Re: [R] Simulating data (stupid question)
Hello Your attachement didn't seem to get through. You can simulate data using rnorm() or any of the r*() functions [1]. You can also use it to add noise to a custom function that you use to generate your specific data. Liviu [1] http://www.statmethods.net/management/functions.html On 8/25/09, KABELI MEFANE kabelimef...@yahoo.co.uk wrote: Dear All I know that you do not have to help me but please do, i am new to R as a CPI compiler, i just need to do a sample to see which sampling method best works in different situations, therefore since this is for practice purposes nobody will finance a real project thats why i need you to help me direct me as to how simulate data (just direct me,not 100% help). See my attachment for problem formulation, you can even suggest a different problem and how i can simulate __ 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. -- Do you know how to read? http://www.alienetworks.com/srtest.cfm Do you know how to write? http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail __ 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] Combining matrices
My prior solution was not correct. If the idea is to combine each row of x with each row of y then convert the matrices to data frames and perform and outer join with SQL like this: library(sqldf) X - as.data.frame(x) Y - as.data.frame(y) as.matrix(sqldf(select * from X, Y, method = raw)) See http://sqdfl.googlecode.com On Mon, Aug 24, 2009 at 1:10 PM, Gabor Grothendieckggrothendi...@gmail.com wrote: Try this: kronecker(cbind(x, y), rep(1, 3)) On Mon, Aug 24, 2009 at 12:16 PM, Daniel Nordlunddjnordl...@verizon.net wrote: If I have two matrices like x - matrix(rep(c(1,2,3),3),3) y - matrix(rep(c(4,5,6),3),3) How can I combine them to get ? 1 1 1 4 4 4 1 1 1 5 5 5 1 1 1 6 6 6 2 2 2 4 4 4 2 2 2 5 5 5 2 2 2 6 6 6 3 3 3 4 4 4 3 3 3 5 5 5 3 3 3 6 6 6 The number of rows and the actual numbers above are unimportant, they are given so as to illustrate how I want to combine the matrices. I.e., I am looking for a general way to combine the first row of x with each row of y, then the second row of x with y, Thanks, Dan Daniel Nordlund Bothell, WA USA Thanks for __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R with MPI
Hi Polemon -- polemon wrote: On Mon, Aug 24, 2009 at 2:20 PM, polemon pole...@gmail.com wrote: Hello, I plan to use R with my cluster with OpenMPI. I need the packaged 'snow' and 'Rmpi' for that, however, I get an error while downloading and installing them: When I do a: install.packages(Rmpi, dependencies=T) I get this error: checking for mpi.h... no Try to find libmpi.so or libmpich.a checking for main in -lmpi... no libmpi not found. exiting... However, mpi.h is present via the openmpi-devel package on my RHEL 5.3. This is a configuration error, try install.pacakges(Rmpi, dependencies=TRUE, configure.args=--with-mpi=/folder/with/mpiheaders). More below Some of those packages need sprng 2.0 (rsprng, for instance, which is a dependency for another MPI-related package). Sprng 2.0, however, isn't in developement for years, I wonder how I am supposed to keep my software up to date... rsprng is listed as 'Suggests' for both Rmpi and snow, so don't need to be installed for successful Rmpi or snow installation. Your message in the next post is that library(Rmpi) Error in dyn.load(file, DLLpath = DLLpath, ...) : unable to load shared library '/opt/R/lib64/R/library/Rmpi/libs/Rmpi.so': libmpi.so.0: cannot open shared object file: No such file or directory Error in library(Rmpi) : .First.lib failed for 'Rmpi' Error in dyn.unload(file.path(libpath, libs, paste(Rmpi, .Platform$dynlib.ext, : dynamic/shared library '/opt/R/lib64/R/library/Rmpi/libs/Rmpi.so' was not loaded This is a run-time error, and is because your mpi installation has not registered the location of the shared object libraries (or has not been compiled as a shared object?) correctly. See the system command 'ldconfig', or as a workaround launch R with something like LD_LIBRARY_PATH=/path/to/libmpifolder R I'm guessing that path is /usr/lib64/openmpi/1.2.7-gcc/lib, but I didn't look carefully enough at your solution to installing Rmpi to know for sure. For more help, try the R-sig-HPC mailing list, for instance Hao Yu's response to this https://stat.ethz.ch/pipermail/r-sig-hpc/2009-August/000329.html thread. Martin Any ideas on how to workaround that mpi.h problem? Please help, --polemon I did as described here: http://www.cybaea.net/Blogs/Data/R-tips-Installing-Rmpi-on-Fedora-Linux.html Since Fedora and RHEL are pretty equal, I gave that installation a shot, and from what I can tell, I got pretty far. The package installed well, but when I try to load it with library(Rmpi): library(Rmpi) Error in dyn.load(file, DLLpath = DLLpath, ...) : unable to load shared library '/opt/R/lib64/R/library/Rmpi/libs/Rmpi.so': libmpi.so.0: cannot open shared object file: No such file or directory Error in library(Rmpi) : .First.lib failed for 'Rmpi' Error in dyn.unload(file.path(libpath, libs, paste(Rmpi, .Platform$dynlib.ext, : dynamic/shared library '/opt/R/lib64/R/library/Rmpi/libs/Rmpi.so' was not loaded As you can see, R is installed in /opt/R, libmpi.so.0 is available: /usr/lib/lam/lib/libmpi.so.0 /usr/lib/lam/lib/libmpi.so.0.0.0 /usr/lib/openmpi/1.2.7-gcc/lib/libmpi.so /usr/lib/openmpi/1.2.7-gcc/lib/libmpi.so.0 /usr/lib/openmpi/1.2.7-gcc/lib/libmpi.so.0.0.0 /usr/lib64/lam/lib/libmpi.so.0 /usr/lib64/lam/lib/libmpi.so.0.0.0 /usr/lib64/openmpi/1.2.7-gcc/lib/libmpi.so /usr/lib64/openmpi/1.2.7-gcc/lib/libmpi.so.0 /usr/lib64/openmpi/1.2.7-gcc/lib/libmpi.so.0.0.0 What should I do, to make Rmpi available in R? Cheers, --polemon [[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] how to apply a date format for data frame
Hi Everyone, i have a data frame like this labels starts ends priorities 1 firsttask 37987 38049 1 2 secondtask 38019 38112 2 3 thirdtask 38049 38144 3 4 fourthtask 38081 38207 4 5 fifthtask 38112 38239 5 now i want to apply a date format for the two variables they are starts and ends. please help in this aspect it would be appreciable. Thanks in Advance. -- View this message in context: http://www.nabble.com/how-to-apply-a-date-format-for-data-frame-tp25129192p25129192.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] comparing tables from replicated data
Dear friends, I'm examining the characteristics of two models that both
fit the sodium concentration in 16 pigs quite well under treatment or
control conditions. The more complicated model is by anova better than
the less complicated model. To take it further I have generated
replicate data using the independent variables and parameter estimates
under the two models. A clinically important criterion is the change in
sodium concentration during the experiment, and as expected due to the
character of the treatment this is larger in all the treated animals
(n=10) as compared to the controls (n=6). This is also the case for 1000
replicated sets under the more complex model while quite a few of
misclassifications (control animal change treated animal change)
occurs under the less complex model. To understand (a bit at least) what
goes on I have tried to see the observed data under random group
assignment in the hope to be able to compare directly and formally the
results from the replicates under the two models.
Here are the observed changes in the 16 pigs and grp1 is treated and 2
is control.
grp - as.factor(c(rep(1,10),rep(2,6)))
val - c(6,12,11,11,11,13,15,13,11,11,2,3,1,1,1,2)
test - sum(val[grp==1]max(val[grp==2])) # 0
#Now under random perturbations of group assignments,
#what would occur???
TT - NULL
for (i in 1:1000){
ind - sample(c(1:16),16,replace=FALSE)
grp1 - grp[ind]
TT[i] - sum(val1[grp1==1]max(val1[grp1==2]))
}
hist(TT)
table(TT)
TT
0123456789 10
25 126 407 408 1179 1026 3171 879 3850
For the less complex model, the results on 1000 replicates are
evidently better than the TT default
table(test11b)
test11b
0 1 2 3 4 5 6 7 8 9 10
279 294 191 114 53 40 17 6 3 2 1
and for the more advanced model I get even more convincing
table(test11d)
test11d
0
1000
Clinically I can say that it is bad to have 1 in 16 misclassified and
therefore judge the complicated model better, but others might disagree.
Also it is not too good that the method here is insensitive to the size
of the changes.
I hope some of you will have remarks on this problem.
Best wishes
Troels
--
Troels Ring - -
Department of nephrology - -
Aalborg Hospital 9100 Aalborg, Denmark - -
+45 99326629 - -
tr...@gvdnet.dk
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Re: [R] vector size: Fixed
Hi Uwe, I am pleased to inform you that this problem has, following your hint, been resolved. I installed 64-bit version of R and I was alright. Thanks again for your time. Ogbos 2009/6/24 Uwe Ligges lig...@statistik.tu-dortmund.de ogbos okike wrote: Hi, I have a data of size 981.1MB(1028707715) and I intend to calculate the length of the data using tapply function in R. I was able to read the data into R but when I tried to use the factor function, I had an error message Error: cannot allocate vector of size 2.0 Gb. Can anybody tell me what to do? I have tried to increase the size of the memory but I am not sure if R recognizes it as the error message persists. Is there any other thing I need to do for R to recognize the memory increase? My guess is that this is a 32-bit version of R? You need some 64-bit OS and 64-bit R or try to decrease the amount/size of objects that are in the same time in your memory. Best, Uwe Ligges Thank you so much for any help. 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. [[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] robust method to obtain a correlation coeff?
Hi, Thanks for all the answers. Think with your help I got it now. It was apparently a typical newbie question. I'll try omitting NAs globally in this case. Thanks, Christian __ 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 apply a date format for data frame
rajclinasia wrote: Hi Everyone, i have a data frame like this labels starts ends priorities 1 firsttask 37987 38049 1 2 secondtask 38019 38112 2 3 thirdtask 38049 38144 3 4 fourthtask 38081 38207 4 5 fifthtask 38112 38239 5 now i want to apply a date format for the two variables they are starts and ends. Would be helpful to know which date format and what the coding means. Uwe Ligges please help in this aspect it would be appreciable. 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.
Re: [R] Filling in empty arrays/lists from using paste function
I highly suggest to rethink your problem in a way that you store the things labelled, e.g., TA1 to TA5 (as well as all the others) as the elements of a list TA. This way you have just one object TA that can easily be accessed by index operations and the code looks much cleaner in the end. Best, Uwe Ligges Steven Kang wrote: Dear R users, I am trying to fill in arrays (5 different according to distinct id) from objects produced from arbitrary data set below. a - data.frame(id=rep(c(idA1,idA2,idA3,idA4,idA5),2),pro=c(bb,uu,ee,tt,uu,gg,tt,bb,gg,ee),sal=rpois(10,2)) idpro sal 1 idA1 bb 2 2 idA2 uu 0 3 idA3 ee 3 4 idA4 tt 2 5 idA5 uu 4 6 idA1 gg 3 7 idA2 tt 0 8 idA3 bb 1 9 idA4 gg 0 10 idA5 ee 5 My desired outputs (5 arrays/lists classified according to distinct id field) are as follow: TA1 bb ee gg tt uu 2 0 3 0 0 TA2 bb ee gg tt uu 0 0 0 0 0 TA3 bb ee gg tt uu 1 3 0 0 0 .. similarly for TA4 TA5. The above results were produced using TA1 - c(bb=TA1.bb,ee=TA1.ee,gg=TA1.gg,tt=TA1.tt,uu=TA1.uu), TA2 - c(bb=TA2.bb,ee=TA2.ee,gg=TA2.gg,tt=TA2.tt,uu=TA2.uu*)* etc for TA3~TA5. Although these generate the output I desire, I would like to use a single statement for producing 5 different arrays (instead of 5 different statements) I have tried the following codes, however the last statement (paste(T, substring(i,3,4), sep=) - c(bb ..) gives error message reading Error in paste(T, substring(i, 3, 4), sep = ) - c(bb = paste(paste(T, : target of assignment expands to non-language object for (i in unique(a$id)) for (j in unique(a$pro)) assign(paste(paste(T, substring(i,3,4), sep=), j, sep=.), sum(subset(a, a$id == i a$pro == j)$sal)) paste(T, substring(i,3,4), sep=) - c(bb=paste(paste(T, substring(i,3,4), sep=),j, sep=.),ee=paste(paste(T, substring(i,3,4), sep=),j, sep=.), gg=paste(paste(T, substring(i,3,4), sep=),j, sep=.),tt=paste(paste(T, substring(i,3,4), sep=),j, sep=.), uu=paste(paste(T, substring(i,3,4), sep=),j, sep=.)) Your solution to this problem would be highly appreciated. Steve [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Between-group variance from ANOVA
Hi Emma, ...from this I can read the within-group variance. can anyone tell me how i may find out the between-group variance? But it's in the table, above the within-group variance. Remember that F is the ratio of these two quantities, i.e. the mean of the group variances divided by the mean of the within-group variances . I will work with my example since you never set seed so your answers are different from mine (which really does not help matters). set.seed(7) TDat - data.frame(response = c(rnorm(100, 5, 2), rnorm(100, 20, 2))) TDat$group - gl(2, 100, labels=c(A,B)) summary(aov(response ~ group, data=TDat)) 11225.25/3.64 [1] 3083.86 There is some rounding error on the mean squares (i.e. mean variances) but F is correct. Using estimates calculated by a different route we have: 11225.249057/3.639801 [1] 3084.028 Does this answer your question? Regards, Mark. emj83 wrote: I have done this in R and this is the following ANOVA table I get: summary(aov(response ~ group, data=TDat)) Df Sum Sq Mean Sq F valuePr(F) group 1 11203.5 11203.5 2505.0 2.2e-16 *** Residuals 198 885.5 4.5 The model is response(i,j)= group(i)+ error(i,j), we assume that group~N(0,P^2) and error~N(0,sigma^2) I know that sigma^2 is equal to 4.5, how do I find out P^2? In the problem that I am trying to apply this to, I have more than 2 groups. I was hoping there would be a function that helps you do this that I don't know about. Thanks for your help Emma Mark Difford wrote: Hi Emma, R gives you the tools to work this out. ## Example set.seed(7) TDat - data.frame(response = c(rnorm(100, 5, 2), rnorm(100, 20, 2))) TDat$group - gl(2, 100, labels=c(A,B)) with(TDat, boxplot(split(response, group))) summary(aov(response ~ group, data=TDat)) Regards, Mark. emj83 wrote: can anyone advise me please? emj83 wrote: I have done some ANOVA tables for some data that I have, from this I can read the within-group variance. can anyone tell me how i may find out the between-group variance? Thanks Emma -- View this message in context: http://www.nabble.com/Between-group-variance-from-ANOVA-tp24954045p25129942.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to apply a date format for data frame
On Mon, 24 Aug 2009 23:42:11 -0700 (PDT) rajclinasia r...@clinasia.com wrote: R labels starts ends priorities R 1 firsttask 37987 38049 1 It is easier with the other data.frame you had with: labels starts ends 1 first task 1-Jan-04 3-Mar-04 and as.Date see ?as.Date e.g. as.Date(1-Jan-04,%d-%b-%y) or for the whole column: myframe$starts-as.Date(myframe$starts,%d-%b-%y) now I hope you have learnt how to get the data into a list for the gantt.chart... hth Stefan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with BRugs
R Heberto Ghezzo, Dr wrote:
Hello, I am sorry, I have this problem before and Uwe send me the answer but I misplaced it
Oh dear! But it is not lost, since the answer to the second part of your
problem was: Please read the documentation!
and can not find it.
writing a model for BRugs
library(BRugs)
Loading required package: coda
Loading required package: lattice
Welcome to BRugs running on OpenBUGS version 3.0.3
setwd(c:/tmp)
Error in setwd(c:/tmp) : cannot change working directory
So c:/tmp does not exist or you don't have permissions there?
mo - function(){
+ for (k in 1:p){
+ delta[1,k] ~ dnorm(0,0.1)I(,delta[2,k])
Error: unexpected symbol in:
The answer was: Please read the documentation!
?writeModel (which you want to use on this function) tells you:
As a difference, BUGS syntax allows truncation specification like this:
dnorm(...) I(...) but this is illegal in R. To overcome this
incompatibility, use %_% before I(...): dnorm(...) %_% I(...). The dummy
operator %_% will be removed before the BUGS code is saved.
for (k in 1:p){
delta[1,k] ~ dnorm(0,0.1)I
delta[2,k] ~ dnorm(0,0.1)I(delta[1,k],delta[3,k])
Error: unexpected symbol in delta[2,k] ~ dnorm(0,0.1)I
delta[3,k] ~ dnorm(0,0.1)I(delta[2,k],)}
Error: unexpected symbol in delta[3,k] ~ dnorm(0,0.1)I
}
Error: unexpected '}' in }
so R parser does not like the I(,) construct
which is *not* the problem, see above.
Best wishes,
Uwe
, What is the alternative way of propgramming the
constrain I(lower,upper)
Thanks
Heberto Ghezzo
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[R] Re : table function
Hi Mark, Thank you for your answer !! it works but if i have NA in the vector z what i shoud do to count its number in Z? x y z 1 0 100 5 1 1500 6 1 NA 2 2 500 1 1 NA 5 2 2000 8 5 4500 i did the same but it gives me this error message: [0 - 1000] [1000 - 3000] 3000 0 0 0 Warning message: In inherits(x, factor) : NAs introduced by coercion Thank you De : Marc Schwartz marc_schwa...@me.com Cc : r-help@r-project.org Envoyé le : Lundi, 24 Août 2009, 18h33mn 52s Objet : Re: [R] table function On Aug 24, 2009, at 10:59 AM, Inchallah Yarab wrote: hi, i want to use the function table to build a table not of frequence (number of time the vareable is repeated in a list or a data frame!!) but in function of classes [[elided Yahoo spam]] example x y z 1 0 100 5 1 1500 6 1 1200 2 2 500 1 1 3500 5 2 2000 8 5 4500 i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] thank you very much for your help See ?cut, which bins a continuous variable. DF x y z 1 1 0 100 2 5 1 1500 3 6 1 1200 4 2 2 500 5 1 1 3500 6 5 2 2000 7 8 5 4500 table(cut(DF$z, breaks = c(-Inf, 1000, 3000, Inf), labels = c(0 - 1000, 1000 - 3000, 3000))) 0 - 1000 1000 - 3000 3000 2 3 2 HTH, Marc Schwartz [[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] Between-group variance from ANOVA
Hi Emma, ... I forgot to add the tabular ouput, which doesn't help either: T.sum - summary(aov(response ~ group, data=TDat)) print(T.sum) Df Sum Sq Mean Sq F valuePr(F) group 1 11225.2 11225.23084 2.2e-16 *** Residuals 198 720.7 3.6 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 unlist(T.sum) unlist(T.sum)[5]/unlist(T.sum)[6] Mean Sq1 3084.028 Regards, Mark. Mark Difford wrote: Hi Emma, ...from this I can read the within-group variance. can anyone tell me how i may find out the between-group variance? But it's in the table, above the within-group variance. Remember that F is the ratio of these two quantities, i.e. the mean of the group variances divided by the mean of the within-group variances . I will work with my example since you never set seed so your answers are different from mine (which really does not help matters). set.seed(7) TDat - data.frame(response = c(rnorm(100, 5, 2), rnorm(100, 20, 2))) TDat$group - gl(2, 100, labels=c(A,B)) summary(aov(response ~ group, data=TDat)) 11225.25/3.64 [1] 3083.86 There is some rounding error on the mean squares (i.e. mean variances) but F is correct. Using estimates calculated by a different route we have: 11225.249057/3.639801 [1] 3084.028 Does this answer your question? Regards, Mark. emj83 wrote: I have done this in R and this is the following ANOVA table I get: summary(aov(response ~ group, data=TDat)) Df Sum Sq Mean Sq F valuePr(F) group 1 11203.5 11203.5 2505.0 2.2e-16 *** Residuals 198 885.5 4.5 The model is response(i,j)= group(i)+ error(i,j), we assume that group~N(0,P^2) and error~N(0,sigma^2) I know that sigma^2 is equal to 4.5, how do I find out P^2? In the problem that I am trying to apply this to, I have more than 2 groups. I was hoping there would be a function that helps you do this that I don't know about. Thanks for your help Emma Mark Difford wrote: Hi Emma, R gives you the tools to work this out. ## Example set.seed(7) TDat - data.frame(response = c(rnorm(100, 5, 2), rnorm(100, 20, 2))) TDat$group - gl(2, 100, labels=c(A,B)) with(TDat, boxplot(split(response, group))) summary(aov(response ~ group, data=TDat)) Regards, Mark. emj83 wrote: can anyone advise me please? emj83 wrote: I have done some ANOVA tables for some data that I have, from this I can read the within-group variance. can anyone tell me how i may find out the between-group variance? Thanks Emma -- View this message in context: http://www.nabble.com/Between-group-variance-from-ANOVA-tp24954045p25130266.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] table function
Hi Mark, Thank you for your answer !! it works but if i have NA in the vector z what i shoud do to count its number in Z? x y z 1 0 100 5 1 1500 6 1 NA 2 2 500 1 1 NA 5 2 2000 8 5 4500 i did the same but it gives me this error message: [0 - 1000] [1000 - 3000] 3000 0 0 0 Warning message: In inherits(x, factor) : NAs introduced by coercion Thank you De : Marc Schwartz marc_schwa...@me.com Cc : r-help@r-project.org Envoyé le : Lundi, 24 Août 2009, 18h33mn 52s Objet : Re: [R] table function On Aug 24, 2009, at 10:59 AM, Inchallah Yarab wrote: hi, i want to use the function table to build a table not of frequence (number of time the vareable is repeated in a list or a data frame!!) but in function of classes [[elided Yahoo spam]] example x y z 1 0 100 5 1 1500 6 1 1200 2 2 500 1 1 3500 5 2 2000 8 5 4500 i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] thank you very much for your help See ?cut, which bins a continuous variable. DF x y z 1 1 0 100 2 5 1 1500 3 6 1 1200 4 2 2 500 5 1 1 3500 6 5 2 2000 7 8 5 4500 table(cut(DF$z, breaks = c(-Inf, 1000, 3000, Inf), labels = c(0 - 1000, 1000 - 3000, 3000))) 0 - 1000 1000 - 3000 3000 2 3 2 HTH, Marc Schwartz [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] MSM package and qmatrix
Hi Ross, For an eleven state model you could define transitions as follows: transitions_allowed - matrix(c( 0,1,1,1,1,1,1,1,1,1,1, 1,0,1,1,1,1,1,1,1,1,1, 1,1,0,1,1,1,1,1,1,1,1, 1,1,1,0,1,1,1,1,1,1,1, 1,1,1,1,0,1,1,1,1,1,1, 1,1,1,1,1,0,1,1,1,1,1, 1,1,1,1,1,1,0,1,1,1,1, 1,1,1,1,1,1,1,0,1,1,1, 1,1,1,1,1,1,1,1,0,1,1, 1,1,1,1,1,1,1,1,1,0,1, 1,1,1,1,1,1,1,1,1,1,0 ), nrow=11, ncol=11, byrow=TRUE, dimnames=list(from=1:11,to=1:11)) In your case - I think you may be getting the error message as you have 23 200 201 203 999 as states and not 1 2 3 4 5 6 7 8 9 10. Try recoding 23 200 201 203 999 to 6 7 8 9 10? HTH, Peter Ross Culloch wrote: Hi R-helpers, I am having a problem understanding how to construct the qmatrix, i have read the help menu for msm but I am still a bit lost. Below is an example of one of my transition matrices: statetable.msm(BEH, ID, data = A1) to from 1 2 3 4 5 23 200 201 203 999 1 86 11 2 20 1 9 3 11 1 22 2 18 4 4 4 0 1 1 1 0 1 35 2 0 5 0 0 2 0 0 3 4 17 5 5 13 0 4 3 3 0 7 51 0 1 1 0 0 0 0 0 0 23 7 3 0 4 0 3 1 3 0 3 200 5 0 2 3 1 2 1 3 0 3 201 5 5 3 2 1 0 2 6 1 5 203 3 0 0 0 0 0 0 0 0 0 999 18 4 0 5 0 5 7 3 1 0 Instantaneous transition IS allowed from state r to state s. But what i don't understand is what is the logical way to create the initial transition matrix - i've tried to just give even numbers to all transition possibilities e.g. every transition given 0.1 or 1, but i get the following error message: Error in msm.check.state(nrow(qmatrix), state, cens$censor) : State vector contains elements not in 1, 2, ... , 10 Does anyone have any suggestions, i can only assume that this is a simple problem that i have overlooked because nobody else seems to have asked it! Any suggestions on how to construct a qmatrix would be much appreciated, Berst wishes, Ross -- View this message in context: http://www.nabble.com/MSM-package-and-qmatrix-tp24796485p25117814.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] hdf5 package segfault when processing large data
Hi Budi and Bill,
As a workaround, you could try setting your stack limit to a larger
limit (ulimit -s). I'll change that to a heap allocation.
Thanks,
Marcus
William Dunlap wrote:
This is probably due to the code in hdf5.c allocating a huge
matrix, buf, on the stack with
883 unsigned char buf[rowcount][size];
It dies with the segmentatio fault (stack overflow, in particular)
at line 898, where it tries to access this buf.
885 for (ri = 0; ri rowcount; ri++)
886 for (pos = 0; pos colcount; pos++)
887 {
888 SEXP item = VECTOR_ELT (val, pos);
889 SEXPTYPE type = TYPEOF (item);
890 void *ptr = buf[ri][offsets[pos]];
891
892 switch (type)
893 {
894 case REALSXP:
895 memcpy (ptr, REAL (item)[ri], sizeof
(double));
896 break;
897 case INTSXP:
898 memcpy (ptr, INTEGER (item)[ri], sizeof
(int));
899 break;
The code should use one of the allocators in the R API instead
of putting the big memory block on the stack.
The data example, the list continue for more than 250,000
rows: sample.txt
DateTimef1 f2 f3 f4 f5
2007032807:56 463 463.07 462.9 463.01 1100
2007032807:57 463.01 463.01 463.01 463.01 200
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[R] order of effects plot panels
I can't figure out how to customize the order of the panels for effects plots. Any help would be appreciated. Here's my code: library(effects) lm.hour3eff-allEffects(lm.hour3, xlevels=list(road=c(0, 1000, 2000, 3000), trail=c(0, 1000, 2000, 3000))) plot(lm.hour3eff,road:trail, xlab=Trail Distance (m), ylab=Step Length (m), main=NULL) Thanks, McCrea -- View this message in context: http://www.nabble.com/order-of-effects-plot-panels-tp25123680p25123680.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] Monotone Smoothing specifically I splines
Hello I am looking for a function to create an Integrated (I) spline basis, somehting similar to the likes of 'bs' and 'ns'. I have come across the funcitons, fda::eval.monfd Values of a Monotone Functional Data Object fda::/.fd FDA internal functions fda::monfnEvaluates a monotone function fda::smooth.monotone Monotone Smoothing of Data mgcv::mono.conMonotonicity constraints for a cubic regression spline splines::backSpline Monotone Inverse Spline stats::isoreg Isotonic / Monotone Regression but none of these are specifically for an I-spline. Any help would be much appreciated. Helen Powell (Glsagow) __ 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] Between-group variance from ANOVA
can anyone advise me please? emj83 wrote: I have done some ANOVA tables for some data that I have, from this I can read the within-group variance. can anyone tell me how i may find out the between-group variance? Thanks Emma -- View this message in context: http://www.nabble.com/Between-group-variance-from-ANOVA-tp24954045p25120522.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] allowing line wrap for long strip text in xyplot (lattice)
Hi. Am brand new to R and to mailing lists - have never posted anywhere
before, so hope I do this right.
Am using R 2.9.1 with lattice graphics (just installed, fully up to date).
Am doing trellis xyplot with y (emp=employment), x (yearmo=a time measure)
and conditioning variable (indf - factor describing industry) -- i.e., (emp
~ yearmo | indf), where all three variables are in a dataframe. The
conditioning variable is a factor with a long text description (e.g.,
Offices of bank holding companies and of other holding companies
(60551112) constructed from a much-less-descriptive industry code - in this
case, 60551112). That long text goes into the strip text above each panel.
With default settings the text for many of the factor levels is too long for
the strip area and I simply see some of it but not all of it. I would like
to be able to see the full text description, which means I would like to
allow it to wrap over several lines in the strip area - say 2 or 3 lines.
Note that the text itself does NOT have any embedded line-wrap characters
(\n) and I would like to avoid writing something to guess where to insert
them into each of more than 1,000 levels of the factor. I suspect that
xyplot, perhaps through par.strip.text, has some simple way of allowing it
to do intelligent line wrap. I have searched high and low, but have not
figured this out:
- I have read the lattice documentation but haven't quite figured out from
that what I should do
- I have tried using lines=2 in par.strip.text and lineheight=2 but the
former appears to change the height of the strip area without forcing line
wrap, and the latter as I understand it controls spacing between lines
- I have tried workarounds like using cex=0.2 in par.strip.text to make the
text really small, but it gets hard to read; and I have tried
abbreviate=TRUE in par.strip.text but I would rather have full titles
I am sure there is a really simple solution to this but I can't figure it
out. If anyone could tell me how I can get long strip titles in xyplot to
wrap lines (without inserting \ns in them) I would appreciate it greatly.
Thanks!
Don Boyd
To be more concrete, here is sample code, stripped to its essentials. indf
is the factor that has very long text for some of its levels:
xyplot(emp ~ yearmo | indf,
data = subset(empall, subset=(yearmo=as.Date(2006/12/01))),
scales = list(relation=free, x = list(rot = 45)),
type = b,
par.strip.text = list(lines=2, cex=.6),
layout=c(4,3),
xlab=NULL,
# put reference lines on the plot
panel=function(...){
panel.xyplot(...)
panel.abline(h=seq(-10,5,by=1))
}
)
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[R] Number of CPU's
Any way to get access to the number of CPU's, optionally their type, from within R? In linux I can just read /proc/cpuinfo but for win/mac ? Thanks! Håvard -- Håvard Rue Department of Mathematical Sciences Norwegian University of Science and Technology N-7491 Trondheim, Norway Voice: +47-7359-3533URL : http://www.math.ntnu.no/~hrue Fax : +47-7359-3524Email: havard@math.ntnu.no This message was created in a Microsoft-free computing environment. __ 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] Fw: Re: Simulating data for sampling (stupid question)
Dear All  I know that you do not have to help me (as this is not a pure R problem) but please do, i am new to R as a CPI compiler, i just need to do a sample to see which sampling method best works in different situations, therefore since this is for practice purposes nobody will finance a real project thats why i need you to help me direct me as to how simulate data (just direct me,not 100% help). See my attachment for problem formulation, you can even suggest a different problem and how i can simulate it.  Problem Formulation:  If I want to measure customer satisfaction on say 5000 business outlets that I supply with soft drink. The rating is on a 10 scale point where 1 is lowest and 10 highest. The outlets range from 1 to 5 (1= big supermarkets, 2= medium, 3= Small, 4 = Mini markets, 5 = corner shops).  Data: I have to simulate this sort of data for example  Outlet type Population N Average Buying Power (L/M) Combined Average Buying Proportions of total buying 1 50 31000 155 19.9100835 2 200 13500 270 34.6820809 3 350 4500 1575000 20.2312139 4 1000 600 60 7.70712909 5 3400 400 136 17.4694926 Total 5000 5 7785000 100        L/M =litres per month    This means I have to simulate 50 outlets of type 1 who buy between say 2 and 4 L/M, 200 outlets of type two buying 12000 â 1 L/M,⦠etc. Also I have to simulate ratings randomly from 1 to 10.  I really do not know how to simulate data, after simulating I am going to use dollar stratification to sample this data to get info. I want to compare different sampling techniques to see which one is best.  My objective is to sample from this data in such a way that my company will benefit from this survey. If I use SRS my survey results may show that customers are satisfied with average rating of 8, but this sample may not include my most valued customers who buy 19 to 54 percent of my stock  Best Regards R novice [[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] Filling matrix secondary diagonal
Hi, how do i fill the secondary diagonals of a matrix? Is there an funktion like the diag funktion in matlab, where i can specify the diagonal i´d like to fill? -- View this message in context: http://www.nabble.com/Filling-matrix-secondary-diagonal-tp25121745p25121745.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] Between-group variance from ANOVA
I have done this in R and this is the following ANOVA table I get: summary(aov(response ~ group, data=TDat)) Df Sum Sq Mean Sq F valuePr(F) group 1 11203.5 11203.5 2505.0 2.2e-16 *** Residuals 198 885.5 4.5 The model is response(i,j)= group(i)+ error(i,j), we assume that group~N(0,P^2) and error~N(0,sigma^2) I know that sigma^2 is equal to 4.5, how do I find out P^2? In the problem that I am trying to apply this to, I have more than 2 groups. I was hoping there would be a function that helps you do this that I don't know about. Thanks for your help Emma Mark Difford wrote: Hi Emma, R gives you the tools to work this out. ## Example set.seed(7) TDat - data.frame(response = c(rnorm(100, 5, 2), rnorm(100, 20, 2))) TDat$group - gl(2, 100, labels=c(A,B)) with(TDat, boxplot(split(response, group))) summary(aov(response ~ group, data=TDat)) Regards, Mark. emj83 wrote: can anyone advise me please? emj83 wrote: I have done some ANOVA tables for some data that I have, from this I can read the within-group variance. can anyone tell me how i may find out the between-group variance? Thanks Emma -- View this message in context: http://www.nabble.com/Between-group-variance-from-ANOVA-tp24954045p25122960.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] Creating a simple line graph
Hey everyone, Sorry for yet another simple question but hopefully it makes whoever comes up with the answer feel good about helping others. I would like to simply plot the following two sets of data in a line graph. The one set is an observed set of points and the latter is the predicted. I have looked through the documentation (which makes any graphing very complicated to me) but i havent found what i need. So for: Sz= c(h1,h2,h3,h4) Pred=c(34790.0 ,47559.8, 21197.8, 28198.6) Obs=c(34740 ,48615 ,20420, 26840) MeanEst2000.Sz=cbind(Sz,Pred) LaneCo2000HH.Sz =cbind(Sz,Obs) I would like the x-axis to display the labels(Sz) and the y-axis to be the vlaues I am currently using the below (wont work with sample data) which gives me the proportions of the observed versus the predicted in four different graphs in histogram format. panelHist(DataMatrix=t(apply(Hh2000.SnSz, 1, 4)), ObsMeans=proportion(rowSums(LaneCo2000HH.SzWk),4), Bounds=c(0.95, 1.05) ) Also, if there is additional documentation for these operations i would appreciate any insights./ Thanks -- View this message in context: http://www.nabble.com/Creating-a-simple-line-graph-tp25123681p25123681.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] Saving heatmaps as PDFs
Hi,
I'm trying to save heatmaps as PDFs. However, the PDF version of the
heatmaps (Heatmap_CAFvsTNF_run2.pdf) is blurred when compared to its
counterpart, which was saved manually by using the software
Grab (Heatmap_CAFvsTNF_run2.tiff).
-R code
sample_output - stroma_run2
filename - exp_limma_results_final_probesets_Pbelowpoint05.txt
# read in data
my.data - read.delim(filename, sep=\t)
my.dataM -my.data[,2:ncol(my.data)]
hr - hclust(as.dist(1-cor(t(my.dataM), method=pearson)),
method=complete)
hc - hclust(as.dist(1-cor(my.dataM, method=spearman)),
method=complete)
# generate heat map
library(gplots)
if (is.null(sample_output)) {
pdf( Heatmap.pdf)
} else {
pdf(file=paste(Heatmap_, sample_output, .pdf, sep=)) }
heatmap.2(as.matrix(my.dataM),Rowv=as.dendrogram(hr),
Colv=as.dendrogram(hc),cexRow=0.01,
cexCol=0.8, dendrogram=none, col=greenred(100),
scale=row,key=TRUE, trace=none,density.info=none,
symkey=FALSE, main=paste(Heatmap_, sample_output))
dev.off()
--
Thanks,
Dorothy
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[R] Fw: Re: Simulating data (stupid question)
Dear All  I know that you do not have to help me (as this is not a pure R problem) but please do, i am new to R as a CPI compiler, i just need to do a sample to see which sampling method best works in different situations, therefore since this is for practice purposes nobody will finance a real project thats why i need you to help me direct me as to how simulate data (just direct me,not 100% help). See my attachment for problem formulation, you can even suggest a different problem and how i can simulate it.  Problem Formulation:  If I want to measure customer satisfaction on say 5000 business outlets that I supply with soft drink. The rating is on a 10 scale point where 1 is lowest and 10 highest. The outlets range from 1 to 5 (1= big supermarkets, 2= medium, 3= Small, 4 = Mini markets, 5 = corner shops).  Data: I have to simulate this sort of data for example  Outlet type Population N Average Buying Power (L/M) Combined Average Buying Proportions of total buying 1 50 31000 155 19.9100835 2 200 13500 270 34.6820809 3 350 4500 1575000 20.2312139 4 1000 600 60 7.70712909 5 3400 400 136 17.4694926 Total 5000 5 7785000 100        L/M =litres per month    This means I have to simulate 50 outlets of type 1 who buy between say 2 and 4 L/M, 200 outlets of type two buying 12000 â 1 L/M,⦠etc. Also I have to simulate ratings randomly from 1 to 10.  I really do not know how to simulate data, after simulating I am going to use dollar stratification to sample this data to get info. I want to compare different sampling techniques to see which one is best.  My objective is to sample from this data in such a way that my company will benefit from this survey. If I use SRS my survey results may show that customers are satisfied with average rating of 8, but this sample may not include my most valued customers who buy 19 to 54 percent of my stock  Best Regards R novice [[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] How to create a random matrix
Here is another nice way of doing it:
replicate(10, rnorm(20))
# this will give you 10 columns of vectors with 20 random variables taken
from the normal distribution.
If any one got a faster way of doing this, please let me know.
Tal galili
On Tue, Aug 25, 2009 at 1:30 AM, Rolf Turner r.tur...@auckland.ac.nzwrote:
On 25/08/2009, at 10:17 AM, Peng Yu wrote:
Hi,
I did a search but I was able to find how to generate a random matrix.
Can somebody let me know how to do it?
Uhhh, generate some random numbers and then arrange them in a matrix?
?matrix
?runif
?rnorm
?rgamma
.
.
.
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
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--
--
My contact information:
Tal Galili
Phone number: 972-50-3373767
FaceBook: Tal Galili
My Blogs:
http://www.r-statistics.com/
http://www.talgalili.com
http://www.biostatistics.co.il
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[R] [] for R
I am assuming the variable out is the output parameter.
However, I don't understand what is out[1+xi*xx0]?
Can someone explain this to me?
Thanks in advance,
Chen
dGEV - function(x, xi, mu = 0, sigma = 1, logvalue=FALSE)
{
xx - (x-mu)/sigma
#use the new dGumbel which passes mu and sigma:
#if (xi==0) out - dGumbel(xx,logvalue=TRUE)-log(sigma)
if (xi==0) {
return(out - dGumbel(x, mu, sigma, logvalue));
}
else
{ out - rep(-Inf,length(x))
out[1+xi*xx0] - (-1/xi-1)*log(1+xi*xx[1+xi*xx0]) -
(1+xi*xx[1+xi*xx0])^(-1/xi) -log(sigma)
}
if (!(logvalue))
out - exp(out)
out
}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating a simple line graph
Hello I couldn't quite understand exactly what graph you are looking for, so I'd be unable to provide code. Couple of pointers, though. - Put your data into a data frame and look at Rcmdr Graphs menu. It can do many types of them, and it should give you enough examples to get started with. - Try plot(x, y) and see if it satisfies your needs - Once you get working the command, try say playwith(plot(x, y)) - Quick-R [1] contains many good examples for beginners. Best Liviu [1] http://www.statmethods.net/graphs/index.html On 8/24/09, PDXRugger j_r...@hotmail.com wrote: Hey everyone, Sorry for yet another simple question but hopefully it makes whoever comes up with the answer feel good about helping others. I would like to simply plot the following two sets of data in a line graph. The one set is an observed set of points and the latter is the predicted. I have looked through the documentation (which makes any graphing very complicated to me) but i havent found what i need. So for: Sz= c(h1,h2,h3,h4) Pred=c(34790.0 ,47559.8, 21197.8, 28198.6) Obs=c(34740 ,48615 ,20420, 26840) MeanEst2000.Sz=cbind(Sz,Pred) LaneCo2000HH.Sz =cbind(Sz,Obs) I would like the x-axis to display the labels(Sz) and the y-axis to be the vlaues I am currently using the below (wont work with sample data) which gives me the proportions of the observed versus the predicted in four different graphs in histogram format. panelHist(DataMatrix=t(apply(Hh2000.SnSz, 1, 4)), ObsMeans=proportion(rowSums(LaneCo2000HH.SzWk),4), Bounds=c(0.95, 1.05) ) Also, if there is additional documentation for these operations i would appreciate any insights./ Thanks -- View this message in context: http://www.nabble.com/Creating-a-simple-line-graph-tp25123681p25123681.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. -- Do you know how to read? http://www.alienetworks.com/srtest.cfm Do you know how to write? http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail __ 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] [Rd] Formulas in gam function of mgcv package
Dear Gavin / Rlings, thanks for your kind answer and sorry for posting to the dev mailing list. Concerning the specific of your answer: I am working with 6 to 36 covariates, and they are all centred and scaled. I represented the problem with two variables to simplify the question. So ideally, the situation is: 1) y ~ s(x1) + + s(x36) vs. 2) y~s(x1, ,x36) I am trying to build a predictive model. Since the the variables are centred and scaled, I think I need an isotropic smooth. I am also interested in having the interactions between the variables included, that is not a purely additive model. It is not clear to me when should I give preference to tensor smooths, possibly because I have not understood well how they work. I am reading Wood(2003) as recommended and I have also read rather extensively Simon N. Wood. Generalized Additive Models: An Introduction, 2006, but still I am stuck. Any additional suggestion or reading recommendation would be greatly appreciated. I have also some difficulties in understanding the values you have chosen for k in the first example (why 60?). Thanks Best, On Monday 24 August 2009 17:33:55 Gavin Simpson wrote: [Note R-Devel is the wrong list for such questions. R-Help is where this should have been directed - redirected there now] On Mon, 2009-08-24 at 17:02 +0100, Corrado wrote: Dear R-experts, I have a question on the formulas used in the gam function of the mgcv package. I am trying to understand the relationships between: y~s(x1)+s(x2)+s(x3)+s(x4) and y~s(x1,x2,x3,x4) Does the latter contain the former? what about the smoothers of all interaction terms? I'm not 100% certain how this scales to smooths of more than 2 variables, but Sections 4.10.2 and 5.2.2 of Simon Wood's book GAM: An Introduction with R (2006, Chapman Hall/CRC) discuss this for smooths of 2 variables. Strictly y ~ s(x1) + s(x2) is not nested in y ~ s(x1, x2) as the bases used to produce the smoothers in the two models may not be the same in both models. One option to ensure nestedness is to fit the more complicated model as something like this: ## if simpler model were: y ~ s(x1, k=20) + s(x2, k = 20) y ~ s(x1, k=20) + s(x2, k = 20) + s(x1, x2, k = 60) ^ where the last term (^^^ above) has the same k as used in s(x1, x2) Note that these are isotropic smooths; are x1 and x2 measured in the same units etc.? Tensor product smooths may be more appropriate if not, and if we specify the bases when fitting models s(x1) + s(x2) *is* strictly nested in te(x1, x2), eg. y ~ s(x1, bs = cr, k = 10) + s(x2, bs = cr, k = 10) is strictly nested within y ~ te(x1, x2, k = 10) ## is the same as y ~ te(x1, x2, bs = cr, k = 10) [Note that bs = cr is the default basis in te() smooths, hence we don't need to specify it, and k = 10 refers to each individual smooth in the te().] HTH G I have (tried to) read the manual pages of gam, formula.gam, smooth.terms, linear.functional.terms but could not understand properly. Regards -- Corrado Topi Global Climate Change Biodiversity Indicators Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Re : table function
Hi r-help-boun...@r-project.org napsal dne 25.08.2009 10:08:36: Hi Mark, Thank you for your answer !! it works but if i have NA in the vector z what i shoud do to count its number in Z? You do not have NA in z, you manage to convert it somehow to factor. Please try to read about data types and its behaviour. Start with this chapter 2.8 Other types of objects in R intro manual which I suppose you have in doc folder of R program directory. You possibly can convert it back to numeric by e.g. DF$z - as.numeric(as.character(DF$z)) but I presume you need to check your original data maybe by str(your.data) what mode they are and why they are factor if you expect them numeric. Regards Petr x yz 10 100 51 1500 61 NA 22 500 11 NA 522000 854500 i did the same but it gives me this error message: [0 - 1000] [1000 - 3000] 3000 0 0 0 Warning message: In inherits(x, factor) : NAs introduced by coercion Thank you De : Marc Schwartz marc_schwa...@me.com Cc : r-help@r-project.org Envoyé le : Lundi, 24 Aoűt 2009, 18h33mn 52s Objet : Re: [R] table function On Aug 24, 2009, at 10:59 AM, Inchallah Yarab wrote: hi, i want to use the function table to build a table not of frequence (number of time the vareable is repeated in a list or a data frame!!) but in function of classes [[elided Yahoo spam]] example x yz 10 100 51 1500 61 1200 22 500 11 3500 522000 854500 i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] thank you very much for your help See ?cut, which bins a continuous variable. DF x yz 1 1 0 100 2 5 1 1500 3 6 1 1200 4 2 2 500 5 1 1 3500 6 5 2 2000 7 8 5 4500 table(cut(DF$z, breaks = c(-Inf, 1000, 3000, Inf), labels = c(0 - 1000, 1000 - 3000, 3000))) 0 - 1000 1000 - 30003000 232 HTH, Marc Schwartz [[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] Saving heatmaps as PDFs
Try playing with the
width, height
of the pdf()
Tal
On Tue, Aug 25, 2009 at 2:04 AM, dorothyc dorot...@bcgsc.ca wrote:
Hi,
I'm trying to save heatmaps as PDFs. However, the PDF version of the
heatmaps (Heatmap_CAFvsTNF_run2.pdf) is blurred when compared to its
counterpart, which was saved manually by using the software Grab
(Heatmap_CAFvsTNF_run2.tiff).
-R code
sample_output - stroma_run2
filename - exp_limma_results_final_probesets_Pbelowpoint05.txt
# read in data
my.data - read.delim(filename, sep=\t)
my.dataM -my.data[,2:ncol(my.data)]
hr - hclust(as.dist(1-cor(t(my.dataM), method=pearson)),
method=complete)
hc - hclust(as.dist(1-cor(my.dataM, method=spearman)),
method=complete)
# generate heat map
library(gplots)
if (is.null(sample_output)) {
pdf( Heatmap.pdf)
} else {
pdf(file=paste(Heatmap_, sample_output, .pdf, sep=)) }
heatmap.2(as.matrix(my.dataM),Rowv=as.dendrogram(hr),
Colv=as.dendrogram(hc),cexRow=0.01,
cexCol=0.8, dendrogram=none, col=greenred(100),
scale=row,key=TRUE, trace=none,density.info=none,
symkey=FALSE, main=paste(Heatmap_, sample_output))
dev.off()
--
Thanks,
Dorothy
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--
--
My contact information:
Tal Galili
Phone number: 972-50-3373767
FaceBook: Tal Galili
My Blogs:
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[R] Odp: Fw: Re: Simulating data for sampling (stupid question)
Well you shall consult (surprisingly :-) ?sample function and for rating ?runif function as was pointed out earlier Regards Petr r-help-boun...@r-project.org napsal dne 25.08.2009 05:14:00: Dear All  I know that you do not have to help me (as this is not a pure R problem) but please do, i am new to R as a CPI compiler, i just need to do a sample to see which sampling method best works in different situations, therefore since this is for practice purposes nobody will finance a real project thats why i need you to help me direct me as to how simulate data (just direct me,not 100% help). See my attachment for problem formulation, you can even suggest a different problem and how i can simulate it.  Problem Formulation:  If I want to measure customer satisfaction on say 5000 business outlets that I supply with soft drink. The rating is on a 10 scale point where 1 is lowest and 10 highest. The outlets range from 1 to 5 (1= big supermarkets, 2= medium, 3= Small, 4 = Mini markets, 5 = corner shops).  Data: I have to simulate this sort of data for example  Outlet type Population N Average Buying Power (L/M) Combined Average Buying Proportions of total buying 1 50 31000 155 19.9100835 2 200 13500 270 34.6820809 3 350 4500 1575000 20.2312139 4 1000 600 60 7.70712909 5 3400 400 136 17.4694926 Total 5000 5 7785000 100        L/M =litres per month    This means I have to simulate 50 outlets of type 1 who buy between say 2 and 4 L/M, 200 outlets of type two buying 12000 – 1 L/M,… etc. Also I have to simulate ratings randomly from 1 to 10.  I really do not know how to simulate data, after simulating I am going to use dollar stratification to sample this data to get info. I want to compare different sampling techniques to see which one is best.  My objective is to sample from this data in such a way that my company will benefit from this survey. If I use SRS my survey results may show that customers are satisfied with average rating of 8, but this sample may not include my most valued customers who buy 19 to 54 percent of my stock  Best Regards R novice [[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] Re : Odp: Re : table function
 Thank you Peter, in my vector Z i have missing value NA and i want to count its number in the vector i had did alla this i know the difference between a numeirc and a factor the goal of this exercice that to count the number of missing value, number betwwen 0-1000 , 1000-3000, 3000. Thank you again for your help De : Petr PIKAL petr.pi...@precheza.cz Cc : r-help@r-project.org Envoyé le : Mardi, 25 Août 2009, 11h15mn 23s Objet : Odp: [R] Re : table function Hi r-help-boun...@r-project.org napsal dne 25.08.2009 10:08:36: Hi Mark, Thank you for your answer !! it works but if i have NA in the vector z what i shoud do to count its number in Z? You do not have NA in z, you manage to convert it somehow to factor. Please try to read about data types and its behaviour. Start with this chapter 2.8 Other types of objects in R intro manual which I suppose you have in doc folder of R program directory. You possibly can convert it back to numeric by e.g. DF$z - as.numeric(as.character(DF$z)) but I presume you need to check your original data maybe by str(your.data) what mode they are and why they are factor if you expect them numeric. Regards Petr x   y    z 1  0   100 5  1   1500 6  1   NA 2  2   500 1  1   NA 5  2  2000 8  5  4500 i did the same but it gives me this error message:  [0 - 1000] [1000 - 3000]    3000       0      0      0 Warning message: In inherits(x, factor) : NAs introduced by coercion Thank you De : Marc Schwartz marc_schwa...@me.com Cc : r-help@r-project.org Envoyé le : Lundi, 24 Aoűt 2009, 18h33mn 52s Objet : Re: [R] table function On Aug 24, 2009, at 10:59 AM, Inchallah Yarab wrote: hi, i want to use the function table to build a table not of frequence (number of time the vareable is repeated in a list or a data frame!!) but in function of classes [[elided Yahoo spam]] example x   y    z 1  0   100 5  1   1500 6  1   1200 2  2   500 1  1   3500 5  2  2000 8  5  4500 i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] thank you very much for your help See ?cut, which bins a continuous variable. DF  x y  z 1 1 0 100 2 5 1 1500 3 6 1 1200 4 2 2 500 5 1 1 3500 6 5 2 2000 7 8 5 4500 table(cut(DF$z, breaks = c(-Inf, 1000, 3000, Inf),       labels = c(0 - 1000, 1000 - 3000, 3000)))   0 - 1000 1000 - 3000    3000      2      3      2 HTH, Marc Schwartz   [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating a simple line graph
Hi Do not use cbind as it results in matrix and it can have only data of one type, in your case character. Use MeanEst2000.Sz=data.frame(Sz,Pred) instead. And you probably could go step further to put everything into one data frame DF - data.frame(Sz,Pred, Obs) then with(DF, plot(Pred, Obs, axes=F)) axis(2) axis(1, at=DF$Pred, DF$Sz) box() Regards Petr r-help-boun...@r-project.org napsal dne 25.08.2009 10:51:16: Hello I couldn't quite understand exactly what graph you are looking for, so I'd be unable to provide code. Couple of pointers, though. - Put your data into a data frame and look at Rcmdr Graphs menu. It can do many types of them, and it should give you enough examples to get started with. - Try plot(x, y) and see if it satisfies your needs - Once you get working the command, try say playwith(plot(x, y)) - Quick-R [1] contains many good examples for beginners. Best Liviu [1] http://www.statmethods.net/graphs/index.html On 8/24/09, PDXRugger j_r...@hotmail.com wrote: Hey everyone, Sorry for yet another simple question but hopefully it makes whoever comes up with the answer feel good about helping others. I would like to simply plot the following two sets of data in a line graph. The one set is an observed set of points and the latter is the predicted. I have looked through the documentation (which makes any graphing very complicated to me) but i havent found what i need. So for: Sz= c(h1,h2,h3,h4) Pred=c(34790.0 ,47559.8, 21197.8, 28198.6) Obs=c(34740 ,48615 ,20420, 26840) MeanEst2000.Sz=cbind(Sz,Pred) LaneCo2000HH.Sz =cbind(Sz,Obs) I would like the x-axis to display the labels(Sz) and the y-axis to be the vlaues I am currently using the below (wont work with sample data) which gives me the proportions of the observed versus the predicted in four different graphs in histogram format. panelHist(DataMatrix=t(apply(Hh2000.SnSz, 1, 4)), ObsMeans=proportion(rowSums(LaneCo2000HH.SzWk),4), Bounds=c(0.95, 1.05) ) Also, if there is additional documentation for these operations i would appreciate any insights./ Thanks -- View this message in context: http://www.nabble.com/Creating-a-simple-line- graph-tp25123681p25123681.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. -- Do you know how to read? http://www.alienetworks.com/srtest.cfm Do you know how to write? http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on comparing two matrices
David: Ah, so that was the reason! I didn't realize that. :) Ok, so I try to go through the code and understand it: The last line seems to bring the rows into the order given by the order command. But how does the order command get the order? Lets look into the order function: the Reduce function is applied to all seven rows by the sapply command. Which itself does what? Reduce(paste, sm[x,]) seems to paste all rows while paste means just write next to each other as string. So we get one big string consisting of all the 7 rows inside the order function. But how does the order function get this order [1] 7 6 1 3 5 2 4 out of this big string? Sorry, I'm doing my best to understand it... :( Daniel: I think I understand your idea (though not the code yet). So if the matrix dimensions are x=x3+x4+x5... and y=y3+y4+y5... (where x_i is the number of rows with the sum i) then x3!x4!x5!...y3!y4!y5!... combinations have to be checked in the worst case if the matrices are equivalent (or in each case if they're not). I think it's a bit computationally intensive (since the comparison algorithm has to be used many times on many matrices) but if everything else fails, it will do. :) Thanks to all for your help! Michael David Winsemius schrieb: On Aug 24, 2009, at 4:01 PM, Michael Kogan wrote: David: Well, e.g. the first row has 2 ones in your output while there were no rows with 2 ones in the original matrix. Since the row and column sums can't be changed by sorting them, the output matrix can't be equivalent to the original one. But that means nothing, maybe it's intended and just for comparison reasons? :) But I don't get how the ones can get lost by making a string out of the row values... OK, so shoot me. I screwed up and forgot to use byrow=TRUE in my scan operation. So I ended up with a different starting matrix than you. This is what it should have looked like: sm - matrix(scan(textConnection( + 01110110 + 11000101 + 10100011 + 11001000 + 10111000 + 01011000 + 00000111)), 7, 8, byrow=TRUE) Read 56 items sm [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,]01110110 [2,]11000101 [3,]10100011 [4,]11001000 [5,]10111000 [6,]01011000 [7,]00000111 order(sapply(1:7, function(x) Reduce(paste, sm[x,])) ) [1] 7 6 1 3 5 2 4 sm[order(sapply(1:7, function(x) Reduce(paste, sm[x,])) ), ] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,]00000111 [2,]01011000 [3,]01110110 [4,]10100011 [5,]10111000 [6,]11000101 [7,]11001000 The process creates a sorted index and then just outputs rows from the original matrix, so there cannot be any row that was not there at the start. Gabor's solution will do the same operation and certainly looks more elegant than mine. (His input operation did the same mutilation on your input string as did mine.) __ 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] latitude and longitude distribution
Good day to you all, I have lightning data containing date, time, latitude and longitude. I hope that distribution of latitude and longitude will give number of lightning occurrence in a region. I have used factor function to sum up the number of events on latitude and longitude axis and saved as x and y. But when I tried to plot the two, I had and error message ( Error in image.default(x, y, z) : increasing 'x' and 'y' values expected). Every other effort I made failed to be successful - please I am a new user of R. Part of the data I am using is: latlon -5.1821 147.4462 -5.1826 147.3733 -5.1680 147.3855 31.4489 -63.2175 28.3199 -62.1831 -18.2495 -44.6471 -11.3654 -42.8249 -2.6652 -73.2344 -13.3543 -56.5338 27.9594 -157.9413 29.4454 -62.0620 -5.1953 116.0678 31.7057 -62.7211 -15.6194 124.6710 -7.1434 -72.2230 -19.1066 -45.6655 I would be extremely grateful should anyone be king enough as to guide me on how to handle this plot. Thank you for any help. 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] Odp: Re : Odp: Re : table function
Hi r-help-boun...@r-project.org napsal dne 25.08.2009 11:28:31:  Thank you Peter, in my vector Z i have missing value NA and i want to count its number in the vector i had did alla this i know the difference between a numeirc and a factor OK. If you know difference between factor and numeric you probably have seen ?factor where is note about how to make NA an extra level #Here is z z-c(10,100, 1000, 1200, 2000, 2200, 3000, 3200, 5000, 6000) #let's put some NA values into it z[c(2,5)] -NA #let's make a cut z.c-cut(z, breaks = c(-Inf, 1000, 3000, Inf), labels = c(0 - 1000, 1000 - 3000, 3000)) #as you see there are NA values but they are not extra level z.c [1] 0 - 1000 NA 0 - 1000 1000 - 3000 NA [6] 1000 - 3000 1000 - 3000 300030003000 Levels: 0 - 1000 1000 - 3000 3000 is.na(z.c) [1] FALSE TRUE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE # so let's try what help page says factor(z.c, exclude=NULL) [1] 0 - 1000 NA 0 - 1000 1000 - 3000 NA [6] 1000 - 3000 1000 - 3000 300030003000 Levels: 0 - 1000 1000 - 3000 3000 NA # wow we have NA as extra level, let's do table table(factor(z.c, exclude=NULL)) 0 - 1000 1000 - 30003000 NA 2332 Regards Petr the goal of this exercice that to count the number of missing value, number betwwen 0-1000 , 1000-3000, 3000. Thank you again for your help De : Petr PIKAL petr.pi...@precheza.cz Cc : r-help@r-project.org Envoyé le : Mardi, 25 Août 2009, 11h15mn 23s Objet : Odp: [R] Re : table function Hi r-help-boun...@r-project.org napsal dne 25.08.2009 10:08:36: Hi Mark, Thank you for your answer !! it works but if i have NA in the vector z what i shoud do to count its number in Z? You do not have NA in z, you manage to convert it somehow to factor. Please try to read about data types and its behaviour. Start with this chapter 2.8 Other types of objects in R intro manual which I suppose you have in doc folder of R program directory. You possibly can convert it back to numeric by e.g. DF$z - as.numeric(as.character(DF$z)) but I presume you need to check your original data maybe by str(your.data) what mode they are and why they are factor if you expect them numeric. Regards Petr x   y    z 1  0   100 5  1   1500 6  1   NA 2  2   500 1  1   NA 5  2  2000 8  5  4500 i did the same but it gives me this error message:  [0 - 1000] [1000 - 3000]    3000       0      0      0 Warning message: In inherits(x, factor) : NAs introduced by coercion Thank you De : Marc Schwartz marc_schwa...@me.com Cc : r-help@r-project.org Envoyé le : Lundi, 24 Aoűt 2009, 18h33mn 52s Objet : Re: [R] table function On Aug 24, 2009, at 10:59 AM, Inchallah Yarab wrote: hi, i want to use the function table to build a table not of frequence (number of time the vareable is repeated in a list or a data frame!!) but in function of classes [[elided Yahoo spam]] example x   y    z 1  0   100 5  1   1500 6  1   1200 2  2   500 1  1   3500 5  2  2000 8  5  4500 i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] thank you very much for your help See ?cut, which bins a continuous variable. DF  x y  z 1 1 0 100 2 5 1 1500 3 6 1 1200 4 2 2 500 5 1 1 3500 6 5 2 2000 7 8 5 4500 table(cut(DF$z, breaks = c(-Inf, 1000, 3000, Inf),       labels = c(0 - 1000, 1000 - 3000, 3000)))   0 - 1000 1000 - 3000    3000      2      3      2 HTH, Marc Schwartz   [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-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
Re: [R] problem with BRugs
Hello Uwe,
Just a related question, OpenBugs is using C and T for Censoring and
Truncation. But this does not seem to work with BRugs.
I am using this document
http://mathstat.helsinki.fi/openbugs/Manuals/ModelSpecification.html#TheBUGSLanguageStochasticNodes
Thanks,
Vitalie.
On Tue, 25 Aug 2009 10:00:44 +0200, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
R Heberto Ghezzo, Dr wrote:
Hello, I am sorry, I have this problem before and Uwe send me the
answer but I misplaced it
Oh dear! But it is not lost, since the answer to the second part of your
problem was: Please read the documentation!
and can not find it.
writing a model for BRugs
library(BRugs)
Loading required package: coda
Loading required package: lattice
Welcome to BRugs running on OpenBUGS version 3.0.3
setwd(c:/tmp)
Error in setwd(c:/tmp) : cannot change working directory
So c:/tmp does not exist or you don't have permissions there?
mo - function(){
+ for (k in 1:p){
+ delta[1,k] ~ dnorm(0,0.1)I(,delta[2,k])
Error: unexpected symbol in:
The answer was: Please read the documentation!
?writeModel (which you want to use on this function) tells you:
As a difference, BUGS syntax allows truncation specification like this:
dnorm(...) I(...) but this is illegal in R. To overcome this
incompatibility, use %_% before I(...): dnorm(...) %_% I(...). The dummy
operator %_% will be removed before the BUGS code is saved.
for (k in 1:p){
delta[1,k] ~ dnorm(0,0.1)I
delta[2,k] ~ dnorm(0,0.1)I(delta[1,k],delta[3,k])
Error: unexpected symbol in delta[2,k] ~ dnorm(0,0.1)I
delta[3,k] ~ dnorm(0,0.1)I(delta[2,k],)}
Error: unexpected symbol in delta[3,k] ~ dnorm(0,0.1)I
}
Error: unexpected '}' in }
so R parser does not like the I(,) construct
which is *not* the problem, see above.
Best wishes,
Uwe
, What is the alternative way of propgramming the
constrain I(lower,upper)
Thanks
Heberto Ghezzo
__
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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
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
--
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] [] for R
see ?[
For '['-indexing only: 'i, j, ...' can be logical vectors,
indicating elements/slices to select. Such vectors are
recycled if necessary to match the corresponding extent. 'i,
j, ...' can also be negative integers, indicating
elements/slices to leave out of the selection.
so 1+xi*xx0 is a logical vector.
On Mon, 24 Aug 2009 17:33:36 +0200, kfcnhl zhengchenj...@hotmail.com
wrote:
I am assuming the variable out is the output parameter.
However, I don't understand what is out[1+xi*xx0]?
Can someone explain this to me?
Thanks in advance,
Chen
dGEV - function(x, xi, mu = 0, sigma = 1, logvalue=FALSE)
{
xx - (x-mu)/sigma
#use the new dGumbel which passes mu and sigma:
#if (xi==0) out - dGumbel(xx,logvalue=TRUE)-log(sigma)
if (xi==0) {
return(out - dGumbel(x, mu, sigma, logvalue));
}
else
{ out - rep(-Inf,length(x))
out[1+xi*xx0] - (-1/xi-1)*log(1+xi*xx[1+xi*xx0]) -
(1+xi*xx[1+xi*xx0])^(-1/xi) -log(sigma)
}
if (!(logvalue))
out - exp(out)
out
}
--
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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] robust method to obtain a correlation coeff?
JK == John Kane jrkrid...@yahoo.ca on Mon, 24 Aug 2009 10:01:56 -0700 (PDT) writes: JK I may be misunderstanding the question but would cor(d1, JK use='complete.obs') or some other variant of use help? Yes! I've been a bit appalled to read that nobody else gave this (in my view *the correct* !) answer. Leaving away NA's via na.omit() or similar is more general, but for cor() definitely less flexible! Martin Maechler, ETH Zurich JK --- On Mon, 8/24/09, Christian Meesters JK meest...@imbie.uni-bonn.de wrote: From: Christian Meesters meest...@imbie.uni-bonn.de Subject: [R] robust method to obtain a correlation coeff? To: r-help@r-project.org Help r-help@r-project.org Received: Monday, August 24, 2009, 10:47 AM Hi, Being a R-newbie I am wondering how to calculate a correlation coefficient (preferably with an associated p-value) for data like: d[,1] [1] 25.5 25.3 25.1 NA 23.3 21.5 23.8 23.2 24.2 22.7 27.6 24.2 ... d[,2] [1] 0.0 11.1 0.0 NA 0.0 10.1 10.6 9.5 0.0 57.9 0.0 0.0 ... Apparently corr(d) from the boot-library fails with NAs in the data, also cor.test cannot cope with a different number of NAs. Is there a solution to this problem (calculating a correlation coefficient and ignoring different number of NAs), e.g. Pearson's corr coeff? If so, please point me to the relevant piece of documentation. TIA Christian __ 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. __ JK Be smarter than spam. See how smart SpamGuard is ons in JK Mail and switch to New Mail today or register for free JK at http://mail.yahoo.ca JK __ JK R-help@r-project.org mailing list JK https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do JK read the posting guide JK http://www.R-project.org/posting-guide.html and provide JK 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] problem with BRugs
Vitalie S. wrote:
Hello Uwe,
Just a related question, OpenBugs is using C and T for Censoring and
Truncation. But this does not seem to work with BRugs.
I am using this document
http://mathstat.helsinki.fi/openbugs/Manuals/ModelSpecification.html#TheBUGSLanguageStochasticNodes
What's the problem? Error message? Reproducible code?
Specifying, e.g.
model - function()
x ~ dnorm(mu, tau)%_%T(lower, upper)
writeModel(model)
works for me.
Best,
Uwe
Thanks,
Vitalie.
On Tue, 25 Aug 2009 10:00:44 +0200, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
R Heberto Ghezzo, Dr wrote:
Hello, I am sorry, I have this problem before and Uwe send me the
answer but I misplaced it
Oh dear! But it is not lost, since the answer to the second part of
your problem was: Please read the documentation!
and can not find it.
writing a model for BRugs
library(BRugs)
Loading required package: coda
Loading required package: lattice
Welcome to BRugs running on OpenBUGS version 3.0.3
setwd(c:/tmp)
Error in setwd(c:/tmp) : cannot change working directory
So c:/tmp does not exist or you don't have permissions there?
mo - function(){
+ for (k in 1:p){
+ delta[1,k] ~ dnorm(0,0.1)I(,delta[2,k])
Error: unexpected symbol in:
The answer was: Please read the documentation!
?writeModel (which you want to use on this function) tells you:
As a difference, BUGS syntax allows truncation specification like
this: dnorm(...) I(...) but this is illegal in R. To overcome this
incompatibility, use %_% before I(...): dnorm(...) %_% I(...). The
dummy operator %_% will be removed before the BUGS code is saved.
for (k in 1:p){
delta[1,k] ~ dnorm(0,0.1)I
delta[2,k] ~ dnorm(0,0.1)I(delta[1,k],delta[3,k])
Error: unexpected symbol in delta[2,k] ~ dnorm(0,0.1)I
delta[3,k] ~ dnorm(0,0.1)I(delta[2,k],)}
Error: unexpected symbol in delta[3,k] ~ dnorm(0,0.1)I
}
Error: unexpected '}' in }
so R parser does not like the I(,) construct
which is *not* the problem, see above.
Best wishes,
Uwe
, What is the alternative way of propgramming the
constrain I(lower,upper)
Thanks
Heberto Ghezzo
__
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.
[R] table, sum, cat function
Hi, the second step in my exercice is to calculate the sum of the amout for each class et not the frequency i have this vector x y 1 100 2 1500 3 3250 4 6250 5 2000 6 450 i want to use the function table and cat to calculate the sum of the amount in each class [0-1000], [1000-3000],[ 3000] Thank you for your help?? inchallah [[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 : Odp: Re : Odp: Re : table function
Thank you very much Peter it works!! the second step is to calculate the sum of variable in each class!!! how can i do this!!! thank you in advance inchallah De : Petr PIKAL petr.pi...@precheza.cz Cc : r-help@r-project.org Envoyé le : Mardi, 25 Août 2009, 11h53mn 21s Objet : Odp: [R] Re : Odp: Re : table function Hi r-help-boun...@r-project.org napsal dne 25.08.2009 11:28:31: à Thank you Peter, in my vector Z i have missing value NA and i want to count its number in the vector i had did alla this i know the difference between a numeirc and a factor OK. If you know difference between factor and numeric you probably have seen ?factor where is note about how to make NA an extra level #Here is z z-c(10,100, 1000, 1200, 2000, 2200, 3000, 3200, 5000, 6000) #let's put some NA values into it z[c(2,5)] -NA #let's make a cut z.c-cut(z, breaks = c(-Inf, 1000, 3000, Inf), labels = c(0 - 1000, 1000 - 3000, 3000)) #as you see there are NA values but they are not extra level z.c [1] 0 - 1000  NA    0 - 1000  1000 - 3000 NA [6] 1000 - 3000 1000 - 3000 3000    3000    3000 Levels: 0 - 1000 1000 - 3000 3000 is.na(z.c) [1] FALSE TRUE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE # so let's try what help page says factor(z.c, exclude=NULL) [1] 0 - 1000  NA    0 - 1000  1000 - 3000 NA [6] 1000 - 3000 1000 - 3000 3000    3000    3000 Levels: 0 - 1000 1000 - 3000 3000 NA # wow we have NA as extra level, let's do table table(factor(z.c, exclude=NULL))   0 - 1000 1000 - 3000    3000    NA      2      3      3      2 Regards Petr the goal of this exercice that to count the number of missing value, number betwwen 0-1000 , 1000-3000, 3000. Thank you again for your help De : Petr PIKAL petr.pi...@precheza.cz Cc : r-help@r-project.org EnvoyÄ© le : Mardi, 25 AoÄ»t 2009, 11h15mn 23s Objetà : Odp: [R] Re : table function Hi r-help-boun...@r-project.org napsal dne 25.08.2009 10:08:36: Hi Mark, Thank you for your answer !! it works but if i have NA in the vector z what i shoud do to count its number in Z? You do not have NA in z, you manage to convert it somehow to factor. Please try to read about data types and its behaviour. Start with this chapter 2.8 Other types of objects in R intro manual which I suppose you have in doc folder of R program directory. You possibly can convert it back to numeric by e.g. DF$z - as.numeric(as.character(DF$z)) but I presume you need to check your original data maybe by str(your.data) what mode they are and why they are factor if you expect them numeric. Regards Petr xàààyààààz 1àà0ààà100 5àà1ààà1500 6àà1àààNA 2àà2ààà500 1àà1àààNA 5àà2àà2000 8àà5àà4500 i did the same but it gives me this erroràmessage: à[0 - 1000] [1000 - 3000]àààà3000 àààààà0àààààà0àààààà0 Warning message: In inherits(x, factor) : NAs introduced by coercion Thank you De : Marc Schwartz marc_schwa...@me.com Cc : r-help@r-project.org EnvoyÄ© le : Lundi, 24 Aoűt 2009, 18h33mn 52s Objet : Re: [R] table function On Aug 24, 2009, at 10:59 AM, Inchallah Yarab wrote: hi, i want to use the function table to build a table not of frequence (number of time the vareable is repeated in a list or a data frame!!) but in function of classes [[elided Yahoo spam]] example xàààyààààz 1àà0ààà100 5àà1ààà1500 6àà1ààà1200 2àà2ààà500 1àà1ààà3500 5àà2àà2000 8àà5àà4500 i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] thank you very much for your help See ?cut, which bins a continuous variable. DF àx yààz 1 1 0à100 2 5 1 1500 3 6 1 1200 4 2 2à500 5 1 1 3500 6 5 2 2000 7 8 5 4500 table(cut(DF$z, breaks = c(-Inf, 1000, 3000, Inf), ààààààlabels = c(0 - 1000, 1000 - 3000, 3000))) àà0 - 1000 1000 - 3000àààà3000 ààààà2àààààà3àààààà2 HTH, Marc Schwartz àà[[alternative HTML version deleted]] __ R-help@r-project.org mailing list
Re: [R] error in creating gantt chart.
rajclinasia wrote: hi every one, i have a excel sheet like this labels starts ends 1 first task 1-Jan-04 3-Mar-04 2 second task 2-Feb-04 5-May-04 3 third task 3-Mar-04 6-Jun-04 4 fourth task 4-Apr-04 8-Aug-04 5 fifth task 5-May-04 9-Sep-04 now i converted this excel sheet into csv file and i read the csv file into R with the below code. my.gantt.info-read.csv(C:/Documents and Settings/balakrishna/Desktop/one.csv). and for create gantt chart i used below code. gantt.chart(my.gantt.info). if i run this above code i am getting the error like this Error in x$starts : $ operator is invalid for atomic vectors. can anybody help in this aspect it would be very appreciable. Hi Raj, My apologies for not testing the solution I sent. The dates in the spreadsheet were converted to character strings, then factors in the input stage. If you reconvert them as follows: my.gantt.info$starts-as.POSIXct(strptime(my.gantt.info$starts,%d-%b-%y)) my.gantt.info$ends-as.POSIXct(strptime(my.gantt.info$ends,%d-%b-%y)) things should work properly. 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] image plot
ogbos okike wrote: Hi, I am trying to use the image function to do a color plot. My matrix columns are labeled y and x. I tried image(y, x) but I had error message (Error in image.default(y, x) : increasing 'x' and 'y' values expected). Could anybody please tell me how to add these increasing 'x' and 'y' values. Hi Ogbos, I think you may be passing the values that you want displayed rather than the positions of those values on a 2D grid. Try sending your matrix to color2D.matplot in the plotrix package. That just converts the values into colors and may be more like what you want. 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] table, sum, cat function
the second step in my exercice is to calculate the sum of the amout for each class et not the frequency i have this vector x y 1 100 2 1500 3 3250 4 6250 5 2000 6 450 i want to use the function table and cat to calculate the sum of the amount in each class [0-1000], [1000-3000],[ 3000] You want to use cut, not cat. Take a look at the examples on the cut help page - they will help you. Regards, Richie. Mathematical Sciences Unit [1]HSL _ ATTENTION: This message contains privileged and confidential information intended for the addressee(s) only. If this message was sent to you in error, you must not disseminate, copy or take any action in reliance on it and we request that you notify the sender immediately by return email. Opinions expressed in this message and any attachments are not necessarily those held by the [2]Health and Safety Laboratory or any person connected with the organisation, save those by whom the opinions were expressed. Please note that any messages sent or received by the [3]Health and Safety Laboratory email system may be monitored and stored in an information retrieval system. _ Think before you print - do you really need to print this email? _ _ Scanned by MailMarshal - Marshal's comprehensive email content security solution. Download a free evaluation of MailMarshal at [4]www.marshal.com _ References 1. http://www.hsl.gov.uk/contact-us.htm 2. http://www.hsl.gov.uk/ 3. http://www.hsl.gov.uk/ 4. http://www.marshal.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] Passing a list object to lapply
Fowler, Mark wrote: Hello, I'm having difficulty passing an object name to a lapply function. Can somebody tell me the trick to make this work? lapply() works on objects rather than names of objects, hence you need get(paste(...)). Uwe Ligges #Works T13702 - TRACKDATA[[13702.xls]][[data]] min(unlist(lapply(list(T13702), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3] 16553 #Works d-2 assign(paste(T,substr(names(TRACKDATA)[d],1,(nchar(names(TRACKDATA)[d] )-4)),sep=),TRACKDATA[[d]][[data]],pos=1) min(unlist(lapply(list(T13702), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3] 16553 #Fails. d-2 assign(paste(T,substr(names(TRACKDATA)[d],1,(nchar(names(TRACKDATA)[d] )-4)),sep=),TRACKDATA[[d]][[data]],pos=1) min(unlist(lapply(list(paste(T,substr(names(TRACKDATA)[d],1,(nchar(nam es(TRACKDATA)[d])-4)),sep=)), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3] Error in x[1, 2] : incorrect number of dimensions traceback() 4: mdy.date(x[1, 2], x[1, 1], x[1, 3]) 3: FUN(X[[1L]], ...) 2: lapply(list(paste(T, substr(names(TRACKDATA)[d], 1, (nchar(names(TRACKDATA)[d]) - 4)), sep = )), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3])) 1: unlist(lapply(list(paste(T, substr(names(TRACKDATA)[d], 1, (nchar(names(TRACKDATA)[d]) - 4)), sep = )), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3]))) #Fails (trying noquote). min(unlist(lapply(list(noquote(paste(T,substr(names(TRACKDATA)[d],1,(n char(names(TRACKDATA)[d])-4)),sep=))), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3] Error in unclass(x)[...] : incorrect number of dimensions traceback() 6: `[.noquote`(x, 1, 2) 5: x[1, 2] 4: mdy.date(x[1, 2], x[1, 1], x[1, 3]) 3: FUN(X[[1L]], ...) 2: lapply(list(noquote(paste(T, substr(names(TRACKDATA)[d], 1, (nchar(names(TRACKDATA)[d]) - 4)), sep = ))), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3])) 1: unlist(lapply(list(noquote(paste(T, substr(names(TRACKDATA)[d], 1, (nchar(names(TRACKDATA)[d]) - 4)), sep = ))), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3]))) Mark Fowler Population Ecology Division Bedford Inst of Oceanography Dept Fisheries Oceans Dartmouth NS Canada B2Y 4A2 Tel. (902) 426-3529 Fax (902) 426-9710 Email fowl...@mar.dfo-mpo.gc.ca Home Tel. (902) 461-0708 Home Email mark.fow...@ns.sympatico.ca [[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] latitude and longitude distribution
Hi there, How about you send a minimum reproducible code as suggested on posting guide :-) bests milton On Tue, Aug 25, 2009 at 5:48 AM, ogbos okike ogbos.ok...@gmail.com wrote: Good day to you all, I have lightning data containing date, time, latitude and longitude. I hope that distribution of latitude and longitude will give number of lightning occurrence in a region. I have used factor function to sum up the number of events on latitude and longitude axis and saved as x and y. But when I tried to plot the two, I had and error message ( Error in image.default(x, y, z) : increasing 'x' and 'y' values expected). Every other effort I made failed to be successful - please I am a new user of R. Part of the data I am using is: latlon -5.1821 147.4462 -5.1826 147.3733 -5.1680 147.3855 31.4489 -63.2175 28.3199 -62.1831 -18.2495 -44.6471 -11.3654 -42.8249 -2.6652 -73.2344 -13.3543 -56.5338 27.9594 -157.9413 29.4454 -62.0620 -5.1953 116.0678 31.7057 -62.7211 -15.6194 124.6710 -7.1434 -72.2230 -19.1066 -45.6655 I would be extremely grateful should anyone be king enough as to guide me on how to handle this plot. Thank you for any help. 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.htmlhttp://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] How to create a random matrix
Like tris: NCols=5 NRows=7 myMat-matrix(runif(NCols*NRows), ncol=NCols) myMat bests milton On Mon, Aug 24, 2009 at 6:17 PM, Peng Yu pengyu...@gmail.com wrote: Hi, I did a search but I was able to find how to generate a random matrix. Can somebody let me know how to do it? 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.htmlhttp://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] [] for R
On 8/24/09, kfcnhl zhengchenj...@hotmail.com wrote: I am assuming the variable out is the output parameter. However, I don't understand what is out[1+xi*xx0]? Can someone explain this to me? http://www.statmethods.net/management/subset.html Liviu __ 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] Horisontal line in xtable
Late answer, but still.. On 6/17/09, christiaan pauw cjp...@gmail.com wrote: I want a horisontal line that separates the results from the sample information i.e. a horisontal line on the bottom of the line second from last ?print.xtable hline.after: When 'type=latex', a vector of numbers between -1 and 'nrow(x)', inclusive, indicating the rows after which a horizontal line should appear. If 'NULL' is used no lines are produced. Default value is 'c(-1,0,nrow(x))' which means draw a line before and after the columns names and at the end of the table. Repeated values are allowed. Liviu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with BRugs
On Tue, 25 Aug 2009 12:13:21 +0200, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
Vitalie S. wrote:
Hello Uwe,
Just a related question, OpenBugs is using C and T for Censoring and
Truncation. But this does not seem to work with BRugs.
I am using this document
http://mathstat.helsinki.fi/openbugs/Manuals/ModelSpecification.html#TheBUGSLanguageStochasticNodes
What's the problem? Error message? Reproducible code?
Oh, sorry,I justs assumed it would be an easy answer, like not
implemented yet.
Here is the code:
fModel- function()
{
beta ~ dnorm(0,1) %_% T(0, 1)
}
writeModel(fModel,Model.txt);
modelCheck(Model.txt);
#ERROR: this density cannot be truncated error pos 33 (error on line 3)
Note that C and I work fine in the code above.
My info:
sessionInfo(BRugs)
R version 2.9.0 (2009-04-17)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
character(0)
other attached packages:
[1] BRugs_0.5-1
Vitalie.
Specifying, e.g.
model - function()
x ~ dnorm(mu, tau)%_%T(lower, upper)
writeModel(model)
works for me.
Best,
Uwe
Thanks,
Vitalie.
On Tue, 25 Aug 2009 10:00:44 +0200, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
R Heberto Ghezzo, Dr wrote:
Hello, I am sorry, I have this problem before and Uwe send me the
answer but I misplaced it
Oh dear! But it is not lost, since the answer to the second part of
your problem was: Please read the documentation!
and can not find it.
writing a model for BRugs
library(BRugs)
Loading required package: coda
Loading required package: lattice
Welcome to BRugs running on OpenBUGS version 3.0.3
setwd(c:/tmp)
Error in setwd(c:/tmp) : cannot change working directory
So c:/tmp does not exist or you don't have permissions there?
mo - function(){
+ for (k in 1:p){
+ delta[1,k] ~ dnorm(0,0.1)I(,delta[2,k])
Error: unexpected symbol in:
The answer was: Please read the documentation!
?writeModel (which you want to use on this function) tells you:
As a difference, BUGS syntax allows truncation specification like
this: dnorm(...) I(...) but this is illegal in R. To overcome this
incompatibility, use %_% before I(...): dnorm(...) %_% I(...). The
dummy operator %_% will be removed before the BUGS code is saved.
for (k in 1:p){
delta[1,k] ~ dnorm(0,0.1)I
delta[2,k] ~ dnorm(0,0.1)I(delta[1,k],delta[3,k])
Error: unexpected symbol in delta[2,k] ~ dnorm(0,0.1)I
delta[3,k] ~ dnorm(0,0.1)I(delta[2,k],)}
Error: unexpected symbol in delta[3,k] ~ dnorm(0,0.1)I
}
Error: unexpected '}' in }
so R parser does not like the I(,) construct
which is *not* the problem, see above.
Best wishes,
Uwe
, What is the alternative way of propgramming the
constrain I(lower,upper)
Thanks
Heberto Ghezzo
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with BRugs
Vitalie S. wrote:
On Tue, 25 Aug 2009 12:13:21 +0200, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
Vitalie S. wrote:
Hello Uwe,
Just a related question, OpenBugs is using C and T for Censoring and
Truncation. But this does not seem to work with BRugs.
I am using this document
http://mathstat.helsinki.fi/openbugs/Manuals/ModelSpecification.html#TheBUGSLanguageStochasticNodes
What's the problem? Error message? Reproducible code?
Oh, sorry,I justs assumed it would be an easy answer, like not
implemented yet.
Here is the code:
fModel- function()
{
beta ~ dnorm(0,1) %_% T(0, 1)
}
writeModel(fModel,Model.txt);
modelCheck(Model.txt);
#ERROR: this density cannot be truncated error pos 33 (error on line 3)
This is a BUGS error message that indeed tells you that BUGS cannot
truncate that density - not related to R at all.
Best,
Uwe Ligges
Note that C and I work fine in the code above.
My info:
sessionInfo(BRugs)
R version 2.9.0 (2009-04-17)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
character(0)
other attached packages:
[1] BRugs_0.5-1
Vitalie.
Specifying, e.g.
model - function()
x ~ dnorm(mu, tau)%_%T(lower, upper)
writeModel(model)
works for me.
Best,
Uwe
Thanks,
Vitalie.
On Tue, 25 Aug 2009 10:00:44 +0200, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
R Heberto Ghezzo, Dr wrote:
Hello, I am sorry, I have this problem before and Uwe send me the
answer but I misplaced it
Oh dear! But it is not lost, since the answer to the second part of
your problem was: Please read the documentation!
and can not find it.
writing a model for BRugs
library(BRugs)
Loading required package: coda
Loading required package: lattice
Welcome to BRugs running on OpenBUGS version 3.0.3
setwd(c:/tmp)
Error in setwd(c:/tmp) : cannot change working directory
So c:/tmp does not exist or you don't have permissions there?
mo - function(){
+ for (k in 1:p){
+ delta[1,k] ~ dnorm(0,0.1)I(,delta[2,k])
Error: unexpected symbol in:
The answer was: Please read the documentation!
?writeModel (which you want to use on this function) tells you:
As a difference, BUGS syntax allows truncation specification like
this: dnorm(...) I(...) but this is illegal in R. To overcome this
incompatibility, use %_% before I(...): dnorm(...) %_% I(...). The
dummy operator %_% will be removed before the BUGS code is saved.
for (k in 1:p){
delta[1,k] ~ dnorm(0,0.1)I
delta[2,k] ~ dnorm(0,0.1)I(delta[1,k],delta[3,k])
Error: unexpected symbol in delta[2,k] ~ dnorm(0,0.1)I
delta[3,k] ~ dnorm(0,0.1)I(delta[2,k],)}
Error: unexpected symbol in delta[3,k] ~ dnorm(0,0.1)I
}
Error: unexpected '}' in }
so R parser does not like the I(,) construct
which is *not* the problem, see above.
Best wishes,
Uwe
, What is the alternative way of propgramming the
constrain I(lower,upper)
Thanks
Heberto Ghezzo
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] xtable longtable with caption on top
Dear all
I would like to export a matrix to a LaTeX longtable environment
that would have its caption on top of the table. However xtable()
seems to bar me from doing this:
print(xtable(mtcars[1:10,c(1,2)], caption=CARS),
+ tabular.environment=longtable, caption.placement=top
+ , floating=F)
% latex table generated in R 2.9.1 by xtable 1.5-5 package
% Tue Aug 25 13:48:01 2009
\begin{longtable}{rrr}
\hline
mpg cyl \\
\hline
Mazda RX4 21.00 6.00 \\
Mazda RX4 Wag 21.00 6.00 \\
Datsun 710 22.80 4.00 \\
Hornet 4 Drive 21.40 6.00 \\
Hornet Sportabout 18.70 8.00 \\
Valiant 18.10 6.00 \\
Duster 360 14.30 8.00 \\
Merc 240D 24.40 4.00 \\
Merc 230 22.80 4.00 \\
Merc 280 19.20 6.00 \\
\hline
\hline
\caption{CARS}
\end{longtable}
Warning message:
In print.xtable(xtable(mtcars[1:10, c(1, 2)], caption = CARS), :
Attempt to use longtable with caption.placement=top. Changing to bottom.
The error seems strange since I would assume that longtable tables
would naturally have the caption at the beginning of the table,
especially if it streches over multiple pages. There is no such error
when caption.placement=bottom. Would this be a bug in xtable()?
Thank you
Liviu
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Re: [R] latitude and longitude distribution
ogbos okike wrote: Good day to you all, I have lightning data containing date, time, latitude and longitude. I hope that distribution of latitude and longitude will give number of lightning occurrence in a region. I have used factor function to sum up the number of events on latitude and longitude axis and saved as x and y. But when I tried to plot the two, I had and error message ( Error in image.default(x, y, z) : increasing 'x' and 'y' values expected). Every other effort I made failed to be successful - please I am a new user of R. Part of the data I am using is: Hi Ogbos, That's better. What may be more to your liking is this: library(maps) data(worldMapEnv) map(world) points(lightning$lon,lightning$lat,col=red) 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.
[R] Partykit Document
Hi All, I'm a newcomer to the R community, just started learning R. I need to use partykit for developing CART. But I'm not getting any document/help for using partykit/rpart. http://www.agrocampus-ouest.fr/math/useR-2009/slides/Hothorn+Zeileis.pdfgives some example, but it doesn't tell how my own data should be formatted so that it can be accepted by partykit. Example, data(GlaucomaM, package = ipred) is accepted. Now instead of GlaucomaM, I need to give my own data. But the data format for GlaucomaM is not given. So how can I know that? Please help me in this. Thanks and Regards. - Indranil Basu, Bangalore, India [[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] Filling matrix secondary diagonal
Try this: m - matrix(sample(9), 3) diag(m) diag(m[-1,]) diag(m[,-1]) On Mon, Aug 24, 2009 at 4:03 PM, Vaupo m-va...@web.de wrote: Hi, how do i fill the secondary diagonals of a matrix? Is there an funktion like the diag funktion in matlab, where i can specify the diagonal i´d like to fill? -- View this message in context: http://www.nabble.com/Filling-matrix-secondary-diagonal-tp25121745p25121745.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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: table, sum, cat function
Hi r-help-boun...@r-project.org napsal dne 25.08.2009 12:07:11: Hi, the second step in my exercice is to calculate the sum of the amout for each class et not the frequency i have this vector x y 1 100 2 1500 3 3250 4 6250 5 2000 6 450 i want to use the function table and cat to calculate the sum of the amount in each class [0-1000], [1000-3000],[ 3000] It starts to look like some kind of homework. If yes you shall seek the help from your tutor. If you want to do it yourself than tapply, by, aggregate or package doBy could be good starting point. Regards Petr Thank you for your help?? inchallah [[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] FW: Pec Function in R - syntax
Hello everyone - sorry, I forgot to add the syntax for my program in the last message: My syntax is as follows: library(Hmisc) library(survival) library(pec) AL-read.table(PredErr2.txt,header=TRUE) Models-list(Kaplan.Meier=survfit(Surv(OVS,dead)~1,data=AL),EBMT+SN Pgroup=coxph(Surv(OVS,dead)~GroupMod,method=efron,data=AL),EBMTgro up=coxph(Surv(OVS,dead)~GroupGrat, method=efron,data=AL),EBMToriginal=coxph(Surv(OVS,dead)~Gratwohl, method=efron,data=AL),EBMT+SNPoriginal= coxph(Surv(OVS,dead)~Gratwohl+PIL1rAnyC+PIL4AnyT, method=efron,data=AL)) PredError.632plus-pec(object=Models,formula=Surv(OVS,dead)~1,data=AL, exact=TRUE,cens.model=marginal,replan=boot632plus,B=100,verbose=TR UE) Thankyou so much. Kim Dr Kim Pearce CStat Industrial Statistics Research Unit (ISRU) School of Mathematics and Statistics Herschel Building University of Newcastle Newcastle upon Tyne United Kingdom NE1 7RU Tel. 0044 (0)191 222 6244 (direct) Fax. 0044 (0)191 222 8020 Email: k.f.pea...@ncl.ac.uk __ 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] mixed normal distriburtion
I'm trying to draw the density function of a mixed normal distribution in the form of: .6*N(.4,.1)+ .4*N(.8,.1) At first I generate a random sample with size 200 by the below code: means = c(.4,.8) sds = sqrt(c(.1,.1)) ind = sample(1:2, n, replace=TRUE, prob=c(.6,.4)) x=rnorm(n,mean=means[ind],sd=sds[ind]) Then I use the below code for drawing the graph: plot(density(x)) The plot doesn't seem to be belonging to the desired distribution, because there is just one mode in it (I've seen the real graph of this mixed normal in a paper, it has two clear distinct modes). Even the hist() doesn't draw a plot similar to the real graph. I think the generation code isn't correct. Is it? (I've asked the generation code here!) -- Free e-mail accounts at http://zworg.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] Pec function in R
Hello everyone, These are some questions about the 'pec' function in R. These questions deal with prediction error curves and their derivation. Prediction error curves are documented in, for example, Efron-type measures of prediction error for survival analysis by Gerds and Schumacher. I have detailed some syntax that I have used at the bottom of this email. The associated data is available upon request. In the 'pec' function I have formula=Surv(OVS,dead)~1 ...apparently for right censored data, the RH side of the formula is used to specify conditional censoring models when there are no covariates (as in our case)...I understand that we are assuming that censoring occurs totally at random (for all models). Say we have a data set containing potential predictor variables X1,X2Xp. Our survival time variable (time) is measured in months and our status variable (status) has 0=alive and 1=dead. Firstly, I would like to ask about how to derive conditional censoring models. From past conversations it seems that to establish the form of our censoring model(s), we would use status=event=alive =1 and status=event=death=0. Is this correct?.. Now, say we are assuming that the model for censoring is of the cox regression type, do we assess the model for censoring using the usual variable selection procedures where candidate variables for variable selection are X1,X2Xp and we code *alive=1 and dead=0*? Say we found that X1 and X5 should be included in our cox regression model for censoring, then we would enter: formula=Surv(time,status)~X1+X5 and cens.model=cox in function pec. Am I thinking about this correctly? I think it could be more difficult than this, so I'd appreciate some guidance. Replan is the method for estimating prediction error curves. I understand that replan=none would be used when we don't cross validate i.e. we would create a model using a specific bootstrap sample and then evaluate its performance on the same sample. If we are looking at bootstrap crossvalidation (OutOfBag) method...say we have 500 patients.it says in the paper - Gerds and Schumacher - that bootstrap samples Q*_1Q*_B each of size n are drawn with replacement from the original data (we have chosen 100 bootstrap samples i.e. B=100 in our work). So here I assume that n is equal to the total number of patients i.e. 500 in this example? When we decide to sample with replacement, as here,...I am assuming that, for each of the bootstrap samples, there are 500 patients but some of these patients could occur in this bootstrap sample more than once. The documentation says M is the size of the bootstrap samples for sampling *without* replacement. Hence am I correct in thinking that for sampling *with* replacement, M is equal to n ? i.e. M=n by default. However, if we choose Mn then each of our bootstrap samples will have 'sampling without replacement' i.e. the training set would be comprised of n-M patients. Am I correct? [Each of the bootstrap samples acts as a 'training data set' to generate a modelthe model is then validated using the patients which weren't in the bootstrap sample. ] In a study I did using a 286 case data set, I noticed that my prediction error curves seem to terminate at around 35 months when the actual last survival time was 248 months. I checked the survival times and corresponding alive/dead for this data set and noticed that the number of 'deaths' gets very sparse after month 35but I'm still a bit puzzled as to why the curves end at around 35 months. Perhaps the small number of cases in the original data set leads to small training data sets and hence to a termination of the curves at low times? Has anybody else encountered this? In our study, we wanted to develop a model on a specific bootstrap sample and then test it out using observations which aren't in the bootstrap sample (this would be repeated B times)...we chose the 0.632+ bootstrap estimator. There are many types of estimators for prediction error curves. How should we decide which is 'best'? Thanks for any advice on these questions, Kindest Regards, Kim Dr Kim Pearce CStat Industrial Statistics Research Unit (ISRU) School of Mathematics and Statistics Herschel Building University of Newcastle Newcastle upon Tyne United Kingdom NE1 7RU Tel. 0044 (0)191 222 6244 (direct) Fax. 0044 (0)191 222 8020 Email: k.f.pea...@ncl.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with BRugs
This is a BUGS error message that indeed tells you that BUGS cannot truncate that density - not related to R at all. Best, Uwe Ligges Yes, indeed, that does not work in OpenBugs as well. Means that documentation of Open bugs is incorrect (they are using dnorm there). Also I works and T does not work for any distribution I've tried. The problem is that BRugs documentation is not mentioning T and C operators. OpenBugs tels about T and C but not I. And obviously nothing about what T does not work and I should be used instead! If I didn't know from previous versions about I operator I would have been stuck with BRugs code. So I believe this is also a problem of BRugs' documentation. T and C should be mentioned there along side with I. The guys at OpenBugs obviously now the problem - just new users are utterly confused, that's all. Sorry for taking your time. And thanks for the port; indeed, great tool. Vitalie. __ 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] multiple expressions in a legend
Hi, Trying to do something fairly simple. I'm trying to get a legend that combines superscripts with a sequence, like this: r^2 = 1 r^2 = 2 r^2 = 3 ... Except that r^2 is properly formatted as a superscript. I've been playing with substitute and expression and can get an individual line to appear, but I can't seem to get the syntax correct so that I can get several of these to appear using legend. this works: i-1 substitute(r^2==k,list(k=i)) but not this i-seq(1,10) substitute(r^2==k,list(k=i)) Can anyone help? Thanks Michael __ 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] converting a matrix into a fn
Dear All, I have a matrix m of the form m [,1] [,2] [1,] 9072.302 0.0004462366 [2,] 9086.811 0.0005628169 [3,] 9101.320 0.0007126347 [4,] 9115.830 0.0008986976 [5,] 9130.339 0.001126 [6,] 9144.848 0.0014018405 [7,] 9159.357 0.0017344229 [8,] 9173.866 0.0021363563 [9,] 9188.375 0.0026389996 [10,] 9202.884 0.0032406678 where the first col is my variable x and the second is f(x). I've obtained this from the result of a kernel density estimate so I don't know the explict form of f(x) . For some reason i want to integerate f(x) over a certain range using integrate( ). So i want R to consider the above matrix as a function where the first col is the argument x and the second is the realization f(x). I've tried as.function and vectorize but it didn't work out. Could u help me do it? Thanks Maram [[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] formats
Hi every one, what do you mean by %d-%b-%y. is it reading format or writing format. Thanks in Advance. -- View this message in context: http://www.nabble.com/formats-tp25133503p25133503.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] multiple expressions in a legend
Try this: plot(0) ix - as.double(1:3) legend(top, as.expression(lapply(ix, function(i) bquote(r^.(i) On Tue, Aug 25, 2009 at 8:45 AM, Michael Chore...@channing.harvard.edu wrote: Hi, Trying to do something fairly simple. I'm trying to get a legend that combines superscripts with a sequence, like this: r^2 = 1 r^2 = 2 r^2 = 3 ... Except that r^2 is properly formatted as a superscript. I've been playing with substitute and expression and can get an individual line to appear, but I can't seem to get the syntax correct so that I can get several of these to appear using legend. this works: i-1 substitute(r^2==k,list(k=i)) but not this i-seq(1,10) substitute(r^2==k,list(k=i)) Can anyone help? Thanks Michael __ 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] mixed normal distriburtion
I'm trying to draw the density function of a mixed normal distribution
in the form of:
.6*N(.4,.1)+ .4*N(.8,.1)
At first I generate a random sample with size 200 by the below code:
means = c(.4,.8)
sds = sqrt(c(.1,.1))
ind = sample(1:2, n, replace=TRUE, prob=c(.6,.4))
x=rnorm(n,mean=means[ind],sd=sds[ind])
Then I use the below code for drawing the graph:
plot(density(x))
The plot doesn't seem to be belonging to the desired distribution,
because there is just one mode in it (I've seen the real graph of this
mixed normal in a paper, it has two clear distinct modes). Even the
hist() doesn't draw a plot similar to the real graph. I think the
generation code isn't correct. Is it? (I've asked the generation code
here!)
The code is fine - the reason you can't see two peaks is that the two
distributions overlap a lot. Set
means - c(.4, 10)
to see double peaks more clearly.
Regards,
Richie.
Mathematical Sciences Unit
HSL
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[R] Elastic net in R (enet package)
Dear R users, I am using enet package in R for applying elastic net method. In elastic net, two penalities are applied one is lambda1 for LASSO and lambda2 for ridge ( zou, 2005) penalty. But while running the analysis, I realised tht, I optimised only one lambda. ( even when I looked at the example in R, they used only one penality) So, I am wandering which penalty they are referring to? Is it a combination of penalties or one of them. I read the paper of zou and hastie but still in doubt. Thanks in advance Alex [[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] formats
what do you mean by %d-%b-%y. is it reading format or writing format.
%d-%b-%y is a date format - see the help page for strptime.
Example usage:
strptime(01-Jan-84, %d-%b-%y)
strftime(Sys.time(), %d-%b-%y)
Regards,
Richie.
Mathematical Sciences Unit
HSL
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Re: [R] problem with BRugs
Vitalie S. wrote: This is a BUGS error message that indeed tells you that BUGS cannot truncate that density - not related to R at all. Best, Uwe Ligges Yes, indeed, that does not work in OpenBugs as well. Means that documentation of Open bugs is incorrect (they are using dnorm there). Also I works and T does not work for any distribution I've tried. The problem is that BRugs documentation is not mentioning T and C operators. OpenBugs tels about T and C but not I. And obviously nothing about what T does not work and I should be used instead! If I didn't know from previous versions about I operator I would have been stuck with BRugs code. So I believe this is also a problem of BRugs' documentation. T and C should be mentioned there along side with I. The guys at OpenBugs obviously now the problem - just new users are utterly confused, that's all. Please report OpenBUGS documentation issues at the OpenBUGS list. I do not have write access there. Thank you very much, Uwe Sorry for taking your time. And thanks for the port; indeed, great tool. Vitalie. __ 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] Re : Odp: Re : Odp: Re : table function
On Aug 25, 2009, at 6:25 AM, Inchallah Yarab wrote: Thank you very much Peter it works!! the second step is to calculate the sum of variable in each class!!! how can i do this!!! ?table ?xtab De : Petr PIKAL petr.pi...@precheza.cz Cc : r-help@r-project.org Envoyé le : Mardi, 25 Août 2009, 11h53mn 21s Objet : Odp: [R] Re : Odp: Re : table function Hi r-help-boun...@r-project.org napsal dne 25.08.2009 11:28:31:  Thank you Peter, in my vector Z i have missing value NA and i want to count its number in the vector i had did alla this i know the difference between a numeirc and a factor OK. If you know difference between factor and numeric you probably have seen ?factor where is note about how to make NA an extra level #Here is z z-c(10,100, 1000, 1200, 2000, 2200, 3000, 3200, 5000, 6000) #let's put some NA values into it z[c(2,5)] -NA #let's make a cut z.c-cut(z, breaks = c(-Inf, 1000, 3000, Inf), labels = c(0 - 1000, 1000 - 3000, 3000)) #as you see there are NA values but they are not extra level z.c [1] 0 - 1000NA0 - 10001000 - 3000 NA [6] 1000 - 3000 1000 - 3000 300030003000 Levels: 0 - 1000 1000 - 3000 3000 is.na(z.c) [1] FALSE TRUE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE # so let's try what help page says factor(z.c, exclude=NULL) [1] 0 - 1000NA0 - 10001000 - 3000 NA [6] 1000 - 3000 1000 - 3000 300030003000 Levels: 0 - 1000 1000 - 3000 3000 NA # wow we have NA as extra level, let's do table table(factor(z.c, exclude=NULL)) 0 - 1000 1000 - 30003000NA 2332 Regards Petr the goal of this exercice that to count the number of missing value, number betwwen 0-1000 , 1000-3000, 3000. Thank you again for your help De : Petr PIKAL petr.pi...@precheza.cz Cc : r-help@r-project.org EnvoyĂ© le : Mardi, 25 AoĂ»t 2009, 11h15mn 23s Objet : Odp: [R] Re : table function Hi r-help-boun...@r-project.org napsal dne 25.08.2009 10:08:36: Hi Mark, Thank you for your answer !! it works but if i have NA in the vector z what i shoud do to count its number in Z? You do not have NA in z, you manage to convert it somehow to factor. Please try to read about data types and its behaviour. Start with this chapter 2.8 Other types of objects in R intro manual which I suppose you have in doc folder of R program directory. You possibly can convert it back to numeric by e.g. DF$z - as.numeric(as.character(DF$z)) but I presume you need to check your original data maybe by str(your.data) what mode they are and why they are factor if you expect them numeric. Regards Petr x   y    z 1  0   100 5  1   1500 6  1   NA 2  2   500 1  1   NA 5  2  2000 8  5  4500 i did the same but it gives me this error message:  [0 - 1000] [1000 - 3000]    3000       0      0      0 Warning message: In inherits(x, factor) : NAs introduced by coercion Thank you De : Marc Schwartz marc_schwa...@me.com Cc : r-help@r-project.org EnvoyĂ© le : Lundi, 24 Aoűt 2009, 18h33mn 52s Objet : Re: [R] table function On Aug 24, 2009, at 10:59 AM, Inchallah Yarab wrote: hi, i want to use the function table to build a table not of frequence (number of time the vareable is repeated in a list or a data frame!!) but in function of classes [[elided Yahoo spam]] example x   y    z 1  0   100 5  1   1500 6  1   1200 2  2   500 1  1   3500 5  2  2000 8  5  4500 i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] thank you very much for your help See ?cut, which bins a continuous variable. DF  x y  z 1 1 0 100 2 5 1 1500 3 6 1 1200 4 2 2 500 5 1 1 3500 6 5 2 2000 7 8 5 4500 table(cut(DF$z, breaks = c(-Inf, 1000, 3000, Inf),       labels = c(0 - 1000, 1000 - 3000, 3000)))   0 - 1000 1000 - 3000    3000      2      3      2 HTH, Marc Schwartz   [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]]
Re: [R] Help on comparing two matrices
They can be regarded as incidence matrices rather than adjacency
matrices and in that case it follows:
library(igraph)
# incidence matrix to canonical edge list
inc2canel - function(m) {
g - graph.incidence(m)
cp - canonical.permutation(g)
can - permute.vertices(g, cp$labeling)
el - get.edgelist(can)
el[ order(el[,1], el[,2]), ]
}
# test it out
m1 - matrix(c(0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1,
0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0,
0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1), 7)
m2 - m1[7:1, 8:1]
m3 - m2; m3[1, ] - 1
identical(inc2canel(m1), inc2canel(m2)) # TRUE
identical(inc2canel(m1), inc2canel(m3)) # FALSE
On Sun, Aug 23, 2009 at 6:09 PM, Steve
Lianogloumailinglist.honey...@gmail.com wrote:
Hi,
On Sun, Aug 23, 2009 at 4:14 PM, Michael Koganmichael.ko...@gmx.net wrote:
Thanks for all the replies!
Steve: I don't know whether my suggestion is a good one. I'm quite new to
programming, have absolutely no experience and this was the only one I could
think of. :-) I'm not sure whether I'm able to put your tips into practice,
unfortunately I had no time for much reading today but I'll dive into it
tomorrow.
Ok, yeah. I'm not sure what the best way to do this myself, I would at
first see if one could reduce these matrices by some principled manner
and then do a comparison, which might jump to:
Ted: Wow, that's heavy reading. In fact the matrices that I need to compare
are incidence matrices so I suppose it's exactly the thing I need, but I
don't know if I have the basics knowledge to understand this paper within
the next months.
Ted's sol'n. I haven't read the paper, but its title gives me an idea.
Perhaps you can assume the two matrices you are comparing are
adjacency matrices for a graph then use the igraph library to do a
graph isomorphism test between the two graphs represented by your
adjacency matrices and see if they are the same.
This is probably not the most efficient (computationally) way to do
it, but it might be the quickest way out coding-wise.
I see your original example isn't using square matrices, and an
adjacency matrix has to be square. Maybe you can pad your matrices
with zero rows or columns (depending on what's deficient) as an easy
way out.
Just an idea.
Of course, if David's solution is what you need, then no need to
bother with any of this.
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
__
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Help on comparing two matrices
Hi, It looks like you're getting more good stuff, but just to follow up: On Aug 24, 2009, at 4:01 PM, Michael Kogan wrote: Steve: The two matrices I want to compare really are graph matrices, just not adjacency but incidence matrices. There should be a way to get an adjacency matrix of a graph out of its incidence matrix but I don't know it... If you're working with graph data, do yourself a favor and install igraph (no matter what solution you end up using for this particular problem). http://cran.r-project.org/web/packages/igraph/ http://igraph.sourceforge.net/ In there, you'll find the `graph.incidence` function which creates a graph from its incidence matrix. You can then test if the two graphs are isomorphic. That would look like so: library(igraph) g1 - graph.incidence(matrix.1) g2 - graph.incidence(matrix.2) is.iso - graph.isomorphic(g1, g2) # Or, using the (somehow fast) vf2 algorithm is.iso - graph.isomorphic.vf2(g1, g2) HTH, -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on comparing two matrices
Gabor: Wow, that seems to be exactly what I need! Does it matter that
my incidence matrices represent neighborhood relations between
vertices and faces rather than between vertices and edges?
Steve: Yep, I realize that this package is exactly what I'm searching
for. :)
Gabor Grothendieck schrieb:
They can be regarded as incidence matrices rather than adjacency
matrices and in that case it follows:
library(igraph)
# incidence matrix to canonical edge list
inc2canel - function(m) {
g - graph.incidence(m)
cp - canonical.permutation(g)
can - permute.vertices(g, cp$labeling)
el - get.edgelist(can)
el[ order(el[,1], el[,2]), ]
}
# test it out
m1 - matrix(c(0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1,
0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0,
0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1), 7)
m2 - m1[7:1, 8:1]
m3 - m2; m3[1, ] - 1
identical(inc2canel(m1), inc2canel(m2)) # TRUE
identical(inc2canel(m1), inc2canel(m3)) # FALSE
On Sun, Aug 23, 2009 at 6:09 PM, Steve
Lianogloumailinglist.honey...@gmail.com wrote:
Hi,
On Sun, Aug 23, 2009 at 4:14 PM, Michael Koganmichael.ko...@gmx.net wrote:
Thanks for all the replies!
Steve: I don't know whether my suggestion is a good one. I'm quite new to
programming, have absolutely no experience and this was the only one I could
think of. :-) I'm not sure whether I'm able to put your tips into practice,
unfortunately I had no time for much reading today but I'll dive into it
tomorrow.
Ok, yeah. I'm not sure what the best way to do this myself, I would at
first see if one could reduce these matrices by some principled manner
and then do a comparison, which might jump to:
Ted: Wow, that's heavy reading. In fact the matrices that I need to compare
are incidence matrices so I suppose it's exactly the thing I need, but I
don't know if I have the basics knowledge to understand this paper within
the next months.
Ted's sol'n. I haven't read the paper, but its title gives me an idea.
Perhaps you can assume the two matrices you are comparing are
adjacency matrices for a graph then use the igraph library to do a
graph isomorphism test between the two graphs represented by your
adjacency matrices and see if they are the same.
This is probably not the most efficient (computationally) way to do
it, but it might be the quickest way out coding-wise.
I see your original example isn't using square matrices, and an
adjacency matrix has to be square. Maybe you can pad your matrices
with zero rows or columns (depending on what's deficient) as an easy
way out.
Just an idea.
Of course, if David's solution is what you need, then no need to
bother with any of this.
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
__
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] Lattice graph tweaking
To: silwood-r Subject: Removing lattice graph gridlines and editing label box colour Hi, Is it possible to remove the background gridlines from a lattice graph (ie graph made up of multiple individual graphs with annoying blue grid in the backgroun)? Also, Is it possible to change the colour of the individual graph label boxes? - ie the default pink boxes above the individual graphs Thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice graph tweaking
A reproducible example would be nice. Try grid = FALSE for the first question, though I'm unaware which lattice plot you are using where the default is TRUE. So I can't guarantee that will even work. For your second question, add par.settings = list(strip.background = list(col = white)) to your call (e.g. xyplot(..., par.settings = ...)). To see what other parameters you can set in par.settings, try: str(trellis.par.get()) HTH, --sundar On Tue, Aug 25, 2009 at 6:14 AM, Wallis, Daviddavid.walli...@imperial.ac.uk wrote: To: silwood-r Subject: Removing lattice graph gridlines and editing label box colour Hi, Is it possible to remove the background gridlines from a lattice graph (ie graph made up of multiple individual graphs with annoying blue grid in the backgroun)? Also, Is it possible to change the colour of the individual graph label boxes? - ie the default pink boxes above the individual graphs Thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ 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] Plotting the Log-Likelihood
Dear Users,
I would like to plot the log-likelihood (depending on two
parameters).
I wrote the following code:
library(mvtnorm)
sigma-
matrix(c(4,2,2,3),ncol=2)
x-rmvnorm(n=500,mean=c(1,2),sigma=sigma)
likel-
function(param,data){
pos1-which(data[,1]0)
pos2-which(data[,1]=0)
#density
dens-rep(0,nrow(data))
dens[pos1]-dmvnorm(data[pos1,])
*param[1]
dens[pos2]-dmvnorm(data[pos2,])*param[2]
#return log-
likelihood
sum(log(dens))
}
#plotting the log-likelihood
param1-
seq(from=0,to=1,length=200)
param2-seq(from=0,to=1,length=200)
combin-
as.matrix(expand.grid(param1,param2,KEEP.OUT.ATTRS=FALSE))
likel.values-
apply(combin,1,likel,data=x)
matr.likel.values-matrix(likel.values,200,200)
persp(param1,param2,matr.likel.values,
xlab=param1,ylab=param2,
zlab=Likelihood,
theta=5,phi=20,expand=0.5,
col=lightblue,
ltheta=120,shade=0.75,ticktype=detailed)
It is all ok, but I am searching
for a new lighting way in order to earn time.
I need an alternative function
for apply(), when I calculate the log-likelihood for every single couple of
values (param1,param2).
Do you have any ideas?
Thank you in advance,
Enrico Foscolo
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[R] MASS: lda collinearity
I am trying to understand what the lda function in the MASS package calculates when there are more dimensions than examples. It is my understanding that the Fisher Linear Discriminant is not applicable in this case, because the inverse of the covariance matrix cannot be calculated. My question is how the output is calculated when the lda function is sent a dataset with more dimensions than examples. I had expected it to fail. Is a pseudoinverse (or some other technique) used internally to allow computation of the lda in this case? Thank you in advance for your assistance. __ 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] Problems with loading 'wordnet' in a standalone pc
Thanks Uwe. The wordnet is working now. My previous configuration that causes the error was: R - version 2.9.1 wordnet - version 0.1-4 rJava 0.6-2 After updating rJava to version 0.7-0 everything works now. Uwe Ligges-3 wrote: Looks like nobody answered so far: Kelvin Lam wrote: Hi group, I have the following error code after submitting library(wordnet) in a standalone pc. rJava is already in place. Please read the posting guide and tell us: - Which version of R? - Which version of wordnet? - Which version of rJava? - Which version of Java? If anywhere not the most recent one: Please upgrade and report again. Error in .jpackage(pkgname,lib.loc=libname) unused argument(s) (lib.loc=libname) Error: .onLoad failed in 'loadNamespace' for 'wordnet' Error: package/namespace load failed for 'wordnet' My guess is I miss something from the namespace but unfortunately I can't get to the internet. Does anyone know what I'm missing (from the namespace?). Thank you very much! No, probably nothing is missing from the namespace, but probably too old versions or strange libraries in use. Best, Uwe Ligges Kelvin __ 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. -- View this message in context: http://www.nabble.com/Problems-with-loading-%27wordnet%27-in-a-standalone-pc-tp25062412p25136409.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] Removing objects from workspace
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Ronggui Huang Sent: Monday, August 24, 2009 9:26 PM To: Steven Kang Cc: r-help@r-project.org Subject: Re: [R] Removing objects from workspace It depends. If there are patterns in the names, you can make use of pattern argument of ls(). For example, x1=a;x2=b;x3=c ls(pattern=x[1-2]) [1] x1 x2 ls(pattern=x[^1-2]) [1] x3 # to remove x1-x3 rm(list=ls(pattern=x[1-2])) More generally, if you want to remove all butx,xx,xxx, you can use rm(list=ls()[! ls() %in% c(x,xx,xxx)]) You need to quote the x, xx, and xxx. Also, ls()[ ! ls() %in% c(x,xx,xxx) ] is the same as the more direct setdiff( ls(), c(x,xx,xxx) ) Bill Dunlap TIBCO Software Inc - Spotfire Division wdunlap tibco.com Hope this helps. 2009/8/25 Steven Kang stochastick...@gmail.com: Hi all, I am currently woking with hundreds of objects in workspace and whenever I invoke ls() to observe the names of the objects, there are too much of unnecessary variables. For example, if I only require say 3 or 4 objects from hundreds of objects in workspace, are there any methods that may do the job? I have tried rm(-c(x,xx,xxx)), but no luck.. Your feedback in this problem would be highly appreciated. Steve [[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. -- HUANG Ronggui, Wincent PhD Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html __ 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 in plotting a legend
Hello, List, I am a new user of the R project, and I need some help in plotting a legend. I am using the PBSmapping library to plot map of Ohio and heat color it with the count of employees in each county. As a guide, I am using Data Mashups in R. I am able to plot the map with the colors; however, I would like to put a legend with a single box full of these colors, but only show the max and the min value on the left corner and the right corner respectively. Here's a part of the code I am using: #empdata has the employee data with the latitude and longitude for each employee library(PBSmapping) addressEvents-as.EventData(empdataRC,projection=NA) #myShapeFile is a shape file of Ohio imported using importShapefile addressPolys-findPolys(addressEvents,myShapeFile) myTrtFC- table(factor(addressPolys$PID,levels=levels(as.factor(myShapeFile$PID log(myTrtFC)-lTrt mapColors-heat.colors(max(lTrt)+1,alpha=.6)[max(lTrt)-lTrt+1] mapColors[ is.na(mapColors) ] - white pdf(RC-Employees.pdf,version=1.4) plotPolys(myShapeFile,axes=FALSE,bg=white,main= Employees ,xlab=,ylab=,col=mapColors) #here is the current legend, but as you can see, the colors and the box count doesn't match #i would like to see only one box with the gradients of the colors that I used, and the max and the min value on top. I even tried a rectangle but I could not plot it legend(topleft,legend=seq(max(myTrtFC), 0, length.out=5), fill=heat.colors(max(myTrtFC)+1,alpha=.6),title=Headcount, horiz=TRUE,cex=.5,bty=n) Thanks. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in building new function
It worked perfectly. Thanks a lot. On Mon, Aug 24, 2009 at 4:16 AM, Gabor Grothendieck ggrothendi...@gmail.com wrote: 1. Just enter arima at the R console to see its source code (without comments). The source tar.gz for R is found by googling for R, clicking on CRAN in left column and choosing mirror. Or to view it online or get it via svn: https://svn.r-project.org/R/ 2. You want myarima's free variables to be found in the stats package so set the environment of myarima like this: environment(myarima) - asNamespace(stats) On Mon, Aug 24, 2009 at 2:18 AM, Ana Paula Moraanamor...@gmail.com wrote: Hi: I've installed the precompiled binary for Windows. I need to use an existing function, but I want to introduce some slight changes to it. 1. Is there a way for me to find the source files through windows explorer? I know I can see it using edit(object name) but I want to know if I can see it via explorer in some location under the R directory. 2. I don´t want to modify the code of that function until I'm sure that my changes are not causing any harm. So, I got the R-2.9.1.tar.gz file and open the arima.R file. I change the name of the function to myarima, the name of the file to myarima.R, save it and the loaded it using the source command. So far, so good. When I try to execute my function I get the error message Error in Delta %+% c(1, -1) : object 'R_TSconv' not found. So far, the only change I made is add a Hello world in the first line, so my change is not the source of the problem. Looks like my function (although it is exactly the same) is not being able to see this object. Can someone help me out? Am I missing something? Thanks a lot in advance. Regards, Ana [[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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] seeking tutor on statistical learning and data-mining
(posting for my friend) Hi all, We are looking for a tutor who could teach me statistical learning and data-mining using the book: The Elements of Statistical Learning: Data Mining, Inference, and Prediction Please drop me a line if you are interested. Thank you! [[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] Lattice graph tweaking
Sorry, That worked. However, is it possible to single out specific boxes and change their colour? I want the four graph headers in the top right of the trellice to be different colour to the rest. Thanks very much From: Sundar Dorai-Raj [sdorai...@gmail.com] Sent: Tuesday, August 25, 2009 3:19 PM To: Wallis, David Cc: silwood-r; r-help@r-project.org Subject: Re: [R] Lattice graph tweaking A reproducible example would be nice. Try grid = FALSE for the first question, though I'm unaware which lattice plot you are using where the default is TRUE. So I can't guarantee that will even work. For your second question, add par.settings = list(strip.background = list(col = white)) to your call (e.g. xyplot(..., par.settings = ...)). To see what other parameters you can set in par.settings, try: str(trellis.par.get()) HTH, --sundar On Tue, Aug 25, 2009 at 6:14 AM, Wallis, Daviddavid.walli...@imperial.ac.uk wrote: To: silwood-r Subject: Removing lattice graph gridlines and editing label box colour Hi, Is it possible to remove the background gridlines from a lattice graph (ie graph made up of multiple individual graphs with annoying blue grid in the backgroun)? Also, Is it possible to change the colour of the individual graph label boxes? - ie the default pink boxes above the individual graphs Thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ 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] Appending strings at the beginning of a text file
Dear All, I have a piece of text that I want to append to a text file at the beginning of the text file. I have thought about using cat() with the option 'append=T', but the appending, in this case, is done at the bottom of the text file. Any ideas? Thanks in advance, Paul __ 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] Covariates in NLS (Multiple nonlinear regression)
Dear R-users, I am trying to create a model using the NLS function, such that: Y = f(X) + q + e Where f is a nonlinear (Weibull: a*(1-exp(-b*X^c)) function of X and q is a covariate (continous variable) and e is an error term. I know that you can create multiple nonlinear regressions where x is polynomial for example, but is it possible to do this kind of thing when x is a function with unknown coefficients (a,b,c)? Ultimately, I am expecting the output to give individual regression models for each coefficient (a,b,c) with q as a covariate. I have tried the following code, and get the resultant error messages: weib.nls - nls(Y ~ (a*(1-exp(-b*X^c)))|q, + data=DATA, + start=c(a=75,b=0.05,c=0.7)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates summary(weib.nls) weib.nls2 - nls(Y~ (a*(1-exp(-b*X^c)))+q, + data=DATA, + start=c(a=75,b=0.05,c=0.7)) Error in numericDeriv(form[[3L]], names(ind), env) : Missing value or an infinity produced when evaluating the model Many thanks in advance! Lindsay __ 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] converting a matrix into a fn
On Aug 25, 2009, at 8:45 AM, maram salem wrote: Dear All, I have a matrix m of the form m [,1] [,2] [1,] 9072.302 0.0004462366 [2,] 9086.811 0.0005628169 [3,] 9101.320 0.0007126347 [4,] 9115.830 0.0008986976 [5,] 9130.339 0.001126 [6,] 9144.848 0.0014018405 [7,] 9159.357 0.0017344229 [8,] 9173.866 0.0021363563 [9,] 9188.375 0.0026389996 [10,] 9202.884 0.0032406678 where the first col is my variable x and the second is f(x). I've obtained this from the result of a kernel density estimate so I don't know the explict form of f(x) . For some reason i want to integerate f(x) over a certain range using integrate( ). So i want R to consider the above matrix as a function where the first col is the argument x and the second is the realization f(x). I've tried as.function and vectorize but it didn't work out. Could u help me do it? Thanks Maram http://finzi.psych.upenn.edu/Rhelp08/2008-July/169121.html [[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. 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] Appending strings at the beginning of a text file
Try this;
fConn - file('test.txt', 'r+')
Lines - readLines(fConn)
writeLines(paste(Text at beginning of file, Lines, sep = \n),
con = fConn)
On Tue, Aug 25, 2009 at 12:54 PM, Paul Smith phh...@gmail.com wrote:
Dear All,
I have a piece of text that I want to append to a text file at the
beginning of the text file.
I have thought about using cat() with the option 'append=T', but the
appending, in this case, is done at the bottom of the text file. Any
ideas?
Thanks in advance,
Paul
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
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[R] Help with nls and error messages singular gradient
Hi All, I'm trying to run nls on the data from the study by Marske (Biochemical Oxygen Demand Interpretation Using Sum of Squares Surface. M.S. thesis, University of Wisconsin, Madison, 1967) and was reported in Bates and Watts (1988). Data is as follows, (stored as mydata) time bod 11 0.47 22 0.74 33 1.17 44 1.42 55 1.60 67 1.84 79 2.19 8 11 2.17 I then run the following; #Plot initial curve plot(mydata$time, mydata$bod,xlab=Time (in days),ylab=biochemical oxygen demand (mg/l) ) model - nls(bod ~ beta1/(1 - exp(beta2*time)), data = mydata, start=list(beta1 = 3, beta2 = -0.1),trace=T) The start values are recommended, (I have used these values in SPSS without any problems, SPSS returns values of Beta1 = 2.4979 and Beta2 = -2.02 456) but return the error message, Error in nls(bod ~ beta1/(1 - exp(beta2 * time)), data = mydata, start = list(beta1 = 3, : singular gradient Can anyone offer any advice? Thanks in advance Mike -- Michael Pearmain Senior Analytics Research Specialist Statistics are like women; mirrors of purest virtue and truth, or like whores to use as one pleases Google UK Ltd Belgrave House 76 Buckingham Palace Road London SW1W 9TQ United Kingdom t +44 (0) 2032191684 mpearm...@google.com If 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. [[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] lme, lmer, gls, and spatial autocorrelation
Manuel,
Thanks for the reference. I printed it out and read through it this
morning.
I think I'm going to take a gls approach. I've spent the last couple weeks
reading about spatial autocorrelation, and found that the world of SAC is
large, complex, and requires more time than I currently have. Using gls
seems a reasonable compromise between statistical rigour, and the
unfortunate but real constraint of my limited time to work on this project.
According to Dorman et al, in their (admittedly limited) tests, GLS worked
reasonably well with Poisson distributed synthetic data.
Also, I've come to think that the ability to do model comparison would be
useful. While I would like to be able to confidently choose a model for
spatial autocorrelation a priori, based on biological knowledge, I don't
have enough information to do this. Even after some data exploration, using
variograms and plots of Moran's I, it still seems like there's insufficient
information. Using a fitness score such as AIC, I could compare a small
number of reasonable models to find the most appropriate error structure.
Additionally, I could compare the SAC-informed and SAC-ignorant models to
get a holistic assessment of the importance of SAC in my data.
Tim Handley
Fire Effects Monitor
Santa Monica Mountains National Recreation Area
401 W. Hillcrest Dr.
Thousand Oaks, CA 91360
805-370-2347
Manuel Morales
Manuel.A.Morales
@williams.edu To
timothy_hand...@nps.gov
08/24/2009 05:31 cc
PMBert Gunter
gunter.ber...@gene.com,
r-help@r-project.org
Subject
Re: [R] lme, lmer, gls, and spatial
autocorrelation
Hi Tim,
I don't believe there is a satisfactory solution in R - at least yet -
for non-normal models. Ultimately, this should be possible using lmer()
but not in the near-term. One possibility is to use glmPQL as described
in:
Dormann, F. C., McPherson, J. M., Araújo, M. B., Bivand, R., Bolliger,
J., Carl, G., Davies, R. G., Hirzel, A., Jetz, W., Kissling, W. D.,
Kühn, I., Ohlemüller, R., Peres-Neto, P. R., Reineking, B., Schröder,
B., Schurr, F. M. and Wilson, R. 2007. Methods to account for spatial
autocorrelation in the analysis of species distributional data: a
review. – Ecography 30: 609–628.
However, note the caution:
This is an inofficial abuse of a Generalized Linear Mixed Model
function (glmmPQL {MASS}), which is a wrapper function for lme {nlme},
which in turn internally calls gls {nlme}.
If all you need are parameter estimates, fine. If you want to do model
comparison, though, no luck.
Manuel
On Mon, 2009-08-24 at 12:10 -0700, timothy_hand...@nps.gov wrote:
Bert -
I took a look at that page just now, and I'd classify my problem as
spatial regression. Unfortunately, I don't think the spdep library fits
my
needs. Or at least, I can't figure out how to use it for this problem.
The
examples I have seen all use spdep with networks. They build a graph,
connecting each location to something like the nearest N neighbors,
attach
some set of weights, and then do an analysis. The plots in my data have a
very irregular, semi-random, yet somewhat clumped (several isolated
islands), spatial distribution. Honestly, it's quite weird looking. I
don't
know how to cleanly turn this into a network, and even if I did, I don't
know that I ought to. To me (and please feel free to disagree) it seems
more natural to use a matrix of distances and associated correlations,
which is what the gls function appears to do.
In the ecological analysis section, it looks like both 'ade4' and 'vegan'
may have helpful tools. I'll explore that some more. However, I still
think
that one of lme or gls already has the functionality I need, and I just
need to learn how to use them properly.
Tim Handley
Fire Effects Monitor
Santa Monica Mountains National Recreation Area
401 W. Hillcrest Dr.
Thousand Oaks, CA
[R] draw n-1 lines with n X and n Y
I want to draw lines using a matrix with the X-axe on the first column, and the the Y-axe on the second column. These lines have to link the n points of a graph, so they are n-1, but the resout using these two commands: X - matrix(scan(MaxInterEdges.R,0), ncol=2); lines(X[,1], X[,2], type = l, col = red); is wrong, becose the lines are many more. What can I do? -- View this message in context: http://www.nabble.com/draw-n-1-lines-with-n-X-and-n-Y-tp25137514p25137514.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] scatterplot3d bug??
I finally found some time to look at it: Yes, it does not work properly for angles 180 degrees. Will try to find a fix and make a new release soon. Best wishes, Uwe Ligges Vivek Ayer wrote: Hey guys, Not sure if I encountered a bug with the scatterplot3d function. Here's the calls I made: s3d1 - scatterplot3d(TotLogDisttenp,TotDifftenp,TotMeasuredRSLtenp,pch=16,highlight.3d=TRUE,angle=40,type=h,main=MRSL ~ LogDist + Diff); s3d1$plane3d(fitols); s3d1 - scatterplot3d(TotLogDisttenp,TotDifftenp,TotMeasuredRSLtenp,pch=16,highlight.3d=TRUE,angle=130,type=h,main=MRSL ~ LogDist + Diff); s3d1$plane3d(fitols); s3d1 - scatterplot3d(TotLogDisttenp,TotDifftenp,TotMeasuredRSLtenp,pch=16,highlight.3d=TRUE,angle=210,type=h,main=MRSL ~ LogDist + Diff); s3d1$plane3d(fitols); s3d1 - scatterplot3d(TotLogDisttenp,TotDifftenp,TotMeasuredRSLtenp,pch=16,highlight.3d=TRUE,angle=310,type=h,main=MRSL ~ LogDist + Diff); s3d1$plane3d(fitols); Essentially four plots showing the data from different angles. This includes the fit plane. The first two graphs make sense, but for the latter two, the fit plane is not making sense. Take a look at the attached png. Is it a bug? Thanks, Vivek __ 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] Covariates in NLS (Multiple nonlinear regression)
I think you need to consult a local statistician, as you appear to be so far out of your depth statistically that even the best-intentioned help from this list may not suffice. For example, if I were that local statistician, I would ask: what is the context? -- what is the underlying scientific issue that led to this model? -- what are the data? -- etc. etc. before I would even hazard advice on the statistical details(Please do not answer these questions either publicly or privately, as they are just fyi's). Bert Gunter Genentech Nonclinical Biostatisics -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Lindsay Banin Sent: Tuesday, August 25, 2009 9:02 AM To: 'r-help@r-project.org' Subject: [R] Covariates in NLS (Multiple nonlinear regression) Dear R-users, I am trying to create a model using the NLS function, such that: Y = f(X) + q + e Where f is a nonlinear (Weibull: a*(1-exp(-b*X^c)) function of X and q is a covariate (continous variable) and e is an error term. I know that you can create multiple nonlinear regressions where x is polynomial for example, but is it possible to do this kind of thing when x is a function with unknown coefficients (a,b,c)? Ultimately, I am expecting the output to give individual regression models for each coefficient (a,b,c) with q as a covariate. I have tried the following code, and get the resultant error messages: weib.nls - nls(Y ~ (a*(1-exp(-b*X^c)))|q, + data=DATA, + start=c(a=75,b=0.05,c=0.7)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates summary(weib.nls) weib.nls2 - nls(Y~ (a*(1-exp(-b*X^c)))+q, + data=DATA, + start=c(a=75,b=0.05,c=0.7)) Error in numericDeriv(form[[3L]], names(ind), env) : Missing value or an infinity produced when evaluating the model Many thanks in advance! Lindsay __ 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] Decision boundaries for lda function?
I got hold of 'Modern Applied Statistics with S' by Venables and Ripley (p.
335-336) and I was able to answer my own question:
###
AA-read.table(http://www.natursyn.dk/online/fingerprinting.txt,header=T)
aa.grp - factor(c(rep('f',13),rep('b',10),rep('p',10)))
aa.lda-lda(as.matrix(AA[3:9]),AA$group)
aa.ld-predict(aa.lda,dimen=2)$x
eqscplot(aa.ld,type=n,xlab=LD1, ylab=LD2,tol=0.25, las=1)
text(aa.ld,labels = as.character(aa.grp))
aa.lda2 - lda(aa.ld, aa.grp)
x - seq(-7, 5.5, 0.25)
y - seq(-4.5, 6.5, 0.25)
Xcon - matrix(c(rep(x,length(y)),
rep(y, rep(length(x), length(y,,2)
aa.pr1 - predict(aa.lda2, Xcon)$post[, c(f,b)] %*% c(1,1)
contour(x, y, matrix(aa.pr1, length(x), length(y)),
levels=0.5, add=T, lty=3,method=simple)
aa.pr2 - predict(aa.lda2, Xcon)$post[, c(p,b)] %*% c(1,1)
contour(x, y, matrix(aa.pr2, length(x), length(y)),
levels=0.5, add=T, lty=3,method=simple)
aa.pr3 - predict(aa.lda2, Xcon)$post[, c(f,p)] %*% c(1,1)
contour(x, y, matrix(aa.pr3, length(x), length(y)),
levels=0.5, add=T, lty=3,drawlabels=F)
###
Thomas Larsen wrote:
Hi,
I am using the lda function from the MASS library. I would to find the
decision boundaries of each class and subsequently plot them. I wonder if
anybody can offer any help on this topic?
Below I applied the lda function on a small dataset of mine.
Any help will be much appreciated.
library(MASS)
AA-read.table(http://www.natursyn.dk/online/fingerprinting.txt,header=T)
aa.lda-lda(as.matrix[3:9],AA$group)
aa.ld-predict(aa.lda,dimen=2)$x
eqscplot(aa.ld,type=n,xlab=LD1, ylab=LD2,las=1)
text(aa.ld,c(rep('f',13),rep('b',10),rep('p',10)))
aa.mean-lda(aa.ld,AA$group)$means
points(aa.mean,pch=3)
Best,
Thomas Larsen
Leibniz-Laboratory for Stable Isotope Research
Max-Eyth-Str. 11-13, 24118 Kiel, Germany
Work: +49-431-880-3896
--
View this message in context:
http://www.nabble.com/Decision-boundaries-for-lda-function--tp24823213p25137633.html
Sent from the R help mailing list archive at Nabble.com.
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[R] how to pass user input to a function?
Hi everyone,
I'm building a website (http://yourpsyche.org) using Jeffrey Horner's
awesome Rapache module. I want to take user input, and pass it to an R
script. At first I was simply using if else statements, but after a
while I had so many nested if else's in my code that my head was
spinning. So then I started using cat() and source() to write
temporary files and read them back in (see example below). I've
searched around, and I think there might be a better way to do it with
substitute(), but I can't seem to figure it out (see attempt below).
Here is a minimal example:
###set up simple example###
GET - list(pass.var=b)
a - 1:10
b - 11:20
###using if else works but becomes confusing when I have a lot of variables
to pass###
if(GET$pass.var==a)
+ {
+ mean(a)
+ } else if(GET$pass.var==b)
+ {
+ mean(b)
+ }
[1] 15.5
###writing to a temporary file works but feels like a hack and results in
many temp
files###
cat('print(mean(', GET$pass.var,'))', file=tmp.R,sep=)
source(tmp.R)
[1] 15.5
###this seems promising but I can't figure it out###
substitute(mean(x), list(x=GET$pass.var))
mean(b)
###is there a better way?###
Thanks!
--
Ista Zahn
Graduate student
University of Rochester
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Re: [R] scatterplot3d bug??
Great! Thanks again! Vivek 2009/8/25 Uwe Ligges lig...@statistik.tu-dortmund.de: I finally found some time to look at it: Yes, it does not work properly for angles 180 degrees. Will try to find a fix and make a new release soon. Best wishes, Uwe Ligges Vivek Ayer wrote: Hey guys, Not sure if I encountered a bug with the scatterplot3d function. Here's the calls I made: s3d1 - scatterplot3d(TotLogDisttenp,TotDifftenp,TotMeasuredRSLtenp,pch=16,highlight.3d=TRUE,angle=40,type=h,main=MRSL ~ LogDist + Diff); s3d1$plane3d(fitols); s3d1 - scatterplot3d(TotLogDisttenp,TotDifftenp,TotMeasuredRSLtenp,pch=16,highlight.3d=TRUE,angle=130,type=h,main=MRSL ~ LogDist + Diff); s3d1$plane3d(fitols); s3d1 - scatterplot3d(TotLogDisttenp,TotDifftenp,TotMeasuredRSLtenp,pch=16,highlight.3d=TRUE,angle=210,type=h,main=MRSL ~ LogDist + Diff); s3d1$plane3d(fitols); s3d1 - scatterplot3d(TotLogDisttenp,TotDifftenp,TotMeasuredRSLtenp,pch=16,highlight.3d=TRUE,angle=310,type=h,main=MRSL ~ LogDist + Diff); s3d1$plane3d(fitols); Essentially four plots showing the data from different angles. This includes the fit plane. The first two graphs make sense, but for the latter two, the fit plane is not making sense. Take a look at the attached png. Is it a bug? Thanks, Vivek __ 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 pass user input to a function?
Hi Ista - I've had success using the do.call function with RApache do.call(func, arglst) where 'func' is the function you desire to use, 'arglst' is a list of arguments. You may need to do some cleanup of the arguments using evalq() function (so that numbers are treated as numbers, for example). HTH, Mike -- p: 415.860.4347 b: www.dataspora.com/blog t: www.twitter.com/dataspora __ 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.
