Re: [R] Function try and Results of a program
On Sep 4, 2010, at 6:10 AM, Evgenia wrote:
Hello, users.
Dear users,
***I have a function f to simulate data from a model (example
below used
only to show my problems)
f-function(n,mean1){
a-matrix(rnorm(n, mean1 , sd = 1),ncol=5)
b-matrix(runif(n),ncol=5)
data-rbind(a,b)
out-data
out}
*I want to simulate 1000 datasets (here only 5) so I use
S-list()
for (i in 1:5){
S[[i]]-f(n=10,mean1=0)}
**I have a very complicated function for estimation of a model
which I
want to apply to Each one of the above simulated datasets
fun-function(data){data-as.matrix(data)
sink(' Example.txt',append=TRUE)
cat(\n***\nEstimation
\n\nDataset Sim : ,
i )
d-data%*%t(data)
s-solve(d)
print(s)
out-list (s,d)
out
}
results-list()
for(i in 1:5){
tmp - try(fun(data=S[[i]]))
results[[i]] - ifelse(is(tmp,try-error),NA,tmp)
}
My problem is that results have only the 1st element of the
result lists
of fun (i.e. only although tmp gives me both s and d.
Two problems:
One: is the misguided use of unmatched sink calls resulting in an
accumulation of diversions of the R output. If your run that at the
console you need to type sink() five times to get any response back
from the console.
Two: the misguided use of ifelse when you should be using if ()
{}else{} to test a single condition and execute conditional
assignment. ifelse if for working with vectors, not with lists.
Suggestions:
use the append = TRUE parameter to sink and unsink at the end of that
function
I'm not sure about how you are using the test for error but since you
did not construct any errors I cannot really be too sure. If it is
working for you then use this instead:
if (is(tmp,try-error) ){results[[i]] - NA} else{results[[i]] - tmp}
--
David.
Thanks
Evgenia
--
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Re: [R] Function try and Results of a program
On Sep 4, 2010, at 12:41 PM, Evgenia wrote:
David,
your suggestion about try works perfect for me.
I still have a problem with sink. Could you explain me better your
suggestion?
When you sink to a file, you will continue sending console output to
that file until you issue sink(). And every time you do it it creates
an extra layer of redirection (the help page calls these diversions)
that will need to be undone to get back to regular console behavior.
?sink # yes, one needs to R~all~TM
If you wanted a record of what that function was doing you would need
to:
a) initialize the file with append=FALSE outside the loop (not sure if
you need to do that, but it does help to get rid of earlier failed
efforts as well
b) open the sink file with append=TRUE inside the function
c) cat() the two matrices separately since lists cannot be cat()-
ted,,, and
d)unsink with sink() at the end of the function.
sink(example.txt, append=FALSE); cat(\n ); sink() #blank line to
initialize
fun-function(data){ data-as.matrix(data)
sink(example.txt, append=TRUE); cat(\nEstimate : , i, \n )
d-data%*%t(data); cat(d= \n,d, \n)
s-solve(d); cat(s= \n,s, \n)
out-list(s=s,d=d); sink()
return(out)
}
View this message in context:
http://r.789695.n4.nabble.com/Function-try-and-Results-of-a-program-tp2526621p2526822.html
Sent from the R help mailing list archive at Nabble.com.
--
David Winsemius, MD
West Hartford, CT
#
An unfortunate effect of Nabble use is that it leads one to believe
that the entire world sees your earlier postings:
#-
f-function(n,mean1){
a-matrix(rnorm(n, mean1 , sd = 1),ncol=5)
b-matrix(runif(n),ncol=5)
data-rbind(a,b)
out-data
out}
*I want to simulate 1000 datasets (here only 5) so I use
S-list()
for (i in 1:5){
S[[i]]-f(n=10,mean1=0)}
**I have a very complicated function for estimation of a model
which I
want to apply to Each one of the above simulated datasets
fun-function(data){data-as.matrix(data)
sink(' Example.txt',append=TRUE)
cat(\n***\nEstimation
\n\nDataset Sim : ,
i )
d-data%*%t(data)
s-solve(d)
print(s)
out-list (s,d)
out
}
results-list()
for(i in 1:5){
tmp - try(fun(data=S[[i]]))
results[[i]] - ifelse(is(tmp,try-error),NA,tmp)
}
My problem is that results have only the 1st element of the result
lists
of fun (i.e. only although tmp gives me both s and d.
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Re: [R] What solve() does?
On Sep 4, 2010, at 2:29 PM, Petar Milin wrote: Thank you so much! This is very useful! Any thoughts about how to run Gaussian elimination? Do some searching? RSiteSearch(gaussian elimination, restrict = c(Rhelp10, Rhelp08, Rhelp02, functions ) ) returns (among other things) a link to a John Fox post from 2005: http://finzi.psych.upenn.edu/R/Rhelp02/archive/49950.html -- David. Best, PM On 04/09/10 20:23, Paul Johnson wrote: On Wed, Sep 1, 2010 at 5:36 AM, Petar Milinpmi...@ff.uns.ac.rs wrote: Hello! Can anyone explain me what solve() function does: Gaussian elimination or iterative, numeric solve? In addition, I would need both the Gaussian elimination and iterative solution for the course. Are the two built in R? Thanks! PM Hello, Petar: I think you are assuming that solve uses an elementary linear algebra paper and pencil procedure, but I don't think it does. In a digital computer, those things are not precise, and I think the folks here will even say you shouldn't use solve to get an inverse, but I can't remember all of the details. To see how solve works ... Let me show you a trick I just learned. Read ?solve notice it is a generic method, meaning it does not actually do the calculations for you. Rather, there are specific implementations for different types of cases. To find the implementations, run methods(solve) I get: methods(solve) [1] solve.default solve.qr Then if you want to read HOW solve does what it does (which I think was your question), run this: solve.default or solve.qr In that code, you will see the chosen procedure depends on the linear algebra libraries you make available. I'm no expert on the details, but it appears QR decomposition is the preferred method. You can read about that online or in numerical algebra books. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] non-zero exit status error when install GenomeGraphs
(Caveat: I am not a bioc user.) The error messages suggest that you are missing dependencies. I looked at the documentation for GenomeGraphs and it does not list any dependencies, but I have no way of knowing how careful or knowledgeable the authors may or may not have bben when they composed that document. The fact that you are posting to the wrong mailing list and are not including what version of linux (although there is a hint it may be RedHat5) you are running suggests you could be fairly new at this. Is biocLite the correct function for installing a bioc package? It appears it may be, but I'm wondering if there is an argument for dependencies as there is in install.packages() that you need to set to TRUE? If biocLite has a ,... in its argument list (and the error message suggests that it does) then you may get better results with the same call with an addition of dependencies=TRUE. Or you could first install the packages that are reported missing: XML and biomaRt, and then try again as you did before. Links to the bioc mailing lists can be found here: http://www.bioconductor.org/help/index.html -- David. On Sep 4, 2010, at 4:07 PM, chen chao wrote: Hi, I am trying to install GenomeGraphs package from bioconductor, but failed by a non-zero exit error. From the error message, it seems that there is a shared library problem. Any suggestion on fixing it? Thanks so much. sessionInfo() R version 2.10.1 (2009-12-14) x86_64-unknown-linux-gnu locale: [1] LC_CTYPE=en_US.iso885915 LC_NUMERIC=C [3] LC_TIME=en_US.iso885915LC_COLLATE=en_US.iso885915 [5] LC_MONETARY=C LC_MESSAGES=en_US.iso885915 [7] LC_PAPER=en_US.iso885915 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.iso885915 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.10.1 source(http://bioconductor.org/biocLite.R;) biocLite(GenomeGraphs)Warning messages: 1: In safeSource() : Redefining 'biocinstall' 2: In safeSource() : Redefining 'biocinstallPkgGroups' 3: In safeSource() : Redefining 'biocinstallRepos' biocLite(GenomeGraphs) Using R version 2.10.1, biocinstall version 2.5.11. Installing Bioconductor version 2.5 packages: [1] GenomeGraphs Please wait... Warning in install.packages(pkgs = pkgs, repos = repos, ...) : argument 'lib' is missing: using '/cchome/cchen1/R/x86_64-unknown-linux-gnu-li brary/2.10' trying URL ' http://www.bioconductor.org/packages/2.5/bioc/src/contrib/GenomeGrap hs_1.6.0.tar.gz' Content type 'application/x-gzip' length 585078 bytes (571 Kb) opened URL == downloaded 571 Kb * installing *source* package 'GenomeGraphs' ... ** R ** data ** inst ** preparing package for lazy loading Error in dyn.load(file, DLLpath = DLLpath, ...) : unable to load shared library '/apps/rhel5/x86_64/R/R-2.10.1//lib64/R/library/ XML/libs/XML.so': libxmlsec1.so.1: cannot open shared object file: No such file or directory Error : .onLoad failed in 'loadNamespace' for 'XML' Error : package 'biomaRt' could not be loaded ERROR: lazy loading failed for package 'GenomeGraphs' * removing '/userhom2/3/cchen1/R/x86_64-unknown-linux-gnu-library/2.10/GenomeGra phs' The downloaded packages are in '/tmp/Rtmp3wsJxw/downloaded_packages' Warning message: In install.packages(pkgs = pkgs, repos = repos, ...) : installation of package 'GenomeGraphs' had non-zero exit status -- Chen, Chao Psychiatry University of Chicago 924 E 57th St, Chicago, IL 60637 U. S. A. MOE Key Laboratory of Contemporary Anthropology and Center for Evolutionary Biology, School of Life Sciences and Institutes of Biomedical Sciences, Fudan University 220# Handan Road, Shanghai (200433) P.R.China [[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 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] Please explain do.call in this context, or critique to stack this list faster
:
## http://stackoverflow.com/questions/tagged/r
## I started to type in the question and 3 plausible answers
## popped up before I could finish.
## The terse answer is:
shortAnswer - do.call(rbind,mylist)
## That's the right answer, see:
shortAnswer == dataComplete
## But I don't understand why it works.
## More importantly, I don't know if it is fastest, or best.
## It is certainly less error prone than dataComplete
## First, make a bigger test case and use system.time to evaluate
phony - function(i){
data.frame(w=rnorm(1000), x=rnorm(1000),y=rnorm(1000),z=rnorm(1000))
}
mylist - lapply(1:1000, phony)
### First, try the terse way
system.time( shortAnswer - do.call(rbind, mylist) )
### Second, try the complete way:
m - 1000
nr - nrow(df1)
nc - ncol(df1)
system.time(
dataComplete - as.data.frame(matrix(0, nrow = nr*m, ncol = nc))
)
system.time(
for (j in 1:m) dataComplete[(((j-1)*nr) + 1):(j*nr), ] -
mylist[[j]]
)
## On my Thinkpad T62 dual core, the shortAnswer approach takes
about
## three times as long:
## system.time( bestAnswer - do.call(rbind,mylist) )
##user system elapsed
## 14.270 1.170 15.433
## system.time(
## +dataComplete - as.data.frame(matrix(0, nrow = nr*m, ncol =
nc))
## + )
##user system elapsed
## 0.000 0.000 0.006
## system.time(
## + for (j in 1:m) dataComplete[(((j-1)*nr) + 1):(j*nr), ] -
mylist[[j]]
## + )
##user system elapsed
## 4.940 0.050 4.989
## That makes the do.call way look slow, and I said hey,
## our stupid for loop at the beginning may not be so bad.
## Wrong. It is a disaster. Check this out:
## resultDF - phony(1)
## system.time(
## + for (i in 2:1000) resultDF - rbind(resultDF, mylist[[i]])
## +)
##user system elapsed
## 159.740 4.150 163.996
--
Paul E. Johnson
Professor, Political Science
1541 Lilac Lane, Room 504
University of Kansas
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--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
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Re: [R] Linear Logistic Regression - Understanding the output (and possibly the test to use!)
On Sep 4, 2010, at 6:53 PM, st...@wittongilbert.free-online.co.uk wrote: Hi I know asking which test to use is frowned upon on this list... so please do read on for at least a couple on sentences... I have some multivariate data slit as follows Tumour Site (one of 5 categories) # Chemo Schedule (one of 3 cats) ## Cycle (one of 3 cats*) ## Dose (one of 3 cats*) # *These are actually integers but for all our other analysis so far we have grouped them into logical bands of categories. The dependant variable is Reaction or No Reaction I have individually analysed each of the independant variables against Reaction/No Reaction using ChiSq and Fisher Tests. Those marked ## produced p values less than 0.05, and those marked # produce p values close to 0.05. We believe that Cycle is the crucial piece of data - the others just appear to be different because there are more early cycles in certain groups than others. SO - I believe what I need to do is a Linear Logistic Regression on the 4 independant variables. And I'm expecting it to show that the tumour site, schedule and dose don't matter, only the cycle matters. Done a lot of reading and I'm clueless!! I think I want to do something like: glm (reaction ~ site + sched + cycle + dose, data=mydata, family=poisson) I am then expecting to see some very long output with lots of numbers... ...my question is TWO fold - 1. is glm the right thing to use before I waste my time Yes, but if your outcome variable is binomial then the family argument should be binomial. (And if you thought it should be poisson, then why below did you use gaussian??? and 2. how do I interpret the result! Result? What result? I do see any description of your data, nor any code. (I'm kind of expect a lecture here as I'm really looking for a nice snappy 'p0.05 means this variable is the one having the influence' type answer and I suspect I'm going to be told thats not possible...! I think you need to consult a statistician or someone who has taken the time to read that statistical mumbo jumbo you don't want to learn. This mailing list is not set up to be a tutorial site. (Re your request below: Some years ago I saw one of those programmed learning texts by Kleinbaum on logistic regression. Maybe you could read it and see if it makes your consulting sessions go more smoothly.) http://www.bookfinder.com/search/?author=kleinbaumtitle=logistic+regressionlang=enisbn=submit=Begin+searchnew_used=*destination=uscurrency=USDmode=basicst=srac=qr I have a couple of Kleinbaum's (et al) other texts and find them to be well written and reasoned, so I suspect the citation above would be as accessible as any. To be clear the example given in the docs is: library(MASS) snipped an example that was not relevant to logistic regression --- Either can someone point me to a decent place that would explain what the means or provide me some pointers? i.e. which of the variables has the influence on the outcome in the anorexia data? Please don't shout!! happy to be pointed to a reference but would prefer one in common english not some stats mumbo jumbo! Calum -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How can I fixe convergence=1 in optim
On Sep 4, 2010, at 4:18 PM, Sally Luo wrote: Hi R users, I am using the optim funciton to maximize a log likelihood function. My code is as follows: p-optim(c(-0.2392925,0.4653128,-0.8332286, 0.0657, -0.0031, -0.00245, 3.366, 0.5885, -0.8, 0.0786,-0.00292,-0.00081, 3.266, -0.3632, -0.49, 0.1856, 0.00394, -0.00193, -0.889, 0.5379, -0.63, 0.213, 0.00338, -0.00026, -0.8912, -0.3023, -0.56), f, method =BFGS, hessian =TRUE, y=y,X=X,W=W) After I ran the code, I got the following results: ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~ p $par [1] 2.235834e-02 1.282826e-01 -3.786014e-01 7.422526e-02 3.037931e-02 -2.570156e-03 3.365872e+00 2.618893e-01 -1.987859e-06 [10] 7.970083e-02 2.878574e-03 -1.391019e-03 3.265966e+00 -4.153697e-01 -3.185684e-03 1.833200e-01 -7.247683e-03 -3.156813e-03 [19] -8.889219e-01 6.208612e-01 2.678643e-04 2.183787e-01 2.715062e-02 2.943905e-04 -8.913260e-01 -5.100482e-01 -3.477559e-04 $value [1] -932.1423 $counts function gradient 1439 100 $convergence [1] 1 $message NULL $hessian ( I omitted the approximation results for the hessian here to save space) ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~ The error code 1 for convergence shown above means that the iteration limit maxit had been reached. How can I fix this problem and achieve convergence for my optimization problem? Can I increase the number of maxit so that convergence might occur? I am wondering how you expect us to guess at the answer? You are the one who know what f is and you are the one who has the option of increasing maxit. If the question is how to increase maxit, then the answer is perhaps as easy as: ?optim -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Linear Logistic Regression - Understanding the output (and possibly the test to use!)
On Sep 5, 2010, at 6:06 AM, st...@wittongilbert.free-online.co.uk wrote: David Winsemius wrote: 1. is glm the right thing to use before I waste my time Yes, but if your outcome variable is binomial then the family argument should be binomial. (And if you thought it should be poisson, then why below did you use gaussian??? Used gaussian below because it was the example from the docs. Thats not my data, its example data which was not binomial. and 2. how do I interpret the result! Result? What result? I do see any description of your data, nor any code. I didn't provide MY DATA because I thought that would complicate things even further. So I was hoping for some advice on how to interpret the result of the example data so that I could then apply that to my data. I haven't even tried to run my data as I couldn't see what the output of the examples was trying to tell me. I didn't think that providing commentary on ols regression results was going to be that germane to setting up and running logistic regression. Why haven't you tried a Google search for tutorials. When I did that I found: http://www.ats.ucla.edu/stat/r/dae/logit.htm Surely there are others. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] converting string vector to integer/numeric vector
On Sep 5, 2010, at 8:48 AM, rajesh j wrote: Hi, Is it possible to convert a string vector to integer or numeric vector? In my situation I receive data in a string vector and have to convert it based on a given type. Can you give an example? I don't understand either what sort of conversion you desire or what you mean by convert it based on a given type. There are a couple of function you may want to consider but I am having difficulty convincing myself they answer the problem posed: ?charToRaw ?stroi strtoi(charToRaw(123 this is a string), base=16) # convert to decimal ASCII [1] 49 50 51 32 116 104 105 115 32 105 115 32 97 32 115 116 114 105 110 103 -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] converting string vector to integer/numeric vector
On Sep 5, 2010, at 9:22 AM, rajesh j wrote: for e.g., I get the following as a string vector int 4 5 6 after reading the first element, I have to convert this to a integer vector But what is the right answer? And what number of items are possble per line? And what are the other possible type identifiers? We need an example that has enough complexity to allow testing. -- David. On Sun, Sep 5, 2010 at 6:44 PM, David Winsemius dwinsem...@comcast.net wrote: On Sep 5, 2010, at 8:48 AM, rajesh j wrote: Hi, Is it possible to convert a string vector to integer or numeric vector? In my situation I receive data in a string vector and have to convert it based on a given type. Can you give an example? I don't understand either what sort of conversion you desire or what you mean by convert it based on a given type. There are a couple of function you may want to consider but I am having difficulty convincing myself they answer the problem posed: ?charToRaw ?stroi strtoi(charToRaw(123 this is a string), base=16) # convert to decimal ASCII [1] 49 50 51 32 116 104 105 115 32 105 115 32 97 32 115 116 114 105 110 103 -- David Winsemius, MD West Hartford, CT -- Rajesh.J David Winsemius, MD West Hartford, CT [[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] converting string vector to integer/numeric vector
So there is one item per line and the task is to recognize the strings
INT and NUM and create variables with numeric type and
INT=c(2,3,4)
NUM=c(2.37, 4.56) # ???
I worry that is not a full description of the task if there be
more than just two variable names and if all the INTs have the same
name, then they will get overwritten but perhaps you have
specified the problem completely, so here goes:
txt - textConnection('INT
+ 2
+ 3
+ 4
+
+ NUM
+ 2.37
+ 4.56')
indat - read.table(txt, stringsAsFactors=FALSE)
indat$nflag - as.numeric(indat$V1)
cumsum(is.na(indat$nflag))
[1] 1 1 1 1 2 2 2
by(indat$V1, cumsum(is.na(indat$nflag)), function(x)
assign(as.character(x[1]), as.numeric(x[-1]) ,envir = .GlobalEnv) )
cumsum(is.na(indat$nflag)): 1
[1] 1 3 4
-
cumsum(is.na(indat$nflag)): 2
[1] 2 5
INT
[1] 1 3 4
NUM
[1] 2 5
.
On Sep 5, 2010, at 9:33 AM, rajesh j wrote:
The string vector actually comes as a part of a list, and the vector
is named int, and the numbers are strings. I then have to make it
a vector that is still called int and has 4,5,6 etc. the types are
either integer or numeric. The number of items in the vector is
unknown.
here's an example,
a list has vectors
INT
2
3
4
NUM
2.37
4.56
On Sun, Sep 5, 2010 at 6:56 PM, David Winsemius dwinsem...@comcast.net
wrote:
On Sep 5, 2010, at 9:22 AM, rajesh j wrote:
for e.g., I get the following as a string vector
int 4 5 6
after reading the first element, I have to convert this to a
integer vector
But what is the right answer? And what number of items are possble
per line? And what are the other possible type identifiers? We need
an example that has enough complexity to allow testing.
--
David.
On Sun, Sep 5, 2010 at 6:44 PM, David Winsemius dwinsem...@comcast.net
wrote:
On Sep 5, 2010, at 8:48 AM, rajesh j wrote:
Hi,
Is it possible to convert a string vector to integer or numeric
vector? In
my situation I receive data in a string vector and have to convert
it based
on a given type.
Can you give an example? I don't understand either what sort of
conversion you desire or what you mean by convert it based on a
given type.
There are a couple of function you may want to consider but I am
having difficulty convincing myself they answer the problem posed:
?charToRaw
?stroi
strtoi(charToRaw(123 this is a string), base=16) # convert to
decimal ASCII
[1] 49 50 51 32 116 104 105 115 32 105 115 32 97 32 115
116 114 105 110 103
--
David Winsemius, MD
West Hartford, CT
--
Rajesh.J
David Winsemius, MD
West Hartford, CT
--
Rajesh.J
David Winsemius, MD
West Hartford, CT
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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] converting string vector to integer/numeric vector
On Sep 5, 2010, at 10:47 AM, rajesh j wrote:
I'm sorry. you seem to have misunderstood my data representation for
input.
Because you did not follow the Posting Guide's advice regarding
producing examples using valid R code. You give me text; I work on text.
#-
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
#
I am going to change the name of the variable to cc because c is a
unfortunate name for a variable, since it is also the name of a
crucial function.
Here's what I have
cc-list(INT=c(1,2,3),NUM=c(2.34,4.56,6.78))
I need
c-list(INT=c(1,2,3),NUM=c(2.34,4.56,6.78))
cc - lapply(cc, as.numeric)
cc
$INT
[1] 1 2 3
$NUM
[1] 2.34 4.56 6.78
--
David.
On Sun, Sep 5, 2010 at 7:57 PM, David Winsemius dwinsem...@comcast.net
wrote:
So there is one item per line and the task is to recognize the
strings INT and NUM and create variables with numeric type and
INT=c(2,3,4)
NUM=c(2.37, 4.56) # ???
I worry that is not a full description of the task if there be
more than just two variable names and if all the INTs have the same
name, then they will get overwritten but perhaps you have
specified the problem completely, so here goes:
txt - textConnection('INT
+ 2
+ 3
+ 4
+
+ NUM
+ 2.37
+ 4.56')
indat - read.table(txt, stringsAsFactors=FALSE)
indat$nflag - as.numeric(indat$V1)
cumsum(is.na(indat$nflag))
[1] 1 1 1 1 2 2 2
by(indat$V1, cumsum(is.na(indat$nflag)), function(x)
assign(as.character(x[1]), as.numeric(x[-1]) ,envir = .GlobalEnv) )
cumsum(is.na(indat$nflag)): 1
[1] 1 3 4
-
cumsum(is.na(indat$nflag)): 2
[1] 2 5
INT
[1] 1 3 4
NUM
[1] 2 5
.
On Sep 5, 2010, at 9:33 AM, rajesh j wrote:
The string vector actually comes as a part of a list, and the vector
is named int, and the numbers are strings. I then have to make it
a vector that is still called int and has 4,5,6 etc. the types are
either integer or numeric. The number of items in the vector is
unknown.
here's an example,
a list has vectors
INT
2
3
4
NUM
2.37
4.56
On Sun, Sep 5, 2010 at 6:56 PM, David Winsemius dwinsem...@comcast.net
wrote:
On Sep 5, 2010, at 9:22 AM, rajesh j wrote:
for e.g., I get the following as a string vector
int 4 5 6
after reading the first element, I have to convert this to a integer
vector
But what is the right answer? And what number of items are possble
per line? And what are the other possible type identifiers? We need
an example that has enough complexity to allow testing.
--
David.
On Sun, Sep 5, 2010 at 6:44 PM, David Winsemius dwinsem...@comcast.net
wrote:
On Sep 5, 2010, at 8:48 AM, rajesh j wrote:
Hi,
Is it possible to convert a string vector to integer or numeric
vector? In
my situation I receive data in a string vector and have to convert
it based
on a given type.
Can you give an example? I don't understand either what sort of
conversion you desire or what you mean by convert it based on a
given type.
There are a couple of function you may want to consider but I am
having difficulty convincing myself they answer the problem posed:
?charToRaw
?stroi
strtoi(charToRaw(123 this is a string), base=16) # convert to
decimal ASCII
[1] 49 50 51 32 116 104 105 115 32 105 115 32 97 32 115 116
114 105 110 103
--
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] dirichlet models
On Sep 5, 2010, at 10:53 AM, Donald Braman wrote: Does anyone know of a package (or workaround) for fitting a dirichlet distribution by maximum likelihood? Searching: RSiteSearch(fitting dirichlet distribution maximum likelihood) (Rapidly) produces this candidate: http://finzi.psych.upenn.edu/R/library/VGAM/html/dirichlet.html (I'm looking for something like this: http://repec.org/bocode/d/dirifit.html , that allows for both dependent variables summing to 1 predictive variables of any sort.) Don -- Donald Braman -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Greek symbols (again but more complicated)
On Sep 5, 2010, at 11:41 AM, John Helly wrote: Hi. I'm trying to get 'mu' to show up as a Greek symbol but, despite trying every example I could find, can't get it to work. Any insights would be welcome. This is what I'm using that works, but displays mu with the letter u. plotTimeXMastPAR - qplot(DT,MastPAR, data=A, xlab = , ylab = quote(PAR (uE ~m^-2 ~s^-1)), geom=line) + opts(legend.position=none) I suppose most readers will know by now that qplot is part of ggplot2, but it would be better manners to include the require() statement that would identify the non-base package, as would inclusion of a working example: require(ggplot2) qplot(mpg, wt, data=mtcars, ylab=expression(PAR*(mu*E~m^-2~s^-1))) or qplot(mpg, wt, data=mtcars, ylab=quote(PAR*(mu*E~m^-2~s^-1))) This is an attempt to get mu to look right but it does not work. It doesn't fail but nothing inside the expression statement gets displayed. plotTimeXMastPAR - qplot(DT,MastPAR, data=A, xlab = , ylab = c(PAR, expression(mu, quote(E ~m^-2 ~s^-1))), geom=line) + opts(legend.position=none) Cheers. -- John Helly, UCSD / San Diego Supercomputer Center / Scripps Institution of Oceanography, Climate, Atmospheric Science, and Physical Oceanography / +01 760 840 8660 mobile / stonesteps (Skype) / stonesteps7 (iChat) / /www.sdsc.edu/~hellyj __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] appending to a list
On Sep 5, 2010, at 1:21 PM, Aks Ism wrote: Hi, I've looked at previous discussions and did not get anything. I want to be able to append to a list in a loop. Is this possible? Of course: ?c ?[[ -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Need Help .RData in Mac OS X, Please
On Sep 5, 2010, at 12:50 PM, Chunhao wrote: Hi R Users, I was accidentally save R file when I quit R program. I try to delete but I am not sure the R saved file was successfuly delete or not. Now everyt time when I open the R, it always appears [R.app GUI 1.34 (5589) i386-apple-darwin9.8.0] [Workspace restored from /Users/ctu/.RData] I have read google, R help and Nabble but I still can't solve this problem. Search on either the R_SIG-MAC list or in Google for show dotted files. Apply that method to your working directory, delete the invisible .RData file, and then decide whether you want to revert to the default Finder behavior. One citation: http://artofgeek.com/2009/09/16/toggle-display-of-hidden-files-in-finder-with-keyboard-shortcut/ I have globally changed my system to show dotted files, but not everyone feels safe doing so. At the terminal window I have entered: defaults write com.apple.Finder AppleShowAllFiles YES . and then pt-click-hold on Dock-Finder-icon, choose relaunch Or you could open a Terminal window which will by default open in / Users/ctu/ and type: rm .RData -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] appending to a list
On Sep 5, 2010, at 10:11 PM, Aks Ism wrote:
No, I mean dynamically. Like list.push_back
Not sure what you mean by dynamically and my Google search for the
unspecified function list.push_back uncovered a C++ function that is
at least as complex as what I believe would be the equivalent R
function. I think you need to expand your discussion and your
executable R examples before people are going to understand what you
are thinking (but not writing).
mylist - c(mylist, my.element)
c() simply adds an element and R requires that you assign it if it is
to be preserved.
Perhaps you mean something like this:
ll - list();
for (i in 1:20) {
my.element - scan();
ll - c(ll, my.element);
if (is.na(my.element)){return(ll);break}}
--
David.
On Mon, Sep 6, 2010 at 1:01 AM, David Winsemius dwinsem...@comcast.net
wrote:
On Sep 5, 2010, at 1:21 PM, Aks Ism wrote:
Hi,
I've looked at previous discussions and did not get anything. I want
to be
able to append to a list in a loop. Is this possible?
Of course:
?c
?[[
--
David Winsemius, MD
West Hartford, CT
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] extracting x,y coordinates from a contour plot
On Sep 5, 2010, at 11:48 PM, Charles Annis, P.E. wrote: Requisite info: R version 2.11.1 (2010-05-31) running on a 64 bit HP Windows 7 machine. Doubt that makes much of a difference here. I have used contour() for several years. Now I would like to extract from a contour plot the x, y coordinates of a contour z=constant. This seems as though it would be straight-forward but I've been unsuccessful in my searches of CRAN. Suggest you read the help page for contour and the pages to which it links as well as working the examples. The answer is illustrated in the examples on that page. Can anyone provide a hint? x - 10*1:nrow(volcano) y - 10*1:ncol(volcano) xy160 - contourLines(x, y, volcano, nlevels=1, levels=160) str(xy160) List of 2 $ :List of 3 ..$ level: num 160 ..$ x: num [1:165] 110 108 105 102 103 ... ..$ y: num [1:165] 295 300 310 320 330 ... $ :List of 3 ..$ level: num 160 ..$ x: num [1:31] 270 263 262 260 260 ... ..$ y: num [1:31] 310 320 330 340 350 ... -- David. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Correct coefficients from treatment contrasts?
On Sep 6, 2010, at 4:03 AM, B W wrote: Snipped out formatting detritus and added back many missing speces. -Hello,I am trying to take the information from the summary of my best fit logisticregression model for the occurrence of a high elevation plant spp. and create the appropriate equation that will calculate probability of occurrence, given the data. My predictors include both continuous variables (slope and a second orderpolynomial of elevation) and a discrete variable for aspect (warm and cool). I have left unchanged the default contrasts option, so I believe that thefollowing coefficients were created using treatment contrasts. My question how can I take this summary output and create the logistic equation that will allow me to calculate probability of occurrence. My interests are touse this to spatially display this info in a GIS environment. I think you should: -- Read the Posting Guide where you should learn that this is a plain text mailing list and that you need to change the configuration of your mail client. -- Read the help page and read other documentation regarding the use of the predict function. I have made adraft equation (shown below) that uses the coefficients from this summaryoutput, but this appears to be incorrect values always return zeroprobabilities. Presumably I need to adjust the values in some way but I am unclear as to how to proceed. Anyguidance would be appreciated! summary ( Call:glm(formula= Po ~ Slope + poly(Elevation, 2) + Aspect_2, family = quasibinomial) DevianceResiduals: Min 1Q Median 3Q Max -1.0532 -0.4167 -0.2760 -0.1823 3.3376 Coefficients: Estimate Std. Error t valuePr(| t|)(Intercept) -4.577707 0.222406 -20.583 2e-16 *** Slope 0.039959 0.003593 11.121 2e-16 *** poly(Elevation,2)1 8.050898 5.601956 1.437 0.1508 poly(Elevation,2)2 -37.694521 6.297806 -5.985 2.39e-09 *** Aspect_2w 0.429229 0.174760 2.456 0.0141 * --- You may get predictions at the original data points with: pred predict(model.Slope.Elevation.Aspect) (1/ (1 + exp(-1 * (-4.577707 + 0.039959*Slope + 8.050898 * poly(Elevation, 2)1 + -37.694521 * poly(Elevation, 2)2 + 0.429229* Aspect_2w) Brendan Wilson 2530 Alexis Road Shoreacres BC Canada V1N 4P6 Ph: 1.250.359.5905 [[alternative HTML version deleted]] David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Aggregating the matrices
On Sep 6, 2010, at 9:56 AM, Sergey Goriatchev wrote: Hello everyone. Say we have the following: a - matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list(06092010, c(ES, PT, Z , CF, GX, ST, EO))) b - matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list(06092010, c(PT, CF, AT, EM, ST))) d - cbind(a, b) I want to calculate sums of the columns that have similar column names and then output this summary What I want to have is an array that looks like: ES PT Z CF... -75 -2 5 11... I tried the following, but it did not work: aggregate(d, list(colnames(d)), sum) ES is not in the duplicated column names so perhaps your English specification is not what you meant: d ES PT Z CF GX ST EO PT CF AT EM ST 06092010 -75 3 5 9 2 3 5 -5 2 4 12 5 dupled - colnames(d)[duplicated(colnames(d))] sapply(dupled, function(x) sum( d[, x])) PT CF ST 3 9 3 If you wanted simple a sum over unique column names then it would have been somewhat simpler (no need to construct a duplicated set): sapply(unique(colnames(d)), function(x) sum( d[, x])) ES PT Z CF GX ST EO AT EM -75 3 5 9 2 3 5 4 12 How can I achieve my objective? Thank you in advance. Sergey __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mac: lib/gtk.pkg
On Sep 5, 2010, at 10:32 AM, Daniele Sluijters wrote: Hello, I'm sorry to just pop-up on the mailing list like this and ask a relatively non-R related question but I had no idea whom else to contact on this matter. I'm working on a completely different port of an application to OS X which requires GTK and through Google'ing stumbled on a rather recent GTK installer for Mac at: http://r.research.att.com/ I was wondering if anyone here knows how GTK and its dependencies were packaged into that pkg? It'd be a lifesaver if someone could point me in the right direction. Again, sorry for the non-R related question but this place seemed like the only option. The is an R-SIG-Mac mailing list. It's webpage is at the top of the SIG entries on: http://www.r-project.org/mail.html I suspect that Simon Urbanek, who maintains the ATT webpages and very probably created that package, sometimes reads rhelp but I'm not sure on what schedule. You might see if he makes his email address available on those pages. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Aggregate certain rows in a matrix
On Sep 6, 2010, at 10:47 AM, Dimitris Rizopoulos wrote: one way is the following: M - cbind(c(1,1,1,1,2,2,3,3,3,3), c(2,2,2,3,4,4,4,5,5,6), c(1,2,3,4,5,6,7,8,9,10)) ind - do.call(paste, c(as.data.frame(M[, 1:2], sep = \r))) M[, 3] - ave(M[, 3], ind, FUN = sum) unique(M) I had been working on a similar approach with ave( ,paste(), sum) inside a datafrmae, but I liked your approach of setting up the results of the paste operation as a vector outside of M. (Skips the dataframe operation I was using.) The above solution is destructive, so I constructed this similar alternative that returns the results without altering M: cbind(M, ave(M[ , 3], list(M[,1], M[,2]), FUN=sum))[ !duplicated(M[,1:2]), c(1,2,4)] [,1] [,2] [,3] [1,]126 [2,]134 [3,]24 11 [4,]347 [5,]35 17 [6,]36 10 I hope it helps. Best, Dimitris On 9/6/2010 4:29 PM, Kennedy wrote: Hi, I have a matrix that looks like this a- c(1,1,1,1,2,2,3,3,3,3) b- c(2,2,2,3,4,4,4,5,5,6) c- c(1,2,3,4,5,6,7,8,9,10) M- matrix(nr=10,nc=3) M[,1]- a M[,2]- b M[,3]- c M [,1] [,2] [,3] [1,]121 [2,]122 [3,]123 [4,]134 [5,]245 [6,]246 [7,]347 [8,]358 [9,]359 [10,]36 10 I want to reduce the matrix according to the following: If the values of the two first columns are the same in two or more rows the values in the third column of the corresponding rows should be added and only one of the rows should be keept. Hence the matrix M above should look like this 1 2 6 1 3 4 2 4 11 3 4 7 3 5 17 3 6 10 Thank you Henrik -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] WriteXLS problem
On Sep 6, 2010, at 12:25 PM, Kenneth Roy Cabrera Torres wrote: Thank you Ivan for you answer: El lun, 06-09-2010 a las 18:11 +0200, Ivan Calandra escribió: Hi, Are you sure you used the correct syntax and object names? It might just be because of that...(reading the error messages) Im sure, because it works with write.csv or write.table. Sure? You are making the incorrect assumption that those write functions have the same syntax. At least for WriteXLS that assumption is false. The help page clearly states that the objects need to be quoted rather than being referred to by their naked names. The error you are getting with your second option suggests to me that you offered an unquoted name of an object. You can offer a vector of quoted names of dataframes to WriteXLS and each named dataframe will be converted to a worksheet within the workbook. -- David. There is another function, xlsReadWrite::write.xls(), that I like a lot: it is really easy to use and does not require Perl or Python. Unfortunately it works on windows, and I am in a non windows platform (ubuntu). Thank you for you advice and help. Kenneth HTH, Ivan Le 9/6/2010 18:03, Kenneth Roy Cabrera Torres a crit : Hi R users: I don't know if you have had the following problem trying to export to an xls format file in a non windows platform. I try to use the following packages: 1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1) 2. WriteXLS (version 1.9.0) (with perl and testPerl working) Even xlsx package that take too long and do not finish. The data frame I try to export has 269363 row and 116 columns. In the first one (dataframe2xls) I get this message: Traceback (most recent call last): File C:/PROGRA~2/R/R-211~1.1PA/library/dataframes2xls/python/ csv2xls.py, line 18, inmodule import pyexcelerator File C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python \pyexcelerator \__init__.py, line 12, inmodule from Workbook import Workbook File C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python \pyexcelerator \Workbook.py, line 526 boundsheets_len += len(BIFFRecords.BoundSheetRecord(0x00L, sheet.hidden, sheet.name).get()) ^ SyntaxError: invalid syntax Using the second option I get this message: Error en get(as.character(i)),envr=envir) : objeto '089' no encontrado Object '089' not found. Im using this R platform: sessionInfo() R version 2.11.1 Patched (2010-08-30 r52848) Platform: x86_64-unknown-linux-gnu (64-bit) Locale: LC_CTYPE=es_CO.UTF-8 Is the only solution to export to .csv and then to .xls format with other program like openoffice? Thank you for your help and advice. Kenneth __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] poisson distribution
On Sep 6, 2010, at 1:13 PM, tamas barjak wrote: Hello! I need some help. How I know it to draw the formula of the poisson distribution? expr-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^- lambda) --- not good ?plotmath (Do not see factorial as a plotmath function Try: expr-expression(P(xi == k) == frac(lambda^k, k*!)*e^-lambda) on the screen the k! not the Poisson Formula, but factorial(k) Thanx! -- David. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Failure to aggregate
On Sep 6, 2010, at 12:15 PM, Dimitri Shvorob wrote: I have a (very big - 1.5 rows) dataframe with a (POSIXt POSIXlt) column h (hour). Surprisingly, I cannot calculate a simple aggregate over the dataframe. n.h1 = sqldf(select distinct h, count(*) from x group by h) Error in sqliteExecStatement(con, statement, bind.data) : RS-DBI driver: (error in statement: no such table: x) In addition: Warning message: In value[[3L]](cond) : RAW() can only be applied to a 'raw', not a 'double' n.h2 = aggregate(x$price, by = x$h, FUN = nrow) A vector argument (x$price) would only have one row (at most). nrow(c(1,2) NULL Error in names(y) - c(names(by), names(x)) : 'names' attribute [10] must be the same length as the vector [2] Try: tapply(x$price, by = x$h, FUN = length) -- David. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 change the xlab name?
On Sep 6, 2010, at 2:07 PM, tooblue wrote: I simply put, plot(density(), main=, + xlab = XXX), it says that I have an unexpected = in it. It may be a case of a confused parser. You have an extraneous + in there: = rnorm(100) plot(density(), main=, xlab = XXX) # works If on the other hand you wanted to construct a more complex title then you will probably need to read the expression and bquote help pages and submit a more descriptive problem statement. -- David. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] two questions
The usual least-squares methods are fairly robust to departures from normality. Furthermore, it is the residuals that are assumed to be normally distributed (not the marginal distributions that you are probably looking at) , so it does not sound as though you have yet examined the data properly. Tell us what the descriptive stats (say the means, variance, 10th and 90th percentiles) are on the residuals within cells cross-classified by the gender and city-of-birth variables (say the means, variance, 10th and 90th percentiles). On Sep 6, 2010, at 4:34 PM, Iasonas Lamprianou wrote: Dear friends, two questions (1) does anyone know if there are any non-parametric equivalents of the two-way ANOVA in R? I have an ordinal non-normally distributed dependent variable and two factors (gender and city of birth). Normally, one would try a two-way anova, but if R has any non- parametric equivalents, that might be great. There is an entire task view page on robust methods if you decide to press on with this quest. (2) Also, if the interaction of gender and city of birth is statistically significant, which post-hoc tests should I run? How many cities are we talking about? Thanks Jason Dr. Iasonas Lamprianou -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] WriteXLS problem
On Sep 6, 2010, at 8:09 PM, Dejian Zhao wrote: The maximum number of rows in excel 2003 or below is 65535, less than your number of rows, so if you export your data into xls files, probably you cannot see all your data in excel. Exel 2007 can hold as many as 1048575 lines, thus xlsx file is a better choice. The maximum number of rows in more editions of Excel.2003 was increased to a million. You may be correct about the Perl module that underlies WriteXLS, however. Here is an extract from the CPAN page for that module: http://search.cpan.org/~jmcnamara/Spreadsheet-WriteExcel/lib/Spreadsheet/WriteExcel.pm#LIMITATIONS LIMITATIONS The following limits are imposed by Excel: Description Limit --- -- Maximum number of chars in a string 32767 Maximum number of columns 256 Maximum number of rows 65536 Maximum chars in a sheet name 31 Maximum chars in a header/footer 254 The minimum file size is 6K due to the OLE overhead. The maximum file size is approximately 7MB (7087104 bytes) of BIFF data. This can be extended by installing Takanori Kawai's OLE::Storage_Lite module http://search.cpan.org/search?dist=OLE-Storage_Lite see the bigfile.pl example in the examples directory of the distro. -- David On 2010-9-7 0:03, Kenneth Roy Cabrera Torres wrote: Hi R users: I don't know if you have had the following problem trying to export to an xls format file in a non windows platform. I try to use the following packages: 1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1) 2. WriteXLS (version 1.9.0) (with perl and testPerl working) Even xlsx package that take too long and do not finish. The data frame I try to export has 269363 row and 116 columns. In the first one (dataframe2xls) I get this message: Traceback (most recent call last): File C:/PROGRA~2/R/R-211~1.1PA/library/dataframes2xls/python/csv2xls.py, line 18, inmodule import pyexcelerator File C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python \pyexcelerator \__init__.py, line 12, inmodule from Workbook import Workbook File C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python \pyexcelerator \Workbook.py, line 526 boundsheets_len += len(BIFFRecords.BoundSheetRecord(0x00L, sheet.hidden, sheet.name).get()) ^ SyntaxError: invalid syntax Using the second option I get this message: Error en get(as.character(i)),envr=envir) : objeto '089' no encontrado Object '089' not found. Im using this R platform: sessionInfo() R version 2.11.1 Patched (2010-08-30 r52848) Platform: x86_64-unknown-linux-gnu (64-bit) Locale: LC_CTYPE=es_CO.UTF-8 Is the only solution to export to .csv and then to .xls format with other program like openoffice? Thank you for your help and advice. Kenneth __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to combine several subsets?
On Sep 6, 2010, at 9:22 PM, tooblue wrote: I simply put, NEVER=subset(infants$bwt,ISNO1) UNTILPREGNANT=subset(infants$bwt, ISNO2) ONCENOTNOW=subset(infants$bwt, ISNO3) and I wanna combine those three. I do it like ISNO=NEVERUNTILPREGNANTONCENOTNOW The operator does not do concatenation, but rather returns a logical vector. You may have had your mind adversely affected by excessive exposure to Excel. Hopefully you did not do just that at the command line. I could imagine thinking that might work as part of the subset argument to subset, but it would be through the use of the or operator, |: ALL - subset(infants$bwt, ISNO1| ISNO2| ISNO3) If they were dataframes, you could also have done: ALL - rbind(NEVER, UNTILPREGNANT, ONCENOTNOW) But below you suggested they might be vectors; if so, why not: ALL - c(NEVER, UNTILPREGNANT, ONCENOTNOW) and R tells me 1: In NEVER UNTILPREGNANT : longer object length is not a multiple of shorter object length 2: In NEVER UNTILPREGNANT ONCENOTNOW : longer object length is not a multiple of shorter object length I'm confused coz these are not objects, but a list of sets of numbers. They really _must_ be objects since you assigned a result to those names. (Greater clarity would occur if you offered at least str(NEVER) -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Prediction and confidence intervals from predict.drc
On Sep 6, 2010, at 7:54 PM, Brant Inman wrote:
R-helpers,
I am using the package drc to fit a 4 parameter logistic model.
When I
use the predict function to get prediction on a new dataset, I am not
getting the requested confidence or prediction intervals. Any idea
what
is going on? Here is code to reproduce the problem:
---
library(drc)
# Fit model to existing dataset in package
spinach.model - drm(SLOPE~DOSE, data = spinach, fct = LL.4())
(Comparing to the predict example code.) You did not specify a CURVE
argument. Not sure what effect that would have.
#Generate new fake dataset
newdt - data.frame(matrix(c(seq(0, 150, 0.1), rep(NA, 1501)), ncol=2,
byrow=F))
colnames(newdt) - c('DOSE', 'SLOPE')
You did not include a CURVE variable. But this provokes nary a
complaint. I wondered if the estimates may be an unlabeled mixture
from the 5 CURVEs
#Use predict function to get prediction and confidence intervals
pred - predict(spinach.model, interval='prediction', newdata=newdt)
You did not specify CURVE. The example uses one:
predict(spinach.model1, data.frame(dose=2, CURVE=c(1, 2, 3)),
interval = prediction)
With your object:
head(predict(spinach.model, data.frame(dose=2, CURVE=c(1)),
+ interval = prediction))
Prediction Lower Upper
0.3500492 -0.2790351 0.9791336
With the original example:
predict(spinach.model1, data.frame(dose=2, CURVE=c(1, 2, 3)),
+ interval = confidence)
Prediction Lower Upper
[1,] 0.9048476 0.8552178 0.9544775
[2,] 0.4208307 0.3626741 0.4789873
[3,] 0.5581673 0.4971838 0.6191509
With your object
predict(spinach.model, data.frame(dose=2, CURVE=c(1)),
+ interval = confidence)
Prediction Lower Upper
0.3500492 0.2673464 0.4327521
conf - predict(spinach.model, interval='confidence', newdata=newdt)
head(pred); head(conf)
---
Examining the output shows the point estimates but not the intervals.
Would like the intervals.
I am using the most recent versions of R and drc on Windows XP.
Thanks,
Brant
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Re: [R] Saving fits (glm, nls) without data
On Sep 7, 2010, at 11:02 AM, Johann Hibschman wrote:
Is there any package that assists in saving and reconstituting glm and
nls fits without bringing along the accompanying data? A quick search
on CRAN didn't turn up anything.
If not, how do other people deal with saving the coefficients of model
fits?
For example, I've run a glm fit that has 23 coefficents on data set
that
had 193,008 rows, by the time the fit was called. When I save the
resulting fit object, I get a 491 MB object, which suggests that it's
pulling along all sorts of junk in the environment, as 23*193k*8 is
only
34 MB. Even so, I would prefer to only save the coefficients
Have you read through the Value section of glm's help page?
...and
?coef
and the
Hessian, not the fit data set.
I'm not sure about whether there will be a Hessian in a glm object.
Have you run str() on your objects. It's likely that the residuals,
fitted.values, weights, prior.weights, and linear.predictors are going
to be fairly large. You could use lapply to run object.size to see
whether I have missed any. When I do that on hte first help page
example, it is the model component that is the second largest, but its
inclusion is optional. The largest compenent is family but I suspect
that is a family of functions and would not increase in size with
larger models.
Is there anything I can do? If I want to save several fits, 490 MB a
shot starts to add up very quickly. If I just save the
coefficients, I
have to manually hack up an object that I can then run 'predict' on
when
I want to evaluate the model, and that feels very error-prone.
The predict.glm function is visible so you can just type its name to
see the code. It appears that the section of the code that does the
work is fairly short. This is my nomination for what happens in most
cases:
if (!se.fit) {# not generally invoked with se.fit=TRUE
}
else {
pred - predict.lm(object, newdata, se.fit, scale = 1,
type = ifelse(type == link, response, type),
terms = terms, na.action = na.action)
switch(type, response = {
pred - family(object)$linkinv(pred)
}, link = , terms = )
}
So maybe you should write a predict function that would work on a
reduced glm object that has a class name of your choosing.
--
David Winsemius, MD
West Hartford, CT
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Re: [R] boundary correction - univariate kernel density estimation
On Sep 7, 2010, at 12:04 PM, sbillin2 wrote: Hey, Does anyone know of a package in R that provides univariate kernel density estimation with boundary correction ? What? you don't believe that tunneling occurs at finite barriers? or how to easily extend an existing bivariate kernel density estimation function (e.g. lambdahat in the spatialkernel package) with boundary corrections to allow univariate density estimation? When this question has been (multiply) posed in the past, the suggested answer has been to use package logspline. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] a^c(1:3)
On Sep 7, 2010, at 12:35 PM, Feng Li wrote: Dear R, I have two small questions confused me recently. Now assume I have a matrix a, like this, a - matrix(1:6, 2, 3) a [,1] [,2] [,3] [1,]135 [2,]246 I sometimes need each row of a raised to a different exponent. So I do a trick like this, a^c(2, 3) [,1] [,2] [,3] [1,]19 25 [2,]8 64 216 My first question is that if it is possible to do this trick column wise? Most questions of this sort are answerable by thinking of R matrices as folded vectors. The folding occurs columnwise (unlike Matlab), so for this problem: a^rep(c(2, 3, 4), each=nrow(a)) # the exponents become 2,2,3,3,4,4 [,1] [,2] [,3] [1,]1 27 625 [2,]4 64 1296 or: a^matrix(c(2, 3, 4), byrow=TRUE, nrow=2, ncol=3) [,1] [,2] [,3] [1,]1 27 625 [2,]4 64 1296 Just out of curiosity, of course I know there are other ways of doing this. And the second question is why I get such result when I put another element in the exponent part like this, Because argument recycling makes the exponents 2,3,4,2,3,4 and they are applied folded column wise a^c(2, 3, 4) [,1] [,2] [,3] [1,]1 81 125 [2,]8 16 1296 BTW, I have a 64bit R version (2.11) for Linux. Any advice would be appreciated. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] remove accents in strings
On Sep 7, 2010, at 1:35 PM, Matt Shotwell wrote: If you know the encoding of the string, or if its encoding is the current locale encoding, then you can use the iconv function to convert the string to ASCII. Something like: iconv(accented.string, to=ASCII//TRANSLIT) While 7-bit ASCII does not permit accented characters, extended (8- bit) ASCII does. Hence, I'm not sure this will work. But it's worth a try. tst - c(à, è, ì, ò, ù , À, È, Ì, Ò, Ù, á, é, í, ó, ú, ý , Á, É, Í, Ó, Ú, Ý) iconv(tst, to=ASCII//TRANSLIT) [1] `a `e `i `o `u `A `E `I `O `U 'a 'e 'i 'o 'u 'y [17] 'A 'E 'I 'O 'U 'Y gsub(`|\\', , iconv(tst, to=ASCII//TRANSLIT)) [1] a e i o u A E I O U a e i o u y A E I O [21] U Y Notice that the accent acute gets converted to a single quote and therefore needs to be dbl-\-ed to get recognized in an R regex pattern. On a Mac with: locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 -- David. -Matt On Tue, 2010-09-07 at 13:04 -0400, lamack lamack wrote: Dear all, there is a R function to remove all accents in strings? best regards. JL [[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. -- Matthew S. Shotwell Graduate Student Division of Biostatistics and Epidemiology Medical University of South Carolina __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] remove accents in strings
On Sep 7, 2010, at 2:29 PM, Matt Shotwell wrote: Weird, my (Ubuntu, s don't tell Dirk) iconv doesn't add the backticks or single quotes. I don't see any promise in the help page that iconv should substitute anything for the accents. It just says each OS may have its own behavior and suggest that you are accessing glibc while I am using libiconv and warns to expect different results. -- David. tst - c(à, è, ì, ò, ù , À, È, Ì, Ò, Ù, á, + é, í, ó, ú, ý , Á, É, Í, Ó, Ú, Ý) iconv(tst, to=ASCII//TRANSLIT) [1] a e i o u A E I O U a e i o u y A E I [20] O U Y By the way, I'll take this moment to remind anyone interested that R still has trouble with embedded zeros in character strings. I may be abusing terminology, but I think that makes R 8-bit dirty. -Matt On Tue, 2010-09-07 at 14:01 -0400, David Winsemius wrote: On Sep 7, 2010, at 1:35 PM, Matt Shotwell wrote: If you know the encoding of the string, or if its encoding is the current locale encoding, then you can use the iconv function to convert the string to ASCII. Something like: iconv(accented.string, to=ASCII//TRANSLIT) While 7-bit ASCII does not permit accented characters, extended (8- bit) ASCII does. Hence, I'm not sure this will work. But it's worth a try. tst - c(à, è, ì, ò, ù , À, È, Ì, Ò, Ù, á, é, í, ó, ú, ý , Á, É, Í, Ó, Ú, Ý) iconv(tst, to=ASCII//TRANSLIT) [1] `a `e `i `o `u `A `E `I `O `U 'a 'e 'i 'o 'u 'y [17] 'A 'E 'I 'O 'U 'Y gsub(`|\\', , iconv(tst, to=ASCII//TRANSLIT)) [1] a e i o u A E I O U a e i o u y A E I O [21] U Y Notice that the accent acute gets converted to a single quote and therefore needs to be dbl-\-ed to get recognized in an R regex pattern. On a Mac with: locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 -- Matthew S. Shotwell Graduate Student Division of Biostatistics and Epidemiology Medical University of South Carolina David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Saving fits (glm, nls) without data
On Sep 7, 2010, at 2:53 PM, Johann Hibschman wrote:
David Winsemius dwinsem...@comcast.net writes:
On Sep 7, 2010, at 11:02 AM, Johann Hibschman wrote:
Even so, I would prefer to only save the coefficients
Have you read through the Value section of glm's help page?
...and
?coef
I have; it's easy to get the coefficients. The part I'm struggling
with
is reconstituting an operational glm object given the coefficient
vector
and the formula.
Really, I was hoping that someone had already done this work, so I
could
stop trying re-create the right kind of term object, while making sure
it doesn't hold on to a pointer to an environment with a lot of data
in
it, etc., etc..
I was assuming you could take all the code work that was already
tested and trim out the non essential code and arguments let it work
on a new class.
The predict.glm function is visible so you can just type its name to
see the code. It appears that the section of the code that does the
work is fairly short. This is my nomination for what happens in most
cases:
if (!se.fit) {# not generally invoked with se.fit=TRUE
if (missing(newdata)) { # forgot this clause in first post
}
else {
pred - predict.lm(object, newdata, se.fit, scale = 1,
type = ifelse(type == link, response, type),
terms = terms, na.action = na.action)
switch(type, response = {
pred - family(object)$linkinv(pred)
}, link = , terms = )
}
I agree. That reduces the problem to confecting a working lm object,
given a formula and coefficients. Unfortunately, I haven't yet
figured
out how to do that.
And I was thinking one would start with the glm object and just set
the unnecessary leaves of the list to NULL.
So maybe you should write a predict function that would work on a
reduced glm object that has a class name of your choosing.
I'm trying to get this to work, but I haven't figured out yet how to
generate the X matrix properly from the formula and the coefficients.
I'm sure I can eventually get it, but it's annoying.
I don't think you need to do anything other than construct a proper
newdata argument and feed it and your stripped down object to a
modified predict.sml_glm function. And it could very well be that all
you need to do is rename the predict.glm code and give it the proper
arguments. I do not see a need to recreate an X matrix for the
newdata, since the code to do that is already in predict.lm(). I
suppose I could be wrong, since I have not done it myself. I just got
my employer to buy more memory.
The whole model whereby fit objects keep around their data so you
don't
have to provide it on a few calls just seems like a mistake.
-Johann
David Winsemius, MD
West Hartford, CT
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Re: [R] Saving fits (glm, nls) without data
On Sep 7, 2010, at 3:16 PM, David Winsemius wrote: On Sep 7, 2010, at 2:53 PM, Johann Hibschman wrote: David Winsemius dwinsem...@comcast.net writes: On Sep 7, 2010, at 11:02 AM, Johann Hibschman wrote: Even so, I would prefer to only save the coefficients Have you read through the Value section of glm's help page? ...and ?coef I have; it's easy to get the coefficients. The part I'm struggling with is reconstituting an operational glm object given the coefficient vector and the formula. Really, I was hoping that someone had already done this work, so I could stop trying re-create the right kind of term object, while making sure it doesn't hold on to a pointer to an environment with a lot of data in it, etc., etc.. I was assuming you could take all the code work that was already tested and trim out the non essential code and arguments let it work on a new class. Just tested my theory and it seems to be holding up. Took the example on the predict help page, set three of the variable length components not needed in the predict operations to NULL and the code still runs fine. It does not appear that either predict.glm or predict.lm check to see if there are any missing components: ldose - rep(0:5, 2) numdead - c(1, 4, 9, 13, 18, 20, 0, 2, 6, 10, 12, 16) sex - factor(rep(c(M, F), c(6, 6))) SF - cbind(numdead, numalive=20-numdead) budworm.lg - glm(SF ~ sex*ldose, family=binomial) budworm.lg[residuals] - NULL budworm.lg[linear.predictors] - NULL budworm.lg[fitted.values] - NULL plot(c(1,32), c(0,1), type = n, xlab = dose, + ylab = prob, log = x) text(2^ldose, numdead/20, as.character(sex)) ld - seq(0, 5, 0.1) lines(2^ld, predict(budworm.lg, data.frame(ldose=ld, +sex=factor(rep(M, length(ld)), levels=levels(sex))), +type = response)) Also took out y, qr, weights, and prior.weights budworm.lg[y] - NULL budworm.lg[weights] - NULL budworm.lg[prior.weights] - NULL budworm.lg[qr] - NULL ... And it continue to perform without throwing an error. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with decimal points
On Sep 7, 2010, at 4:38 PM, Amit Patel wrote: I have found a little problem with an R script. I am trying to merge some data and am finding something unusual going on. As shown below I am trying to assign (MatchedValues[Value2,Value]) to (ClusteredData[k,Value]) which are two separate dataframes. 1) By the following command you can see that the value im transferring is 481844.03 MatchedValues[Value2,Value] [1] 481844.03 6618 Levels: 1.00E+07 1.01E+07 1.02E+07 1.04E+07 1.05E+07 1.06E +07 ... Raw This shows you that MatchValues' column eval(Value) is a factor represented internally by one of 6618 integers with various level labels. The label of the item in row eval(Value2) is 481844.03. The label is a character object. 2) But when I try to replace the values using the command i get a value of 4420 ClusteredData[k,Value] - MatchedValues[Value2,Value] ClusteredData[k,Value] [1] 4420 3) So what am I not doing. How can I keep that same value of 481844.03 I have tried Read FAQ 7.10: http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-do-I-convert-factors-to-numeric_003f -- David. as.double(MatchedValues[Value2,Value]) [1] 4420 as.numeric(MatchedValues[Value2,Value]) [1] 4420 -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] change the for loops with lapply
On Sep 7, 2010, at 5:43 PM, Changbin Du wrote:
cv.fold-function(i, size=3, rang=0.3){
cat('Fold ', i, '\n')
out.fold.c -((i-1)*c.each.part +1):(i*c.each.part)
out.fold.n -((i-1)*n.each.part +1):(i*n.each.part)
train.cv - n.cc[-out.fold.c, c(2:2401, 2417)]
train.nv - n.nn[-out.fold.n, c(2:2401, 2417)]
train.v-rbind(train.cv, train.nv) #training data for feature
selection
# grow tree
fit.dimer - rpart(as.factor(out) ~ ., method=class, data=train.v)
at-grep(leaf, fit.dimer$frame[, var], value=FALSE,
ignore.case=TRUE)
varr-as.character(unique(fit.dimer$frame[-at, var]))
train.cc - n.cc[-out.fold.c,]
valid.cc - n.cc[out.fold.c,]
train.nn - n.nn[-out.fold.n,]
valid.nn - n.nn[out.fold.n,]
train-rbind(train.cc, train.nn) #training data
valid-rbind(valid.cc, valid.nn) # validation data
#creat data set contains the following variables
myvar-names(gh9_h) %in% c(varr, out)
train-train[myvar] # update training set
valid-valid[myvar]
nnet.fit-nnet(as.factor(out) ~ ., data=train, size=size, rang=rang,
decay=5e-4, maxit=500) # model fitting
#get the validation error
mc-table(valid$out, predict(nnet.fit, valid, type=class))
#confusion
matrix
fp-mc[1,2]/sum(mc[1,]) #false positive
fn- mc[2,1]/sum(mc[2,]) #false negative
accuracy.r-1-(mc[1,2]+mc[2,1])/sum(mc) #total accuracy rate
return(c(fp, fn, accuracy.r))
}
result.fun - lapply(1:2, cv.fold(i, size=5, rang=0.3))
I got the following error message:
*Error in match.fun(FUN) :
'cv.fold(i, size = 5, rang = 0.3)' is not a function, character or
symbol
Generally when one is passing an atomic vector argument to a function
one would use sapply (but it may be a distinction withou a difference
here.) ... and furthermore the additional arguments would be given as
named constants:
?sapply
Perhaps (untested):
result.fun - sapply(1:2, cv.fold, size=5, rang=0.3))
or perhaps:
result.fun - sapply(1:2, function(i) cv.fold(i, size=5, rang=0.3))
As always the provision of a working example, perhaps even from one of
the help pages, would allow testing, and it's always good manners to
specify which package has non-base functions:
?n.cc
No documentation for 'n.cc' in specified packages and libraries:
you could try '??n.cc'
?rpart
No documentation for 'rpart' in specified packages and libraries:
you could try '??rpart'
?train
No documentation for 'train' in specified packages and libraries:
you could try '??train'
(I have suspicions which packages they come from, but one never
knows)
--
David.
I do want to change the size and rang parameters some time.
*
Can anyone help me this this? Thanks so much!
--
Sincerely,
Changbin
David Winsemius, MD
West Hartford, CT
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Re: [R] problem with max in a function
On Sep 7, 2010, at 9:06 PM, stephen sefick wrote: s - 1.00 max(s) sprintf(%.2f, max(s)) [1] 1.00 @ as a string/character object returns 1 is there anyway that I can get it to return 1.00. I am using the results of this max statement in a grep statement and it returns the wrong numbers, I will provide more information and code if it would make more sense in context. -- Stephen Sefick | Auburn University | | Department of Biological Sciences | | 331 Funchess Hall | | Auburn, Alabama | | 36849| |___| | sas0...@auburn.edu | | http://www.auburn.edu/~sas0025 | |___| Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R time series analysis
On Sep 7, 2010, at 7:51 PM, lord12 wrote: For each arima model, can you output an associated confidence interval for the predicted value at each time point? ?arima0 arima0 will return ... a list with components pred, the predictions, and se, the estimated standard errors as time series when se.fit = TRUE. -- View this message in context: http://r.789695.n4.nabble.com/R-time-series-analysis-tp2527513p2530595.html -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] multiple graphs
On Sep 7, 2010, at 8:02 PM, moleps wrote:
Dear all,
I´m trying to create multiple graphs on the same page, but they are
all stacked on top of each other.
My code:
par(mfrow=c(2,2))
a-list(levels(bar$h.r)[c(1,3,6)])
print(a)
lapply(a,function(x){
a-subset(bar,h.r==x)
with(a, cdplot(wh~Age,ylab=x))
#plot.new()
})
The plot.new command doesnt help...
Any ideas??
?layout # assuming that the undescribed plotting function is base
graphics. Some plotting functions are hard coded and are able to
defeat the usual formatting options.
--
David Winsemius, MD
West Hartford, CT
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] R time series analysis
On Sep 7, 2010, at 9:33 PM, David Winsemius wrote:
On Sep 7, 2010, at 7:51 PM, lord12 wrote:
For each arima model, can you output an associated confidence
interval for
the predicted value at each time point?
?arima0
arima0 will return ... a list with components pred, the
predictions, and se, the estimated standard errors as time series
when se.fit = TRUE.
See also:
predict.Arima {stats}
--
View this message in context:
http://r.789695.n4.nabble.com/R-time-series-analysis-tp2527513p2530595.html
--
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with max in a function
On Sep 7, 2010, at 9:37 PM, stephen sefick wrote: Here is a striped down example that is not working That dreadful phrase... is not working. When the ESP package comes to fruition, life will be so easy. Until then ... the English language is necessary. Where am we supposed to be looking. Did I miss you saying which of those (unprinted) objects we should be fixing. because of the 1.00 to 1. Any help would be greatly appreciated. measure_bkf - (structure(list(measurment_num = c(0, 0.2, 0.4, 0.6, 0.8, 1, 1.2, 1.4, 1.6, 1.8, 2, 2.2, 2.2, 2.4, 2.6, 2.8), bankfull_depths_m = c(-0.15, -0.09, -0.00998, 0.06, 0.13, 0.26, 0.36, 0.46, 0.56, 0.61, 0.85, 0.93, 0.93, 0.97, 1, 1)), .Names = c(measurment_num, bankfull_depths_m), row.names = c(32L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 29L, 12L, 13L, 14L), class = data.frame)) measure_bkf_not_zero - measure_bkf[grep([^0], measure_bkf[,bankfull_depths_m]),] bkf_min - grep(min(measure_bkf_not_zero[,bankfull_depths_m]), measure_bkf_not_zero[,bankfull_depths_m]) bkf_max - grep(max(measure_bkf_not_zero[,bankfull_depths_m]), measure_bkf_not_zero[,bankfull_depths_m]) bkf_min - ifelse(length(bkf_min)1, bkf_min[1], bkf_min) bkf_max - ifelse(length(bkf_max)1, bkf_max[1], bkf_max) #s - with(measure_bkf_not_zero, approx(measurment_num, bankfull_depths_m, xout=seq(measure_bkf_not_zero[bkf_min,measurment_num], measure_bkf_not_zero[bkf_max,measurment_num], length=2000))) #int_bkf - with(s, x[which.min(y[y0])]) s - with(measure_bkf_not_zero[bkf_min:bkf_max,], approxfun(bankfull_depths_m, measurment_num), ties=mean) int_bkf - s(0) On Tue, Sep 7, 2010 at 8:28 PM, David Winsemius dwinsem...@comcast.net wrote: On Sep 7, 2010, at 9:06 PM, stephen sefick wrote: s - 1.00 max(s) sprintf(%.2f, max(s)) [1] 1.00 @ as a string/character object returns 1 is there anyway that I can get it to return 1.00. I am using the results of this max statement in a grep statement and it returns the wrong numbers, I will provide more information and code if it would make more sense in context. -- Stephen Sefick | Auburn University | | Department of Biological Sciences | | 331 Funchess Hall | | Auburn, Alabama | | 36849| |___| | sas0...@auburn.edu | | http://www.auburn.edu/~sas0025 | |___| Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT -- Stephen Sefick | Auburn University | | Department of Biological Sciences | | 331 Funchess Hall | | Auburn, Alabama | | 36849| |___| | sas0...@auburn.edu | | http://www.auburn.edu/~sas0025 | |___| Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with max in a function
On Sep 7, 2010, at 10:05 PM, stephen sefick wrote: I'm sorry. In the bkf_max - grep(max(measure_bkf_not_zero[,bankfull_depths_m]), measure_bkf_not_zero[,bankfull_depths_m]) it is giving the indexes for 1 5 10 15 16 And it was supposed to yield ... what? I think this is because grep encounters a 1 with either 0. Huh? or nothing in front of it. I would like to find the max and then then the closest (leftmost) which is why the ifelse statment follows. grep is for working with character vectors. It seems to work when given numerics as patterns, but with floating point representations it seems really dangerous. I'm rather amazed you got anything vaguely useful. (And see further negative comments on that strategy below.) If you want the max of a numeric vector, then use max or which.max if you want the index. If you want the next largest then use max( vector[- which.max(vector)] ). So as Jim Holtman's tag line says: what problem are you trying to solve? Again, I am sorry for being vague. I get wrapped up in a problem and forget that I need to communicate. kindest regards, Stephen On Tue, Sep 7, 2010 at 8:48 PM, David Winsemius dwinsem...@comcast.net wrote: On Sep 7, 2010, at 9:37 PM, stephen sefick wrote: Here is a striped down example that is not working That dreadful phrase... is not working. When the ESP package comes to fruition, life will be so easy. Until then ... the English language is necessary. Where am we supposed to be looking. Did I miss you saying which of those (unprinted) objects we should be fixing. because of the 1.00 to 1. Any help would be greatly appreciated. measure_bkf - (structure(list(measurment_num = c(0, 0.2, 0.4, 0.6, 0.8, 1, 1.2, 1.4, 1.6, 1.8, 2, 2.2, 2.2, 2.4, 2.6, 2.8), bankfull_depths_m = c(-0.15, -0.09, -0.00998, 0.06, 0.13, 0.26, 0.36, 0.46, 0.56, 0.61, 0.85, 0.93, 0.93, 0.97, 1, 1)), .Names = c(measurment_num, bankfull_depths_m), row.names = c(32L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 29L, 12L, 13L, 14L), class = data.frame)) measure_bkf_not_zero - measure_bkf[grep([^0], measure_bkf[,bankfull_depths_m]),] You have constructed an odd pattern with the square brackets around caret-zero. Per the regex page: A character class is a list of characters enclosed between [ and ] which matches any single character in that list; unless the first character of the list is the caret ^, when it matches any character not in the list You are trying to find any number with a non-zero in a numeric vector? That seems to be what happened. bkf_min - grep(min(measure_bkf_not_zero[,bankfull_depths_m]), measure_bkf_not_zero[,bankfull_depths_m]) bkf_max - grep(max(measure_bkf_not_zero[,bankfull_depths_m]), measure_bkf_not_zero[,bankfull_depths_m]) bkf_min - ifelse(length(bkf_min)1, bkf_min[1], bkf_min) bkf_max - ifelse(length(bkf_max)1, bkf_max[1], bkf_max) #s - with(measure_bkf_not_zero, approx(measurment_num, bankfull_depths_m, xout=seq(measure_bkf_not_zero[bkf_min,measurment_num], measure_bkf_not_zero[bkf_max,measurment_num], length=2000))) #int_bkf - with(s, x[which.min(y[y0])]) s - with(measure_bkf_not_zero[bkf_min:bkf_max,], approxfun(bankfull_depths_m, measurment_num), ties=mean) int_bkf - s(0) On Tue, Sep 7, 2010 at 8:28 PM, David Winsemius dwinsem...@comcast.net wrote: On Sep 7, 2010, at 9:06 PM, stephen sefick wrote: s - 1.00 max(s) sprintf(%.2f, max(s)) [1] 1.00 @ as a string/character object returns 1 is there anyway that I can get it to return 1.00. I am using the results of this max statement in a grep statement and it returns the wrong numbers, I will provide more information and code if it would make more sense in context. -- Stephen Sefick | Auburn University | | Department of Biological Sciences | | 331 Funchess Hall | | Auburn, Alabama | | 36849| |___| | sas0...@auburn.edu | | http://www.auburn.edu/~sas0025 | |___| Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT
Re: [R] Regression using mapply?
On Sep 8, 2010, at 7:34 AM, Philipp Kunze wrote:
Hi,
I have huge matrices in which the response variable is in the first
column and the regressors are in the other columns. What I wanted to
do
now is something like this:
#this is just to get an example-matrix
DataMatrix - rep(1,1000);
Disturbance - rnorm(900);
DataMatrix[101:1000] - DataMatrix[101:1000]+Disturbance;
DataMatrix - matrix(DataMatrix,ncol=10,nrow=100);
#estimate univariate linear model with each regressor-column, response
in the first column
for(i in 2:10){
result - lm(DataMatrix[,1]~DataMatrix[,i])
}
result - apply(DataMatrix[,2:10], 2, function (x) lm(DataMatrix[,
1]~x) )
Which would have the added advantage that result would not be
overwritten for iterations 3:10, which is what your code would have
done. result will be a list of 9 models which might be a bit
unweildy, so you might consider something like
result - apply(DataMatrix[,2:10], 2, function (x)
coef( lm(DataMatrix[,1]~x) ) )
result
When you do so, you uncover a fatal flaw in your strategy, which
suggests you have not even done this once on your data or simulations.
--
David.
Is there any way to get rid of the for-loop using mapply (or some
other
function)?
--
David Winsemius, MD
West Hartford, CT
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Re: [R] How to change font size in plot() function
That is one method, but he should also review: ?par where he will discover that cex.main is a parameter that could be used from within the plot function. FMH had written, by Peng, C had failed to include context: Could someone please advice me the way to change the size of the title and x and y-label in plot() function. I've tried to use 'font' and 'font.main' call in plot() function, but it didn't make any changes in terms of the size. On Sep 8, 2010, at 7:46 AM, Peng, C wrote: try: ?title -- View this message in context: http://r.789695.n4.nabble.com/How-to-change-font-size-in-plot-function-tp2531127p2531161.html Sent from the R help mailing list archive at Nabble.com. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Checking if the distribution follow a power law
On Sep 8, 2010, at 10:34 AM, NatsumiYotsumoto wrote: Dear all. I'm using igraph package, and do a research about network analysis. With power.law.fit from igraph package, it seems that we can fit a power law distribution to some data. But, I want to know how to judge whether the network distribution follows a power law or not. In order to determine whether something is from distribution A or not- A, one needs to have a sensible way of characterizing or considering what would be in the range of distributions in the not-A. Unfortunately for your question, the range of possible distributions is infinite. That means it would always be possible to have a better fitting distribution than what ever is distribution A. If you have alternatives to the power-law that you want to put to the test, then now is the time to offer them. My guess is that you do not, so I will offer alternatives: Alt A: a) read the citations in the email you cited, especially Newman then ... b) set up a histogram of your data using hist with logarithmic or geometric progression of the breaks argument. c) as a check on you exponent estimate, calculate alpha and se(alpha) as on pg 4-5 of that citation. Alt B: require(sos) ???fitting pareto ???fitting power network # and proceed from there -- David. Does anyone know the way to do this? Thanks for any help. Daigo p.s. Also, I tried several ways such as http://www.mail-archive.com/r-h...@stat.math.ethz.ch/msg62520.html and I got results like this: Profiling... 2.5 % 97.5 % 2.393297 2.412650 What do these suggest? please tell me about this if someone knows. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] try-error can not be test. Why?
On Sep 8, 2010, at 1:18 PM, telm8 wrote: Hi, I am having some strange problem with detecting try-error. From what I have read so far the following statement: try( log(a) ) == try-error should yield TRUE, however, it yields FALSE. I can not figure out why. Can someone help? class(try( log(a), silent=TRUE )) == try-error [1] TRUE Many thanks -- View this message in context: http://r.789695.n4.nabble.com/try-error-can-not-be-test-Why-tp2531675p2531675.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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Replace NAs in one column with data from another column
On Sep 8, 2010, at 2:24 PM, Joshua Wiley wrote:
Hi Jakob,
You can use is.na() to create an index of which rows in column 3 are
missing data, and then select these from column 1. Here is a simple
example:
dat - data.frame(V1 = 1:5, V3 = c(1, NA, 3, 4, NA))
dat$new - dat$V3
my.na - is.na(dat$V3)
dat$new[my.na] - dat$V1[my.na]
dat
This should be quite fast. I broke the steps up to be explicit, but
you can readily simplify them.
I was about to post something similar except I was going to avoid the
$ operator thinking, incorrectly as it turned out, that it would be
faster. I also include the Holtman/Rizopoulos suggestion of ifelse().
I was also surprised that ifelse is the winning strategy:
dat[4] - dat[3]; idx -is.na(dat[, 3])
dat[is.na(dat[, 3]), 4] - dat[is.na(dat[, 3]), 1]
benchmark(meth.ifelse = {dat$z.new - ifelse(is.na(dat$V3), dat$V1,
dat$V3)},
+ meth.dlr.sign={dat$new - dat$V3
+ my.na - is.na(dat$V3)
+ dat$new[my.na] - dat$V1[my.na]},
+ meth.index ={dat[4] - dat[3]; idx -is.na(dat[, 3])
+ dat[idx, 4] - dat[idx, 1]},
+ meth.forloop ={for (i in 1:nrow(dat)){
+ if (is.na(dat[i,3])==TRUE){
+ dat[i,4]- dat[i,1]}
+ else{
+ dat[i,4]- dat[i,3]} }
+ },
+ replications=5000, columns = c(test, replications, elapsed,
+ relative, user.self) )
test replications elapsed relative user.self
2 meth.dlr.sign 5000 0.502 1.081897 0.501
4 meth.forloop 5000 6.419 13.834052 6.409
1 meth.ifelse 5000 0.464 1.00 0.463
3meth.index 5000 2.908 6.267241 2.904
--
David.
HTH,
Josh
On Wed, Sep 8, 2010 at 11:17 AM, Jakob Hedegaard
jakob.hedega...@agrsci.dk wrote:
Hi list,
I have a data frame (m) with 169221 rows and 10 columns and would
like to make a new column containing the content of column 3 but
replace the NAs in column 3 with the data in column 1 (from the
same row as the NA in column 3). Column 1 has data in all rows.
My first attempt was:
for (i in 1:169221){
if (is.na(m[i,3])==TRUE){
m[i,11] - as.character(m[i,1])}
else{
m[i,11] - as.character(m[i,3])}
}
Works - but takes too long time.
I would appreciate alternative solutions.
Best regards, Jakob
--
David Winsemius, MD
West Hartford, CT
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] subbing a string vector for another string vector
On Sep 8, 2010, at 4:45 PM, jlemaitre wrote: I have a data frame with two columns: url.img patt.url 1http://$IMAGE_ID$www.url.com/image.jpg 2$IMAGE_ID$ http://www.blah.com/image.gif ... I want to replace $IMAGE_ID$ with the corresponding entry in the pattern column such that the result would appear as follows: A) Drop the use of the name pattern because it is the argument name for regex functions. Name it something else like patt.url. Furthermore image is also a function name so use somehting more specific there too, say url.img B) use grep(grep-pattern, df$url.img) to identify the rows of patt.url you want to assemble. Perhaps: df[ grep(\\$IMAGE\\_ID\\$, df$url.img), patt.url] url http://www.url.com/image.jpg http://www.blah.com/image.gif Using something like gsub(,image,pattern) doesn't work because it only takes uses the first entry in pattern as the replacement for all image entries: url http://www.url.com/image.jpg www.url.com/image.jpg Please help. Thanks. -- View this message in context: http://r.789695.n4.nabble.com/subbing-a-string-vector-for-another-string-vector-tp2532005p2532005.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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie cross tabulation issue
On Sep 8, 2010, at 6:40 PM, Jonathan Finlay wrote: hi, i'm new in R and i need some help. Please, ¿do you know a function how can process cross tables for many variables and show the result in one table who look like this?: ++ |-- |X variable | |- | Xop1 | Xop2 | Xop3|.| ++ |Yvar1 | Total | %row..| | | Op1 | %row..| | | Op2 | %row..| |+---+ |Yvar2 | Op1 | %row..| | | Op2 | %row...| ++ |Yvar3 | Op1 | %row..| | | Op2 | %row...| | | Op3 | %row...| |+---+ Like a pivot table! ?table ?xtabs ... and if you want all those dashes, dots, pluses and pipes cluttering up your output a la SAS, then there is: gmodels::Crosstable -- David. thanks a lot. -- Jonathan. [[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 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] coxph and ordinal variables?
On Sep 8, 2010, at 6:43 PM, Min-Han Tan wrote: Dear R-help members, Apologies - I am posting on behalf of a colleague, who is a little puzzled as STATA and R seem to be yielding different survival estimates for the same dataset when treating a variable as ordinal. Ordered() is used to represent an ordinal variable) I understand that R's coxph (by default) uses the Efron approximation, whereas STATA uses (by default) the Breslow. but we did compare using the same approximations. I am wondering if this is a result of how coxph manages an ordered factor? Essentially, this is a survival dataset using tumor grade (1, 2, 3 and 4) as the risk factor. This is more of an 'ordinal' variable, rather than a continuous variable. For the same data set of 399 patients, when treating the vector of tumor grade as a continuous variable (range of 1 to 4), testing the Efron and the Breslow approximations yield the same result in both R and STATA. However, when Hist_Grade_4 grp is converted into an ordered factor using ordered(), and the same scripts are applied, rather different results are obtained, relative to the STATA output. This is tested across the different approximations, with consistent results. The comparison using Efron approximation and ordinal data is is below. Are you sure you want an ordered factor? In R this means you will be creating linear, quadratic and cubic contrasts. Notice the L, Q and C designations on the coefficients. That certainly does not look to be comparable to what you are getting from Stata. My suggestion would be to create an un-ordered factor in R and see whether you get results more in line with Stata's output when applied to your data. -- David. Your advice is very much appreciated! Min-Han Apologies below for the slightly malaligned output. STATA output . xi:stcox i.Hist_Grade_4grp, efr i.Hist_Grade_~p _IHist_Grad_1-4 (naturally coded; _IHist_Grad_1 omitted) failure _d: FFR_censor analysis time _t: FFR_month Iteration 0: log likelihood = -1133.369 Iteration 1: log likelihood = -1129.4686 Iteration 2: log likelihood = -1129.3196 Iteration 3: log likelihood = -1129.3191 Refining estimates: Iteration 0: log likelihood = -1129.3191 Cox regression -- Efron method for ties No. of subjects = 399 Number of obs = 399 No. of failures = 218 Time at risk= 9004.484606 LR chi2(3) = 8.10 Log likelihood = -1129.3191 Prob chi2 = 0.0440 -- _t | Haz. Ratio Std. Err. zP|z| [95% Conf. Interval] - + _IHist_Gra~2 | 1.408166 .3166876 1.52 0.128 .9062001 2.188183 _IHist_Gra~3 |1.69506 .3886792 2.30 0.021 1.081443 2.656847 _IHist_Gra~4 | 2.540278 .9997843 2.37 0.018 1.17455 5.49403 R Output using summary ( coxph( Surv(FFR_month,FFR_censor) ~ Hist_Grade_4grp, method=c(breslow))) summary ( coxph( Surv(FFR_month,FFR_censor) ~ Hist_Grade_4grp, method=c(exact))) summary ( coxph( Surv(FFR_month,FFR_censor) ~ Hist_Grade_4grp, method=c(efron))) n=399 (21 observations deleted due to missingness) coef exp(coef) se(coef) z Pr(|z|) Hist_Grade_4grp.L 0.66685 1.94809 0.26644 2.503 0.0123 * Hist_Grade_4grp.Q 0.03113 1.03162 0.20842 0.149 0.8813 Hist_Grade_4grp.C 0.08407 1.08771 0.13233 0.635 0.5252 --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 exp(coef) exp(-coef) lower .95 upper .95 Hist_Grade_4grp.L 1.948 0.51331.1556 3.284 Hist_Grade_4grp.Q 1.032 0.96930.6857 1.552 Hist_Grade_4grp.C 1.088 0.91940.8392 1.410 Rsquare= 0.02 (max possible= 0.997 ) Likelihood ratio test= 8.1 on 3 df, p=0.044 Wald test= 8.02 on 3 df, p=0.0455 Score (logrank) test = 8.2 on 3 df, p=0.04202 [[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 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] Newbie cross tabulation issue
On Sep 8, 2010, at 7:32 PM, Jonathan Finlay wrote: Thanks David, gmodels::Crosstable partially work because can show only 1 x 1 tablen CrossTable(x,y,...) I need something how can process at less 1 variable in X an 10 in Y. I hope you mean only two factors and an n x m table. Thanks for your help. -- Jonathan. [[alternative HTML version deleted]] -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with outer
On Sep 8, 2010, at 6:59 PM, tuggi wrote:
hello,
tank you very very much. it solves my first problem. i hope you can
help me
also with the second problem.
it was like this.
p_11=seq(0,1,0.1)
p_12=seq(0,1,0.1)
then i get also this error message:
Error in rmultinom(n - q + 1, size = 1, prob = rbind(p_11, p_12, (1
- :
non-positive probability
It seems to be a very informative error message. The first element of
seq(0,1,0.1) is going to be zero and rmultinom doesn't accept non-zero
probabilities. If you would explain what you _were_ trying to
accomplish it might be possible to say more, but without some
explication we would be stumbling around in the dark. Furthermore I
wonder whether you really meant to offer a matrix to rmultinom for
probabilities? The help page says prob should be a vector.
OK, I will stumble around:
If you want one random draw one one item from into n-q+1 equally
probable bins, then wouldn't you use:
rmultinom(n - q + 1, size = 1, prob = rep(0.1, 10) )
rmultinom(1, size = 1, prob = rep(0.1, 10) )
[,1]
[1,]0
[2,]0
[3,]0
[4,]0
[5,]0
[6,]0
[7,]0
[8,]0
[9,]1
[10,]0
If you want 10 draws:
rmultinom(10, size = 1, prob = rep(0.1, 10) )
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]100000000 1
[2,]000000000 0
[3,]011000000 0
[4,]000100000 0
[5,]000000001 0
[6,]000000000 0
[7,]000011000 0
[8,]000000100 0
[9,]000000010 0
[10,]000000000 0
And if you want 1 sample of 100 items:
rmultinom(1, size = 100, prob = rep(0.1, 10) )
[,1]
[1,] 10
[2,] 10
[3,] 12
[4,] 12
[5,]8
[6,] 11
[7,]9
[8,]6
[9,]7
[10,] 15
But as I said that's a WAG at where you might be having problems.
--
David.
i try to solve this problem with a if order like this:
p_11=seq(0,1,0.1)
p_12=seq(0,1,0.1)
guete = function(p_11,p_12) { if(p_11+p_121)
set.seed(1000)
S_vek=matrix(0,nrow=N,ncol=1)
for(i in 1:N) {
X_0=rmultinom(q-1,size=1,prob=p_0)
X_1=rmultinom(n-q+1,size=1,prob=cbind(p_11,p_12,(1-p_11-p_12)))
N_0=apply(X_0[,(n-2*k-L+1):(n-k-L)],1,sum)
N_1=apply(X_1[,(n-q-k+2):(n-q+1)],1,sum)
S_vek[i]=((sum(((N_1-k*cbind(p_11,p_12,(1-p_11-p_12)))^2)/
k*cbind(p_11,p_12,(1-p_11-p_12/(sum(((N_0-k*p_0)^2)/k*p_0)))-1
}
1-mean(f_1=S_vek S_vek =f_2)
}
f=outer(p_11,p_12,Vectorize(guete))
but i get the error message:
Error in rmultinom(n - q + 1, size = 1, prob = cbind(p_11, p_12, (1
- :
non-positive probability.
thank for your helps.
Tuggi
--
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Re: [R] on error execute:
On Sep 8, 2010, at 7:37 PM, jcress410 wrote: So, i've been writing code in R interfacing with mysql, its been fun and the documentation has been useful, learning quite a bit. the only annoying thing i've been encountering is while coding/ debugging, my session gets clogged, especially with sql connections. At the end of the code i dbDisconnect, but when the script is stopped by an error, the session stays open. Usually i'll remember to ls() and dbDisconnect at console manually, though, sometimes I forget and wind up with a bunch of extra connections. A couple of questions: First: is there a way to tell R to execute some commands before dumping back to console on error, ?options error: either a function or an expression governing the handling of non- catastrophic errors snipped rest of paragraph and/or second: is there a way to capture error messages and continue executing? ?try Sorry if this has been addressed already, but, even aided by the new awesome google i haven't been able to find it. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] try-error can not be test. Why?
On Sep 8, 2010, at 11:46 PM, Philippe Grosjean wrote:
On 08/09/10 19:25, David Winsemius wrote:
On Sep 8, 2010, at 1:18 PM, telm8 wrote:
Hi,
I am having some strange problem with detecting try-error. From
what I
have read so far the following statement:
try( log(a) ) == try-error
should yield TRUE, however, it yields FALSE. I can not figure out
why.
Can
someone help?
class(try( log(a), silent=TRUE )) == try-error
[1] TRUE
This is perfectly correct in this case, but while we are mentioning
a test on the class of an object, the better syntax is:
inherits(try(log(a)), try-error)
In a more general context, class may be defined with multiple
strings (R way of subclassing S3 objects). For instance, this does
not work:
if (class(Sys.time()) == POSIXct) ok else not ok
... because the class of a `POSIXct' object is defined as:
c(POSIXt, POSIXct). This works:
if (inherits(Sys.time(), POSIXct)) ok else not ok
Alternate valid tests would be (but a little bit less readable):
if (any(class(Sys.time()) == POSIXct)) ok else not ok
or, by installing the operators package, a less conventional, but
cleaner code:
install.packages(operators)
library(operators)
if (Sys.time() %of% POSIXct) ok else not ok
I have also used if (try-error %in% class(try(log(a))) ) { } else
{ }, but the inherits() form looks at the very least less clunky.
Best,
Philippe Grosjean
Many thanks
--
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Re: [R] Determine Bounds of Current Graph
On Sep 9, 2010, at 10:07 AM, Isamoor wrote:
I'm having trouble determining the bounds of my current graph. I
know how
to set the bounds up front (ylim xlim in most cases), but I would
rather
be able to dynamically see what was chosen to use in later code.
Example:
library(maps)
map('state','Indiana')
map.axes()
?par
bounds - par(usr)
bounds
[1] -88.12964 -84.77184 37.74583 41.82082
??Something that lets me know the y-axis is from ~38 to ~42 and
store this
information into a vector
Is there some way to query what the bounds of the current graph are?
Thanks!
David Winsemius, MD
West Hartford, CT
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Re: [R] Calculating with tolerances (error propagation)
On Sep 9, 2010, at 6:50 AM, Jan private wrote: Hello Bernardo, - If I understood your problem this script solve your problem: q-0.15 + c(-.1,0,.1) h-10 + c(-.1,0,.1) 5*q*h [1] 2.475 7.500 12.625 - OK, this solves the simple example. But what if the example is not that simple. E.g. P = 5 * q/h Here, to get the maximum tolerances for P, we need to divide the maximum value for q by the minimum value for h, and vice versa. Is there any way to do this automatically, without thinking about every single step? There is a thing called interval arithmetic (I saw it as an Octave package) which would do something like this. I would have thought that tracking how a (measuring) error propagates through a complex calculation would be a standard problem of statistics?? In other words, I am looking for a data type which is a number with a deviation +- somehow attached to it, with binary operators that automatically knows how to handle the deviation. Thank you, Jan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculating with tolerances (error propagation)
On Sep 9, 2010, at 6:50 AM, Jan private wrote: Hello Bernardo, - If I understood your problem this script solve your problem: q-0.15 + c(-.1,0,.1) h-10 + c(-.1,0,.1) 5*q*h [1] 2.475 7.500 12.625 - OK, this solves the simple example. But what if the example is not that simple. E.g. P = 5 * q/h Here, to get the maximum tolerances for P, we need to divide the maximum value for q by the minimum value for h, and vice versa. Have you considered the division by zero problems? Is there any way to do this automatically, without thinking about every single step? There is a thing called interval arithmetic (I saw it as an Octave package) which would do something like this. (Sorry for the blank reply posting. Serum caffeine has not yet reached optimal levels.) Is it possible that interval arithmetic would produce statistically incorrect tolerance calculation, and that be why it has not been added to R? Those tolerance intervals are presumably some sort of (unspecified) prediction intervals (i.e. contain 95% or 63% or some fraction of a large sample) and combinations under mathematical operations are not going to be properly derived by c( min(XY), max(XY) ) since those are not calculated with any understanding of combining variances of functions on random variables. -- David. I would have thought that tracking how a (measuring) error propagates through a complex calculation would be a standard problem of statistics?? In probability theory, anyway. In other words, I am looking for a data type which is a number with a deviation +- somehow attached to it, with binary operators that automatically knows how to handle the deviation. There is the suite of packages that represent theoretic random variables and support mathematical operations on them. See distrDoc and the rest of that suite. -- David. Thank you, Jan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie cross tabulation issue
On Sep 8, 2010, at 7:32 PM, Jonathan Finlay wrote: Thanks David, gmodels::Crosstable partially work because can show only 1 x 1 tablen CrossTable(x,y,...) I need something how can process at less 1 variable in X an 10 in Y. A further thought (despite a lack of clarification on what your data situation really is.). The strong tendency in R is not to attempt replication of formats in SAS that were developed in an era of dot- matrix printers, but to target modern output devices. As such most of the table output facilities with any degree of sophistication have LaTeX or HTML as targets. RSiteSearch(html tables) produces over 1000 links although they have many that are not for multiway tables where multi is greater than R x C. RSiteSearch(latex tables) produces many fewer. You may want to look at xtable, Sweave, odfWeave, the various HTML utilities, and Harrell's Hmisc::summary.formula -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie cross tabulation issue
On Sep 9, 2010, at 11:20 AM, David Winsemius wrote: On Sep 8, 2010, at 7:32 PM, Jonathan Finlay wrote: Thanks David, gmodels::Crosstable partially work because can show only 1 x 1 tablen CrossTable(x,y,...) I need something how can process at less 1 variable in X an 10 in Y. A further thought (despite a lack of clarification on what your data situation really is.). The strong tendency in R is not to attempt replication of formats in SAS that were developed in an era of dot- matrix printers, but to target modern output devices. As such most of the table output facilities with any degree of sophistication have LaTeX or HTML as targets. RSiteSearch(html tables) produces over 1000 links although they have many that are not for multiway tables where multi is greater than R x C. RSiteSearch(latex tables) produces many fewer. You may want to look at xtable, Sweave, odfWeave, the various HTML utilities, and Harrell's Hmisc::summary.formula Perhaps my final thought. It has none of the dividing lines, but ftable is the standard method for displaying flat contingency tables for greater than two dimensions. Try the examples on the help page. If you wanted to add all that +---+-ing window dressing you could concievable start with its code using: getAnywhere(ftable.default) and stick in the needed cat() statements. You can see how the author of CrossTables proceeded by just typing the function name without quotes: CrossTables -- David. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie cross tabulation issue
On Sep 9, 2010, at 2:14 PM, Marc Schwartz wrote: On Sep 9, 2010, at 12:59 PM, Jonathan Finlay wrote: Ok friends, I tried but I not know! I'm a Linux SysAdmin and Stadistical and i working to migrate all the software in my workplace to free or open software. The OS was easy, ofimatic suite too, multimedia and graphics you know, everything was relatively easy. But i work with SPSS and I produce tables from polls with CTABLES and FRECUENCIES tables. In my migration party I think R is the best option but need replace SPSS functions and procedures. I know that sucks but it's. In the attach you will see an example with a short output like I need, please I'm starting with R and i 'm study but i need to find something to replace this frecuently output while grows my experience in R. This is the syntaxis used: FREQUENCIES VARIABLES= REGION /ORDER= ANALYSIS . CTABLES /VLABELS VARIABLES=zona V1 V3 V4 V5 V6 V7 V8 v51 DISPLAY=DEFAULT /TABLE ZONA [C] +V1 [C] +V3 [C] +V4 [C] +V5 [C] +V6 [C] +V7 [C] +V8 [C] BY v51 [C][ROWPCT.COUNT COMMA40.1, TOTALS[COLPCT.COUNT COMMA40.1]] /SLABELS VISIBLE=NO /CATEGORIES VARIABLES=ZONA ORDER=A KEY=VALUE EMPTY=INCLUDE TOTAL=YES LABEL='Total' POSITION=BEFORE /CATEGORIES VARIABLES= V1 V3 V4 V5 V6 V7 V8 ORDER=A KEY=VALUE EMPTY=EXCLUDE /CATEGORIES VARIABLES=v51 ORDER=D KEY=VALUE EMPTY=INCLUDE TOTAL=YES LABEL='Frecuencia' POSITION=AFTER /TITLES TITLE=' '. R is grand, great and biggest. Thanks for all. I would recommend getting a copy of Bob Muenchen's book: R for SAS and SPSS Users http://sites.google.com/site/r4statistics/the-books/r4sas-spss There is a free smaller version here: http://sites.google.com/site/r4statistics/free-version and you can get a full copy from Amazon.com: http://www.amazon.com/SAS-SPSS-Users-Statistics-Computing/dp/0387094172 That would be the best place to start, relative to moving from SPSS to R. Part of the challenge is not just replicating SPSS code and output in R, but understanding the conceptual differences between the two, so that you can take advantage of R's approach/philosophy in conducting data analysis. Another place to get useful code is the UCLA Statistical Computing website. At one time that group was very negative about using R, but I think the feedback or demand has reversed that attitude and there are now quite a few worked examples of common task in which SAS, SPSS, Stata and R are applied using the same data. http://www.ats.ucla.edu/stat/dae/ -- David. HTH, Marc Schwartz David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sequeeze a data frame
On Sep 9, 2010, at 5:47 PM, array chip wrote:
Hi, suppose I have a data frame as below:
dat-
cbind
(expand
.grid
(id=c(1,2,3),time=c(0,3,6),mode=c('R','L'),rep=(1:3)),y=rnorm(54))
I kind of want to squeeze the data frame into a new one with
averaged y over
rep for the same id, time and mode. taking average is easy with
tapply:
tapply(dat$y, list(dat$id, dat$time, dat$mode), mean)
Try:
dat$avg.y - ave(dat$y, dat$id, dat$time, dat$mode, FUN=mean)
?ave
The syntax of ave is different than tapply or by. The grouping
factors are not presented as a list and the FUN argument comes after
the ,..., so it needs to be named if different than the default of
mean.
But I want the result to be in the same format as dat. Certainly,
we always
can transform the result (array) into the data frame using a lot of
codes. But
is there a simple way to do this?
Thanks
David Winsemius, MD
West Hartford, CT
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] URL error when trying to use help function in R [Sec: UNOFFICIAL]
On Sep 9, 2010, at 6:34 PM, Gosse, Michelle wrote: Greetings, I am using R version 2.11.1 on a Dell computer, via a VMware connection to a remote server. My browser version is IE 8.0.6001.18702 and the OS is some corporate version of Microsoft XP. I'm trying to learn more about the tapply function , so I typed ? tapply into the command line. This opened up a browser window with url http://127.0.0.1:28138/library/base/html/tapply.html which is giving me an error message. I receive the same problem when trying for help on other commands, e.g. ?table http://127.0.0.1:28138/library/base/html/table.html and ? log http://127.0.0.1:28138/library/base/html/Log.html I did a whois on 127.0.0.1 That should always be your own computer. The browser is trying to reach a server on itself over port 28138 and either the port is blocked or you don't have the documentation at that location. via www.geektools.comhttp://www.geektools.com and got the information boxed below, which suggests I'm not going where I am supposed to be going when I ask for help. I suppose you could use that terminology, but you should check the specification of the browser entry in your options(). Does the remote server have the documentation in a library directory? Can you get the sysadmin for that device to give you the locations of the files? ?options options()$browser # would at least tell you where you system thinks it should be going. If you restrict Baron's help search to only this year's questions containing help server browser you get: http://search.r-project.org/cgi-bin/namazu.cgi?query=help+server+browsermax=100result=normalsort=scoreidxname=functionsidxname=Rhelp10 -- David. I'm assuming this problem is associated with my VMware link to the server. Is anyone else running this type of connection through to R and, if so, could they please advise how to remove the issue? I've been able to bring in my data table fine, and do some commands on it, the issue so far appears related only to help files. I'm assuming this is because the help files call a URL that, in my case, appears to be incorrect (whereas the R commands are not calling a URL and therefore are working fine). I found this earlier thread: https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=14155 however, the commentary says the bug was fixed. Final results obtained from whois.arin.net. Results: # # The following results may also be obtained via: # http://whois.arin.net/rest/nets;q=127.0.0.1?showDetails=trueshowARIN=false # NetRange: 127.0.0.0 - 127.255.255.255 CIDR: 127.0.0.0/8 OriginAS: NetName: SPECIAL-IPV4-LOOPBACK-IANA-RESERVED NetHandle: NET-127-0-0-0-1 Parent: NetType: IANA Special Use Comment: This block is assigned for use as the Internet Comment: host loopback address. Datagrams sent to Comment: addresses anywhere within this block loops back Comment: inside the host. Many implementation only Comment: support this for 127.0.0.1. This block was Comment: assigned by the IETF in the Standard document, Comment: RFC 1122 and is further documented in the Best Comment: Current Practice document RFC 5735. These Comment: documents can be found at: Comment: http://www.rfc-editor.org/rfc/rfc1122.txt Comment: http://www.rfc-editor.org/rfc/rfc5735.txt RegDate: Updated: 2010-04-14 Ref: http://whois.arin.net/rest/net/NET-127-0-0-0-1 [extraneous detail here deleted] Thanks in advance, Michelle Michelle Gosse Consumer and Social Sciences Food Standards Australia New Zealand 108 The Terrace Wellington New Zealand ph: 0064-4-978-5652 email: michelle.go...@foodstandards.govt.nzmailto:michelle.go...@foodstandards.govt.nz ** This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. If you have received this email in error please notify the system manager. This footnote also confirms that this email message has been swept by MIMEsweeper for the presence of computer viruses. www.clearswift.com ** [[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 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] survfit question
On Sep 9, 2010, at 8:50 PM, andre bedon wrote: I am attempting to graph a Kaplan Meier estimate for some claims using the survfit function. However, I was wondering if it is possible to plot a cdf of the kaplan meier rather than the survival function. Here is some of my code: It's not really the cdf of the KM since the KM is just an estimator. Yeah, I know, picky, picky. library(survival) Surv(claimj,censorj==0) I'm reasonably sure you need to assign that to something (unless its purpose is just to test the syntax.) survfit(Surv(claimj,censorj==0)~1) surv.all-survfit(Surv(claimj,censorj==0)~1) summary(surv.all) plot(surv.all) I would really appreciate any assistance. Thank you. The survival function is just 1 minus the CDF, (and vice versa). You didn't provide any data, but we can use the aml dataframe in survival: library(survival) surv.all-survfit(Surv(time,status)~1, data=aml) str(surv.all) # x-coord is time and S_KM(t) is surv plot(surv.all$time, 1-surv.all$surv, type=s, ylim=c(0,1)) So that's the KM estimator of the CDF. Doesn't inherit the nice features of the plot.survfit function, though. It's also going to be more messy if you have two/+ groups -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculating with tolerances (error propagation)
On Sep 9, 2010, at 10:57 AM, David Winsemius wrote: On Sep 9, 2010, at 6:50 AM, Jan private wrote: Hello Bernardo, - If I understood your problem this script solve your problem: q-0.15 + c(-.1,0,.1) h-10 + c(-.1,0,.1) 5*q*h [1] 2.475 7.500 12.625 - OK, this solves the simple example. But what if the example is not that simple. E.g. P = 5 * q/h Here, to get the maximum tolerances for P, we need to divide the maximum value for q by the minimum value for h, and vice versa. Is there any way to do this automatically, without thinking about every single step? There is a thing called interval arithmetic (I saw it as an Octave package) which would do something like this. (Sorry for the blank reply posting. Serum caffeine has not yet reached optimal levels.) Is it possible that interval arithmetic would produce statistically incorrect tolerance calculation, and that be why it has not been added to R? Those tolerance intervals are presumably some sort of (unspecified) prediction intervals (i.e. contain 95% or 63% or some fraction of a large sample) and combinations under mathematical operations are not going to be properly derived by c( min(XY), max(XY) ) since those are not calculated with any understanding of combining variances of functions on random variables. There is a function, propagate, in the qpcR package that does incorporate statistical principles in handling error propagation. Thanks to the author, Dr. rer. nat. Andrej-Nikolai Spiess, for drawing it to my attention (and of course for writing it.). It appears that it should handle the data situation offered with only minor modifications. The first example is vary similar to your (more difficult) ratio problem. install.packages(pkgs=qpcR, type=source) # binary install did not succeed on my Mac, but installing from source produced no errors or warnings. require(qpcR) q- c( 0.15 , .1) h-c( 10 , .1) EXPR - expression(5*q/h) DF - cbind(q, h) res - propagate(expr = EXPR, data = DF, type = stat, + do.sim = TRUE, verbose = TRUE) res$summary Sim PermProp Mean0.07500751 NaN 0.0750 s.d.0.05001970 NA 0.05000562 Median 0.07445724 NA NA MAD 0.04922935 NA NA Conf.lower -0.02332498 NA -0.02300922 Conf.upper 0.17475818 NA 0.17300922 (My only suggestion for enhancement would be a print or summary method that did not output every single simulated value.) Three methods for error propagation estimation are included: a) Monte- Carlo simulation using sampling from the data, b) Permutation, and c) Gaussian errors calculated via a Taylor series expansion. There is a rich set of worked examples. It appears to be capable of meeting a wide variety of challenges. -- David. -- David. I would have thought that tracking how a (measuring) error propagates through a complex calculation would be a standard problem of statistics?? In probability theory, anyway. In other words, I am looking for a data type which is a number with a deviation +- somehow attached to it, with binary operators that automatically knows how to handle the deviation. There is the suite of packages that represent theoretic random variables and support mathematical operations on them. See distrDoc and the rest of that suite. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data.frames : difference between x$a and x[, a] ? - How set new values on x$a with a as variable ?
On Sep 10, 2010, at 9:42 AM, Hadley Wickham wrote: I'm having trouble parsing this. What exactly do you want to do? 1 - Put a list as an element of a data.frame. That's quite convenient for my pricing function. I think this is a really bad idea. data.frames are not meant to be used in this way. Why not use a list of lists? It can be very convenient, but I suspect the original poster is confused about the different between vectors and lists. I wouldn't be surprised if someone were confused, since my reading of some (but not all) of the help documents has led me to think that lists _were_ vectors, just not vectors of atomic mode. And one oft- illustrated method for creating a list is: alist - vector(mode=list, length=10). I am perhaps less confused than I was two years ago but my confusion about all the possible permutations of mode, typeof, expression, formula, and class and the extraction methods therefrom definitely persists. I think the authors of the documentation are of divided opinion or usage on this topic. Best; David. Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] convert 1, 10, and 100 to 0001, 0010, 0100 etc.
On Sep 10, 2010, at 4:05 PM, Nick Matzke wrote: Hi, Is there an easy way to convert numbers into a form such that they all have the same number of digits? e.g.: 1, 10, and 100 ...become... 0001, 0010, 0100 etc. sprintf(%05.0f, 100) [1] 00100 sprintf(%04.0f, 100) [1] 0100 I ask because I am producing a large number of files that need to sort consistently by filename. Currently I get this kind of sorting: filename1 filename10 filename11 filename12 filename13 filename14 filename15 filename16 filename17 filename18 filename19 filename2 filename20 filename21 ...etc.. which is annoying. Ideally I'd have: filename0001 filename0002 filename0003 filename0004 filename0005 filename0006 filename0007 filename0008 filename0009 filename0010 filename0011 ...etc.. Basically I want to produce strings like 0010 without an elaborate hack. Thanks! Nick -- Nicholas J. Matzke Ph.D. Candidate, Graduate Student Researcher Huelsenbeck Lab Center for Theoretical Evolutionary Genomics 4151 VLSB (Valley Life Sciences Building) Department of Integrative Biology University of California, Berkeley Graduate Student Instructor, IB200A Principles of Phylogenetics: Systematics http://ib.berkeley.edu/courses/ib200a/index.shtml Lab websites: http://ib.berkeley.edu/people/lab_detail.php?lab=54 http://fisher.berkeley.edu/cteg/hlab.html Dept. personal page: http://ib.berkeley.edu/people/students/person_detail.php?person=370 Lab personal page: http://fisher.berkeley.edu/cteg/members/matzke.html Lab phone: 510-643-6299 Dept. fax: 510-643-6264 Cell phone: 510-301-0179 Email: mat...@berkeley.edu Mailing address: Department of Integrative Biology 3060 VLSB #3140 Berkeley, CA 94720-3140 - [W]hen people thought the earth was flat, they were wrong. When people thought the earth was spherical, they were wrong. But if you think that thinking the earth is spherical is just as wrong as thinking the earth is flat, then your view is wronger than both of them put together. Isaac Asimov (1989). The Relativity of Wrong. The Skeptical Inquirer, 14(1), 35-44. Fall 1989. http://chem.tufts.edu/AnswersInScience/RelativityofWrong.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Solver in R
On Sep 10, 2010, at 5:35 PM, Chien-Pang Chin wrote: Hi all: I'm looking for a package that similar to solver in MS. All I need is find a r to satisfy R0=sum( , where t are from 1 to n and Xt are come from another formula. The most package I found were to max or min the obj. function. Is there any package can do it? If you minimize R0 - sum(r * Xt ), you should get your answer. -- David. Thanks. Peter David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Solver in R
On Sep 10, 2010, at 5:44 PM, David Winsemius wrote: On Sep 10, 2010, at 5:35 PM, Chien-Pang Chin wrote: Hi all: I'm looking for a package that similar to solver in MS. All I need is find a r to satisfy R0=sum( , where t are from 1 to n and Xt are come from another formula. The most package I found were to max or min the obj. function. Is there any package can do it? If you minimize , you should get your answer. Ooops, sorry. Make that: abs( R0 - sum(r * Xt ) ) # or ( R0 - sum(r * Xt ) )^2 -- David. Thanks. Peter David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Where to find R-help options web page
On Sep 10, 2010, at 6:10 PM, Frank Harrell wrote: Thanks for your note Ista. I had seen that page. I don't see on it where you can modify your subscription characteristics. I'd like to remain subscribed but not have e-mail delivered; that way I can just see messages in nabble. It may be that I'm misunderstanding how to use nabble (and to be able to post on it). I will admit that it often takes two or three looks for me each time I do it, but it _is_ on that page. Search for: To unsubscribe from R-help, get a password reminder, or change your subscription options enter your subscription email address: The entry box is just below that sentence. The button gets you to a page where you enter your pwd. When I use Nabble I try to remember to include context with the quotes button. It's kind of a clunky interface. Frank - Frank Harrell Department of Biostatistics, Vanderbilt University -- View this message in context: http://r.789695.n4.nabble.com/Where-to-find-R-help-options-web-page-tp2535123p2535148.html Sent from the R help mailing list archive at Nabble.com. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Formatting of time strings
On Sep 10, 2010, at 6:41 PM, Dennis Fisher wrote: Colleagues, (OS X, R: 11.1) I am stuck on what should be a simple problem. I have time elements in numeric form (e.g., 14050.5). I want to convert these to HH:MM (e.g., 12:00). I can do this with brute force (7 commands) but I suspect that it can be accomplished with a single command. Any help would be appreciated. ?DateTimeClasses # POSIXct objects are internally the number of seconds past the origin as.POSIXct(14050.5, origin=1970-01-01) [1] 1970-01-01 03:54:10 EST # see ?strptime for format specs format(as.POSIXct(14050.5, origin=1970-01-01), format=%H:%M) # which returns character vector [1] 03:54 str(format(as.POSIXct(14050.5 +(1:10)*3600, origin=1970-01-01), format=%H:%M ) ) chr [1:10] 04:54 05:54 06:54 07:54 ... Thanks. Dennis Dennis Fisher MD P (The P Less Than Company) Phone: 1-866-PLessThan (1-866-753-7784) Fax: 1-866-PLessThan (1-866-753-7784) www.PLessThan.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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Greek letter included in a character vector
On Sep 10, 2010, at 7:58 PM, Judith Flores wrote: Hello, In the past I have used expression to include greek letters in axis labels, but this time I need to include the greek letter as part of a legend. Basically, I need to create the following vector to rename the levels of a factor: c(Interferon-gamma, IL-10, IL-5), where gamma obviously needs to be printed as the greek letter gamma. I have tried expression(IFN-*gramma), but it only works when it is isolated, not as part of a vector. An example would have been nice. You should note that expressions are vectors of mode expression and that is what text and legend functions expect. The other thing I have learned is that quoting items unnecessarily inside expression() generally causes more harm than good. Learn to use plotmath syntax and separate items with ~ and *. A comma (outside a quoted string) separates individual expression items. x - seq(-pi, pi, len = 65) plot(x, sin(x), type = l, ylim = c(-1.2, 1.8), col = 3, lty = 2) points(x, cos(x), pch = 3, col = 4) lines(x, tan(x), type = b, lty = 1, pch = 4, col = 6) title(legend(..., lty = c(2, -1, 1), pch = c(-1,3,4), merge = TRUE), cex.main = 1.1) legend(-1, 1.9, expression(Interferon-gamma, IL-10, IL-5), col = c(3,4,6), text.col = green4, lty = c(2, -1, 1), pch = c(-1, 3, 4), merge = TRUE, bg = 'gray90') Also look at plotmath-paste: legend(-1, 1.9, expression(paste(Interferon,-,gamma), IL-10, IL-5), col = c(3,4,6), text.col = green4, lty = c(2, -1, 1), pch = c(-1, 3, 4), merge = TRUE, bg = 'gray90') -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] convert 1, 10, and 100 to 0001, 0010, 0100 etc.
On Sep 10, 2010, at 9:25 PM, Peng, C wrote: These are character values. Is there any way to get 001, 010, ..., as actual numeric values? 001 [1] 1 0100 [1] 100 001 == 1 [1] TRUE 0100 == 100 [1] TRUE -- View this message in context: http://r.789695.n4.nabble.com/convert-1-10-and-100-to-0001-0010-0100-etc-tp2535023p2535296.html Sent from the R help mailing list archive at Nabble.com. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there a bisection method in R?
On Sep 10, 2010, at 8:35 PM, huang min wrote: uniroot ... is not a bisection method. On Sat, Sep 11, 2010 at 6:10 AM, Gregory Gentlemen gregory_gentle...@yahoo.ca wrote: Dear fellow R-users, Is there a function that does the bisection method? I was unable to find one. Thanks in advance. Gregory -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] convert 1, 10, and 100 to 0001, 0010, 0100 etc.
On Sep 10, 2010, at 10:05 PM, Peng, C wrote: I mean to display 001,010, ..., as there are. In other words, whether there is a function, say func(), such that func(001,010) displays 001, 010. Not hard to construct one, but does not behave properly in the sub- unity decimal range. Not surprising. It did better in the range of values from the integer class than I thought it would. func4 - function(x) cat(sprintf(%05.0f, x)) func4(10) 00010 func5 - function(x) cat(sprintf(%05.0f, x)) func5(100) 00100 func5(10) 10 func5(1) 1 func5(0.001) 0 func5(-3) -0003 func5(-300) -300 -- David. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] acceptable filetypes was Re: coxph and ordinal variables?
On Sep 10, 2010, at 10:32 PM, Paul Johnson wrote: Hi, everybody On Wed, Sep 8, 2010 at 5:43 PM, Min-Han Tan minhan.scie...@gmail.com wrote: David said my R code text attachment got rejected by the mailing list. Pooh. I don't think that's nice. I don't see anything in the posting guide about a limit on text attachments. Paul: It's not in the Posting Guide but it is in the Mailing list information page: Furthermore, most binary e-mail attachments are not accepted, i.e., they are removed from the posting completely. As an exception, we allow application/pdf, application/postscript, and image/png (and x- tar and gzip on R-devel). You can use text/plain as well, or simply paste text into your message instead. And as I said the filter is very dumb. filename.R files are intercepted but the very same file with the name filename.txt will be passed through intact. Now I would agree if you said those directions were ambiguous and even possibly misleading, but it did lead me to successful interactions with the mailserver in the past, so I decided it wasn't worth a complaint. As I understand it the amount of control that the list maintainers exert over the server behavior is pretty limited. -- David. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Argument lib is missing
On Sep 11, 2010, at 6:29 AM, khush wrote: Dear all, I have installed R using yum install R-2.9. I am able to use R for general functions but when I installed some library say gplots I am getting the below error. install.packages(gplots) Warning in install.packages(gplots) : argument 'lib' is missing: using '/home/fedora/R/i386-redhat-linux-gnu-library/2.11' That warning should not cause much problem. Warning: unable to access index for repository http://cran.csdb.cn/src/contrib From time to time it happens that repositories are unavailable. I just checked and at the moment that repository is available. You could try again. (Or you could have tried another repository at that time. Trying another repository might make sense even now, since the version of gplots at that repository is fairly outdated. The current version is 2.8.0 and the one at csdb.cn is 2.7.1 from May 2009. There have been several revisions since that date.) Warning message: In getDependencies(pkgs, dependencies, available, lib) : package gplots is not available any suggestion will be appreciable. Thank you Khush David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Latex fonts in R graphics
On Sep 11, 2010, at 7:18 AM, KARAVASILIS GEORGE wrote: Hello, R users. I am trying to embed Computer modern fonts to an R plot and I get the following error. CM - Type1Font(CM, + c(paste(cm-lgc/fonts/afm/public/cm-lgc/, + c(fcmr8a.afm, fcmb8a.afm, fcmri8a.afm, fcmbi8a.afm), sep=), + ./cmsyase.afm)) pdf(cm.pdf, width=3, height=3, family=CM) plot(1:length(y), y, xlab=ss, ylab=expression(x[2])) ## for any vector y dev.off() null device 1 embedFonts(cm.pdf, outfile=cmembed.pdf, fontpaths=c(cm-lgc/ fonts/type1/public/cm-lgc, .)) Error in embedFonts(cm.pdf, outfile = cmembed.pdf, fontpaths = c(cm-lgc/fonts/type1/public/cm-lgc, : status -1 in running command 'gswin32c.exe -dNOPAUSE -dBATCH -q - dAutoRotatePages=/None -sDEVICE=pdfwrite -sOutputFile=C: \DOCUME~1\user\LOCALS~1\Temp\RtmpccEtgV\Rembed28163716 -sFONTPATH=cm- lgc/fonts/type1/public/cm-lgc;. cm.pdf' In addition: Warning message: In system(cmd) : gswin32c.exe not found Operating system: Windows XP, SP2. cm-lgc has been unzipped in the working directory C:/Program Files/ R/R-2.10.1 and the AFM, PFB files are also in same directory. The error message is telling us that you do not have Ghostscript installed properly. While you are at it, you should probably updated R as well. -- David Any help? -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] for loop
On Sep 11, 2010, at 8:39 AM, Peng, C wrote:
or:
k=0
for (i in 1:k) if(k0) print(i)
Because of the way the : operator works, I would have tested k =1
k=0.5
for (i in 1:k) if (k0){print(i)}
[1] 1
But Gabor's suggestion to use seq_len(k) is cleaner, anyway.
--
David Winsemius, MD
West Hartford, CT
__
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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.
Re: [R] Argument lib is missing
On Sep 11, 2010, at 8:02 AM, David Winsemius wrote: On Sep 11, 2010, at 6:29 AM, khush wrote: Dear all, I have installed R using yum install R-2.9. I am able to use R for general functions but when I installed some library say gplots I am getting the below error. install.packages(gplots) Warning in install.packages(gplots) : argument 'lib' is missing: using '/home/fedora/R/i386-redhat-linux-gnu-library/2.11' That warning should not cause much problem. Or maybe it would. I just noticed that you said you were using an ancient version of R and that library location is for the current version. You probably need to get advice from a fellow Redhat user if you want to run an outdated version. Installing packages for the current version is not going to be very successful. There is a repository of packages for older systems and you can find the link at the bottom of the package list on CRAN. http://cran.r-project.org/web/packages/ Look for the Archive link. Warning: unable to access index for repository http://cran.csdb.cn/src/contrib From time to time it happens that repositories are unavailable. I just checked and at the moment that repository is available. You could try again. (Or you could have tried another repository at that time. Trying another repository might make sense even now, since the version of gplots at that repository is fairly outdated. The current version is 2.8.0 and the one at csdb.cn is 2.7.1 from May 2009. There have been several revisions since that date.) Warning message: In getDependencies(pkgs, dependencies, available, lib) : package ‘gplots’ is not available any suggestion will be appreciable. Thank you Khush David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bootstrapping
On Sep 11, 2010, at 9:50 AM, Gregory Ryslik wrote: Hi Everyone, I am implementing a special case of Random forests. At one point, I have a list of which I then sample for replacement. So if the list is 100 elements, I get 100 elements some of them duplicates. How can I easily get the elements that were not included in the list? I realize i can do this with a for loop by going through each element and checking if it's in the list but I am wondering if there is a faster way. ?setdiff ?%in% -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] approxfun-problems (yleft and yright ignored)
On Sep 11, 2010, at 10:04 AM, Martin Maechler wrote: SW == Samuel Wuest wue...@tcd.ie on Thu, 26 Aug 2010 14:34:26 +0100 writes: SW Hi Greg, SW thanks for the suggestion: SW I have attached some small dataset that can be used to reproduce the SW odd behavior of the approxfun-function. SW If it gets stripped off my email, it can also be downloaded at: SW http://bioinf.gen.tcd.ie/approx.data.Rdata SW Strangely, the problem seems specific to the data structure in my SW expression set, when I use simulated data, everything worked fine. SW Here is some code that I run and resulted in the strange output that I SW have described in my initial post: ### load the data: a list called approx.data load(file=approx.data.Rdata) ### contains the slots x, y, input names(approx.data) SW [1] x y input ### with y ranging between 0 and 1 range(approx.data$y) SW [1] 0 1 ### compare ranges of x and input-x values (the latter is a small subset of 500 data points): range(approx.data$x) SW [1] 3.098444 7.268812 range(approx.data$input) SW [1] 3.329408 13.026700 ### generate the interpolation function (warning message benign) interp - approxfun(approx.data$x, approx.data$y, yleft=1, yright=0, rule=2) SW Warning message: SW In approxfun(approx.data$x, approx.data$y, yleft = 1, yright = 0, : SW collapsing to unique 'x' values ### apply to input-values y.out - sapply(approx.data$input, interp) ### still I find output values 1, even though yleft=1: range(y.out) SW [1] 0.00 7.207233 #-- on another 64 bit Mac - load(file=approx.data.Rdata) names(approx.data) [1] x y input range(approx.data$y) [1] 0 1 range(approx.data$x) [1] 3.098444 7.268812 interp - approxfun(approx.data$x, approx.data$y, yleft=1, yright=0, rule=2) Warning message: In approxfun(approx.data$x, approx.data$y, yleft = 1, yright = 0, : collapsing to unique 'x' values y.out - sapply(approx.data$input, interp) range(y.out) [1] -5.143958e+284 9.816907e-01 interp function (v) .C(R_approxfun, as.double(x), as.double(y), as.integer(n), xout = as.double(v), as.integer(length(v)), as.integer(method), as.double(yleft), as.double(yright), as.double(f), NAOK = TRUE, PACKAGE = stats)$xout environment: 0x1598c91b8 # attempt to offer yright and yledft in the sapply call failed y.out - sapply(approx.data$input, interp, yright=0, yleft=1) Error in FUN(c(6.99984458535897, 4.85139079147721, 7.58922165833, 10.8135863246057, : unused argument(s) (yright = 0, yleft = 1) #Create yright and yleft in the calling environment: yright=0; yleft=1 y.out - sapply(approx.data$input, interp) range(y.out) [1] 0.000 0.9816907 Now it works as expected. It seems to me that the yleft and yright values may not be properly incorporated into the constructed function. sessionInfo() R version 2.11.1 Patched (2010-06-14 r52281) x86_64-apple-darwin9.8.0 locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] grid splines stats graphics grDevices [6] utils datasets methods base other attached packages: [1] gplots_2.8.0 caTools_1.10 bitops_1.0-4.1 [4] gdata_2.8.0 gtools_2.6.2 sos_1.2-9 [7] brew_0.1-1 qpcR_1.3-2 minqa_1.1.9 [10] Rcpp_0.8.5 nlme_3.1-96 cluster_1.12.3 [13] rgl_0.91 minpack.lm_1.1-4 MASS_7.3-6 [16] chron_2.3-35 maps_2.1-4 gmodels_2.15.0 [19] rbenchmark_0.3 data.table_1.5 rms_3.0-0 [22] Hmisc_3.8-1 survival_2.35-8 plyr_1.1 [25] lattice_0.18-8 loaded via a namespace (and not attached): [1] tools_2.11.1 -- David Winsemius I get completely different (and correct) results, by the way the *same* you have in the bug report you've submitted (https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=14377) and which does *not* show any bug: range(y.out) [1] 0.000 0.9816907 Of course, I do believe that you've seen the above problems, -- on 64-bit Mac ? as you report in sessionInfo() ? -- but I cannot reproduce them. And also, you seem yourself to be able to get different results for the same data... what are the circumstances? Regards, Martin Maechler, ETH Zurich David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] xlab with text and expression
On Sep 11, 2010, at 11:07 AM, threshold wrote:
Dear R users, say
plot(rnorm(1)~rnorm(1), xlab=paste('abc', expression(x=1)),
plot(rnorm(1)~rnorm(1), xlab=expression(abc~x = 1))
# or if you are uncertain whether abc is a plotmath-special then this
also works
plot(rnorm(1)~rnorm(1), xlab=expression(abc~x = 1))
I want proper sign of weak inequality not just '='
will appreciate!
robert
--
David Winsemius, MD
West Hartford, CT
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Re: [R] confidence bands for a quasipoisson glm
On Sep 11, 2010, at 3:15 PM, Maik Rehnus wrote:
Dear all,
I have a quasipoisson glm for which I need confidence bands in a
graphic:
gm6 - glm(num_leaves ~ b_dist_min_new, family = quasipoisson, data
= beva)
summary(gm6)
library('VIM')
b_dist_min_new - as.numeric(prepare(beva$dist_min,
scaling=classical, transformation=logarithm)).
My first steps for the solution are following:
range(b_dist_min_new)
x - seq(-1.496, 1.839, by=0.01)
newdat - data.frame(b_dist_min_new=x)
y - predict(gm6, newdata=newdat, type=response)
plot(x,y, type=l, ylim=c(0,15), lty=2, xlab=Distance [scaled
log.], ylab=Number of used plant, las=1)
ilogit-function(x) exp(x)/(1 + exp(x))
logit -function(x) log(x/(1 - x))
newdat$logitpred - predict(gm6, newdata=newdat, type=link)
I'm puzzled. You specified that model as quasipoisson and are now
treating it as if it were logistic? The link is going to be log(),
nicht wahr?
newdat$sepred - predict(gm6, newdata=newdat, type=link,
se.fit=TRUE)$se.fit
newdat$logitlower - newdat$logitpred-1.96 * newdat$sepred
newdat$logitupper - newdat$logitpred+1.96 * newdat$sepred
I'm not familiar with ilogit (sounds very useful assuming it to be an
inverse logit), but if one were taking a first stab at an inverse
function for quasipoisson wouldn't that be exp()?
newdat$upper - ilogit(newdat$logitupper)
newdat$lower - ilogit(newdat$logitlower)
lines(x, newdat$lower, lty=3)
lines(x, newdat$upper, lty=3).
In this way I could find a positive correlation. But my created
confidence bands on the graph don't touch my regression line. Could
it be a technical problem or is it a mistake in the calculation?
I am new here and I hope you can help to solve my problem. I could
not find any answers for quasipoisson glm on internet.
Best regards
Maik
David Winsemius, MD
West Hartford, CT
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Re: [R] scalable delimiters in plotmath
On Sep 11, 2010, at 9:00 PM, Peter Ehlers wrote: On 2010-09-11 16:14, Dennis Murphy wrote: Hi Baptiste, You need to use the symbol(\nnn) concept, where nnn denotes the octal symbol number. For it's 074 and for it's 076. This little test seemed to work: plot(1, 1, main = expression(symbol(\074)~'x, y'~symbol(\076))) HTH, Dennis It's a matter of taste, but I would use \341 and \361. However, these are still not scalable, AFAICS. Not exactly scalable angles, but you can fake it: plot(1, 1, main = expression(symbol(\341)~scriptstyle( atop(x,y) )~symbol(\361)), cex.main=3) scriptstyle shrinks the inner atop() material, and since I tested on a Mac it should work for Baptiste. -- David. -Peter Ehlers On Sat, Sep 11, 2010 at 10:01 AM, baptiste auguie baptiste.aug...@googlemail.com wrote: What do people use to show angle bracketsin R graphics? Have I missed something obvious? Thanks, baptiste On 9 September 2010 17:57, baptiste auguie baptiste.aug...@googlemail.com wrote: Dear list, I read in ?plotmath that I can use bgroup to draw scalable delimiters such as [ ] and ( ). The same technique fails withhowever, and I cannot find a workaround, grid.text(expression(bgroup(,atop(x,y),))) Error in bgroup(, atop(x, y),) : invalid group delimiter Regards, baptiste sessionInfo() R version 2.11.1 (2010-05-31) x86_64-apple-darwin9.8.0 David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] create a '3D line plot'
On Sep 12, 2010, at 10:12 AM, Karl Brand wrote:
Esteemed useRs and developeRs,
I need to create a '3D line plot' (proper name?) of which an
excellent example can be viewed here:
http://cococubed.asu.edu/images/87a/images/unknown_pleasures.jpg
Set up blank plot region with proper ranges for x and ylim that could
be say 0:60 with the intent of scaling your 50 individual y ranges to
0-10
Scale your y values to be within 0-10
Fill plot area black:
Draw from top down using polygon with white lines at cex=2 and black
fill.
Then something along these lines:
for(ilevel in 0:50){ polygon(x,yscaled+50-ilevel, col=black,
border=white, lwd=2)}
#--- tested code
opar - par(bg=black)
set.seed(1)
Ldens - vector(mode=list, 50)
for(i in 1:50) {
Ldens[[i]] - density(rnorm(100), from=-3,to=3)
with(Ldens[[i]], polygon(x=c(x[c(1, 1:512, 512)]), y= c(0, 50-i
+20*y[1:512], 0),
col=black, border=white))
}
par(opar)
#-
Got any data?
--
David.
I have some experience using the rgl package to create 3D PCA plots,
but have no idea where to start for an image like this.
I'd really appreciate suggestions help on how might achieve this
using R,
Karl
--
Karl Brand
--
David Winsemius, MD
West Hartford, CT
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Re: [R] reshape matrix entities to columns
On Sep 12, 2010, at 10:45 AM, Natasha Asar wrote: Greeting R helpers J I am not familiar with R but I have to use it to analyze data set that I have (30,000 20,000) I want to change the structure of the dataset and I am wondering how that might be possible in R A main data looks like this: some entities are empty AgeNo. AgeNo. AgeNo. Center1 5 2 8 7 Center2 10 7 20 9 4 10 But what I want the data to look like is Age1 2 3 4 5 6 7 8 9 10 … 20 Center1 2 7 Center2 10 7 9 It should read the entities one by one when j is in age column take its value and consider it as the column number for new matrix then go to next entity (j No. columns) and put that entity under the columns number identified in previous step. In other word it should get the each element in No. columns (one by one) and place them in a new matrix under the column number which are equal to entity of age columns of first matrix i have tired ncol, and cbind and things like that but I guess im on the wrong path because it is not working. I am reading this fine with read.csv and writing back the same way. do you know how I can make this work?? Is it even possible to do something like this? Thank you in advance Natasha [[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 West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to print matrices in standard format was ... Re: How to define new matrix based on an elementary row oper
On Sep 12, 2010, at 11:27 AM, Cuckovic Paik wrote:
I appreciate all you help. This is only for instructional purpose:
A = matrix(c(0,1,1,-2,-3,1,2,-1,0,2,2,4,1,-3,-2,1,-4,-7,-1,-19),
ncol=5,
byrow=T)
B
=matrix(sample(c(0,1,1,-2,-3,1,2,-1,0,2,2,4,1,-3,-2,1,-4,-7,-1,-19),),
ncol=5, byrow=T)
Which print func( A, B, A+B) can print the resulting matrices A and
B and
A+B in the following format?
[,1] [,2] [,3] [,4] [,5] [,1] [,2] [,3] [,4] [,5]
[,1] [,2] [,3] [,4] [,5]
[1,]011 -2 -3 [1,] 2 -102
1[1,]
2 0 1 0-2
[2,]12 -102 + [2,] 1 -42 -2-2
= [2,]
2-2 1-2 0
[3,]241 -3 -2 [3,] -31 -71
-1 [3,]
-1 5-6-2-3
[4,]1 -4 -7 -1 -19 [4,] -304 -19 1
[4,] -2
-4-3 -20 -18
for( i in 1:nrow(A) ) { cat(sprintf(%4.0f, A[i, ]), paste(
,if( i==3 ){+}else{ }, , sep=),sprintf(%4.0f,B[i, ]),
paste( ,if( i==3 ){=}else{ }, , sep=), sprintf(%4.0f, (A
+B)[i, ]), \n )}
--
David Winsemius, MD
West Hartford, CT
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Re: [R] scalable delimiters in plotmath
On Sep 12, 2010, at 6:15 AM, baptiste auguie wrote:
Thanks everyone. I've also had a look at plotmath.c where bgroup is
defined for [, {, (, . but not . It seems quite trivial to
add it, at first sight, however there is a part that I don't
understand in the RenderDelim routine,
static BBOX RenderDelim(int which, double dist, int draw,
mathContext *mc,
pGEcontext gc, pGEDevDesc dd)
{
// [... snipped ...]
case '(':
top = 230; ext = 231; bot = 232; mid = 0;
break;
case ')':
top = 246; ext = 247; bot = 248; mid = 0;
break;
These integer codes make no sense to me, I have no clue which ones I
should use for and .
Does this help? (I think they are using Symbol PS fonts with decimal
indexing.)
as.octmode(c(230, 231, 232, 246, 247, 248) )
[1] 346 347 350 366 367 370
plot(1,1, xlab= expression(
symbol(\346)~# upper 1/3 of left paren
symbol(\347)~# to left of center bar
symbol(\350)~# lower 1/3 of left paren
symbol(\366)~# upper 1/3 of right paren
symbol(\367)~# to right of center bar
symbol(\370) ) ) # lower 1/3 of right paren
(caveat: Maybe not standard glyph-names.)
I added octal annotation to the TestChars(font=5) call that the points
help page offers:
TestChars(font=5)
for(j in 1:14) {
for(i in 0:16){
text(i+0.2, j+.6, labels=as.octmode(i+(j+1)*16), cex=.5)}}
I do not see a trio or pair of glyphs that would form an angle bracket.
--
David.
As far as I understand these codes might
correspond to extended ascii characters whose boundaries and positions
we want to borrow. Then again, maybe it's something else entirely.
Any hints?
Best wishes,
baptiste
On 12 September 2010 03:27, David Winsemius dwinsem...@comcast.net
wrote:
On Sep 11, 2010, at 9:00 PM, Peter Ehlers wrote:
On 2010-09-11 16:14, Dennis Murphy wrote:
Hi Baptiste,
You need to use the symbol(\nnn) concept, where nnn denotes the
octal
symbol number. For it's 074 and for it's 076. This little
test seemed
to
work:
plot(1, 1, main = expression(symbol(\074)~'x, y'~symbol(\076)))
HTH,
Dennis
It's a matter of taste, but I would use \341 and \361.
However, these are still not scalable, AFAICS.
Not exactly scalable angles, but you can fake it:
plot(1, 1, main = expression(symbol(\341)~scriptstyle( atop(x,y)
)~symbol(\361)), cex.main=3)
scriptstyle shrinks the inner atop() material, and since I tested
on a Mac
it should work for Baptiste.
--
David.
-Peter Ehlers
On Sat, Sep 11, 2010 at 10:01 AM, baptiste auguie
baptiste.aug...@googlemail.com wrote:
What do people use to show angle bracketsin R graphics?
Have I
missed something obvious?
Thanks,
baptiste
On 9 September 2010 17:57, baptiste auguie
baptiste.aug...@googlemail.com wrote:
Dear list,
I read in ?plotmath that I can use bgroup to draw scalable
delimiters
such as [ ] and ( ). The same technique fails with
however, and I
cannot find a workaround,
grid.text(expression(bgroup(,atop(x,y),)))
Error in bgroup(, atop(x, y),) : invalid group delimiter
Regards,
baptiste
sessionInfo()
R version 2.11.1 (2010-05-31)
x86_64-apple-darwin9.8.0
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] scalable delimiters in plotmath
, grid.text(expression(bgroup(,atop(x,y),))) Error in bgroup(, atop(x, y),) : invalid group delimiter Regards, baptiste sessionInfo() R version 2.11.1 (2010-05-31) x86_64-apple-darwin9.8.0 David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 p...@stat.auckland.ac.nz http://www.stat.auckland.ac.nz/~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.
Re: [R] reshape matrix entities to columns
On Sep 12, 2010, at 3:34 PM, Dennis Murphy wrote:
Hi:
Natasha said:
I changed it so i hope it will look better now
the matrix is like this:
AgeNo. Age No. AgeNo.
Center1 52 8 7
Center210 720 9 4 10
column name = sequence of age-no.
But what I want the data to look like is this
Age
1 2 3 4 5 6 7 8 9 10
20
Center1
27
Center2
10 7 9
column name= age of ppl
entries = number of ppl with that age in on center
*
It's a continuation of the reshape problem, but we have to
change the NAs in the reshaped data frame to zeros first:
df2[is.na(df2)] - 0
xtabs(n ~ center + age, data = df2)
age
center 5 6 7 8 9 10 11 12 13 14
1 0 10 0 13 0 9 0 7 0 10
2 0 0 12 14 0 0 16 0 0 13
3 6 0 0 0 10 0 12 0 9 0
How's that?
You've done all the hard work, but the OP wanted the full range of age
values from 1:max and that pretty easy to do with one further step
that adds entries fo the missing age levels:
df3 - rbind(df2, data.frame(center=1,time=1, age=1:max(df2$age),
n=0))
xtabs(n ~ center + age, data = df3)
age
center 1 2 3 4 5 6 7 8 9 10 11 12 13 14
1 0 0 0 0 0 10 0 13 0 9 0 7 0 10
2 0 0 0 0 0 0 12 14 0 0 16 0 0 13
3 0 0 0 0 6 0 0 0 10 0 12 0 9 0
--
David.
Dennis
On Sun, Sep 12, 2010 at 9:46 AM, Dennis Murphy djmu...@gmail.com
wrote:
Hi:
Here's a made up example using the reshape function:
Input data:
df - structure(list(center = 1:3, age1 = c(6L, 7L, 5L), n1 = c(10L,
12L, 6L), age2 = c(8L, 8L, 8L), n2 = c(13L, 14L, NA), age3 = c(10L,
10L, 9L), n3 = c(9L, NA, 10L), age4 = c(12L, 11L, 11L), n4 = c(7L,
16L, 12L), age5 = c(14L, 14L, 13L), n5 = c(10L, 13L, 9L)), .Names =
c(center,
age1, n1, age2, n2, age3, n3, age4, n4, age5,
n5), class = data.frame, row.names = c(NA, -3L))
df
center age1 n1 age2 n2 age3 n3 age4 n4 age5 n5
1 16 108 13 10 9 12 7 14 10
2 27 128 14 10 NA 11 16 14 13
3 35 68 NA9 10 11 12 13 9
# To reshape more than one variable at a time, you need
# to put the sets of variables into a list, as follows:
df2 - reshape(df, idvar = 'center', varying =
list(c(paste('age', 1:5, sep = '')), c(paste('n', 1:5, sep = ''))),
v.names = c('age', 'n'), times = 1:5, direction = 'long')
df2
center time age n
1.1 11 6 10
2.1 21 7 12
3.1 31 5 6
1.2 12 8 13
2.2 22 8 14
3.2 32 8 NA
1.3 13 10 9
2.3 23 10 NA
3.3 33 9 10
1.4 14 12 7
2.4 24 11 16
3.4 34 11 12
1.5 15 14 10
2.5 25 14 13
3.5 35 13 9
HTH,
Dennis
On Sun, Sep 12, 2010 at 7:45 AM, Natasha Asar natasha.asa...@yahoo.com
wrote:
Greeting R helpers J
I am not familiar with R but I have to use it to analyze data set
that I
have
(30,000 20,000)
I want to change the structure of the dataset and I am wondering
how that
might
be possible in R
A main data looks like this: some entities are empty
AgeNo. AgeNo. AgeNo.
Center15 2 8
7
Center210 7
20
9 4 10
But what I want the data to look like is
Age1 2 3
4 5 6 7 8
9 10
20
Center1
2 7
Center2
10
7 9
It should read the entities one by one
when j is in age column take its value and consider it as the column
number for
new matrix
then go to next entity (j No. columns) and put that entity under the
columns
number identified in previous step.
In other word
it should get the each element in No. columns (one by one) and
place them
in a
new matrix under the column number which are equal to entity of age
columns of
first matrix
i have tired ncol, and cbind and things like that but I guess im
on the
wrong
path because it is not working. I am reading this fine with
read.csv and
writing back the same way.
do you know how I can make this work?? Is it even possible to do
something
like
this?
Thank you in advance
Natasha
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Re: [R] How to print matrices in standard format was ... Re: How to define new matrix based on an elementary row oper
On Sep 12, 2010, at 12:24 PM, David Winsemius wrote:
On Sep 12, 2010, at 11:27 AM, Cuckovic Paik wrote:
I appreciate all you help. This is only for instructional purpose:
A = matrix(c(0,1,1,-2,-3,1,2,-1,0,2,2,4,1,-3,-2,1,-4,-7,-1,-19),
ncol=5,
byrow=T)
B
=
matrix(sample(c(0,1,1,-2,-3,1,2,-1,0,2,2,4,1,-3,-2,1,-4,-7,-1,-19),),
ncol=5, byrow=T)
Which print func( A, B, A+B) can print the resulting matrices A
and B and
A+B in the following format?
[,1] [,2] [,3] [,4] [,5] [,1] [,2] [,3] [,4] [,
5][,1] [,2] [,3] [,4] [,5]
[1,]011 -2 -3 [1,] 2 -102
1[1,] 2 0 1 0-2
[2,]12 -102 + [2,] 1 -42 -2-2
=[2,] 2-2 1-2 0
[3,]241 -3 -2 [3,] -31 -71
-1[3,] -1 5-6-2-3
[4,]1 -4 -7 -1 -19 [4,] -304 -19
1[4,] -2-4-3 -20 -18
for( i in 1:nrow(A) ) { cat(sprintf(%4.0f, A[i, ]), paste(
,if( i==3 ){+}else{ }, , sep=),sprintf(%4.0f,B[i, ]),
paste( ,if( i==3 ){=}else{ }, , sep=), sprintf(%4.0f,
(A+B)[i, ]), \n )}
for( i in 1:nrow(A) ) { cat(sprintf(%4.0f, A[i, ]),
paste( ,if( i==3 ){+}else{ }, ,
sep=),
sprintf(%4.0f,B[i, ]),
paste( ,if( i==3 ){=}else{ }, ,
sep=),
sprintf(%4.0f, (A+B)[i, ]), \n )}
Even with the prettier printing it stilled seemed like a hack, so here
is a grid graphics solution that gives prettier _output_:
require(grid)
grid.newpage()
pushViewport(plotViewport(c(5,4,2,2))) # implicit limits are
c(0,0,1,1) within plot area
for (i in 1:nrow(A)) { for (j in 1:ncol(A)){grid.text(A[i,j], x=i/
20, y=j/20)}} # plot mtx A
for (i in 1:nrow(B)) { for (j in 1:ncol(B)){grid.text(B[i,j], x=(i
+5)/20, y=j/20)}} # B
for (i in 1:nrow(B)) { for (j in 1:ncol(B)){grid.text(A[i,j]
+B[i,j], x=(i+10)/20, y=j/20)}} # A+B
grid.text(=, x=10/20, y=2.5/20)
grid.text(+, x=5/20, y=2.5/20)
--
David Winsemius, MD
West Hartford, CT
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Re: [R] How to do a trig regression
On Sep 12, 2010, at 10:23 PM, Aaditya Nanduri wrote: Hello All, I cant seem to do a trig regression in R. The equation is as follows : y = a+b*(sin((2*pi*x/360) - c))^2 a, b, c are coefs that I want. y, x are input vectors. The equation I put into R: lm(y ~ sin(2*pi*x/360)^2) This equation is missing the c and I dont get the right answer. Take a look at section 2.3 of: http://www.statoek.wiso.uni-goettingen.de/veranstaltungen/zeitreihen/sommer03/ts_r_intro.pdf Or consider using time series or nonlinear modeling. Also, I dont know how to plot the lm over the x values instead of the indices. Without an example of your data and what you tried it is difficult to intuit what problems you may be facing. -- David. Any help is sincerely appreciated. Thank you all very much. -- Aaditya Nanduri aaditya.nand...@gmail.com [[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] Overlay of two graphs of different axes
On Sep 13, 2010, at 1:54 AM, Uwe Dippel wrote: Though I have read quite a bit, and tried quite a bit, I have yet to find a nice way to overlay 2 or more curves in the same plot, with different ranges. The different axes could be a problem but if you construct both with predefined coordinates systems and scale appropriately it should work out. See the twoord.plot function in plotrix. Here is simplified sample code to demonstrate the question: plot(2*(seq(1,5)), type=l, axes=FALSE) ? lines lines(1.5*(seq(2,5)), type=b) curve(2*(seq(1,5)), type=b, add=TRUE) Error in curve(2 * (seq(1, 5)), type = b, add = TRUE) : 'expr' must be a function or an expression containing 'x' axis(2) curve(x^2, 1, 5, type=b, add=TRUE) axis(4) Firstly, as an aside, I am not clear why 'curve' has a different syntax compared to 'plot'. As a still beginner, I'd for one would be happy to add curves to a plot; curves of just different parameters. Though, I guess, there must be a good reason? curve is like abline in that it draws from left x range to right x range. Mostly, however, I wonder how to plot a number of curves into an original plot, that re-defines the min/max from the most recent curve. lines() or segments() In the example that I constructed, axis(2) does exactly the expected thing. What I want to do next, though, with the least effort, is to add another function in a manner that the added function is scaled, not according to the first function (plot), but to fit into the plotting area. (The example above overshoots the range). Plus, how can I subsequently add the axis suitable to the most recent function? That is, how can I render axis(4) to displaying the scale for the second graph, created with 'curve'? There are many worked examples in the archives, as well as canned solutions in widely used packages. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create a time-series from cross-sectional data that has each year as a separate column
On Sep 13, 2010, at 2:31 AM, Gabriel Bergin wrote: Hi, I have a dataset from ILO, originally in csv-format, that I have read into R. It is cross-sectional time-series data, so I have a bunch of variables and dummy variables that I need to extract data from for the entire time period. However, the years are separated by columns instead of rows, as is usually the case in R. This is what it looks like: str(laborstafinMFBA) 'data.frame': 152 obs. of 39 variables: $ COUNTRY: Factor w/ 164 levels Albania,Algeria,..: 2 4 7 8 9 10 11 11 12 13 ... $ CODE.COUNTRY : Factor w/ 163 levels AE,AG,AI,..: 44 7 8 5 12 11 10 10 13 23 ... $ SOURCE : Factor w/ 7 levels Administrative reports,..: 3 3 3 3 3 3 3 3 3 3 ... $ CODE.SOURCE: Factor w/ 7 levels A,B,BA,CA,..: 3 3 3 3 3 3 3 3 3 3 ... etc.. $ D1990 : num NA NA NA NA NA ... $ D1991 : num NA NA 101 NA NA ... $ D1992 : num NA 38.4 111.2 NA NA ... $ D1993 : num NA NA 94.4 NA NA ... $ D1994 : num NA NA 133.69 NA 1.42 ... $ D1995 : num NA NA 121 NA NA ... $ D1996 : num NA NA 176 NA NA ... $ D1997 : num NA NA 195.31 NA 1.51 ... $ D1998 : num NA NA 202 NA NA ... $ D1999 : num NA NA 201 NA NA ... $ D2000 : num NA NA 207 NA NA ... $ D2001 : num 68.1 NA 198.3 NA NA ... $ D2002 : num NA NA 186 NA NA ... $ D2003 : num 67.6 NA 148.8 NA NA ... $ D2004 : num 68.8 NA 143.7 NA NA ... $ D2005 : num NA NA 163 NA NA ... $ D2006 : num NA NA 189 NA NA ... $ D2007 : num NA NA NA 14 1.91 ... How do I transform this into something that I can make a time-series of? ?reshape package reshape(not the same as the function reshape) package sqldf There might also be the possibility of using t() on just the Dyear columns but it might not give desired results on the association of COUNTRY with SOURCE It looks as though there are a mixture of data types and that you might want to pull out all the types of SOURCE separately before transforming or go to a database-oriented solution. Sincerely, Gabriel Bergin gabr...@bergin.se [[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 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] Saveing plot to multiple locations
On Sep 13, 2010, at 5:32 AM, Joel wrote:
Hi
Im trying to save a plot both to a pdf and as just a picture but
without
success so if someone can help me I would be happy :)
Are you trying to embed a png file in a graphic or do you just want
two different files?
If the first (obviously the more difficult) then Paul Murrell has a
bunch of material he has offered over the years:
cran.r-project.org/web/packages/grImport/vignettes/import.pdf
Vector graphics is the focus but citations to packages handling raster
or bitmap graphics are included.
If it is the second, then just do one task, complete it with
dev.off(), which is needed to close the file, and move on to the other.
--
David
my code:
require(party)
irisct - ctree(Species ~ .,data = iris)
data(iris)
attach(iris)
pdf('/home/joel/Skrivbord/mammamu.pdf')
try(png('/home/joel/Skrivbord/mammamu1.png'))
plot(Sepal.Length, Petal.Length, col=unclass(Species))
legend(4.5, 7, levels(Species), col=plot_colors, cex=0.8, fill=1:3)
try(png('/home/joel/Skrivbord/mammamu2.png'))
plot(irisct)
dev.off()
readBin(pic,'raw',1024*1024)
readBin('/home/joel/Skrivbord/mammamu2.png','raw',1024*1024)
Ive tryed to move around the png and pdf part to see if it worked
better if
they where in an other order but as I said without success.
//Joel
--
View this message in context:
http://r.789695.n4.nabble.com/Saveing-plot-to-multiple-locations-tp2537133p2537133.html
Sent from the R help mailing list archive at Nabble.com.
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and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
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Re: [R] lattice: Set x-axis in italics only
On Sep 13, 2010, at 8:50 AM, Alejo C.S. wrote: Dear list, I making some box-and-whisker plots in R with vertebrate data. The x axis are species names that must be in italics. I tried with the axis function but no luck, and it seems that affects both axes. Any tip? In bwplot just add: ..., scales=list(x=list(font=3)), In this case (and I suppose many others) lattice is much more flexible than base. I initially tried what appeared to be valid strategies with boxplot and ended up tied in knots. Then I noticed your subject line and it was trivial. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Axis break with gap.plot()
On Sep 13, 2010, at 10:34 AM, Filoche wrote: Hi again everyone. Anyone know if there's any limitation with gap.plot concerning the fill color of plotted markers? I would like to fill the circles with a color : library(plotrix); gap.plot(c(1,2,3,4,10), c(1,2,3,4,10), c(5,9), pch = 21, col = red); However, it only change the color of the line. I tried bgcol and some argument but without success. Change pch to 19. (=solid circle) With regards, Phil -- View this message in context: http://r.789695.n4.nabble.com/Axis-break-with-gap-plot-tp2533027p2537540.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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
