste(Sample size =, unq)) datos2$parm - factor(datos2$parm, levels = c(Intercept, time, trt, time*trt)) tp1.sim - xyplot(MSE ~ ntimes | fsamplesize + parm, group = group, data = datos2, type = b, lty = 1:2, pch = 1:2, scales = list(x = list(at = c(2, 4, 8, 16)), alternating = 1), as.table = TRUE, key = list(text = list(c(GNA, PNA)), points = list(pch = 1:2)) ) plot(tp1.sim) I have attached my real dataset (called datos) as well. Kind appreciations to your efforts. George On Wed, Oct 7, 2009 at 9:20 AM, Peter Ehlers ehl...@ucalgary.ca wrote: see below George Kalema wrote: Dear R users, I am trying to have an xyplot of a data set which has the following variables: case (n=10,20,30) parameter (parm=a,b) group (grp=g1,g2) y (y values) x (x=2,4,8) My plot should be parameter by case such that I have 2 rows (each row= each parameter) and 3 columns (each column=each case). My R-code is as follows but I am not able to get what I want to: tp1.sim - xyplot(y~ x | case + parm , group=group, data = data, lty = 1:4 , pch = 1:4) print(tp1.sim) How can I have two lines (for g1 and g2) in each plot (each box)? include the type=b argument How do I label the x-axis with only values 2, 4, 8? include the scales= argument or make x a factor How do I label each column with the corresponding case number? make 'case' a factor The following should do what you want: xyplot(y ~ x | factor(case) + parm, group=group, data=data, type='b', lty=1:2, pch=1:2, scales=list(x=list(at=c(2,4,8))) ) I don't understand why you want 4 line types/point chars. -Peter Ehlers My hypothetical data set is as follows: parm x case y group a 2 10 0.03 g1 b 2 10 0.02 g1 a 4 10 0.03 g1 b 4 10 0.02 g1 a 8 10 0.03 g1 b 8 10 0.02 g1 a 2 20 0.03 g1 b 2 20 0.02 g1 a 4 20 0.03 g1 b 4 20 0.02 g1 a 8 20 0.03 g1 b 8 20 0.02 g1 a 2 30 0.03 g1 b 2 30 0.02 g1 a 4 30 0.03 g1 b 4 30 0.02 g1 a 8 30 0.03 g1 b 8 30 0.02 g1 a 2 10 0.13 g2 b 2 10 0.12 g2 a 4 10 0.13 g2 b 4 10 0.12 g2 a 8 10 0.13 g2 b 8 10 0.12 g2 a 2 20 0.13 g2 b 2 20 0.12 g2 a 4 20 0.13 g2 b 4 20 0.12 g2 a 8 20 0.13 g2 b 8 20 0.12 g2 a 2 30 0.13 g2 b 2 30 0.12 g2 a 4 30 0.13 g2 b 4 30 0.12 g2 a 8 30 0.13 g2 b 8 30 0.12 g2 Many thanks in advance for your response. George [[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. -- George Williams KALEMA, Schapenstraat 37/282, 3000 Leuven, Belgium. Cell: +32 495 33 13 02 __ 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 diff, ifelse on zoo object
Hi, I'm having an issue when using diff and ifelse on a zoo object. x.Date - as.Date(2003-02-01) + c(1, 3, 7, 9, 14) - 1 x - zoo(rnorm(5), x.Date) x.POS - c(0,0,0,1,1) x- merge(x,x.POS) x x x.POS 2003-02-01 -0.1858136 0 2003-02-03 -1.3188533 0 2003-02-070.2709794 0 2003-02-09 -1.4915262 1 2003-02-140.5014170 1 When I create this new zoo object using the previous one (x) I don't get exactly what I need..the traded rate is based on the lagged values, rather than the present ones... TradedRate - ifelse(abs(diff(x[,x.POS],lag= 1))0,ifelse(x[,x.POS] !=1,-x[,x],x[,x]),NA) x - merge(x, TradedRate, all=TRUE) x x x.POS TradedRate 2003-02-01 -0.1858136 0 NA 2003-02-03 -1.3188533 0 NA 2003-02-07 0.2709794 0 NA 2003-02-09 -1.4915262 1 -0.2709794 2003-02-14 0.5014170 1 NA The value for TradedRate on the 9th Feb should be -1.4945262 instead of -0.2709794what am I doing wrong? Thanks very much.. -- View this message in context: http://www.nabble.com/Using-diff%2C-ifelse-on-zoo-object-tp25852822p25852822.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] Vector Allocation -problem
When i tried running Zip-sqldf(select a.*,b.s_lv from Zip a inner join lin b on a.S=B.S) Error: cannot allocate vector of size 15.6 Mb and with following warning Warning messages: 1: In as.list.data.frame(X) : Reached total allocation of 1024Mb: see help(memory.size) 2: In as.list.data.frame(X) : Reached total allocation of 1024Mb: see help(memory.size) 3: In sqliteFetch(rs, n = -1, ...) : Reached total allocation of 1024Mb: see help(memory.size) 4: In sqliteFetch(rs, n = -1, ...) : Reached total allocation of 1024Mb: see help(memory.size) 5: In sqliteFetch(rs, n = -1, ...) : Reached total allocation of 1024Mb: see help(memory.size) 6: In sqliteFetch(rs, n = -1, ...) : Reached total allocation of 1024Mb: see help(memory.size) 7: In sqliteFetch(rs, n = -1, ...) : Reached total allocation of 1024Mb: see help(memory.size) 8: In sqliteFetch(rs, n = -1, ...) : Reached total allocation of 1024Mb: see help(memory.size) 9: In sqliteFetch(rs, n = -1, ...) : Reached total allocation of 1024Mb: see help(memory.size) 10: In sqliteFetch(rs, n = -1, ...) : Reached total allocation of 1024Mb: see help(memory.size) 11: In sqliteFetch(rs, n = -1, ...) : Reached total allocation of 1024Mb: see help(memory.size) 12: In sqliteFetch(rs, n = -1, ...) : Reached total allocation of 1024Mb: see help(memory.size) memory.size() [1] 966.67 If i'm right the memory size shows 1Gb but it was not able to allocate 15mb.Help me to resolve this -- View this message in context: http://www.nabble.com/Vector-Allocation--problem-tp25853258p25853258.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] tktext-window smaller than text
Hi there, I need to build up a tktext-widget that contains a longer text than the tktext-widget actually is. So what I mean is, that the tktext window is of width=100 and the text in it has a length greater 100. But I don't want the window to just wrap the line, but to belong to scrollbar that is able to scroll to the end of the text. tt - tktoplevel() txt - tktext(tt, width=100, height=30) scrergVert - tkscrollbar(tt, orient = vertical, repeatinterval = 1, command = function(...) tkyview(txt, ...)) tkconfigure(txt, yscrollcommand = function(...) tkset(scrergVert, ...)) scrergHor - tkscrollbar(tt, orient = horizontal, repeatinterval = 1, command = function(...) tkxview(txt, ...)) tkconfigure(txt, xscrollcommand = function(...) tkset(scrergHor, ...)) tkgrid(txt, column=0, row=0, sticky=nwse) tkgrid(scrergVert, column=1, row=0, sticky=ns) tkgrid(scrergHor, column=0, row=1, sticky=we) tkinsert(txt, 0.0, paste(capture.output(resultKmean), collapse=\n)) Trying the upper code would wrap the lines that are contain more than 100 characters. How can I keep the size of the window, but make the text scrollable? Thanks in advance, -- Anne Skoeries __ 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] Ordinal response model
I have been asked to analyse some questionnaire data- which is not data I'm that used to dealing with. I'm hoping that I can make use of the nabble expertise (again). The questionnaire has a section which contains a particular issue and then questions which are related to this issue (and potentially to each other): 1) importance of the issue (7 ordinal categories from -3 to +3) 2) impact of the impact (7 ordinal categroies from -3 to +3) 3) percentage affected by the issue (11 ordinal categories from 0, 0-10, 20-30, 30-40.90-100) I also have three participant predictive factors: Gender (M/F) Age (continuous scale) Sector (6 nominal categories) So that my data looks like this: genagesector impac importa percen 1 1 59 4 0 -3 2 2 2 64 3 2 -3 2 3 1 83 6 3 -3 2 4 1 66 5 2 -2 2 5 1 79 5 0 -1 2 6 2 63 4 0 -1 2 I have 2 questions I want my analysis to answer 1) does gender/ age/ sector affect importance, impact, reponse 2) are importance/impact/response correlated in some way I'm thinking I need some ordered logistic or probit model (possibly using polr() command). The problem is the multivariate aspect- I need importance, impact and perecentage in the same model to look at the covariance between them and affects of gender, age and sector on these covariances. It would be good to include a latent variable- at least for the perecentage factor. Any help would be very much appreciated. -- View this message in context: http://www.nabble.com/Ordinal-response-model-tp25856728p25856728.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] R commander.
You seem to have gotten an extraneous list item called Chromosome to
which an empty string has been assigned. What happens if you issue
this command:
ann2 - data.frame( Name=ann$Name, Position=ann$Position,
Chr.num=factor(ann$Chr.num) )
(I took the liberty of making Chr.num into a factor.) And then print
out smaller segments of ann2 (since what you have would take many
hundreds of pages.
ann2[ , 1:60]
--
David
On Oct 12, 2009, at 2:29 AM, Ilyas . wrote:
i have two RData files,,i want to print them to check the format of
the
tables in these files,,,i can load both the files and can read it as
well
load('ann.RData')
str(ann)
List of 4
$ Name : chr [1:561466] rs3094315 rs12562034 rs3934834
rs9442372 ...
$ Position : int [1:561466] 742429 758311 995669 1008567 1011278
1011521
1020428 1021403 1038818 1039813 ...
$ Chromosome: chr(0)
$ Chr.num : num [1:561466] 1 1 1 1 1 1 1 1 1 1 ...
but when i try to display all the table by using the R commander.i
have got
the display of 'pheno.RData' file,,but the other file 'ann.RData'
show me an
error i.e.
ann - as.data.frame(ann)
Error in data.frame(Name = c(rs3094315, rs12562034, rs3934834,
rs9442372, :
arguments imply differing number of rows: 561466, 0
i am sending you both files,,,hope u will help me solve this
problm
Ilyas
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David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] field index given name.
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of tdm
Sent: Monday, October 12, 2009 12:22 AM
To: r-help@r-project.org
Subject: Re: [R] field index given name.
Thanks - would never have guessed that. I eventually got the
following to do
what I want...
colprob - array(dim=NCOL(iris))
for(i in 1:NCOL(iris)){
+ colprob[i]=
+ ifelse(names(iris)[i] == 'Species',1,0.5)
+ }
colprob
[1] 0.5 0.5 0.5 0.5 1.0
I think a more direct way to do what that does
is to make vector with names, where the names
are the column names of 'iris' and then subscript
the vector by name instead of by number. E.g.,
colprob - rep(0.5, length=ncol(iris)) # initialize probs
names(colprob) - colnames(iris) # initialize names
colprob[Species] - 1.0 # change prob for Species
colprob
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
0.5 0.5 0.5 0.5 1.0
colprob[Sepal.Width] # example prob for Sepal.Width
Sepal.Width
0.5
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Schalk Heunis-2 wrote:
Hi Phil
Try the following
which(names(iris)=='Species')
[1] 5
HTH
Schalk Heunis
On Mon, Oct 12, 2009 at 8:53 AM, tdm ph...@philbrierley.com wrote:
Hi,
How do I access the index number of a field given I only
know the field
name?
eg - I want to set the probability of the field 'species'
higher than the
other fields to use in sampling.
colprob - array(dim=NCOL(iris))
for(i in 1:NCOL(iris)){colprob[i]=0.5}
colprob[iris$species] = 1 #this doesn't work
colprob
[1] 0.5 0.5 0.5 0.5 0.5
--
View this message in context:
http://www.nabble.com/field-index-given-name.-tp25851216p25851216.html
Sent from the R help mailing list archive at Nabble.com.
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--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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[R] xyplot does not find variable in data
When we call a lattice function such as xyplot, to what extent does the data designation cause the function to look inside the data for variables? In the examples below, the subset argument understands that Variety is a variable in the data. But the scales argument does not understand that nitro is a variable in the data. What principle is at work? library(MEMSS) # The following works fine: xyplot( yield ~ nitro , data=Oats , scales=list( x=list( at=unique(Oats$nitro) ) ) , subset=Variety==Victory ) # But the following returns an error: xyplot( yield ~ nitro , data=Oats , scales=list( x=list( at=unique(nitro) ) ) ) Thanks for any insight Jacob A. Wegelin Assistant Professor Department of Biostatistics Virginia Commonwealth University 730 East Broad Street Room 3006 P. O. Box 980032 Richmond VA 23298-0032 U.S.A. E-mail: jwege...@vcu.edu URL: http://www.people.vcu.edu/~jwegelin __ 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] combining/overlaying boxplot and barplot
Dear all, I would like to visualise when days are rainy or dry in bar/boxplots. Therefore I've tried to combine raingauge data (boxplots) and percentage of raingauges with 0mm measurements (barplot). See attachment mei2004.pdf (if it came through). I've come this far: barplot(dcp0[monthindex], col=gray, border=NA, axes=F) boxplot(t(dcpn[monthindex,]), names=dates$day[monthindex], main=month, outline=F, add=T, ylim=c(0,30), col=lightblue) Both plots use different x-axis spacing, so May 31 of the barplot extents further to the right than May 31 boxplot. So, how do I get the x-axis of both plots aligned?? I've looked at par() but can't see any parameters for plot size. Only for inner and outer margins. The strange thing is, when I plot the boxplot (add=F), it uses more x-axis space (on the same device), the graph becomes wider (with add=T, it does seem to take the numbers from the barplot in to account). Thanks for the help! Ivan Soenario KNMI Royal Netherlands Meteorological Institute ps dcpn: data.frame containing raingauge data for 310 stations (columns) for 1991-2006 (each day is a row) dcp0: calculation of % 0mm, based on dcpn. monthindex: array with rownumbers, in this case representing month May in 2004 (so length(monthindex) is 31) ps 2 I cheated with percentage, I multiplied the fraction (number of 0mm)/(total number raingauges) with 30, to make it fit with the boxplot's ylim of 30. It seldomly rains more than 30mm in 24hours. To overcome this, I'd need to draw the barplot with it's own y-axis, and, the boxplot shouldn't notice/use the ylim from the barplot. __ 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] spectral analysis
On Oct 12, 2009, at 5:24 AM, sdlywjl666 wrote: Dear all£¬ Is there some functions to estimate the spectrum by the fft of autocorrelation? The blindingly obvious search terms: fft spectrum autocorrelation ... in two hops to what appears to be a function to this: Auto And/Or Cross Correlations via FFT: http://finzi.psych.upenn.edu/R/library/timsac/html/fftcor.html Is there Parzen's lag window in R ? Ditto for the search terms: Parzen window http://finzi.psych.upenn.edu/R/library/sapa/html/SDF.html http://finzi.psych.upenn.edu/R/library/cyclones/html/filters.html and perhaps: http://finzi.psych.upenn.edu/R/library/fracdiff/html/fdSperio.html R-search is your friend:(Not to mention a Posting-Guide- recommended step before posting.) http://search.r-project.org/nmz.html -- David Winsemius, MD Heritage Laboratories 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 with the use of mtext to create main title over multiple plots
I'm trying to use mtext to create a main title over multiple plots. Below is
a simple self-contained example and my sessionInfo (I should note I've also
tried this with R-2.8.1 with the same results). When I execute the code
chunk below, I get the plots, but no title. I've tried this using the screen
driver, pdf, and postscript. I've used different sizes of paper. I suspect I
am making an elementary error but searching the help files and help archives
hasn't provided me an answer.
Thanks for any help, Mark
#
setwd(~/Desktop)
pdf(my.test.plots.pdf, paper = letter)
par(mfrow=c(2,2))
for (i in 1:4){
plot(1:6, 1:6)
}
mtext(text = my test plots, side = 3, outer = TRUE)
dev.off()
#
R version 2.10.0 Under development (unstable) (2009-09-21 r49771)
x86_64-unknown-linux-gnu
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] car_1.2-15
loaded via a namespace (and not attached):
[1] tools_2.10.0
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University School of Medicine
15032 Hunter Court, Westfield, IN 46074
(317) 490-5129 Work, Mobile VoiceMail
(317) 399-1219 Skype No Voicemail please
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with the use of mtext to create main title over multiple plots
Try playing around with the oma setting in par() -- it sets the outer
margins, which by default are zero.
The following shows the mtext label for me, using the windows device:
par(mfrow=c(2,2))
par(oma)
[1] 0 0 0 0
par(oma=c(0,0,2,0))
for (i in 1:4) plot(0:1,0:1)
mtext(text = my test plots, side = 3, outer = TRUE)
Mark Kimpel wrote:
I'm trying to use mtext to create a main title over multiple plots. Below is
a simple self-contained example and my sessionInfo (I should note I've also
tried this with R-2.8.1 with the same results). When I execute the code
chunk below, I get the plots, but no title. I've tried this using the screen
driver, pdf, and postscript. I've used different sizes of paper. I suspect I
am making an elementary error but searching the help files and help archives
hasn't provided me an answer.
Thanks for any help, Mark
#
setwd(~/Desktop)
pdf(my.test.plots.pdf, paper = letter)
par(mfrow=c(2,2))
for (i in 1:4){
plot(1:6, 1:6)
}
mtext(text = my test plots, side = 3, outer = TRUE)
dev.off()
#
R version 2.10.0 Under development (unstable) (2009-09-21 r49771)
x86_64-unknown-linux-gnu
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] car_1.2-15
loaded via a namespace (and not attached):
[1] tools_2.10.0
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University School of Medicine
15032 Hunter Court, Westfield, IN 46074
(317) 490-5129 Work, Mobile VoiceMail
(317) 399-1219 Skype No Voicemail please
[[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.
__
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 with the use of mtext to create main title over multiple plots
On Oct 12, 2009, at 1:41 PM, Tony Plate wrote:
Try playing around with the oma setting in par() -- it sets the
outer margins, which by default are zero.
The following shows the mtext label for me, using the windows device:
par(mfrow=c(2,2))
par(oma)
[1] 0 0 0 0
par(oma=c(0,0,2,0))
for (i in 1:4) plot(0:1,0:1)
mtext(text = my test plots, side = 3, outer = TRUE)
Good advice. Works on a Mac, too:
setwd(~/Desktop)
pdf(my.test.plots.pdf, paper = letter)
opar - par(mfrow=c(2,2), oma=c(0,0,2,0))
for (i in 1:4){
plot(1:6, 1:6);
}
mtext(text = my test plots, side = 3, outer = TRUE)
dev.off(); par(opar)
Mark Kimpel wrote:
I'm trying to use mtext to create a main title over multiple plots.
Below is
a simple self-contained example and my sessionInfo (I should note
I've also
tried this with R-2.8.1 with the same results). When I execute the
code
chunk below, I get the plots, but no title. I've tried this using
the screen
driver, pdf, and postscript. I've used different sizes of paper. I
suspect I
am making an elementary error but searching the help files and help
archives
hasn't provided me an answer.
Thanks for any help, Mark
#
setwd(~/Desktop)
pdf(my.test.plots.pdf, paper = letter)
par(mfrow=c(2,2))
for (i in 1:4){
plot(1:6, 1:6)
}
mtext(text = my test plots, side = 3, outer = TRUE)
dev.off()
#
R version 2.10.0 Under development (unstable) (2009-09-21 r49771)
x86_64-unknown-linux-gnu
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] car_1.2-15
loaded via a namespace (and not attached):
[1] tools_2.10.0
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University School of Medicine
15032 Hunter Court, Westfield, IN 46074
(317) 490-5129 Work, Mobile VoiceMail
(317) 399-1219 Skype No Voicemail please
[[alternative HTML version deleted]]
__
--
David Winsemius, MD
Heritage Laboratories
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] help with the use of mtext to create main title over multiple plots
Mark Kimpel wrote:
I'm trying to use mtext to create a main title over multiple plots. Below
is
a simple self-contained example and my sessionInfo (I should note I've
also
tried this with R-2.8.1 with the same results). When I execute the code
...
Thanks for your nice example showing the problem clearly. I normally prefer
to test these things in a window first, it's faster, though.
You had forgotten to give the poor graphics a bit of outer margin. If you
look carefully, you could have seen a few tail in the plot. The example
overreacts a bit, but you get the idea.
You might also reduce your plot margins a bit (see par) to avoid to large
empty space.
Dieter
par(mfrow=c(2,2),oma=c(10,10,10,10)
for (i in 1:4){
plot(1:6, 1:6)
}
mtext(text = my test plots, side = 1, outer = TRUE)
--
View this message in context:
http://www.nabble.com/help-with-the-use-of-mtext-to-create-main-title-over-multiple-plots-tp25859951p25860178.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] help with the use of mtext to create main title over multiple plots
Hey Mark, The text is actually there -- I can just see the bottom of the 'y' and the 'p' in my plotting window. You can move the text down (into the plot) with the argument line. E.g.: mtext(text = my test plots, side = 3, outer = TRUE, line=-2) Hope that helps... - Simon Bonner Post-Doctoral Fellow Department of Statistics, UBC www.simon.bonners.ca On Mon, 2009-10-12 at 13:29 -0400, Mark Kimpel wrote: mtext(text = my test plots, side = 3, outer = TRUE) __ 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] add lines() to 1st plot in layout() after calling 2nd plot()?
Since you are using LaTeX, you might consider creating the plots all in 1 step, but use the tikzdevice, this provides LaTeX (pgf) commands to create the plot that you can then insert LaTeX commands to built up the plot bit by bit. You could also look at using the subplot function in the TeachingDemos package, start with a blank plot without margins, then use subplot in place of layout to position your graphs, the examples in the help file for subplot show how to add to the subplot after creating it. Another option is to create a function that creates each whole plot, but with an argument and some if statements that will skip certain steps, then just do a for loop calling the functions to build the plots in pieces. This means replotting from scratch each step, but save these as graphics files and it is not overly complex. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Marianne Promberger Sent: Saturday, October 03, 2009 1:15 PM To: r-help Subject: [R] add lines() to 1st plot in layout() after calling 2nd plot()? Dear R users, I create a graphic with two plots side by side using layout(), like this: layout(matrix(c(1,2),1)) plot(1:10,main=left plot) lines(c(3:7,7:3),col=red) plot(10:1,main=right plot) The lines() obivously get added to the left plot plot. Now, I'm trying to write a function that builds up a plot bit by bit to then include it in a LaTeX presentation with overlays. I'm using dev.copy(), and it would make my life much easier (because in fact I call all sorts of additional axis() etc after plot) if I could call the above commands in this order: layout(matrix(c(1,2),1)) plot(1:10,main=left plot) plot(10:1,main=right plot) lines(c(3:7,7:3),col=red) but of course now lines() gets added to the right plot. I Is there any way to make the lines() go to the fist plot (left plot)? Marianne -- Marianne Promberger PhD, King's College London http://promberger.info R version 2.9.2 (2009-08-24) Ubuntu 9.04 __ 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] Ordinal response model
drlucyasher wrote: The questionnaire has a section which contains a particular issue and then questions which are related to this issue (and potentially to each other): 1) importance of the issue (7 ordinal categories from -3 to +3) 2) impact of the impact (7 ordinal categroies from -3 to +3) 3) percentage affected by the issue (11 ordinal categories from 0, 0-10, 20-30, 30-40.90-100) I also have three participant predictive factors: Gender (M/F) Age (continuous scale) Sector (6 nominal categories) Gender and Sector are clear; convert these to factors, preferably giving them meaningful names (m/f, east, west), and everything will be treated correctly by most r function. Age is also clear, leave as is. There will be considerably discussion how to code the scores. If these are not heavily skewed (all -3), in some fields it is accepted to treat these as continuous. Frank Harrell would argue against it. I have revised too many manuscripts in both directions, so my opinion depends on the paper where you publish it. Anyway, Frank Harrel's lrm in Design might give you a starter. There is also a well-known book by him on the subject. Dieter -- View this message in context: http://www.nabble.com/Ordinal-response-model-tp25856728p25860439.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] add lines() to 1st plot in layout() after calling 2nd plot()?
This only works if all the plots are the same size and the defaults are used
for the margins. Try it with different sized figure regions in layout, the
added lines don't match at the end.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of baptiste auguie
Sent: Sunday, October 04, 2009 3:33 AM
To: r-help
Subject: Re: [R] add lines() to 1st plot in layout() after calling 2nd
plot()?
Hi,
Try this,
dev.new()
layout(matrix(1:4,2, by=T))
plot(1:10,main=top left plot)
plot(1:10,main=top right plot)
plot(1:10,main=bottom left plot)
plot(1:10,main=bottom right plot)
for (ii in 1:2){
for (jj in 1:2){
par(mfg=c(ii,jj))
text(5,2, lab=paste(plot #:,ii,,,jj,sep=))
}
}
par(mfg=c(1,1))
lines(c(3:7,7:3),col=red)
HTH,
baptiste
2009/10/4 Marianne Promberger marianne.promber...@kcl.ac.uk:
Thanks for the quick reply. However ...
David Winsemius dwinsem...@comcast.net 03-Oct-09 20:50:
MP layout(matrix(c(1,2),1))
MP plot(1:10,main=left plot)
MP plot(10:1,main=right plot)
MP lines(c(3:7,7:3),col=red)
MP
MP but of course now lines() gets added to the right plot. I
MP
MP Is there any way to make the lines() go to the fist plot (left
MP plot)?
If you look at layout's help page there appears to be a worked
example
of an even more complex task. The answer appears to be assingning
numbers to regions and then inserting par(mar= with an
appropriately
constructed destination arguments prior to each added piece.
Sorry, but I fail to find the solution in the page returned by
?layout, assuming that's what you mean.
Yes, the numbers in the matrix given to layout() give the order of
where plots will be put, so
layout(matrix(c(2,1),1))
then
plot(1:10,main=left plot)
plot(10:1,main=right plot)
lines(c(3:7,7:3),col=red)
puts left plot on the right hand side and right plot on the
left. But the lines() still go to the right plot plot (now on the
left hand side) which gets called last.
The par(mar ... of the scatterplot with marginal histograms example
just set the margins of the histogram plots, then they get plotted to
the region with the next number given in the layout() matrix.
Maybe I'm missing something.
Thanks,
Marianne
--
David
On Oct 3, 2009, at 3:15 PM, Marianne Promberger wrote:
Dear R users,
I create a graphic with two plots side by side using layout(), like
this:
layout(matrix(c(1,2),1))
plot(1:10,main=left plot)
lines(c(3:7,7:3),col=red)
plot(10:1,main=right plot)
The lines() obivously get added to the left plot plot.
Now, I'm trying to write a function that builds up a plot bit by
bit
to
then include it in a LaTeX presentation with overlays. I'm using
dev.copy(), and it would make my life much easier (because in fact
I
call all sorts of additional axis() etc after plot) if I could call
the above commands in this order:
layout(matrix(c(1,2),1))
plot(1:10,main=left plot)
plot(10:1,main=right plot)
lines(c(3:7,7:3),col=red)
but of course now lines() gets added to the right plot. I
Is there any way to make the lines() go to the fist plot (left
plot)?
Marianne
David Winsemius, MD
Heritage Laboratories
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.
--
Marianne Promberger PhD, King's College London
http://promberger.info
R version 2.9.2 (2009-08-24)
Ubuntu 9.04
__
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-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] Speed up and limit memory usage of lm()
gauti wrote: I have been doing series of linear regression models lm(). In this case the execution time and memory usage becomes a huge issue. I have therefore been trying to speed the process and limit the memory usage. Have a look at package biglm. Dieter -- View this message in context: http://www.nabble.com/Speed-up-and-limit-memory-usage-of-lm%28%29-tp25858939p25860485.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] mkdir in R?
Hi, Besides calling shell command mkdir by system(), I'm wondering if there is a buildin command in R to make a new directory. Regards, Peng __ 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] add lines() to 1st plot in layout() after calling 2nd plot()?
Interesting, I hadn't tried this but it probably explains why
navigation to different regions of a layout is neither documented nor
advisable.
Yet another alternative is to use Grid graphics. In particular,
1- lattice or ggplot2 provide ways to arrange several plots in a
rectangular layout, with several options to add output to a specific
panel (but they do require quite a different approach to the creation
of plots)
2- for simple enough plots (or for brave users) you could also use
low-level grid commands, and navigate to different viewports
arbitrarily placed on a page.
3- the gridBase package provides a way to combine (with some
limitations) the power of grid layouts and the output produced with
base graphics.
Best,
baptiste
2009/10/12 Greg Snow greg.s...@imail.org:
This only works if all the plots are the same size and the defaults are used
for the margins. Try it with different sized figure regions in layout, the
added lines don't match at the end.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of baptiste auguie
Sent: Sunday, October 04, 2009 3:33 AM
To: r-help
Subject: Re: [R] add lines() to 1st plot in layout() after calling 2nd
plot()?
Hi,
Try this,
dev.new()
layout(matrix(1:4,2, by=T))
plot(1:10,main=top left plot)
plot(1:10,main=top right plot)
plot(1:10,main=bottom left plot)
plot(1:10,main=bottom right plot)
for (ii in 1:2){
for (jj in 1:2){
par(mfg=c(ii,jj))
text(5,2, lab=paste(plot #:,ii,,,jj,sep=))
}
}
par(mfg=c(1,1))
lines(c(3:7,7:3),col=red)
HTH,
baptiste
2009/10/4 Marianne Promberger marianne.promber...@kcl.ac.uk:
Thanks for the quick reply. However ...
David Winsemius dwinsem...@comcast.net 03-Oct-09 20:50:
MP layout(matrix(c(1,2),1))
MP plot(1:10,main=left plot)
MP plot(10:1,main=right plot)
MP lines(c(3:7,7:3),col=red)
MP
MP but of course now lines() gets added to the right plot. I
MP
MP Is there any way to make the lines() go to the fist plot (left
MP plot)?
If you look at layout's help page there appears to be a worked
example
of an even more complex task. The answer appears to be assingning
numbers to regions and then inserting par(mar= with an
appropriately
constructed destination arguments prior to each added piece.
Sorry, but I fail to find the solution in the page returned by
?layout, assuming that's what you mean.
Yes, the numbers in the matrix given to layout() give the order of
where plots will be put, so
layout(matrix(c(2,1),1))
then
plot(1:10,main=left plot)
plot(10:1,main=right plot)
lines(c(3:7,7:3),col=red)
puts left plot on the right hand side and right plot on the
left. But the lines() still go to the right plot plot (now on the
left hand side) which gets called last.
The par(mar ... of the scatterplot with marginal histograms example
just set the margins of the histogram plots, then they get plotted to
the region with the next number given in the layout() matrix.
Maybe I'm missing something.
Thanks,
Marianne
--
David
On Oct 3, 2009, at 3:15 PM, Marianne Promberger wrote:
Dear R users,
I create a graphic with two plots side by side using layout(), like
this:
layout(matrix(c(1,2),1))
plot(1:10,main=left plot)
lines(c(3:7,7:3),col=red)
plot(10:1,main=right plot)
The lines() obivously get added to the left plot plot.
Now, I'm trying to write a function that builds up a plot bit by
bit
to
then include it in a LaTeX presentation with overlays. I'm using
dev.copy(), and it would make my life much easier (because in fact
I
call all sorts of additional axis() etc after plot) if I could call
the above commands in this order:
layout(matrix(c(1,2),1))
plot(1:10,main=left plot)
plot(10:1,main=right plot)
lines(c(3:7,7:3),col=red)
but of course now lines() gets added to the right plot. I
Is there any way to make the lines() go to the fist plot (left
plot)?
Marianne
David Winsemius, MD
Heritage Laboratories
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.
--
Marianne Promberger PhD, King's College London
http://promberger.info
R version 2.9.2 (2009-08-24)
Ubuntu 9.04
__
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] xyplot does not find variable in data
Deepayan will correct me if I'm wrong, but I'm pretty sure that the answer is that it looks in the frame in the data argument only for variables in the formula argument. Note that the fact that it also works for the subset argument is explicitly mentioned therein: subset: logical or integer indexing vector **(can be specified in terms of variables in data). ** FWIW, I agree that this is not as clearly documented as I think it could be: I kind of intuited it after a bunch of trial and error. Bert Gunter Genentech Nonclinical Biostatistics -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jacob Wegelin Sent: Monday, October 12, 2009 9:33 AM To: r-h...@stat.math.ethz.ch Subject: [R] xyplot does not find variable in data When we call a lattice function such as xyplot, to what extent does the data designation cause the function to look inside the data for variables? In the examples below, the subset argument understands that Variety is a variable in the data. But the scales argument does not understand that nitro is a variable in the data. What principle is at work? library(MEMSS) # The following works fine: xyplot( yield ~ nitro , data=Oats , scales=list( x=list( at=unique(Oats$nitro) ) ) , subset=Variety==Victory ) # But the following returns an error: xyplot( yield ~ nitro , data=Oats , scales=list( x=list( at=unique(nitro) ) ) ) Thanks for any insight Jacob A. Wegelin Assistant Professor Department of Biostatistics Virginia Commonwealth University 730 East Broad Street Room 3006 P. O. Box 980032 Richmond VA 23298-0032 U.S.A. E-mail: jwege...@vcu.edu URL: http://www.people.vcu.edu/~jwegelin __ 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] mkdir in R?
?dir.create -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Peng Yu Sent: Monday, October 12, 2009 1:01 PM To: r-h...@stat.math.ethz.ch Subject: [R] mkdir in R? Hi, Besides calling shell command mkdir by system(), I'm wondering if there is a buildin command in R to make a new directory. Regards, Peng __ 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] mkdir in R?
Try dir.create On Mon, Oct 12, 2009 at 3:00 PM, Peng Yu pengyu...@gmail.com wrote: Hi, Besides calling shell command mkdir by system(), I'm wondering if there is a buildin command in R to make a new directory. Regards, Peng __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ 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] xyplot does not find variable in data
On Mon, Oct 12, 2009 at 9:32 AM, Jacob Wegelin jacob.wege...@gmail.com wrote: When we call a lattice function such as xyplot, to what extent does the data designation cause the function to look inside the data for variables? In the examples below, the subset argument understands that Variety is a variable in the data. But the scales argument does not understand that nitro is a variable in the data. What principle is at work? A strange one called standard non-standard evaluation; see http://developer.r-project.org/nonstandard-eval.pdf for a nice overview by Thomas Lumley. ?xyplot says: data: For the ‘formula’ method, a data frame containing values (or more precisely, anything that is a valid ‘envir’ argument in ‘eval’, e.g. a list or an environment) for any variables in the formula, as well as ‘groups’ and ‘subset’ if applicable. If not found in ‘data’, or if ‘data’ is unspecified, the variables are looked for in the environment of the formula. For other methods (where ‘x’ is not a formula), ‘data’ is usually ignored, often with a warning. so the non-standard evaluation only applies to 'groups' and 'subset'. The list may be different for other functions, e.g., densityplot() also evaluates 'weights' in 'data'. -Deepayan __ 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] crosstabulation and unlist function
Hello R-users, My toy example: aa-c(1:5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) df-data.frame(aa,bb,cc,dd=as.factor(dd)) table(unlist(df[,1:3])) Can anyone point me to what function let's me do a crosstabulation between table(unlist(df[,1:3])) and df$dd? I want to find out when dd==A (or B, or C) how many times do the values 1, 2 ,3,.. appear in df[,1:3]? Thank you very much! Eugen Pircalabelu __ 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 with the use of mtext to create main title over multiple plots
Thanks Tony (and others). Setting oma corrects the problem. Mark
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University School of Medicine
15032 Hunter Court, Westfield, IN 46074
(317) 490-5129 Work, Mobile VoiceMail
(317) 399-1219 Skype No Voicemail please
On Mon, Oct 12, 2009 at 1:41 PM, Tony Plate tpl...@acm.org wrote:
Try playing around with the oma setting in par() -- it sets the outer
margins, which by default are zero.
The following shows the mtext label for me, using the windows device:
par(mfrow=c(2,2))
par(oma)
[1] 0 0 0 0
par(oma=c(0,0,2,0))
for (i in 1:4) plot(0:1,0:1)
mtext(text = my test plots, side = 3, outer = TRUE)
Mark Kimpel wrote:
I'm trying to use mtext to create a main title over multiple plots. Below
is
a simple self-contained example and my sessionInfo (I should note I've
also
tried this with R-2.8.1 with the same results). When I execute the code
chunk below, I get the plots, but no title. I've tried this using the
screen
driver, pdf, and postscript. I've used different sizes of paper. I suspect
I
am making an elementary error but searching the help files and help
archives
hasn't provided me an answer.
Thanks for any help, Mark
#
setwd(~/Desktop)
pdf(my.test.plots.pdf, paper = letter)
par(mfrow=c(2,2))
for (i in 1:4){
plot(1:6, 1:6)
}
mtext(text = my test plots, side = 3, outer = TRUE)
dev.off()
#
R version 2.10.0 Under development (unstable) (2009-09-21 r49771)
x86_64-unknown-linux-gnu
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] car_1.2-15
loaded via a namespace (and not attached):
[1] tools_2.10.0
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University School of Medicine
15032 Hunter Court, Westfield, IN 46074
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[R] Using tabs in output instead of spaces
Is there a way to have xtabs and friends use tabs in their output to separate columns rather than spaces? It would be great for importing into other software. TIA! __ 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] Using tabs in output instead of spaces
?write.table is probably a good starting point. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of ws Sent: Monday, October 12, 2009 1:39 PM To: r-h...@stat.math.ethz.ch Subject: [R] Using tabs in output instead of spaces Is there a way to have xtabs and friends use tabs in their output to separate columns rather than spaces? It would be great for importing into other software. TIA! __ 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] xyplot does not find variable in data
On 10/12/2009 08:30 PM, Deepayan Sarkar wrote: On Mon, Oct 12, 2009 at 9:32 AM, Jacob Wegelinjacob.wege...@gmail.com wrote: When we call a lattice function such as xyplot, to what extent does the data designation cause the function to look inside the data for variables? In the examples below, the subset argument understands that Variety is a variable in the data. But the scales argument does not understand that nitro is a variable in the data. What principle is at work? A strange one called standard non-standard evaluation; see http://developer.r-project.org/nonstandard-eval.pdf for a nice overview by Thomas Lumley. ?xyplot says: data: For the ‘formula’ method, a data frame containing values (or more precisely, anything that is a valid ‘envir’ argument in ‘eval’, e.g. a list or an environment) for any variables in the formula, as well as ‘groups’ and ‘subset’ if applicable. If not found in ‘data’, or if ‘data’ is unspecified, the variables are looked for in the environment of the formula. For other methods (where ‘x’ is not a formula), ‘data’ is usually ignored, often with a warning. so the non-standard evaluation only applies to 'groups' and 'subset'. The list may be different for other functions, e.g., densityplot() also evaluates 'weights' in 'data'. -Deepayan In addition, you might want to use with, as in: with( Oats, xyplot( yield ~ nitro,scales=list(x=list(at=unique(nitro))), subset=Variety==Victory ) ) Romain -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://tr.im/BcPw : celebrating R commit #5 |- http://tr.im/ztCu : RGG #158:161: examples of package IDPmisc `- http://tr.im/yw8E : New R package : sos __ 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] Using diff, ifelse on zoo object
The problem is that ifelse does not work the way you might think (see value section of ?ifelse) and basically should not be used with three zoo objects unless the three arguments to ifelse have the same time index. We can get that effect by using na.pad = TRUE in your diff call: TradedRate - with(x, ifelse(abs(diff(x.POS, na.pad = TRUE)) 0, ifelse(x.POS != 1, -x, x), NA)) or alternately we can merge them together first: xm - merge(x, dif = abs(diff(x$x.POS))) TradedRate - with(xm, ifelse(dif 0, ifelse(x.POS != 1, -x, x), NA)) By the way, to ensure that your output is reproducible be sure to use set.seed(...) before any call to a random number generator when posting. On Mon, Oct 12, 2009 at 5:30 AM, charliegenge charlie.ge...@sc.com wrote: Hi, I'm having an issue when using diff and ifelse on a zoo object. x.Date - as.Date(2003-02-01) + c(1, 3, 7, 9, 14) - 1 x - zoo(rnorm(5), x.Date) x.POS - c(0,0,0,1,1) x- merge(x,x.POS) x x x.POS 2003-02-01 -0.1858136 0 2003-02-03 -1.3188533 0 2003-02-07 0.2709794 0 2003-02-09 -1.4915262 1 2003-02-14 0.5014170 1 When I create this new zoo object using the previous one (x) I don't get exactly what I need..the traded rate is based on the lagged values, rather than the present ones... TradedRate - ifelse(abs(diff(x[,x.POS],lag= 1))0,ifelse(x[,x.POS] !=1,-x[,x],x[,x]),NA) x - merge(x, TradedRate, all=TRUE) x x x.POS TradedRate 2003-02-01 -0.1858136 0 NA 2003-02-03 -1.3188533 0 NA 2003-02-07 0.2709794 0 NA 2003-02-09 -1.4915262 1 -0.2709794 2003-02-14 0.5014170 1 NA The value for TradedRate on the 9th Feb should be -1.4945262 instead of -0.2709794what am I doing wrong? Thanks very much.. -- View this message in context: http://www.nabble.com/Using-diff%2C-ifelse-on-zoo-object-tp25852822p25852822.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.
[R] How to hide tick lines behind the box-and-whisker 's in a boxplot
Dear R people, I wonder how to hide tick lines behind other figures in a plot, e.g. in a boxplot. # Sample code: x- c(rep(4,50),rep(5,20),rep(6,50),rnorm(20,5,1)) boxplot(x) axis(2,tck=1,col.ticks='grey',lty=5 ) # end of sample code The tick lines is put on top of the box-plot, but I would like to put these lines behind the box and whiskers.. Regards Helmer __ 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] Trendline for a subset of data
Others have show how to use the segments function, but this can also be done
using the original abline function along with the clip function. Here is an
example:
plot( iris$Petal.Width, iris$Petal.Length,
col=c('red','green','blue')[iris$Species])
tmp - levels(iris$Species)
tmp2 - par('usr')
for(i in 1:3) {
fit - lm( Petal.Length ~ Petal.Width, data=iris, subset=
Species==tmp[i] )
tmpx - with(iris, Petal.Width[ Species==tmp[i] ] )
clip( min(tmpx), max(tmpx), tmp2[3], tmp2[4] )
abline( fit, col=c('red','green','blue')[i] )
}
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Steve Murray
Sent: Friday, October 09, 2009 3:51 AM
To: r-help@r-project.org
Subject: [R] Trendline for a subset of data
Dear all,
I am using abline(lm ...) to insert a linear trendline through a
portion of my data (e.g. dataset[,36:45]). However, I am finding that
whilst the trendline is correctly displayed and representative of the
data portion I've chosen, the line continues to run beyond this data
segment and continues until it intersects the vertical axes at each
side of the plot.
How do I display the line so that it only runs between point 36 and 45
(as shown in the example above) as doesn't continue to display a line
throughout the rest of the plot space?
Many thanks,
Steve
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Re: [R] crosstabulation and unlist function
On Oct 12, 2009, at 2:36 PM, eugen pircalabelu wrote: Hello R-users, My toy example: aa-c(1:5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) df-data.frame(aa,bb,cc,dd=as.factor(dd)) table(unlist(df[,1:3])) Can anyone point me to what function let's me do a crosstabulation between table(unlist(df[,1:3])) and df$dd? I want to find out when dd==A (or B, or C) how many times do the values 1, 2 ,3,.. appear in df[,1:3]? Thank you very much! One way would be to collect the row sums of those columns first, and then sum by index: tapply(apply(df[,1:3],1,sum, na.rm=TRUE), df$dd, sum) A B C 14 9 10 -- David Winsemius, MD Heritage Laboratories 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] crosstabulation and unlist function
On Oct 12, 2009, at 3:25 PM, David Winsemius wrote: On Oct 12, 2009, at 2:36 PM, eugen pircalabelu wrote: Hello R-users, My toy example: aa-c(1:5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) df-data.frame(aa,bb,cc,dd=as.factor(dd)) table(unlist(df[,1:3])) Can anyone point me to what function let's me do a crosstabulation between table(unlist(df[,1:3])) and df$dd? I want to find out when dd==A (or B, or C) how many times do the values 1, 2 ,3,.. appear in df[,1:3]? Thank you very much! One way would be to collect the row sums of those columns first, and then sum by index: tapply(apply(df[,1:3],1,sum, na.rm=TRUE), df$dd, sum) A B C 14 9 10 This method is safer than working on table(unlist(df[, 1:3]) since it does not break when an entire row is empty. aa-c(1,2,NA,4,5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) df-data.frame(aa,bb,cc,dd=as.factor(dd)) table(unlist(df[,1:3])) 1 2 4 5 2 3 3 2 # missing row willno longer be aligned with dd. tapply(table(unlist(df[,1:3])), df$dd, sum) Error in tapply(table(unlist(df[, 1:3])), df$dd, sum) : arguments must have same length tapply(apply(df[,1:3],1,sum, na.rm=TRUE), df$dd, sum) A B C 14 6 10 -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] crosstabulation and unlist function
Hello, First of all, thank you David for your reply, but sadly this is not what i wanted (i am sorry for not being more specific about my problem!) aa-c(1:5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) table(unlist(df[,1:3])) df aa bb cc dd 1 1 NA 1 A 2 2 2 2 B 3 3 NA NA B 4 4 4 4 A 5 5 5 NA C I do not want to get this: tapply(apply(df[,1:3],1,sum, na.rm=TRUE), df$dd, sum) A B C 14 6 10 but a crosstabulation between table(unlist(df[,1:3])) and df$dd, which should look something like this: 1 2 3 4 5 A 2 0 0 3 0 B 0 3 1 0 0 C 0 0 0 0 2 meaning that when dd is A 1 appears 2 times, 2 doesn't appear, 3 doesn't appear, 4 appears 3times, 5 doesn't appear; when dd is C only 5 appears 2 times (i am not really interested in the NA occurence). Hopefully, this time my question was a lot more clear. Thank you very much ! - Original Message From: David Winsemius dwinsem...@comcast.net To: David Winsemius dwinsem...@comcast.net Cc: eugen pircalabelu eugen_pircalab...@yahoo.com; R-help r-h...@stat.math.ethz.ch Sent: Mon, October 12, 2009 9:36:39 PM Subject: Re: [R] crosstabulation and unlist function On Oct 12, 2009, at 3:25 PM, David Winsemius wrote: On Oct 12, 2009, at 2:36 PM, eugen pircalabelu wrote: Hello R-users, My toy example: aa-c(1:5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) df-data.frame(aa,bb,cc,dd=as.factor(dd)) table(unlist(df[,1:3])) Can anyone point me to what function let's me do a crosstabulation between table(unlist(df[,1:3])) and df$dd? I want to find out when dd==A (or B, or C) how many times do the values 1, 2 ,3,.. appear in df[,1:3]? Thank you very much! One way would be to collect the row sums of those columns first, and then sum by index: tapply(apply(df[,1:3],1,sum, na.rm=TRUE), df$dd, sum) A B C 14 9 10 This method is safer than working on table(unlist(df[, 1:3]) since it does not break when an entire row is empty. aa-c(1,2,NA,4,5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) df-data.frame(aa,bb,cc,dd=as.factor(dd)) table(unlist(df[,1:3])) 1 2 4 5 2 3 3 2 # missing row willno longer be aligned with dd. tapply(table(unlist(df[,1:3])), df$dd, sum) Error in tapply(table(unlist(df[, 1:3])), df$dd, sum) : arguments must have same length tapply(apply(df[,1:3],1,sum, na.rm=TRUE), df$dd, sum) A B C 14 6 10 -- David Winsemius, MD Heritage Laboratories 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. David Winsemius, MD Heritage Laboratories 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] Creating object referant from argument name
Hi all. I'd like to define an object within a function based on an
argument to that function.
Specifically, I've got:
do.something-function(input){
id-substring(input,3,3)
j-list1
if(id==2)j-list2
if(id==3)j-list3
if(id==4)j-list4
...}
Instead of all these if() arguments, I was hoping to use something like:
j-paste(list,substring(input,3,3),sep=)
but this just assigns j the value of listx of _character_ mode,
instead of the actual object 'listx'.
Is there any way to get around this?
Thanks.
__
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.
Re: [R] xyplot does not find variable in data
What principle is at work? A strange one called standard non-standard evaluation; see http://developer.r-project.org/nonstandard-eval.pdf for a nice overview by Thomas Lumley. ?xyplot says: data: For the 'formula' method, a data frame containing values (or more precisely, anything that is a valid 'envir' argument in 'eval', e.g. a list or an environment) for any variables in the formula, as well as 'groups' and 'subset' if applicable. If not found in 'data', or if 'data' is unspecified, the variables are looked for in the environment of the formula. For other methods (where 'x' is not a formula), 'data' is usually ignored, often with a warning. so the non-standard evaluation only applies to 'groups' and 'subset'. The list may be different for other functions, e.g., densityplot() also evaluates 'weights' in 'data'. -Deepayan In addition, you might want to use with, as in: with( Oats, xyplot( yield ~ nitro,scales=list(x=list(at=unique(nitro))), subset=Variety==Victory ) ) Romain -- which standardizes the standard non-standardized evaluation I guess :-). Bert __ 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] Creating object referant from argument name
On 13/10/2009, at 9:14 AM, Maxwell Reback wrote:
Hi all. I'd like to define an object within a function based on an
argument to that function.
Specifically, I've got:
do.something-function(input){
id-substring(input,3,3)
j-list1
if(id==2)j-list2
if(id==3)j-list3
if(id==4)j-list4
...}
Instead of all these if() arguments, I was hoping to use something
like:
j-paste(list,substring(input,3,3),sep=)
but this just assigns j the value of listx of _character_ mode,
instead of the actual object 'listx'.
Is there any way to get around this?
Thanks.
?get
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
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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.
Re: [R] Creating object referant from argument name
try this: (?get)
j- get(paste(list,substring(input,3,3),sep=))
On Mon, Oct 12, 2009 at 4:14 PM, Maxwell Reback mreb...@gmail.com wrote:
Hi all. I'd like to define an object within a function based on an
argument to that function.
Specifically, I've got:
do.something-function(input){
id-substring(input,3,3)
j-list1
if(id==2)j-list2
if(id==3)j-list3
if(id==4)j-list4
...}
Instead of all these if() arguments, I was hoping to use something like:
j-paste(list,substring(input,3,3),sep=)
but this just assigns j the value of listx of _character_ mode,
instead of the actual object 'listx'.
Is there any way to get around this?
Thanks.
__
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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.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] crosstabulation and unlist function
What you're really saying is that you don't care about the distinction between aa, bb and cc. In that case, a different arrangement of the data will be more useful: library (reshape ) df.melt - melt ( df, id.var = dd) with ( df.melt, table ( dd, value ) ) Eric - Original message - From: eugen pircalabelu eugen_pircalab...@yahoo.com To: David Winsemius dwinsem...@comcast.net Cc: R-help r-h...@stat.math.ethz.ch Date: Mon, 12 Oct 2009 13:05:33 -0700 (PDT) Subject: Re: [R] crosstabulation and unlist function Hello, First of all, thank you David for your reply, but sadly this is not what i wanted (i am sorry for not being more specific about my problem!) aa-c(1:5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) table(unlist(df[,1:3])) df aa bb cc dd 1 1 NA 1 A 2 2 2 2 B 3 3 NA NA B 4 4 4 4 A 5 5 5 NA C I do not want to get this: tapply(apply(df[,1:3],1,sum, na.rm=TRUE), df$dd, sum) A B C 14 6 10 but a crosstabulation between table(unlist(df[,1:3])) and df$dd, which should look something like this: 1 2 3 4 5 A 2 0 0 3 0 B 0 3 1 0 0 C 0 0 0 0 2 meaning that when dd is A 1 appears 2 times, 2 doesn't appear, 3 doesn't appear, 4 appears 3times, 5 doesn't appear; when dd is C only 5 appears 2 times (i am not really interested in the NA occurence). Hopefully, this time my question was a lot more clear. Thank you very much ! - Original Message From: David Winsemius dwinsem...@comcast.net To: David Winsemius dwinsem...@comcast.net Cc: eugen pircalabelu eugen_pircalab...@yahoo.com; R-help r-h...@stat.math.ethz.ch Sent: Mon, October 12, 2009 9:36:39 PM Subject: Re: [R] crosstabulation and unlist function On Oct 12, 2009, at 3:25 PM, David Winsemius wrote: On Oct 12, 2009, at 2:36 PM, eugen pircalabelu wrote: Hello R-users, My toy example: aa-c(1:5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) df-data.frame(aa,bb,cc,dd=as.factor(dd)) table(unlist(df[,1:3])) Can anyone point me to what function let's me do a crosstabulation between table(unlist(df[,1:3])) and df$dd? I want to find out when dd==A (or B, or C) how many times do the values 1, 2 ,3,.. appear in df[,1:3]? Thank you very much! One way would be to collect the row sums of those columns first, and then sum by index: tapply(apply(df[,1:3],1,sum, na.rm=TRUE), df$dd, sum) A B C 14 9 10 This method is safer than working on table(unlist(df[, 1:3]) since it does not break when an entire row is empty. aa-c(1,2,NA,4,5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) df-data.frame(aa,bb,cc,dd=as.factor(dd)) table(unlist(df[,1:3])) 1 2 4 5 2 3 3 2 # missing row willno longer be aligned with dd. tapply(table(unlist(df[,1:3])), df$dd, sum) Error in tapply(table(unlist(df[, 1:3])), df$dd, sum) : arguments must have same length tapply(apply(df[,1:3],1,sum, na.rm=TRUE), df$dd, sum) A B C 14 6 10 -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] crosstabulation and unlist function
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of eugen pircalabelu Sent: Monday, October 12, 2009 1:06 PM To: David Winsemius Cc: R-help Subject: Re: [R] crosstabulation and unlist function Hello, First of all, thank you David for your reply, but sadly this is not what i wanted (i am sorry for not being more specific about my problem!) aa-c(1:5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) You forget to say how you made 'df', which I assume is df - data.frame(aa,bb,cc,dd) Having a self-contained way to reproduce your problem makes much easier to solve! table(unlist(df[,1:3])) df aa bb cc dd 1 1 NA 1 A 2 2 2 2 B 3 3 NA NA B 4 4 4 4 A 5 5 5 NA C I do not want to get this: tapply(apply(df[,1:3],1,sum, na.rm=TRUE), df$dd, sum) A B C 14 6 10 but a crosstabulation between table(unlist(df[,1:3])) and df$dd, which should look something like this: 1 2 3 4 5 A 2 0 0 3 0 B 0 3 1 0 0 C 0 0 0 0 2 Try with(df, table(rep(dd,3), c(aa,bb,cc))) 1 2 3 4 5 A 2 0 0 3 0 B 0 3 1 0 0 C 0 0 0 0 2 or table(rep(df$dd, 3), unlist(df[,1:3])) 1 2 3 4 5 A 2 0 0 3 0 B 0 3 1 0 0 C 0 0 0 0 2 You need the rep() to show how the 5 elements of dd should correspond to the 15 elements of aa, bb, and cc. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com meaning that when dd is A 1 appears 2 times, 2 doesn't appear, 3 doesn't appear, 4 appears 3times, 5 doesn't appear; when dd is C only 5 appears 2 times (i am not really interested in the NA occurence). Hopefully, this time my question was a lot more clear. Thank you very much ! - Original Message From: David Winsemius dwinsem...@comcast.net To: David Winsemius dwinsem...@comcast.net Cc: eugen pircalabelu eugen_pircalab...@yahoo.com; R-help r-h...@stat.math.ethz.ch Sent: Mon, October 12, 2009 9:36:39 PM Subject: Re: [R] crosstabulation and unlist function On Oct 12, 2009, at 3:25 PM, David Winsemius wrote: On Oct 12, 2009, at 2:36 PM, eugen pircalabelu wrote: Hello R-users, My toy example: aa-c(1:5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) df-data.frame(aa,bb,cc,dd=as.factor(dd)) table(unlist(df[,1:3])) Can anyone point me to what function let's me do a crosstabulation between table(unlist(df[,1:3])) and df$dd? I want to find out when dd==A (or B, or C) how many times do the values 1, 2 ,3,.. appear in df[,1:3]? Thank you very much! One way would be to collect the row sums of those columns first, and then sum by index: tapply(apply(df[,1:3],1,sum, na.rm=TRUE), df$dd, sum) A B C 14 9 10 This method is safer than working on table(unlist(df[, 1:3]) since it does not break when an entire row is empty. aa-c(1,2,NA,4,5) bb-c(NA,2,NA,4,5) cc-c(1,2,NA,4,NA) dd-c(A,B,B,A,C) df-data.frame(aa,bb,cc,dd=as.factor(dd)) table(unlist(df[,1:3])) 1 2 4 5 2 3 3 2 # missing row willno longer be aligned with dd. tapply(table(unlist(df[,1:3])), df$dd, sum) Error in tapply(table(unlist(df[, 1:3])), df$dd, sum) : arguments must have same length tapply(apply(df[,1:3],1,sum, na.rm=TRUE), df$dd, sum) A B C 14 6 10 -- David Winsemius, MD Heritage Laboratories 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. David Winsemius, MD Heritage Laboratories 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@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 plot a data by different group
Hi Catherine, Assuming your variables are in a dataframe called myData, some variation of the following may be what you want: library(ggplot2) myData.m - melt(myData, measure.vars=c(Y1, Y2)) qplot(X, value, colour=variable, shape=groupf3, facets=groupf1~groupf2, geom=point, data=myData.m) -Ista On Mon, Oct 12, 2009 at 4:36 PM, catherineLF cath...@hotmail.com wrote: Dear everyone, sorry to bother you. I have a big data, suppose it has 200 groups and each group has 100 data. So the data have 2 observations in total. The variables are groupf1 groupf2 groupf3 X Y1 Y2 1 1 1 1 0.5 0.5 groupf1, groupf2 and groupf3 are defining the 200 groups. I want to make 200 graphs for each group. For each group, graph Y1 and Y2 vs X. Is there any easy way to do that? Thank you very much for your help! Catherine -- View this message in context: http://www.nabble.com/how-to-plot-a-data-by-different-group-tp25862739p25862739.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. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.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.
[R] lapply / mapply and assignments
I have a list of isometric structures, and I want to change the same part of each structure in the list, assigning one element of a vector to each of these parts. In other words, I want to achieve the following: l - list( list(a=1,b=2), list(a=3,b=4)) unlist(lapply(l, [[, a)) [1] 1 3 # This will not actually work l[[]][a] - 5:6 unlist(lapply(l, [[, a)) [1] 5 6 I figure mapply is the solution to the problem, and I tried the following: mapply( [-, l, a, 5:6) But the result is not the same shape as the original. Any thoughts? Best, Magnus __ 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] lapply / mapply and assignments
On Mon, 12 Oct 2009, Magnus Torfason wrote: I have a list of isometric structures, and I want to change the same part of each structure in the list, assigning one element of a vector to each of these parts. In other words, I want to achieve the following: l - list( list(a=1,b=2), list(a=3,b=4)) unlist(lapply(l, [[, a)) [1] 1 3 # This will not actually work l[[]][a] - 5:6 unlist(lapply(l, [[, a)) [1] 5 6 I figure mapply is the solution to the problem, and I tried the following: mapply( [-, l, a, 5:6) But the result is not the same shape as the original. Any thoughts? See ?relist something like: relist( unlist( mapply( [-, l, a, 5:6) ), l ) HTH, Chuck Best, Magnus __ 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] lapply / mapply and assignments
Hmmm, after thinking about it, better see ?mapply SIMPLIFY logical; attempt to reduce the result to a vector or matrix? Try mapply( ..., SIMPLIFY=FALSE ) HTH, Chuck On Mon, 12 Oct 2009, Magnus Torfason wrote: I have a list of isometric structures, and I want to change the same part of each structure in the list, assigning one element of a vector to each of these parts. In other words, I want to achieve the following: l - list( list(a=1,b=2), list(a=3,b=4)) unlist(lapply(l, [[, a)) [1] 1 3 # This will not actually work l[[]][a] - 5:6 unlist(lapply(l, [[, a)) [1] 5 6 I figure mapply is the solution to the problem, and I tried the following: mapply( [-, l, a, 5:6) But the result is not the same shape as the original. Any thoughts? Best, Magnus __ 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] lapply / mapply and assignments
Try: mapply(..., SIMPLIFY = FALSE) On Mon, Oct 12, 2009 at 5:25 PM, Magnus Torfason zulutime@gmail.com wrote: I have a list of isometric structures, and I want to change the same part of each structure in the list, assigning one element of a vector to each of these parts. In other words, I want to achieve the following: l - list( list(a=1,b=2), list(a=3,b=4)) unlist(lapply(l, [[, a)) [1] 1 3 # This will not actually work l[[]][a] - 5:6 unlist(lapply(l, [[, a)) [1] 5 6 I figure mapply is the solution to the problem, and I tried the following: mapply( [-, l, a, 5:6) But the result is not the same shape as the original. Any thoughts? Best, Magnus __ 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] xyplot does not find variable in data
One other way that is sometimes useful is to use the tick.number argument as in: xyplot(yield ~ nitro, data=Oats, scales=list(x=list(tick.number=4)), subset=Variety==Victory ) This is especially handy if you want to just tick/label every other value. -Peter Ehlers Jacob Wegelin wrote: When we call a lattice function such as xyplot, to what extent does the data designation cause the function to look inside the data for variables? In the examples below, the subset argument understands that Variety is a variable in the data. But the scales argument does not understand that nitro is a variable in the data. What principle is at work? library(MEMSS) # The following works fine: xyplot( yield ~ nitro , data=Oats , scales=list( x=list( at=unique(Oats$nitro) ) ) , subset=Variety==Victory ) # But the following returns an error: xyplot( yield ~ nitro , data=Oats , scales=list( x=list( at=unique(nitro) ) ) ) Thanks for any insight Jacob A. Wegelin Assistant Professor Department of Biostatistics Virginia Commonwealth University 730 East Broad Street Room 3006 P. O. Box 980032 Richmond VA 23298-0032 U.S.A. E-mail: jwege...@vcu.edu URL: http://www.people.vcu.edu/~jwegelin __ 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] how to output profile plots for groups using lattice package
Hi George,
Your problem is not with xyplot, but with the NA occurrences in
your data. Try adding
subset = {!is.na(MSE)},
to your xyplot call, or (better), subset the data before
calling xyplot.
-Peter Ehlers
George Kalema wrote:
Hi Peter (and anyone else willing to help me out),
Many thanks for your help. Having used your code plus a few other
modifications, I only get the points plotted but without the two lines. I
just cannot figure out what the problem is.
My code is as follows:
library(lattice)
datos2 - subset(datos, samplesize != 10 parm != Theta0)
unq - sort(unique(datos2$samplesize))
datos2$fsamplesize - factor(datos2$samplesize, labels = paste(Sample
size =, unq))
datos2$parm - factor(datos2$parm, levels = c(Intercept, time,
trt, time*trt))
tp1.sim - xyplot(MSE ~ ntimes | fsamplesize + parm, group = group, data
= datos2,
type = b, lty = 1:2, pch = 1:2,
scales = list(x = list(at = c(2, 4, 8, 16)), alternating = 1),
as.table = TRUE, key = list(text = list(c(GNA, PNA)), points =
list(pch = 1:2))
)
plot(tp1.sim)
I have attached my real dataset (called datos) as well.
Kind appreciations to your efforts.
George
On Wed, Oct 7, 2009 at 9:20 AM, Peter Ehlers ehl...@ucalgary.ca wrote:
see below
George Kalema wrote:
Dear R users,
I am trying to have an xyplot of a data set which has the following
variables:
case (n=10,20,30)
parameter (parm=a,b)
group (grp=g1,g2)
y (y values)
x (x=2,4,8)
My plot should be parameter by case such that I have 2 rows (each row=
each
parameter) and 3 columns (each column=each case). My R-code is as follows
but I am not able to get what I want to:
tp1.sim - xyplot(y~ x | case + parm , group=group, data = data, lty = 1:4
,
pch = 1:4)
print(tp1.sim)
How can I have two lines (for g1 and g2) in each plot (each box)?
include the type=b argument
How do I label the x-axis with only values 2, 4, 8?
include the scales= argument or make x a factor
How do I label each column with the corresponding case number?
make 'case' a factor
The following should do what you want:
xyplot(y ~ x | factor(case) + parm, group=group, data=data,
type='b', lty=1:2, pch=1:2,
scales=list(x=list(at=c(2,4,8)))
)
I don't understand why you want 4 line types/point chars.
-Peter Ehlers
My hypothetical data set is as follows:
parm x case y group
a 2 10 0.03 g1
b 2 10 0.02 g1
a 4 10 0.03 g1
b 4 10 0.02 g1
a 8 10 0.03 g1
b 8 10 0.02 g1
a 2 20 0.03 g1
b 2 20 0.02 g1
a 4 20 0.03 g1
b 4 20 0.02 g1
a 8 20 0.03 g1
b 8 20 0.02 g1
a 2 30 0.03 g1
b 2 30 0.02 g1
a 4 30 0.03 g1
b 4 30 0.02 g1
a 8 30 0.03 g1
b 8 30 0.02 g1
a 2 10 0.13 g2
b 2 10 0.12 g2
a 4 10 0.13 g2
b 4 10 0.12 g2
a 8 10 0.13 g2
b 8 10 0.12 g2
a 2 20 0.13 g2
b 2 20 0.12 g2
a 4 20 0.13 g2
b 4 20 0.12 g2
a 8 20 0.13 g2
b 8 20 0.12 g2
a 2 30 0.13 g2
b 2 30 0.12 g2
a 4 30 0.13 g2
b 4 30 0.12 g2
a 8 30 0.13 g2
b 8 30 0.12 g2
Many thanks in advance for your response.
George
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Re: [R] What is the correct way to define __hash__?
On Mon, Oct 12, 2009 at 4:03 PM, Robert Kern robert.k...@gmail.com wrote: On 2009-10-12 15:45 PM, Peng Yu wrote: Hi, I'm wondering what is the general way to define __hash__. I could add up all the members. But I am wondering if this would cause a performance issue for certain classes. Unless if you are very familiar with the math of hash functions, I don't recommend that you try to implement one directly. Instead, make a tuple of the hashable content of your class and return the result of calling hash() on that tuple. Be sure to make your equality comparison do the right thing. class A(object): def __init__(self, a, b): self.a = a self.b = b def _key(self): # I include the name of the class so as to differentiate between other # classes that might also have a _key() method. If you have several classes # or subclasses that are allowed to compare equal to each other, use some # other common string here. return (type(self).__name__, a, b) def __hash__(self): return hash(self._key()) # Coincidentally, the _key() method can usually be reused for comparisons. # I recommend doing this for the equality comparisons, at least, when you can # because of the requirement that two items that compare equal must have the # same hash value. def __eq__(self, other): return self._key() == other._key() def __ne__(self, other): return not (self == other) ... Do I need to define other 4 comparison operators besides __eq__ and __ne__? __ 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] Invoking par(mfrow...) on an already subdivided plot
I'd like to generate on a single device multiple plots, each of which
contains two plots. Essentially, I've got sub-plots which consist of
two tracks, the upper one displaying gene expression data, and the
lower one mapping position. I'd like to display four of these
two-track sub-plots on one device, but I can't seem to invoke the
par(mfrow=...) or layout(matrix(...)) functions at more than one
level.
I've got something like:
plot.gene-function(gene){
par(fig=c(0,1,0.3,1)) #expression: track 1
plot(...)
par(fig=c(0,1,0,0.4),new=TRUE) #position: track 2
plot(...)}
plot.multigene-function(gene,...){
pdf(paste(gene,.pdf,sep=))
par(mfrow=c(2,2))
tapply(gene,gene, plot.gene)
dev.off()}
The par(mfrow=) in plot.multigene, even when 'new=TRUE', is
disregarded and I just get one sub-plot per page. Suggestions?
(Thanks!)
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Re: [R] How to hide tick lines behind the box-and-whisker 's in a boxplot
Helmer, You can just place another boxplot on top of the first like this: boxplot(x) axis(2,tck=1,col.ticks='grey',lty=5 ) boxplot(x, col=white, add=TRUE) -Peter Ehlers Helmer Belbo wrote: Dear R people, I wonder how to hide tick lines behind other figures in a plot, e.g. in a boxplot. # Sample code: x- c(rep(4,50),rep(5,20),rep(6,50),rnorm(20,5,1)) boxplot(x) axis(2,tck=1,col.ticks='grey',lty=5 ) # end of sample code The tick lines is put on top of the box-plot, but I would like to put these lines behind the box and whiskers.. Regards Helmer __ 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] Kolmogorov smirnov test
Hi r-users, I would like to use Kolmogorov smirnov test but in my observed data(xobs) there are ties. I got the warning message. My question is can I do something about it? ks.test(xobs, xsyn) Two-sample Kolmogorov-Smirnov test data: xobs and xsyn D = 0.0502, p-value = 0.924 alternative hypothesis: two-sided Warning message: In ks.test(xobs, xsyn) : cannot compute correct p-values with ties Thank you for all your help. [[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] Re use objects from within a custom made function
Hi everyone,
i'm having a problem extracting objects out of functions i've created, so i
can use them for further analysis. Here's a small example:
# ---
test - function(i, j){
x - i:j
y - i*j
z - i/j
return(x,y,z)
}
# ---
This returns the 3 objects as $x, $y and $z. I cannot, however, access these
objects individually by typing for example:
test$x
I know i can do this by adding an extra arrow head to the assignment arrow
(-), but I am sure that is not how it is done in some of the established R
functions (like when calling lm$coef out of the lm function). Is there a
simple command i've omitted from the function that allows access to objects
inside it?
Thanks in advance,
Steve
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Re: [R] Re use objects from within a custom made function
test$x doesn't evaluate the function, you want something like test(1,2)$x, e.g.:
test - function(i, j){
x - i:j
y - i*j
z - i/j
return(list(x=x,y=y,z=z))
}
test(1,2)$x
[1] 1 2
test(1,2)$y
[1] 2
test(1,2)$z
[1] 0.5
Or if you want to avoid evaluating your function multiple times:
res - test(1,2)
res$x
[1] 1 2
res$y
[1] 2
res$z
[1] 0.5
res
$x
[1] 1 2
$y
[1] 2
$z
[1] 0.5
Stropharia wrote:
Hi everyone,
i'm having a problem extracting objects out of functions i've created, so i
can use them for further analysis. Here's a small example:
# ---
test - function(i, j){
x - i:j
y - i*j
z - i/j
return(x,y,z)
}
# ---
This returns the 3 objects as $x, $y and $z. I cannot, however, access these
objects individually by typing for example:
test$x
I know i can do this by adding an extra arrow head to the assignment arrow
(-), but I am sure that is not how it is done in some of the established R
functions (like when calling lm$coef out of the lm function). Is there a
simple command i've omitted from the function that allows access to objects
inside it?
Thanks in advance,
Steve
__
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Re: [R] Re use objects from within a custom made function
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Stropharia
Sent: Monday, October 12, 2009 4:07 PM
To: r-help@r-project.org
Subject: [R] Re use objects from within a custom made function
Hi everyone,
i'm having a problem extracting objects out of functions i've
created, so i
can use them for further analysis. Here's a small example:
# ---
test - function(i, j){
x - i:j
y - i*j
z - i/j
return(x,y,z)
}
# ---
This returns the 3 objects as $x, $y and $z. I cannot,
however, access these
objects individually by typing for example:
test$x
I know i can do this by adding an extra arrow head to the
assignment arrow
(-), but I am sure that is not how it is done in some of
the established R
functions (like when calling lm$coef out of the lm function).
Is there a
simple command i've omitted from the function that allows
access to objects
inside it?
Thanks in advance,
Steve
--
Steve,
I see a couple of problems (at least from my perspective). If you ever got
your function to run, you would see a warning stating that returning
multiple objects is deprecated. In your example of using the function, you
don't call it with any parameters, and the definition has no default values.
I also would be inclined to assign the results of the function call to a
variable/object and access the values from there.
So I would approach your problem as follows:
test - function(i, j){
x - i:j
y - i*j
z - i/j
list(x=x, y=y, z=z)
}
my.result - test(1,4)
my.result$x
[1] 1 2 3 4
If you don't want to assign the output of test() to an object, then you will
need to do something like
test(1,4)$x
But this is inefficient because each time you access one of the components,
you need to rerun the function.
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
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[R] splitting dataframe, assign to new dataframe, add new rows to new dataframe
Hi, all,
My objective is to split a dataframe named cmbine according to the value of
classes. After the split, I will take the first instance from each class and
bin them into a new dataframe, df1. In the 2nd iteration, I will take the 2nd
available instance and bin them into another new dataframe, df2.
cmbine$names
apple tiger pencil chicken banana pear
cmbine$mass
0.50 100.00 0.01 1.00 0.15 0.30
cmbine$classes
1 2 3 2 1 1
These are the results which I want to obtain:
df1
classes mass
apple 0.50
tiger 100.00
pencil 0.01
df2
classes mass
banana 0.15
chicken 1.00
df3
classes mass
pear 0.30
Below shows what I have tried. The main problem I have = I don't know how to
assign the selected instance into a new dataframe with a name which is
generated 'on-the-fly' based on the value of j (the jth row).
for (i in 1:3) {
same_cell - cmbine[cmbine$classes == i, ]
if (nrow(same_cell)!=0){
for (j in 1:nrow(same_cell)){
picked - same_cell[j, ]
assign(paste(df, j, sep=), picked)
#assign(paste(df,j, sep=), paste(df, j, sep=))
}
}
The problem is that the assign function overwrites the previous value of df and
therefore the I have not been able to insert rows in the three df dataframes
and always end up with only 1 (final) row in df1, df2 and df3. I have tried
using rbind but was not able to assign values back to the on-the-fly variable
names.
I really need your advice and assistance since I have stuck with this for some
time now.
Thank you.
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Re: [R] splitting dataframe, assign to new dataframe, add new rows to new dataframe
wk yeo wrote:
Hi, all,
My objective is to split a dataframe named cmbine according to the value
of classes. After the split, I will take the first instance from each
class and bin them into a new dataframe, df1. In the 2nd iteration, I
will take the 2nd available instance and bin them into another new
dataframe, df2.
cmbine$names
apple tiger pencil chicken banana pear
cmbine$mass
0.50 100.00 0.01 1.00 0.15 0.30
cmbine$classes
1 2 3 2 1 1
If possible, it would be helpful to provide sample data in a form that could
be copied and pasted directly into an R session, like so:
cmbine - data.frame( names = c('apple', 'tiger', 'pencil', 'chicken',
'banana', 'pear' ) )
cmbine['mass'] - c(0.50, 100.00, 0.01, 1.00, 0.15, 0.30)
cmbine['classes'] - factor(c(1, 2, 3, 2,1 ,1))
It saves people on the list a bunch of coping/pasting/quote adding. Another
quick way to do this is to use the dump() which spits out the structure of
your object in a way that can be copied and pasted:
dump( 'cmbine', file='' )
wk yeo wrote:
These are the results which I want to obtain:
df1
classes mass
apple 0.50
tiger 100.00
pencil 0.01
df2
classes mass
banana 0.15
chicken 1.00
df3
classes mass
pear 0.30
Below shows what I have tried. The main problem I have = I don't know how
to assign the selected instance into a new dataframe with a name which is
generated 'on-the-fly' based on the value of j (the jth row).
for (i in 1:3) {
same_cell - cmbine[cmbine$classes == i, ]
if (nrow(same_cell)!=0){
for (j in 1:nrow(same_cell)){
picked - same_cell[j, ]
assign(paste(df, j, sep=), picked)
#assign(paste(df,j, sep=), paste(df, j, sep=))
}
}
I'm assuming you want the results grouped by class, i.e. all the 1s in one
data frame all the 2s in another. This can be done with a slight
modification of your loop:
for (i in 1:3) {
same_cell - cmbine[cmbine$classes == i, ]
if (nrow(same_cell)!=0){
assign(paste(df, i, sep=), same_cell)
}
}
However, the results I get aren't the same as the results you said you
wanted:
df1
names mass classes
1 apple 0.50 1
5 banana 0.15 1
6 pear 0.30 1
df2
names mass classes
2 tiger 100 2
4 chicken1 2
df3
names mass classes
3 pencil 0.01 3
The R way of doing this is to use the by() function, which breaks a data
frame into sub-data frames based on a column of factors-- such as the
classes. For your example, it would be used as:
by( cmbine, cmbine[['classes']], function( df ){
# Lots of stuff can happen inside this function, in this case we are
really
# just returning the subset that got passed in.
return( df )
})
cmbine[[classes]]: 1
names mass classes
1 apple 0.50 1
5 banana 0.15 1
6 pear 0.30 1
---
cmbine[[classes]]: 2
names mass classes
2 tiger 100 2
4 chicken1 2
---
cmbine[[classes]]: 3
names mass classes
3 pencil 0.01 3
The by() function returns a fancy list, each component of which can be
accessed using the [] operator.
Hope this helps!
-Charlie
-
Charlie Sharpsteen
Undergraduate
Environmental Resources Engineering
Humboldt State University
--
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Re: [R] Re use objects from within a custom made function
Thanks a lot Tony and Daniel for making that clear.
best,
Steve
Stropharia wrote:
Hi everyone,
i'm having a problem extracting objects out of functions i've created, so
i can use them for further analysis. Here's a small example:
# ---
test - function(i, j){
x - i:j
y - i*j
z - i/j
return(x,y,z)
}
# ---
This returns the 3 objects as $x, $y and $z. I cannot, however, access
these objects individually by typing for example:
test$x
I know i can do this by adding an extra arrow head to the assignment arrow
(-), but I am sure that is not how it is done in some of the established
R functions (like when calling lm$coef out of the lm function). Is there a
simple command i've omitted from the function that allows access to
objects inside it?
Thanks in advance,
Steve
--
View this message in context:
http://www.nabble.com/Reuse-objects-from-within-a-custom-made-function-tp25864695p25866081.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.
Re: [R] Re use objects from within a custom made function
On Oct 12, 2009, at 7:06 PM, Stropharia wrote:
Hi everyone,
i'm having a problem extracting objects out of functions i've
created, so i
can use them for further analysis. Here's a small example:
# ---
test - function(i, j){
x - i:j
y - i*j
z - i/j
return(x,y,z)
}
# ---
This returns the 3 objects as $x, $y and $z. I cannot, however,
access these
objects individually by typing for example:
test$x
Generally one assigns the resutlts of a function to an object name:
ll - test(1,2)
Warning message:
In return(x, y, z) : multi-argument returns are deprecated
ll
$x
[1] 1 2
$y
[1] 2
$z
[1] 0.5
str(ll)
List of 3
$ x: int [1:2] 1 2
$ y: num 2
$ z: num 0.5
So you would not get the warnings if you instead ended the function
with:
return(list(x,y,z) )
I know i can do this by adding an extra arrow head to the assignment
arrow
(-), but I am sure that is not how it is done in some of the
established R
functions (like when calling lm$coef out of the lm function). Is
there a
simple command i've omitted from the function that allows access to
objects
inside it?
Thanks in advance,
Steve
--
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
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