[R] how to improve summary.lm
Hello, I need a shorter summary.lm, instead of Call: lm(formula = E$t ~ E$cfs) Residuals: Min1QMedian3Q Max -0.239674 -0.007694 0.006430 0.014330 2.496551 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) -1.994e-02 1.419e-04 -140.5 2e-16 *** E$cfs1.675e-05 4.714e-09 3552.7 2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.03238 on 65268 degrees of freedom Multiple R-squared: 0.9949, Adjusted R-squared: 0.9949 F-statistic: 1.262e+07 on 1 and 65268 DF, p-value: 2.2e-16 I need lm(formula = E$t ~ E$cfs) Estimate Std. Error t value Pr(|t|) (Intercept) -1.994e-02 1.419e-04 -140.5 2e-16 *** E$cfs1.675e-05 4.714e-09 3552.7 2e-16 *** Multiple R-squared: 0.9949, Adjusted R-squared: 0.9949 I looked at the code of summary.lm, with the intention to copy parts of it into a new function short_sum_lm, but couldn't find the parts I'm interested in. I hope it's not too complicated to achieve that. (In general it would be great to have more influence on the summary-function -- typically it's too spacious.) Thanks for your attention Oliver __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] AFT model time-dependent with weibull distribution
On Sat, Aug 20, 2011 at 7:33 PM, JPF xpfen...@gmail.com wrote: Göran Broström wrote: Good. Do you still need answers to your other questions? Yes. Could answer the following two questions: 1- Can I use phreg function to estimate a model with time-dependent covariates? In case of a positive answer, how? Yes. You do it in the way you did it with your first example. No 'id' argument is needed (or allowed) in the call to phreg. The reason is that in PH regression, the hazard function at any time depends only on the covariate value at that time point. In the AFT model, on the other hand, the hazard function at time t depends on all covariate values in the interval (0, t). Therefore aftreg needs the 'id' argument, but phreg does not. 2- I could not find any example that clearly explains how to interpret aftreg output. Specially, refering to the diference between survreg and aftreg output (intercept and sign of the estimates). I think you already answered that question. You can read more about it in the vignette about parametric models in eha. G. I include below an example of output of a regression with coxph, survreg, phreg and aftreg and a time-independent variable. I would appreciate if you could explain it or provide an external example that explains how it works. n=26 events=25 time at risk=45 a/ coxph(Surv(time,s) ~ Z1, data=data.frame(data)) coef exp(coef) se(coef) z p Z1 0.0249 1.03 0.00907 2.75 0.006 b/ phreg(Surv(time,s) ~ Z1, data=data.frame(data), dist=weibull) Covariate W.mean Coef Exp(Coef) se(Coef) Wald p Z1 43.689 0.033 1.033 0.009 0.000 log(scale) 0.641 1.899 0.065 0.000 log(shape) 1.172 3.230 0.158 0.000 Max. log. likelihood -22.135 LR test statistic 13.1 Degrees of freedom 1 Overall p-value 0.000302689 c/ aftreg(Surv(time,s) ~ Z1, data=data.frame(data), dist=weibull) Covariate W.mean Coef Exp(Coef) se(Coef) Wald p mas 43.689 0.010 1.010 0.002 0.000 log(scale) 1.147 3.149 0.141 0.000 log(shape) 1.172 3.230 0.158 0.000 Max. log. likelihood -22.135 LR test statistic 13.1 Degrees of freedom 1 Overall p-value 0.000302692 d/ survreg(Surv(time,s) ~ Z1, data=data.frame(data), dist=weibull) Value Std. Error z p (Intercept) 1.1476 0.13498 8.50 1.87e-17 mas -0.0101 0.00232 -4.34 1.45e-05 Log(scale) -1.1724 0.15787 -7.43 1.11e-13 Scale= 0.310 Weibull distribution Loglik(model)= -22.1 Loglik(intercept only)= -28.7 Chisq= 13.05 on 1 degrees of freedom, p= 3e-04 Number of Newton-Raphson Iterations: 5 Thank you very much, J -- View this message in context: http://r.789695.n4.nabble.com/AFT-model-time-dependent-with-weibull-distribution-tp3755079p3757387.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. -- Göran Broström __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 improve summary.lm
Hi Oliver,
For the the output you would like to have, you may take a look at
names(summary(yourmodel))
str(summary(yourmodel))
The above will help you to extract the components you need from the lm
object.
Below is my attempt to do what you want. However, you will need to work
a little more in order to get the *** besides the p-values ;-) Also,
note that it is assumed that the lm function is used to build up the
linear model.
# function
myout - function(lmfit, k = 22){
res - summary(fit)
cat(rep(--, k), \n)
print(res$call)
print(res$coef)
cat('Multiple R-squared:', res$r.squared, ' Adjusted
R-squared:', res$adj.r.squared, \n)
cat(rep(--, k), \n)
}
# some data
set.seed(123)
x1 - rnorm(100)
x2 - rnorm(100)
y - .5 + 1.2*x1 - .2*x2 + rnorm(100)
# linear model
fit - lm(y ~ x1 + x2)
myout(fit)
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
lm(formula = y ~ x1 + x2)
Estimate Std. Error t value Pr(|t|)
(Intercept) 0.6350654 0.09614007 6.605627 2.129617e-09
x1 1.0668285 0.10486949 10.172915 5.677194e-17
x2 -0.1761887 0.09899469 -1.779779 7.824368e-02
Multiple R-squared: 0.5284765Adjusted R-squared: 0.5187544
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
HTH,
Jorge
On 8/21/11 3:16 AM, Oliver Kullmann wrote:
Hello,
I need a shorter summary.lm, instead of
Call:
lm(formula = E$t ~ E$cfs)
Residuals:
Min1QMedian3Q Max
-0.239674 -0.007694 0.006430 0.014330 2.496551
Coefficients:
Estimate Std. Error t value Pr(|t|)
(Intercept) -1.994e-02 1.419e-04 -140.52e-16 ***
E$cfs1.675e-05 4.714e-09 3552.72e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.03238 on 65268 degrees of freedom
Multiple R-squared: 0.9949, Adjusted R-squared: 0.9949
F-statistic: 1.262e+07 on 1 and 65268 DF, p-value: 2.2e-16
I need
lm(formula = E$t ~ E$cfs)
Estimate Std. Error t value Pr(|t|)
(Intercept) -1.994e-02 1.419e-04 -140.52e-16 ***
E$cfs1.675e-05 4.714e-09 3552.72e-16 ***
Multiple R-squared: 0.9949, Adjusted R-squared: 0.9949
I looked at the code of summary.lm, with the intention
to copy parts of it into a new function short_sum_lm,
but couldn't find the parts I'm interested in.
I hope it's not too complicated to achieve that.
(In general it would be great to have more influence
on the summary-function -- typically it's too spacious.)
Thanks for your attention
Oliver
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Re: [R] how to improve summary.lm
On Aug 21, 2011, at 3:16 AM, Oliver Kullmann wrote: Hello, I need a shorter summary.lm, instead of Call: lm(formula = E$t ~ E$cfs) Residuals: Min1QMedian3Q Max -0.239674 -0.007694 0.006430 0.014330 2.496551 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) -1.994e-02 1.419e-04 -140.5 2e-16 *** E$cfs1.675e-05 4.714e-09 3552.7 2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.03238 on 65268 degrees of freedom Multiple R-squared: 0.9949, Adjusted R-squared: 0.9949 F-statistic: 1.262e+07 on 1 and 65268 DF, p-value: 2.2e-16 I need lm(formula = E$t ~ E$cfs) Estimate Std. Error t value Pr(|t|) (Intercept) -1.994e-02 1.419e-04 -140.5 2e-16 *** E$cfs1.675e-05 4.714e-09 3552.7 2e-16 *** Multiple R-squared: 0.9949, Adjusted R-squared: 0.9949 I looked at the code of summary.lm, with the intention to copy parts of it into a new function short_sum_lm, but couldn't find the parts I'm interested in. I hope it's not too complicated to achieve that. (In general it would be great to have more influence on the summary-function -- typically it's too spacious.) When you simply type summary.lm there is an implicit call to print.summary.lm but its code is not visible unless you use getAnywhere(print.summary.lm ). Reading the code you find that the coefficient matrix and the significance stars are handled by a function , so this should give you what you want: printCoefmat(summary(model)$coefficients) -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to improve summary.lm
When you simply type summary.lm there is an implicit call to
print.summary.lm but its code is not visible unless you use
getAnywhere(print.summary.lm ).
Aha, that's great -- now I see all the \n (and can get rid off
them ;-)).
Reading the code you find that the
coefficient matrix and the significance stars are handled by a
function , so this should give you what you want:
printCoefmat(summary(model)$coefficients)
For the record, now I'm using
short_summary_lm = function(L) {
S = summary(L)
printCoefmat(S$coefficients, signif.legend=FALSE)
digits = max(3,getOption(digits)-3)
cat(R-squared:, formatC(S$r.squared, digits = digits), \n)
}
Thanks!
Oliver
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[R] How to navigate (zoom, pan) in a plot/graph
Hello all, I need to zoom in and out and travel(pan) inside a plot, like you can do on a Matlab plot. If possible, I would also like the option to use the mouse to set a marker on the graph and get the (x,y) data for it, again, like in Matlab. Is this possible in R with the regular packages, or do you maybe know a different package that will allow this? Eran. * * [[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] Build a package - check error
On 8/20/2011 3:37 PM, Uwe Ligges wrote: ?.First.lib has an example: ## Suppose a package needs to call a DLL named 'fooEXT', ## where 'EXT' is the system-specific extension. Then you should use .First.lib - function(lib, pkg) library.dynam(foo, pkg, lib) Shouldn't the first line here be ## Suppose a package needs to call a DLL named 'foo.EXT', ^ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 a matrix
At 18:21 20/08/2011, Uwe Ligges wrote: On 20.08.2011 19:04, David Winsemius wrote: On Aug 20, 2011, at 12:32 PM, Uwe Ligges wrote: [snip original problem] David, I think there are some good examples on the help page. What is missing? What is not clearly explained? If a longer tutorial is needed, that may be an article for the R Help Desk in The R Journal. Anybody volunteering? Uwe I think the problem is that those of us who do not understand are in such a state of ignorance that we do not know what it is that we do not understand. I have resigned myself to the realisation that there are a very small number of things about R which I shall never understand and always solve by trial and error and the parameterisation of reshape is the leading example (closely followed by backslashes). I do not think the writers of the documentation are at fault here, it is either just inherently difficult to understand or my cortex is wired up inappropriately. Best, Uwe So Hadley wrote an alternate facility ... the reshape package that does not have a reshape function in it but rather two functions 'melt' and 'cast'. Your data is all ready molten, i.e. it is in the long format (in the terminology of the base reshape function) with identifier values in each row and a single column of values. library(reshape) cast(dataframe,A~B) Using C as value column. Use the value argument to cast to override this choice A control sample 1 d0 1e+05 1e+05 2 d1 2e+02 3e+02 3 d2 4e+02 5e+02 Basically the cast formula keeps the LHS variables in the rows and hte RHD variables get arranges in columns. (For reasons that are unclear to me the dataframe argument was placed first when using positional argument passing, unlike most other formula methods in R.) Michael Dewey i...@aghmed.fsnet.co.uk http://www.aghmed.fsnet.co.uk/home.html __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] val.surv
I've attached a sample of the data sets, this is my full code. #R 2.13 #library(survival) #library(Hmisc) #library(splines) library(rms) train-as.data.frame( train-read.csv(G:\\train.txt, header=T, sep=\t)) test-as.data.frame( test-read.table(G:\\test.txt, header=T, sep=\t)) f.1-cph(Surv(time,event)~age, x=T, y=T,surv=T, data=train) val.surv(f.1, newdata=test, u=10) #plot(calibrate(f.1, u=30, B=20)) From: David Winsemius dwinsem...@comcast.net To: Salvo Mac salvo_...@yahoo.com Cc: r-help@R-project.org r-help@r-project.org Sent: Sunday, August 21, 2011 5:29 AM Subject: Re: [R] val.surv On Aug 20, 2011, at 10:25 PM, Salvo Mac wrote: The test and train are like split data sets, contain similar variables but from different countries so the two sets are somehow independent. And yes it is a data frame. What is a data.frame? test and train may be dataframes, but test[, age] is not a dataframe. So I extracted age, time and event. Code? The code you offered before would have created a newdata object (yes, a data.frame) with a single column bearing the same name as the vector argument , test1. Not named age. Try it. do str on such an object: str(dataframe(test1)) So test is data frame,(age, time, event). does that suffice? It certainly does not allow me to reproduce the error you got (which I still think is probably related to the structure of your argument to newdata.) That's all I can say without data and code. --David. From: David Winsemius dwinsem...@comcast.net Cc: r-help@R-project.org r-help@r-project.org Sent: Sunday, August 21, 2011 3:19 AM Subject: Re: [R] val.surv On Aug 20, 2011, at 8:08 PM, Salvo Mac wrote: Thanks David However, I tried your trick on val.surv with newdata=test['age'] but still didn't work. Still gives the same error message: Error in val.surv(f.1, newdata = test1[age], u = 10) : dims [product 1797] do not match the length of object [2496] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length As I said (and you did not act upon): The fundamental thing you are doing wrong for q1 is failing to unambiguously describe the test object. I said it was a guess. Now stop wasting our time and offer what is needed. --david. Salvo From: David Winsemius dwinsem...@comcast.net Cc: r-help@R-project.org r-help@r-project.org Sent: Sunday, August 21, 2011 12:55 AM Subject: Re: [R] val.surv On Aug 20, 2011, at 3:32 PM, Salvo Mac wrote: Dear R-users, I have two questions regarding validation and calibration of Survival regression models. 1. I am trying to calibrate and validate a cox model using val.surv. here is my code: f.1-cph(Surv(time,event)~age, x=T, y=T, data=train) test1-test[,age] val.surv(f.1, newdata=data.frame(test1), u=10) but I get an error message: Error in val.surv(f.1, newdata = data.frame(testi), u = 10) : dims [product 1797] do not match the length of object [2496] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length I ran the example in the r-documentation but couldn't extract dxy from result. What am I doing wrong? The fundamental thing you are doing wrong for q1 is failing to unambiguously describe the test object. I would think that if test were a dataframe then wrapping data.frame around a vector might not get it named correctly as 'age'. You might try newdata= test['age']. Just a guess. 2. In validate and calibrate cph functions. If it is frailty fit, does the the bootstrap resample clusters or just individuals The code above appears to be dependent on the rms package. The frailty function is part of the underlying survival package and I do not see it mentioned in the index for Harrell's RMS text. You will probably need to wait until Frank comes across this. He is generally very good about correction my errors and knowledge gaps. David Winsemius, MD West Hartford, CTid age weight gender timeevent 1 32 14 0 183 0 3 56 9 0 12 1 4 42 5 0 183 0 5 79 13 0 4 1 6 30 3 0 183 0 7 19 10 0 183 0 8 18 12 0 3 1 9 26 14 0 183 0 10 24 14 0 1 1 11 35 6 0 183 0 12 78 14 0 2 1 13 22 7 0 183 0 14 48 5 0 183 0 15 49 13 0 183 0 16 61 5 0 183 0 19 82 14 0 6 1 20 22 10 0 183 0 21 32 4 1
Re: [R] I have a problem with R!!
On 08/21/2011 05:29 AM, nferr...@fceia.unr.edu.ar wrote:
Dear all
i´m working with a program i´ve made in R (using functions that others
created)
to run my program i need a sample. if i generate the sample using for
example, rnorm(n, mu, sigma) i have no problem
but if i obtain a sample from a column in excel and i copy it, the program
says that there is a mistake: it says Error en `[.data.frame`(data,
indices) : undefined columns selected
my program is:
d- read.delim(clipboard, header = T, dec = ,)
#Para determinar los valores de las componentes del vector de capacidad es
necesario definir primero las especificaciones y el valor objetivo, T, así
como el máximo valor admitido para la proporción de producción no
conforme, a cada lado de los límites de especificaciones#
# Ingrese ahora el valor del límite inferior de especificaciones#
LIE- 13
# Ingrese ahora el valor del límite superior de especificaciones#
LSE- 17
# Ingrese ahora el valor objetivo#
T- 14.5
# Ingrese ahora el máximo valor admitido para la proporción de producción
no conforme a cada lado de los límites de especificaciones#
MA- 0.00135
D- min ((LSE-T), (T-LIE))
compo1- function(data, indices)
{
d- data[indices]
n = length (d)
desvio- sd(d)
y- rep(1:n)
y[x= mean(d)]- 1
y[xmean(d)]- 0
RI1- D/(3*desvio*2*mean(y))
RI2- D/(3*desvio*2*(1-mean(y)))
return (min (RI1, RI2))
}
compo2- function(data, indices)
{
d- data[indices]
c2- (abs(mean(d) - T))/D
return (1-c2)
}
compo3-function(data, indices)
{
d- data[indices]
n- length (d)
y- rep(1:n)
y[d LIE]- 1
y[d= LIE]- 0
INFE- mean (y);
y- rep(1:n)
y[d LSE]- 1
y[d= LSE]- 0
SUPE- mean (y);
PPI- (1 - INFE)/(1-MA)
PPS- (1 - SUPE)/(1-MA)
return (min (PPI, PPS))
}
save(file = compo1.RData)
save(file = compo2.RData)
save(file = compo3.RData)
compos- function(data, indices)
{
d- data[indices]
capacidad- c(compo1(d), compo2(d), compo3(d))
return(capacidad)
}
save(file = compos.RData)
require (boot)
vectorcapacidad- boot (d, compos, R = 3000)
ETC. ETC.
WHEN I START MY PROGRAM WRITING:
d- rnorm (n, mu, sigma)
I HAVE NO PROBLEM. BUT WHEN I READ A VECTOR FROM EXCEL, R TELLS ME
Error en `[.data.frame`(data, indices) : undefined columns selected
CAN YOU HELP ME THANK YOU VERY MUCH!
NOEMI FERRERI, ROSARIO, ARGENTINA
SCHOOL OF INDUSTRIAL ENGINEERING
Hi Noemi,
Without some sample data, I can only guess, but I would first try saving
the Excel spreadsheet in CSV format and then reading the data in with
read.csv. This might solve your problem.
Jim
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Re: [R] reshape a matrix
On 21.08.2011 12:54, Michael Dewey wrote: At 18:21 20/08/2011, Uwe Ligges wrote: On 20.08.2011 19:04, David Winsemius wrote: On Aug 20, 2011, at 12:32 PM, Uwe Ligges wrote: [snip original problem] David, I think there are some good examples on the help page. What is missing? What is not clearly explained? If a longer tutorial is needed, that may be an article for the R Help Desk in The R Journal. Anybody volunteering? Uwe I think the problem is that those of us who do not understand are in such a state of ignorance that we do not know what it is that we do not understand. I have resigned myself to the realisation that there are a very small number of things about R which I shall never understand and always solve by trial and error and the parameterisation of reshape is the leading example (closely followed by backslashes). I do not think the writers of the documentation are at fault here, it is either just inherently difficult to understand or my cortex is wired up inappropriately. In think tha major problem is that stuff like id and time etc. may be misleading, if they do not fit to the context. Don't believe I do things correct right away here. Nevertheless, improving the wording of the reshape() documentation is not easy at all if we want to be precise. Maybe we find another volunteer or I find some time to write a short column for the next issue of The R Journal. Best wishes, Uwe Best, Uwe So Hadley wrote an alternate facility ... the reshape package that does not have a reshape function in it but rather two functions 'melt' and 'cast'. Your data is all ready molten, i.e. it is in the long format (in the terminology of the base reshape function) with identifier values in each row and a single column of values. library(reshape) cast(dataframe,A~B) Using C as value column. Use the value argument to cast to override this choice A control sample 1 d0 1e+05 1e+05 2 d1 2e+02 3e+02 3 d2 4e+02 5e+02 Basically the cast formula keeps the LHS variables in the rows and hte RHD variables get arranges in columns. (For reasons that are unclear to me the dataframe argument was placed first when using positional argument passing, unlike most other formula methods in R.) Michael Dewey i...@aghmed.fsnet.co.uk http://www.aghmed.fsnet.co.uk/home.html __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] adding text to a plot created with strat.plot() from package rioja
On 21.08.2011 01:01, Jason Paul Joines wrote: I completely missed that entering just the function name prints the code for that function making it even easier to copy and create a custom version. Better use the original code as you did before: It preserves the comments. Anyway, best way would be to contribute your improvements back to the package. Best, Uwe Ligges Jason === Original Message Subject: Re: [R] adding text to a plot created with strat.plot() from package rioja From: Jason Paul Joines ja...@joines.org To: r-help@r-project.org Date: 2011.08.20.Sat.14:48:35 I'm pretty new to R and finding the responsible line would not have been obvious to me without your help. I downloaded the source for package Rioja and was surprised to see that each function was supplied in it's own file. That made it pretty straightforward to copy, modify, and use my own version of it. Still, it's going to take quite a while to get familiar with all of the graphical capabilities of R. Thanks, Jason === Original Message Subject: Re: [R] adding text to a plot created with strat.plot() from package rioja From: Uwe Ligges lig...@statistik.tu-dortmund.de To: Jason Paul Joines ja...@joines.org Date: 2011.08.20.Sat.8:43:30 On 19.08.2011 18:40, Jason Paul Joines wrote: I have a plot created with strat.plot() from package rioja. When the plot is created with scale.percent=FALSE, each x axes is labeled at 0 and its maximum. However, when scale.percent=TRUE, the x axes are not labeled. I need to use scale.percent=TRUE and I need labels for the x axes. I have been able to add labels to the x axes with mtext but it is very tedious to find the correct position. Is there a better way to do this or a better way to find the desired coordinates than trial and error? Yes: You can change the code and suggest your improvements to the package maintainer, for example. The relevant line in that function obviously is: axis(side = 1, at = seq(0, colM[i], by = 10), labels = FALSE) Uwe Ligges Jason === __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to navigate (zoom, pan) in a plot/graph
On 21.08.2011 10:17, Eran Eidinger wrote: Hello all, I need to zoom in and out and travel(pan) inside a plot, like you can do on a Matlab plot. If possible, I would also like the option to use the mouse to set a marker on the graph and get the (x,y) data for it, again, like in Matlab. Don't know about that feature in Matlab, but see ?locator and ?identify for regular R graphics. Uwe Ligges Is this possible in R with the regular packages, or do you maybe know a different package that will allow this? Eran. * * [[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] Best way to setClass and setMethod for an R package?
On 21.08.2011 00:48, Jonathan Greenberg wrote: Folks: I'm putting together an R package, and I was wondering where, specifically, I put both class definitions (via setClass) as well as setMethod calls? Is there a particular file name I need to use? No, any file name, e.g. .../name_of_your_package/R/Name_of_your_class.R Uwe Ligges Thanks! --j __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Build a package - check error
On 21.08.2011 11:29, Erich Neuwirth wrote: On 8/20/2011 3:37 PM, Uwe Ligges wrote: ?.First.lib has an example: ## Suppose a package needs to call a DLL named 'fooEXT', ## where 'EXT' is the system-specific extension. Then you should use .First.lib- function(lib, pkg) library.dynam(foo, pkg, lib) Shouldn't the first line here be ## Suppose a package needs to call a DLL named 'foo.EXT', ^ No, since .dll is the extension in this case: See ?.Platform which has an element dynlib.ext containing the current's platform specific extension. That one contains the extension such as .dll, .so and so on. Therefore fooEXT evaluates to foo.so under Linux, for example. Best, Uwe __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to navigate (zoom, pan) in a plot/graph
On 21.08.2011 15:04, Eran Eidinger wrote: Thank you Uwe, That solves the second question. Still looking for some solution to zooming in and out dynamically. That is not implemented in standard R graphics. Packages rgl and iplots may be able to do that. Best, Uwe Eran. 2011/8/21 Uwe Ligges lig...@statistik.tu-dortmund.de mailto:lig...@statistik.tu-dortmund.de On 21.08.2011 10 tel:21.08.2011%2010:17, Eran Eidinger wrote: Hello all, I need to zoom in and out and travel(pan) inside a plot, like you can do on a Matlab plot. If possible, I would also like the option to use the mouse to set a marker on the graph and get the (x,y) data for it, again, like in Matlab. Don't know about that feature in Matlab, but see ?locator and ?identify for regular R graphics. Uwe Ligges Is this possible in R with the regular packages, or do you maybe know a different package that will allow this? Eran. * * [[alternative HTML version deleted]] R-help@r-project.org mailto:R-help@r-project.org mailing list https://stat.ethz.ch/mailman/__listinfo/r-help https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/__posting-guide.html http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- * Eran Eidinger | Taykey Ltd | +972-54-5908077 | www.taykey.com http://www.taykey.com * __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Dot plot with two grouping variables concurrently
Dear R help(ers). I'm an R-learner (about 10 hours now) trying to make a ranked dot plot where the symbols are coded by two variables concurrently. I'm trying to use Deepanyan Sarkar's book 'Lattice' as a guide but get the feeling it is a bit advanced for my level of understanding. I have three questions Q1. Right now I like to know how to get the dual coding working on the points on a dotplot and also annotate these in a key. Specifically I'm attempting to code the fill of the points by a variable called 'Commodity' but would like have different symbols to refelect a second variable called 'Year' (say square, triangle and circle for different years). My code so far is as follows: ## initalise library(lattice) ##read the data to a variable Cal_dat - read.table(Calibration2.dat,header = TRUE,sep = \t,) ## set up plotting colours for the fills col.pat-c(violet,cyan,green,red,blue,black,yellow) ##set up the plot key to be inside the plot plot.key-list( corner=c(1,0), text=list(levels(Cal_dat$Commodity)), title=Ore type, points=list(pch=21,cex=1.3,fill=col.pat,col=black) ) ##set some parameters for the dotplot trellis.par.set( dot.line=list(col = transparent), axis.line=list(col = grey90), axis.text=list(col =grey50, cex=0.8), panel.background=list(col=grey98), par.xlab.text= list(col=grey50) ) ## Create the dot plot dotplot(reorder(Mine, Resc_Gt)~ Resc_Gt,groups=Commodity, data=Cal_dat, cex=1.2, pch=21, aspect=2.0, key=plot.key, col=black, fill=col.pat, origin=0, type = c(p, h), main = Resource Tonnage, xlab= tonnes (billions) ) The dot plot so far is attached as well as the input data. Q2 I'm puzzled by the waring message (below) I get when runnin the current code as follows (What does this mean and why does it occur). Warning message: In trellis.par.set(dot.line = list(col = transparent), axis.line = list(col = grey90), : Note: The default device has been opened to honour attempt to modify trellis settings Q3 Can anybody recommend a good 'starters' book for learning plotting in R that is reasonably clear on the basics. My primary interest is the graphing tools that are in lattice (hence trying the Deepayan Sarkar book), but I'm finding this guide rather tough going. Thanks in advance - MarkM http://r.789695.n4.nabble.com/file/n3758248/Calibration2.dat Calibration2.dat http://r.789695.n4.nabble.com/file/n3758248/Ranked_boxplot_by_commodity.png Ranked_boxplot_by_commodity.png -- View this message in context: http://r.789695.n4.nabble.com/Dot-plot-with-two-grouping-variables-concurrently-tp3758248p3758248.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How can I divide an image and randomly plot the segements
Hi! I would like to divide a 401 by 401 image into 20 equal blocks and then re-plot the image with the segments randomly distributed in it. How can I do this? . Regards, Kenduiywo Benson [[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] Histogram from frequency data in pre-made bins
Dear R user, I am using UK census data on travel to work. The authorities have provided a breakdown in each area by mode (car, bicycle etc.) and distance travelled (0 – 2 km, 2 – 5 km etc). Therefore, after processing, the data for Sheffield look like this https://files.one.ubuntu.com/ej2VtVbJTEaelvMRlsocRg : dshef - read.table(distmodesheff.csv, sep=,, header=TRUE) print(dshef) Dist Tr Bici Met Pas Foot Bus Car 1 245 571 491 2125 16644 4469 13494 2 2 – 5 80 1136 2540 4738 3659 17290 30212 3 5 – 10 217 466 2335 3994 1041 12963 35221 4 10 – 20 191 76 491 1333 332 2439 16322 5 20 – 30 1686 25 23541 175 3711 6 30 – 40 7863 1222074 2179 7 40 – 60 3496 21 26196 333 3501 8 60 332 62 125 369 534 433 3276 9Other 148 40 79 905 388 622 6481 It's interesting to look at the different distributions of different transport modes: attach(dshef) rs - rbind(Tr,Bici,Met,Pas,Foot,Bus,Car) barplot(rs, beside=TRUE, names=Dist, col=rainbow(7), legend=TRUE) http://r.789695.n4.nabble.com/file/n3758198/1.png This is brilliant, and creates output similar to that of OO calc: http://r.789695.n4.nabble.com/file/n3758198/egraphmini.jpg However, as you can see, the pre-made categories (0 – 2 km etc.) are unevenly spaced bins within a continuous variable. This puts the analysis into histogram mode (with frequency determined by the area, not the height). What I would look for for the vector Car, for example, would be something like this: n - c(rep(1.5,Car[1]), rep(3,Car[2]), rep(7.7,Car[3]), rep(15,Car[4]),rep(25,Car[5]), rep(35,Car[6]), rep(50,Car[7]), rep(100,Car[8]) ) hist(n, breaks=c(0,2,5,10,20,30,40,60,200)) http://r.789695.n4.nabble.com/file/n3758198/2.png This produces a histogram, but it's a tedious an ugly way of getting there. Also, this does not allow for trend-line analysis of the likely distribution of the continuous variable distance: lines(density(n)), for example results in peaks around my arbitrary value. Has anyone else encountered similar issues? I've searched high and low but can find no solution other than creating a barplot with variable widths: http://r.789695.n4.nabble.com/Histogram-using-frequency-data-td827927.html Any ideas about how to resolve this issue very greatly appreciated. Eventually I hope to model the distribution of distances travelled in order to estimate the mean distance within each bin. Many thanks, Robin -- View this message in context: http://r.789695.n4.nabble.com/Histogram-from-frequency-data-in-pre-made-bins-tp3758198p3758198.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to navigate (zoom, pan) in a plot/graph
Thank you Uwe, That solves the second question. Still looking for some solution to zooming in and out dynamically. Eran. 2011/8/21 Uwe Ligges lig...@statistik.tu-dortmund.de On 21.08.2011 10:17, Eran Eidinger wrote: Hello all, I need to zoom in and out and travel(pan) inside a plot, like you can do on a Matlab plot. If possible, I would also like the option to use the mouse to set a marker on the graph and get the (x,y) data for it, again, like in Matlab. Don't know about that feature in Matlab, but see ?locator and ?identify for regular R graphics. Uwe Ligges Is this possible in R with the regular packages, or do you maybe know a different package that will allow this? Eran. * * [[alternative HTML version deleted]] __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- * Eran Eidinger | Taykey Ltd | +972-54-5908077 | www.taykey.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.
Re: [R] AFT model time-dependent with weibull distribution
thanks! -- View this message in context: http://r.789695.n4.nabble.com/AFT-model-time-dependent-with-weibull-distribution-tp3755079p3758267.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dot plot with two grouping variables concurrently
Q1. Right now I like to know how to get the dual coding working on the points on a dotplot and also annotate these in a key. Specifically I'm attempting to code the fill of the points by a variable called 'Commodity' but would like have different symbols to refelect a second variable called 'Year' (say square, triangle and circle for different years). You can custumize pch as to reflect the levels of Year in both your dotplot and key my.pch- 1:length(levels(Cal_dat$Year)) ### for example Then, use my.pch in the places of pch in your codes of dotplot and key HTH Weidong Gu My code so far is as follows: ## initalise library(lattice) ##read the data to a variable Cal_dat - read.table(Calibration2.dat,header = TRUE,sep = \t,) ## set up plotting colours for the fills col.pat-c(violet,cyan,green,red,blue,black,yellow) ##set up the plot key to be inside the plot plot.key-list( corner=c(1,0), text=list(levels(Cal_dat$Commodity)), title=Ore type, points=list(pch=21,cex=1.3,fill=col.pat,col=black) ) ##set some parameters for the dotplot trellis.par.set( dot.line=list(col = transparent), axis.line=list(col = grey90), axis.text=list(col =grey50, cex=0.8), panel.background=list(col=grey98), par.xlab.text= list(col=grey50) ) ## Create the dot plot dotplot(reorder(Mine, Resc_Gt)~ Resc_Gt,groups=Commodity, data=Cal_dat, cex=1.2, pch=21, aspect=2.0, key=plot.key, col=black, fill=col.pat, origin=0, type = c(p, h), main = Resource Tonnage, xlab= tonnes (billions) ) The dot plot so far is attached as well as the input data. Q2 I'm puzzled by the waring message (below) I get when runnin the current code as follows (What does this mean and why does it occur). Warning message: In trellis.par.set(dot.line = list(col = transparent), axis.line = list(col = grey90), : Note: The default device has been opened to honour attempt to modify trellis settings Q3 Can anybody recommend a good 'starters' book for learning plotting in R that is reasonably clear on the basics. My primary interest is the graphing tools that are in lattice (hence trying the Deepayan Sarkar book), but I'm finding this guide rather tough going. Thanks in advance - MarkM http://r.789695.n4.nabble.com/file/n3758248/Calibration2.dat Calibration2.dat http://r.789695.n4.nabble.com/file/n3758248/Ranked_boxplot_by_commodity.png Ranked_boxplot_by_commodity.png -- View this message in context: http://r.789695.n4.nabble.com/Dot-plot-with-two-grouping-variables-concurrently-tp3758248p3758248.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Best way to setClass and setMethod for an R package?
On 08/21/2011 05:56 AM, Uwe Ligges wrote: On 21.08.2011 00:48, Jonathan Greenberg wrote: Folks: I'm putting together an R package, and I was wondering where, specifically, I put both class definitions (via setClass) as well as setMethod calls? Is there a particular file name I need to use? Hi Jonathan In principle, S4 methods can dispatch on multiple classes, and an approach has been to have files AllClasses.R and AllGenerics.R to implement class and generic definitions, then foo-methods.R for methods defined on the generic foo. An 'advantage', for the lazy amongst us, is that this usually leads to a collation order that is appropriate -- class definition before use in method dispatch -- so that the Collates: field in the DESCRIPTION file does not have to be maintained. In practice many S4 generics dispatch on a single argument and while there might be the equivalent of AllGenerics.R, the classes and their associated methods are defined in MyClass-methods.R. Martin No, any file name, e.g. .../name_of_your_package/R/Name_of_your_class.R Uwe Ligges Thanks! --j __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Computational Biology Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: M1-B861 Telephone: 206 667-2793 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to navigate (zoom, pan) in a plot/graph
See setGraphicsEventHandlers function. On Sun, Aug 21, 2011 at 10:04 AM, Eran Eidinger e...@taykey.com wrote: Thank you Uwe, That solves the second question. Still looking for some solution to zooming in and out dynamically. Eran. 2011/8/21 Uwe Ligges lig...@statistik.tu-dortmund.de On 21.08.2011 10:17, Eran Eidinger wrote: Hello all, I need to zoom in and out and travel(pan) inside a plot, like you can do on a Matlab plot. If possible, I would also like the option to use the mouse to set a marker on the graph and get the (x,y) data for it, again, like in Matlab. Don't know about that feature in Matlab, but see ?locator and ?identify for regular R graphics. Uwe Ligges Is this possible in R with the regular packages, or do you maybe know a different package that will allow this? Eran. * * [[alternative HTML version deleted]] __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- * Eran Eidinger | Taykey Ltd | +972-54-5908077 | www.taykey.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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to navigate (zoom, pan) in a plot/graph
On Sun, Aug 21, 2011 at 3:04 PM, Eran Eidinger e...@taykey.com wrote: Thank you Uwe, That solves the second question. Still looking for some solution to zooming in and out dynamically. Try package 'playwith'. Liviu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] val.surv
On Aug 21, 2011, at 7:03 AM, Salvo Mac wrote: I've attached a sample of the data sets, this is my full code. #R 2.13 #library(survival) #library(Hmisc) #library(splines) library(rms) train-as.data.frame( train-read.csv(G:\\train.txt, header=T, sep=\t)) test-as.data.frame( test-read.table(G:\\test.txt, header=T, sep=\t)) f.1-cph(Surv(time,event)~age, x=T, y=T,surv=T, data=train) val.surv(f.1, newdata=test, u=10) #plot(calibrate(f.1, u=30, B=20)) Salvo; I'm not sure where the error is coming from (although I will share my speculation below). I modified your code somewhat. I was under the impression that read.csv had sep=, hardcoded and the read.csv( ... seo=\t) would throw and error. The as.data.fame around either of the read.* statements is completely unnecessary: Input code: train-read.table(~/train.txt, header=T, sep=\t) test-read.table(~/test.txt, header=T, sep=\t) I thought the newdata=test argument should succeed, but it does throw the error that you report: f.1-cph(Surv(time,event)~age, x=T, y=T,surv=T, data=train) val.surv(f.1, newdata=test, u=10) Error in val.surv(f.1, newdata = test, u = 10) : dims [product 210] do not match the length of object [314] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length I tried sampling from test with replacement to generate a dataframe of equal extent and with that object I do not get the same errors: extend - test[sample(1:nrow(test), 314, replace=TRUE), ] f.1-cph(Surv(time,event)~age, x=T, y=T,surv=T, data=train) val.surv(f.1, newdata=extend, u=10) Validation of Predicted Survival at Time= 10n= 314 , events= 64 hare fit: dim A/D loglik AICpenalty minmax 1 Add -485.24976.22 271.33 Inf 2 Del -349.57710.645.67 271.33 3 Del -346.74710.720.965.67 4 Del -346.26715.510.010.96 5 Add -346.25721.250.000.01 the present optimal number of dimensions is 2. penalty(AIC) was 5.75, the default (BIC), would have been 5.75. dim1 dim2 betaSE Wald Constant -10 0.55 -18.14 Time43 0.15 0.015 10.30 Function used to transform predictions: function (p) log(-log(p)) Mean absolute error in predicted probabilities: 0.0331 0.9 Quantile of absolute errors : 0.0613 I have also tried looking at the code and adding options(error=utils::recover) to see if I can identify the point where the length mismatch is being generated, (but I am NOT an ace debugger). I can see that est.surv is created. I can also get the predict(fit, newdata, type = lp) call to run with train and give sensible numbers. You did not create (or at least say so) a datadist. I tried that when I saw that lim was dependent on datadist limits. ddT - datadist(train); options(datadist=ddT) Seeing the usehare was set to TRUE; I submitted the code section that was intended for that situation to a browser session and the the first line throws the error that I see at the console: Enter a frame number, or 0 to exit 1: val.surv(f.1, newdata = test, u = 10) Selection: 1 Called from: top level Browse[1] i - !is.na(est.surv + S[, 1] + S[, 2]) Error during wrapup: dims [product 210] do not match the length of object [314] As I read this that line is supposed to create an index of valid rows in newdata and the fitted values but is failing in the situation where the nrows of the newdata does not match that of the fit. At this point one option might be trying to generate a structure that matches the val.surv output by going through the code and building it up bit by bit: w - structure(list(harefit = f, p = est.surv, actual = actual, pseq = pseq, actualseq = actualseq, u = u, fun = fun, n = nrow(S), d = sum(S[, 2]), units = units), class = val.survh) return(w) But a much better option would be to report the error to Frank Harrell. I'm copying him since I think your .txt files probably reached the list as well as my mailbox. -- David. From: David Winsemius dwinsem...@comcast.net To: Salvo Mac salvo_...@yahoo.com Cc: r-help@R-project.org r-help@r-project.org Sent: Sunday, August 21, 2011 5:29 AM Subject: Re: [R] val.surv On Aug 20, 2011, at 10:25 PM, Salvo Mac wrote: The test and train are like split data sets, contain similar variables but from different countries so the two sets are somehow independent. And yes it is a data frame. What is a data.frame? test and train may be dataframes, but test[, age] is not a dataframe. So I extracted age, time and event. Code? The code you offered before would have created a newdata object (yes, a data.frame) with a single column bearing the same name as the vector argument , test1. Not named age.
[R] Sweave doesn't work
Hi R users.
I've got a problem in producing the pdf file from Latex with R code. When I
run the code Sweave(example.Rtex) in R it seems working, but when I run
the Latex file it doesn't. The code error shown to me is below:
*Runaway argument?
{echo=FALSE}
data- read.csv(C:\\Users\\Daniele\\Desktop\\dati\\dati_england
! File ended while scanning use of \FV@BeginScanning.
inserted text
\par
* ...le/Desktop/dati/LaTeX1.Rtex*
The Sweave code in Latex is like this:
*\begin{Scode}{echo=FALSE}
data-
read.csv(c:\\Users\\Daniele\\Desktop\\dati\\dati_england.csv,header=T,sep=,)
\end{Scode}
\begin{figure}[!h]
\centering
\begin{Scode}{fig=TRUE, width=6, height=9, echo=FALSE}
data1=matrix(0,ncol=4,nrow=4)
\end{Scode}
\caption{Data}
\end{figure}*
I don't know how to fix this problem. Any advices?
Thanks very much
--
View this message in context:
http://r.789695.n4.nabble.com/Sweave-doesn-t-work-tp3758658p3758658.html
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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] val.surv
I welcome a trivial self-contained set of code, not using frailties, that generates its own data and which fails. Then I'll debug. Frank David Winsemius wrote: On Aug 21, 2011, at 7:03 AM, Salvo Mac wrote: I've attached a sample of the data sets, this is my full code. #R 2.13 #library(survival) #library(Hmisc) #library(splines) library(rms) train-as.data.frame( train-read.csv(G:\\train.txt, header=T, sep=\t)) test-as.data.frame( test-read.table(G:\\test.txt, header=T, sep=\t)) f.1-cph(Surv(time,event)~age, x=T, y=T,surv=T, data=train) val.surv(f.1, newdata=test, u=10) #plot(calibrate(f.1, u=30, B=20)) Salvo; I'm not sure where the error is coming from (although I will share my speculation below). I modified your code somewhat. I was under the impression that read.csv had sep=, hardcoded and the read.csv( ... seo=\t) would throw and error. The as.data.fame around either of the read.* statements is completely unnecessary: Input code: train-read.table(~/train.txt, header=T, sep=\t) test-read.table(~/test.txt, header=T, sep=\t) I thought the newdata=test argument should succeed, but it does throw the error that you report: f.1-cph(Surv(time,event)~age, x=T, y=T,surv=T, data=train) val.surv(f.1, newdata=test, u=10) Error in val.surv(f.1, newdata = test, u = 10) : dims [product 210] do not match the length of object [314] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length I tried sampling from test with replacement to generate a dataframe of equal extent and with that object I do not get the same errors: extend - test[sample(1:nrow(test), 314, replace=TRUE), ] f.1-cph(Surv(time,event)~age, x=T, y=T,surv=T, data=train) val.surv(f.1, newdata=extend, u=10) Validation of Predicted Survival at Time= 10 n= 314 , events= 64 hare fit: dim A/D loglik AICpenalty minmax 1 Add -485.24976.22 271.33 Inf 2 Del -349.57710.645.67 271.33 3 Del -346.74710.720.965.67 4 Del -346.26715.510.010.96 5 Add -346.25721.250.000.01 the present optimal number of dimensions is 2. penalty(AIC) was 5.75, the default (BIC), would have been 5.75. dim1 dim2 betaSE Wald Constant -10 0.55 -18.14 Time43 0.15 0.015 10.30 Function used to transform predictions: function (p) log(-log(p)) Mean absolute error in predicted probabilities: 0.0331 0.9 Quantile of absolute errors : 0.0613 I have also tried looking at the code and adding options(error=utils::recover) to see if I can identify the point where the length mismatch is being generated, (but I am NOT an ace debugger). I can see that est.surv is created. I can also get the predict(fit, newdata, type = lp) call to run with train and give sensible numbers. You did not create (or at least say so) a datadist. I tried that when I saw that lim was dependent on datadist limits. ddT - datadist(train); options(datadist=ddT) Seeing the usehare was set to TRUE; I submitted the code section that was intended for that situation to a browser session and the the first line throws the error that I see at the console: Enter a frame number, or 0 to exit 1: val.surv(f.1, newdata = test, u = 10) Selection: 1 Called from: top level Browse[1] i - !is.na(est.surv + S[, 1] + S[, 2]) Error during wrapup: dims [product 210] do not match the length of object [314] As I read this that line is supposed to create an index of valid rows in newdata and the fitted values but is failing in the situation where the nrows of the newdata does not match that of the fit. At this point one option might be trying to generate a structure that matches the val.surv output by going through the code and building it up bit by bit: w - structure(list(harefit = f, p = est.surv, actual = actual, pseq = pseq, actualseq = actualseq, u = u, fun = fun, n = nrow(S), d = sum(S[, 2]), units = units), class = val.survh) return(w) But a much better option would be to report the error to Frank Harrell. I'm copying him since I think your .txt files probably reached the list as well as my mailbox. -- David. From: David Winsemius lt;dwinsem...@comcast.netgt; To: Salvo Mac lt;salvo_...@yahoo.comgt; Cc: r-help@R-project.org lt;r-help@r-project.orggt; Sent: Sunday, August 21, 2011 5:29 AM Subject: Re: [R] val.surv On Aug 20, 2011, at 10:25 PM, Salvo Mac wrote: The test and train are like split data sets, contain similar variables but from different countries so the two sets are somehow independent. And yes it is a data frame. What is a
[R] rank analysis - reinventing the wheel?
Hello, I have two data frames. One is my dependent variable and the other is my independent variable. For each row I'd like to split the independent variable into fractiles (25 or more) and calculate the average value of the dependent variable. Then I would like to plot the average of the averages for each row for each fractile. If possible, I'd like to have the option to 1) define the fractiles independently for each row, and 2) define the fractiles based off the independent variable data frame as a whole at the start and keep it fixed. Anyway, is there a pacakge/function in R that does something like this so I don't have to reinvent the wheel? If not, any help on how to do this efficiently in R would be great. Here is a super simple example using 3 fractiles: independent data frame: item1 item2 item3 1181516 2121217 3201318 dependent data frame: item1 item2 item3 1 3 1 6 2 2 2 7 3 1 3 8 fractiles by row by item: (done with independent data frame, redefining fractiles each row) fractiles: 12 3 row 1item2 item3 item1 (because 15 16 18) row 2item1,item2[none] item3 (because 12 17 ... or however ties are handled is fine) row 3item2 item3 item1 (because 13 18 20) Note: in the above fractiles there would almost always be more than one item in each fractile if this example wasn't so simple... dependent variable averages by row by fractile: fractiles: 12 3 row 1 1 63 row 2 2 [none] 7 row 3 3 8 1 Note: obviously this example is so simple there aren't actual averages because there aren't enough items for each fractile, but I hope you get the point. So then the averages of the fractiles would be: fractiles: 12 3 dependent data avg: 2 73.67 Then I would like to plot a histogram of the fractile averages directly above. Thanks! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Raw epoch time from XTS
Josh, Good point about including an example. calling xcoredata() does work, but only for a *single* row of the data at a time. In R, I'm used to passing an entire data structure or vector to a function and automatically getting back a vector of all the results. In this case, it doesn't work that way. The attr() function is probably the best solution. Thanks! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building Los Angeles, CA 90095 On Aug 20, 2011, at 1:03 AM, Joshua Wiley wrote: Hi Noah, This is one of those cases where following the posting guide (particularly the minimal, reproducible example part) would have really helped. Are you saying that calling: xcoredata(your_xts_object) does not give you the internal representation of time that you want? data(sample_matrix) sample.xts - as.xts(sample_matrix, descr='my new xts object') # returns a list, the index is the numeric representation of time displayed in the rows xcoredata(sample.xts) You could also try the more direct: attr(xts_object, index) If this is not what you want or is not working for you, providing us the output of dput() from the first few rows of your dataset and an example of what you do want would be spectacular. Cheers, Josh On Sat, Aug 20, 2011 at 12:44 AM, Noah Silverman noahsilver...@ucla.edu wrote: Hi, I have a very large data set stored as an xts object. xts is very nice about showing row labels as human readable dates and times. I want the actual epoch values that are stored internally. The only way I can find to access them is one-at-a-time using the internal function: xcoredata() Calling this in an entire column, the R way doesn't work. It will only return a single value. Calling it in a loop for each row works but is painfully slow. Since the epoch is stored internally, there must be some way to just grab it as a vector. Does anyone know how? Thanks! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building Los Angeles, CA 90095 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Compile on Fedora
Hi, I want to create a custom build of R, for my particular machine that will run as fast as possible. CPU: Dual cora AMD 64 bit OS: Fedora 15 I know there are a lot of options for compile time flags, lapack, blas, atlas, etc. Not sure what the best combination is. Has someone already worked this out? Thanks! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building Los Angeles, CA 90095 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Compile on Fedora
On Sun, Aug 21, 2011 at 10:18 AM, Noah Silverman noahsilver...@ucla.edu wrote: Hi, I want to create a custom build of R, for my particular machine that will run as fast as possible. CPU: Dual cora AMD 64 bit OS: Fedora 15 I know there are a lot of options for compile time flags, lapack, blas, atlas, etc. Not sure what the best combination is. If you need any kind of matrix manipulations with larger matrices (say more than a few hundred rows or columns), I would definitely suggest a custom BLAS, for example Goto BLAS or ATLAS. When I last compared them a couple of years ago, Goto BLAS was somewhat (on the order of 20%) faster than ATLAS. This may have changed because of better compilers, plus I only used Intel processors, so you may want to look into that. A problem with Goto BLAS was that it would inexplicably slow down, or grind to a halt, if two separate R processes were calling the BLAS routines at the same time, whereas ATLAS never gave us any trouble. My understanding is that Goto BLAS is not maintained anymore but the existing version should work well. HTH, Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] rank analysis - reinventing the wheel?
This is close: require(Hmisc); ?xYplot and see method='quantile' Frank Ben qant wrote: Hello, I have two data frames. One is my dependent variable and the other is my independent variable. For each row I'd like to split the independent variable into fractiles (25 or more) and calculate the average value of the dependent variable. Then I would like to plot the average of the averages for each row for each fractile. If possible, I'd like to have the option to 1) define the fractiles independently for each row, and 2) define the fractiles based off the independent variable data frame as a whole at the start and keep it fixed. Anyway, is there a pacakge/function in R that does something like this so I don't have to reinvent the wheel? If not, any help on how to do this efficiently in R would be great. Here is a super simple example using 3 fractiles: independent data frame: item1 item2 item3 1181516 2121217 3201318 dependent data frame: item1 item2 item3 1 3 1 6 2 2 2 7 3 1 3 8 fractiles by row by item: (done with independent data frame, redefining fractiles each row) fractiles: 12 3 row 1item2 item3 item1 (because 15 16 18) row 2item1,item2[none] item3 (because 12 17 ... or however ties are handled is fine) row 3item2 item3 item1 (because 13 18 20) Note: in the above fractiles there would almost always be more than one item in each fractile if this example wasn't so simple... dependent variable averages by row by fractile: fractiles: 12 3 row 1 1 63 row 2 2 [none] 7 row 3 3 8 1 Note: obviously this example is so simple there aren't actual averages because there aren't enough items for each fractile, but I hope you get the point. So then the averages of the fractiles would be: fractiles: 12 3 dependent data avg: 2 73.67 Then I would like to plot a histogram of the fractile averages directly above. Thanks! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Frank Harrell Department of Biostatistics, Vanderbilt University -- View this message in context: http://r.789695.n4.nabble.com/rank-analysis-reinventing-the-wheel-tp3758729p3758906.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Raw epoch time from XTS
Noah, For me xcoredata() returns a vector of the results, which makes me wonder if you are using an old version of R or your data is somehow stored differently. Cheers, Josh On Sun, Aug 21, 2011 at 10:15 AM, Noah Silverman noahsilver...@ucla.edu wrote: Josh, Good point about including an example. calling xcoredata() does work, but only for a *single* row of the data at a time. In R, I'm used to passing an entire data structure or vector to a function and automatically getting back a vector of all the results. In this case, it doesn't work that way. The attr() function is probably the best solution. Thanks! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building Los Angeles, CA 90095 On Aug 20, 2011, at 1:03 AM, Joshua Wiley wrote: Hi Noah, This is one of those cases where following the posting guide (particularly the minimal, reproducible example part) would have really helped. Are you saying that calling: xcoredata(your_xts_object) does not give you the internal representation of time that you want? data(sample_matrix) sample.xts - as.xts(sample_matrix, descr='my new xts object') # returns a list, the index is the numeric representation of time displayed in the rows xcoredata(sample.xts) You could also try the more direct: attr(xts_object, index) If this is not what you want or is not working for you, providing us the output of dput() from the first few rows of your dataset and an example of what you do want would be spectacular. Cheers, Josh On Sat, Aug 20, 2011 at 12:44 AM, Noah Silverman noahsilver...@ucla.edu wrote: Hi, I have a very large data set stored as an xts object. xts is very nice about showing row labels as human readable dates and times. I want the actual epoch values that are stored internally. The only way I can find to access them is one-at-a-time using the internal function: xcoredata() Calling this in an entire column, the R way doesn't work. It will only return a single value. Calling it in a loop for each row works but is painfully slow. Since the epoch is stored internally, there must be some way to just grab it as a vector. Does anyone know how? Thanks! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building Los Angeles, CA 90095 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sweave doesn't work
Hi,
You would probably be better off providing us with the code you weave
rather than the TeX code. Also, the problem is quite possibly not
near where the error occurred. In my (admittedly limited) experience,
runaway argument issues tend to mean at some earlier point I:
A) did not choose the right environment
B) did not close an environment
C) did not properly escape something leading essentially to A or B
Something seems weird to me with the LaTeX code you show though. In
addition to posting your code (not the TeX output), it would also be
good to post your version of R.
Cheers,
Josh
On Sun, Aug 21, 2011 at 9:18 AM, danielepippo dan...@hotmail.it wrote:
Hi R users.
I've got a problem in producing the pdf file from Latex with R code. When I
run the code Sweave(example.Rtex) in R it seems working, but when I run
the Latex file it doesn't. The code error shown to me is below:
*Runaway argument?
{echo=FALSE}
data- read.csv(C:\\Users\\Daniele\\Desktop\\dati\\dati_england
! File ended while scanning use of \FV@BeginScanning.
inserted text
\par
* ...le/Desktop/dati/LaTeX1.Rtex*
The Sweave code in Latex is like this:
*\begin{Scode}{echo=FALSE}
data-
read.csv(c:\\Users\\Daniele\\Desktop\\dati\\dati_england.csv,header=T,sep=,)
\end{Scode}
\begin{figure}[!h]
\centering
\begin{Scode}{fig=TRUE, width=6, height=9, echo=FALSE}
data1=matrix(0,ncol=4,nrow=4)
\end{Scode}
\caption{Data}
\end{figure}*
I don't know how to fix this problem. Any advices?
Thanks very much
--
View this message in context:
http://r.789695.n4.nabble.com/Sweave-doesn-t-work-tp3758658p3758658.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.
--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, ATS Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.com/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Compile on Fedora
On 21 August 2011 at 10:18, Noah Silverman wrote: | Hi, | | I want to create a custom build of R, for my particular machine that will run as fast as possible. | | CPU: Dual cora AMD 64 bit | | OS: Fedora 15 | | I know there are a lot of options for compile time flags, lapack, blas, atlas, etc. Not sure what the best combination is. | | Has someone already worked this out? Please the fine manual at http://cran.r-project.org/doc/manuals/R-admin.html#Configuration-on-a-Unix_002dalike Dirk -- Two new Rcpp master classes for R and C++ integration scheduled for New York (Sep 24) and San Francisco (Oct 8), more details are at http://dirk.eddelbuettel.com/blog/2011/08/04#rcpp_classes_2011-09_and_2011-10 http://www.revolutionanalytics.com/products/training/public/rcpp-master-class.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] trouble installing packages on OpenSuse 11.4
I have similar problems when trying to install packages while at work. It might have something to do with your firewall settings. On Wed, Aug 17, 2011 at 4:38 AM, Dinesh din...@milkorwater.com wrote: Hi, I am trying to install a bunch of packages via command line and can use some help in getting it right. My env is a freshly setup OpenSuse 11.4 on an amd desktop. I have not yet installed gcc (Will I need gcc to install packages? I have installed make, assuming R might need it.). I have tried it both under R2.12 and R2.13. I have a list of packages to install such as fImport, fGarch, zoo, and several more. I am logged in as root because I want the packages to be visible to all users a) R CMD INSTALL fImport This produces no results. b) sudo R CMD INSTALL fImport This downloads timeDate and gets stuck there. (I did sudo based on my experience on Win7 where I had to start R as administrator to update the common package library regardless of how I was logged in.) c) R then, install.packages( fImport, repo=|http://cran.r-project.org; ); If I executed via command R then it furnishes a listbox of mirrors, I select a mirror and then it hangs. If I executed via command sudo R then it furnishes a 24x80 style (not tk) list of choices, I select a mirror and then it hangs. I will appreciate your help in figuring this out. Ideally, I would ideally to run R CMD INSTALL or R CMD BATCH with a short command file so that I can configure a new machine via a script. I also tried to download the tarballs - but then figuring out all the dependencies gets very messy. Besides, that would be, like, reinventing the wheel! |-- -- Regards, Dinesh K. Somani -- /Improving Your Odds/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Split split plot analysis for unbalanced data using lmer
Hello, I'm attempting to analyze a split-split plot model, currently using lmer. My design is similar to that used in Casella's Experimental Designs Ozone Example (example 5.7, p 197), but I have been unable to find any coding help for split-split plot models through searches of several textbooks and the R list serve. I have three treatments of interest (A,B,C), each with 2 treatment levels. One level of treatment A (Y or N) was applied to 6 blocks (Block, labelled 1,2,3,4,5,6). Each block was subdivided into 4 subplots (Positon, labelled 1,2,3,4, so implicit nesting may be a problem), to which treatment B (Y or N) was randomly assigned (2 replicates of each level of treatment B per block). Each of these subplots was divided in half and randomly assigned treatment C (B or M). My response variable is the number of predetermined spots (out of 132) occupied by a certain species; I have a unbalanced data set since one observation was not recorded for a total of 47 counts. My data is set up as: Block A Position B C Count 1 1 Y4 N B64 2 1 Y4 N M 109 3 1 Y3 Y M88 4 1 Y3 Y B 0 5 1 Y1 Y B47 6 1 Y1 Y M91 with both position and block set to factors, not integers. My current model is: model=lmer(cbind(june$Count, 132-june$Count)~A*B*C+(1|A/B)+(1|Block),june, family=binomial) with the resulting summary model Generalized linear mixed model fit by the Laplace approximation Formula: cbind(june$Count, 132 - june$Count) ~ A * B * C + (1 | A/B) + (1 | Block) Data: june AIC BIC logLik deviance 806.9 827.2 -392.4784.9 Random effects: Groups NameVariance Std.Dev. Block (Intercept) 0.074222 0.27244 B:A(Intercept) 0.00 0.0 A (Intercept) 0.00 0.0 Number of obs: 47, groups: Block, 6; B:A, 4; A, 2 Fixed effects: Estimate Std. Error z value Pr(|z|) (Intercept) -0.094520.17295 -0.547 0.58470 AY 0.237560.24453 0.971 0.33130 BY -0.309490.10259 -3.017 0.00255 ** CM -0.574380.10434 -5.505 3.70e-08 *** AY:BY0.649310.14922 4.351 1.35e-05 *** AY:CM1.122200.14754 7.606 2.83e-14 *** BY:CM0.394170.14769 2.669 0.00761 ** AY:BY:CM-0.450400.21368 -2.108 0.03504 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Correlation of Fixed Effects: (Intr) AY BY CM AY:BY AY:CM BY:CM AY -0.707 BY -0.291 0.206 CM -0.286 0.202 0.483 AY:BY 0.200 -0.283 -0.688 -0.332 AY:CM 0.202 -0.286 -0.342 -0.707 0.469 BY:CM 0.202 -0.143 -0.695 -0.706 0.478 0.500 AY:BY:CM -0.140 0.197 0.480 0.488 -0.698 -0.691 -0.691 I'm not sure if I have specified the model correctly (should Position be included in the model specification as well?) or how to interpret the results. I think the group numbers for blocks and A are correct, but I'm not sure how B:A should be checked or if the implicit nesting of B is a problem. Do I need to include factor C in my nesting structure (even though its the lowest level)? My initial conclusions are that Treatments B and C have a main effect and higher order interactions suggest treatment A is also important. All my random effects also seem extremely small, which seems concerning. Any help (or pertinent references) for model setup or interpretation is greatly appreciated. Stephen Gosnell Department of Ecology, Evolution, and Marine Biology, UCSB __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sweave doesn't work
On Sun, Aug 21, 2011 at 09:18:25AM -0700, danielepippo wrote: Sweave(example.Rtex) in R it seems working [...] * ...le/Desktop/dati/LaTeX1.Rtex* Sounds like you are first running sweave on the file 'example.Rtex' and later LaTex on 'LaTeX1.Rtex' Two points: 1) why do these files have totally differnt names? If the Sweave file was called 'example.Rtex' I'd expect the corresponding LaTex file to be 'example.tex' 2) Why to both files have the same extension? Commonly, the Sweave files are called 'something.Snw' or 'something.Rnw' and the resulting LaTex Files would be 'something.tex' cu Philipp __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] pooled hazard model with aftreg and time-dependent variables
Dear R-users, I have two samples with individuals that are in more than one of the samples and individuals that are only in one sample. I have been trying to do a pooled hazard model, stacking one sample below the other, with aftreg and time-dependent covariates. The idea behind is to see aggregate effects of covariates, but need to control for ther effects of same individuals in both samples and the samples themselves. I created two variables for this purpose: a variable representing individuals, as a factor, and a variable for each sample, and included them in the model. I provide an example below. If I include the factor sample the analysis works. If I include both sample + individual like this: weibullaft-aftreg(Surv(sta,time,S) ~ TDC1 + TIC1 + sample + individual, dist=weibull, data.frame=Data, id=ID) , it appears an error with aftreg saying: Error in solve.default(fit$hessian) : Lapack routine dgesv: system is exactly singular With phreg the error is: fail in [dsytrf]; info = 14 1 - I am probably doing something wrong. Any suggestion to control for individuals too would be highly appreciated. 2 - Is including factor(sample) enough to control for the samples? Data S sta time TDC1 total_time TIC1 ID sample individual A 1 0 1 48.50 1 1 1 1 1 B 0 0 1 65.96 2 1 2 1 2 B 1 1 2 65.08 2 121 2 C 0 0 1 0.002 4 3 1 3 C 1 1 2 0.002 4 3 1 3 D 0 0 1 72.742 5 41 4 D 1 1 2 72.522 5 4 1 4 E 0 0 1 61.84 2 351 5 E 0 1 2 60.562 351 5 F 0 0 1 35.044 261 6 F 0 1 2 36.974 261 6 F 0 2 3 37.924 26 1 6 F 1 3 4 39.014 261 6 G 0 0 1 15.96 2 4 7 2 7 G 1 1 2 15.08 2 47 2 7 H 0 0 1 33.042 282 8 H 0 1 2 39.972 282 8 D 0 0 1 75-342 5 92 4 D 0 1 2 76.992 5 9 2 4 E 0 0 1 70.45 2 3102 5 E 1 1 2 45.012 3102 5 F 0 0 1 35.044 2112 6 F 1 1 2 36.974 2112 6 time - time to event sta - starting time TDC - time dependent covariates TIC - time independent covariate total_time - total time at risk ID - ID Individual - as.factor code for individual Sample - as.factor code for sample thanks, J -- View this message in context: http://r.789695.n4.nabble.com/pooled-hazard-model-with-aftreg-and-time-dependent-variables-tp3758805p3758805.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error message: object of type 'closure' is not subsettable
Thank you for your help, even though there was such an obvious mistake, Im
sorry for that
I have now tried to incorporate your suggested solution, but just as last
time (the other post that you referred to), I get the values of the initial
parameters when I run nlminb.
I have changed the code a bit, see below, due to some error messages.
Can anyone see what I am doing wrong?
Thank you in advance!
http://r.789695.n4.nabble.com/file/n3758834/S%26P_500_calls%2C_jan-jun_2010.csv
S%26P_500_calls%2C_jan-jun_2010.csv
setwd(F:/Data til speciale/)
## Calibration of Heston model parameters
marketdata - read.csv(file=SP 500 calls, jan-jun 2010.csv, header=TRUE,
sep=;)
BS_Call - function(S0, K, T, r, sigma, q)
{
sig - sigma * sqrt(T)
d1 - (log (S0/K) + (r - q + sigma^2/2) * T ) / sig
d2 - d1 - sig
Presentvalue - exp(-r*T)
return (S0 * exp(-q*T) * pnorm(d1) - K*Presentvalue*pnorm(d2))
}
#- Values --
Data imported
S0 - 1136.02
X - marketdata[1:460,9]
t - marketdata[1:460,17]/365 #Notice the T is measured in years now
implvol - marketdata[1:460,12]
k - log(X/(S0*exp(r-q)*t))
## Initial values
kappa - 0.0663227 # Lambda = -kappa
rho - -0.6678461
eta - 0.002124704
theta - 0.0001421415
v0 - 0.0001421415
q - 0.02145608
r - 0.01268737
#-
The price of a Call option (Eq. (5.6) of The Volatility Surface,
Gatheral)
# In terms of log moneyness
Price_call - function(phi, k, t)
{
integrand - function(u) {Re(exp(-1i*u*k)*phi(u - 1i/2, t)/(u^2 +
1/4))}
res - S0*exp(-q*t) -
exp(k/2)/pi*integrate(Vectorize(integrand),lower=0,upper=Inf,
subdivisions=460)$value
return(res)
}
# The characteric formula for the Heston model (Eq. XX)
phiHeston - function(kappa, rho, eta, theta, v0)
{
lambda - -kappa
function(u, t)
{
alpha - -u*u/2 - 1i*u/2
beta - lambda - rho*eta*1i*u
gamma - eta^2/2
d - sqrt(beta*beta - 4*alpha*gamma)
rplus - (beta + d)/(eta^2)
rminus - (beta - d)/(eta^2)
g - rminus / rplus
D - rminus * (1 - exp(-d*t))/ (1 - g*exp(-d*t))
C - lambda* (rminus * t - 2/eta^2 * log( (1 -
g*exp(-(d*t)))/(1 - g)) )
return(exp(C*theta + D*v0))
}
}
## Calculating the Heston model price with fourier
HestonCall-function(k,t)
{
res-Price_call(phiHeston(kappa, rho, eta, theta, v0),k,t)
return(res)
}
# Vectorizing the function to handle vectors of strikes and maturities
HestonCallVec - function(k,t)
{
mapply (HestonCall, k, t)
}
lb - c(0, -0.9, 0, 0, 0)
ub - c(100, 0.9, 0.5, 1, 1)
#BS_Call(S0, exp(k), t, r, implvol, q)
difference - function(k, t, S0, r, implvol, q, kappa, rho, eta, theta, v0)
{
return(HestonCallVec(k,t) - BS_Call(S0, exp(k), t, r, implvol, q))
}
difference(k,t,S0, r, implvol, q, kappa, rho, eta, theta, v0)
y - function(par) {kappahat-par[1]; rhohat-par[2]; etahat-par[3];
thetahat-par[4]; v0hat-par[5]; sum(difference(k, t, S0, r, implvol, q,
kappahat, rhohat, etahat, thetahat, v0hat)^2)}
nlminb(start=c(kappa, rho, eta, theta, v0), objective = y, lower =lb, upper
=ub)
--
View this message in context:
http://r.789695.n4.nabble.com/Error-message-object-of-type-closure-is-not-subsettable-tp3752886p3758834.html
Sent from the R help mailing list archive at Nabble.com.
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Increase the size of the boxes but not the text in a legend
HI there, I want to add a legend to a plot using the density and angle argument, so patterns with lines in different angles are used in the plot and should be referred to. When I use default settings, the filled boxes are too small. With the cex argument I can enlarge the whole legend, but then the text gets too big. How could I just increase the size of the single boxes and not the text. I tried: legend(topright, c(Trefferquote,Falschalarmquote), density=c(20,16), angle=c(45,0),cex=1.5, y.intersp=0.8) The sizes of the boxes are good now, but the text is too large and so also the whole legend box gets too big. Thank you very much for your help Jürgen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Time series plot shift
I have a plot and my x axis is Time in days, I need to shift it by +15 days, how do i do it? Thanks. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Time series plot shift
Depending on the time series object you are using, the lag() function will do it. Read the documentation ?lag for details on your specific ts class. Michael Weylandt On Aug 21, 2011, at 3:04 PM, Andrey A ava...@gmail.com wrote: I have a plot and my x axis is Time in days, I need to shift it by +15 days, how do i do it? Thanks. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Multiple R linear models into one Latex table
Dear community,
I had been looking for an easy way to produce latex tables from R
output. xtable() and the package apsrtable produce good outputs but they are not
exactly what I was looking for.
I wrote this code that generates regression tables from multiple R
linear models. I want to share it because it might be
useful for someone else, and because I would appreciate comments on how to
optimize the code. I'm also open to suggestions about design or packages to
include. Or pointing out bugs. I hope this is the right list to post.
Please consider it is still in a very primitive form, and the generated table
might look strange for some disciplines (I come from political science).
Thanks,
Alex
## Start script
#
R models to Latex table
#
# nice() - Multiple R linear models into one Latex table.
#
#nice(list(fit1,fit2,...,fitn)) for generic table with n models
#
# or
#
# for better tables, create objects 'model.names' and 'final.varnames', then
#
#nice(list(fit1,fit2,...,fitn), model.names, final.varnames)
nice - function(modelos, model.names=NULL, final.varnames=NULL) {
var.names-vector(mode=character)
k-1
for (i in 1:length(modelos)) {
l-length(names(coef(modelos[[i]])))
var.names[c(k:c(k+l-1))]-names(coef(modelos[[i]]))
k-k+l
}
var.names-unique(var.names)
if(is.null(final.varnames)) {
final.varnames-var.names
}
if(is.null(model.names)) {
model.names-paste(Model,1:length(modelos))
}
mat.all-matrix(data=NA, nrow=length(var.names)*2, byrow=FALSE,
ncol=length(model.names),
dimnames=list(c(rep(c(coef,sd),length=length(var.names)*2)),
c(model.names)))
dimnames(mat.all)[[1]][c(seq(1,dim(mat.all)[1], by=2))]-var.names;
dimnames(mat.all)[[1]][c(seq(2,dim(mat.all)[1], by=2))]-
for (j in 1:length(modelos)) {
mat.all[c(match(names(coef(modelos[[j]])),dimnames(mat.all)[[1]])),j]-coef(modelos[[j]])
mat.all[c(match(names(coef(modelos[[j]])),dimnames(mat.all)[[1]])+1),j]-sqrt(diag(vcov(modelos[[j]])))
}
mat.all-signif(mat.all, digits=3);
dimnames(mat.all)[[1]][c(seq(1,dim(mat.all)[1], by=2))]-final.varnames
cat(\\begin{table}[!hbt], \n,\\caption[Short for List of Tables]{Long
for this
title},\n,\\begin{center},\n,\\begin{tabular}{ll},\n)
cat(,paste(model.names[1:length(model.names)-1],),model.names[length(model.names)],,\n,\\hline,\n)
for(i in 1:dim(mat.all)[1]) {
if(i %in% c(seq(1,dim(mat.all)[1], by=2))) { # Odd rows (coefficients)
cat(dimnames(mat.all)[[1]][i],)
cat(
for (j in 1:(length(modelos)-1)) {
if(!is.na(mat.all[i,j])) {
p.val-abs(mat.all[i,j]/mat.all[i+1,j])
if(p.val1.644) {
if(1.645=p.val p.val1.96) cat(\\textbf{,round(mat.all[i,j],
digits=5),*}, sep=) else {
if(1.96p.val p.val2.576) cat(\\textbf{,round(mat.all[i,j],
digits=5),**}, sep=) else {
cat(\\textbf{,round(mat.all[i,j], digits=5),***}, sep=)
}
}
} else cat(round(mat.all[i,j], digits=5),)
} else cat()
})
cat(
for (j in length(modelos) ) {
if(!is.na(mat.all[i,j])) {
p.val-abs(mat.all[i,j]/mat.all[i+1,j])
if(p.val1.644) {
if(1.645=p.val p.val1.96) cat(\\textbf{,round(mat.all[i,j],
digits=5),*}, sep=) else {
if(1.96p.val p.val2.576) cat(\\textbf{,round(mat.all[i,j],
digits=5),**}, sep=) else {
cat(\\textbf{,round(mat.all[i,j], digits=5),***}, sep=)
}
}
} else cat(round(mat.all[i,j], digits=5))
}
})
cat(, \n )
} else {
cat(,ifelse(!is.na(mat.all[i,1:(length(dimnames(mat.all)[[2]])-1)]),paste((\\emph{,round(mat.all[i,1:(length(dimnames(mat.all)[[2]])-1)],
digits=5),}),sep=), paste()),
ifelse(!is.na(mat.all[i,length(dimnames(mat.all)[[2]])]),paste((\\emph{,round(mat.all[i,length(dimnames(mat.all)[[2]])],
digits=5),}),sep=),paste( )),
, sep=, fill=TRUE)}
}
cat(\\hline,\n,\\emph{n},)
cat(
for(j in 1:(length(modelos)-1)) {
cat(length(modelos[[j]]$residuals),,sep=)
},
for(j in length(modelos)) {
cat(length(modelos[[j]]$residuals), , \n, sep=)
}
)
cat(\\emph{Adj-R$^2$})
cat(
for(j in 1:(length(modelos)-1)) {
if (class(modelos[[j]])[1]==lm)
{cat(round(summary(modelos[[j]])$adj.r.squared,digits=2),, sep=)} else {
cat()
}
},
for(j in length(modelos)) {
if (class(modelos[[j]])[1]==lm)
{cat(round(summary(modelos[[j]])$adj.r.squared,digits=2),,\n)} else {
cat(,\n)
}
}
)
cat(\\emph{AIC})
cat(
for(j in 1:(length(modelos)-1)) {
if (class(modelos[[j]])[1]==glm)
{cat(round(summary(modelos[[j]])$aic, digits=0),, sep=)} else {
cat()
}
},
for(j in length(modelos)) {
if
[R] Changing data scales
I have data that ranges from 0.3 to 2 and I want to change the scale to be from 0 to 1. Can this be done in R? -- Thanks, Jim. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Changing data scales
On 8/21/11 8:04 PM, Jim Silverton wrote: I have data that ranges from 0.3 to 2 and I want to change the scale to be from 0 to 1. Can this be done in R? Take a look at ?scale HTH, Jorge __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Quadratic equation
these guys wont help you with your homework. But have you ever heard of Google?... if so, R has plenty of online manuals and cheat sheets. -- View this message in context: http://r.789695.n4.nabble.com/Quadratic-equation-tp3758790p3759239.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How can I divide an image and randomly plot the segements
On 21/08/11 23:06, Benson Kenduiywo wrote:
Hi! I would like to divide a 401 by 401 image into 20 equal blocks and then
re-plot the image with the segments randomly distributed in it.
How can I do this?
For a 401 x 401 pixellation --- I don't know. If you use a 400 x 400
pixellation,
then the class im in the spatstat package can be used via the attached
function shuffleIm(). The number of pixels in each dimension must be
divisible
by the number of blocks into which that dimension is subdivided. E.g.
if you
create your 20 blocks by making 4 divisions in the x-direction and 5
divisions in
the y-direction then the number of pixels in the x-direction must be
divisible by 4
and the number in the y-direction must be divisible by 5. Since 401 is
prime
it's kind of hard to do any kind of subdividing with a 401 x 401 image.
Note that if X is an object of class im with a 401 x 401 pixellation,
you can
convert it to one with a 400 x 400 pixellation via
X - as.im(X,dimyx=400)
This makes the pixel values of the new image equal to the values of the old
image at the closest pixel centre. This shouldn't (???) cause too much
distortion.
Said he, optimistically. (In the one experiment that I did, the images were
visually indistinguishable.)
If you have a 400 x 400 pixel image, say X, of class im then
Y - shuffleIm(X,nx=4,ny=5)
gives you a new image in which the matrix of pixel values has been
subdivided into
a 5 x 4 array of 80 x 100 submatrices and these submatrices randomly
shuffled.
(Note that the rows of the matrix of pixel values in an object of class
im correspond
to the ***y*** direction and the columns correspond to the ***x***
direction.
See fortune(conventions) for a comment. :-) )
Hope this does what you want.
cheers,
Rolf Turner
shuffleIm - function (X,nx,ny) {
verifyclass(X,im)
ddd - dim(X)
if(ddd[1]%%ny != 0) stop(Argument \ny\ must divide first dimension of
image.\n)
if(ddd[2]%%nx != 0) stop(Argument \nx\ must divide second dimension of
image.\n)
ky - ddd[1]/ny
kx - ddd[2]/nx
n - nx*ny
ind - sample(1:n,n)
V - matrix(NA,ddd[1],ddd[2])
for(k in 1:n) {
j - k%%nx
if(j==0) j - nx
i - (k-j)/nx + 1
j1 - (j-1)*kx + 1:kx
i1 - (i-1)*ky + 1:ky
kr - ind[k]
j - kr%%nx
if(j==0) j - nx
i - (kr-j)/nx + 1
j2 - (j-1)*kx + 1:kx
i2 - (i-1)*ky + 1:ky
V[i1,j1] - X$v[i2,j2]
}
X$v - V
X}
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Re: [R] Changing data scales
On Sun, Aug 21, 2011 at 5:50 PM, Jorge I Velez jorgeivanve...@gmail.com wrote: On 8/21/11 8:04 PM, Jim Silverton wrote: I have data that ranges from 0.3 to 2 and I want to change the scale to be from 0 to 1. Can this be done in R? Take a look at ?scale Which leads one to: scale(x, .3, 1.7) but this is far from the only possible way. Another possibility would be: f - function(x) ifelse(x 0, (1/(2 - x)), (1 - 1/(2 + x))) finv - function(y) ifelse((0 y) (y .5), (2 - (1/y)), (1/(1-y) - 2)) ## use function to move to the (0, 1) set f(seq(.3, 2, .1)) ## show the inverse works finv(f(seq(.3, 2, .1))) ## Graph the two functions curve(f, from = -100, to = 100, n = 1000) curve(finv, from = .001, to = .999, n = 1000) although this technically puts you into the (0, 1) set rather than [0, 1]. For more information on this general sort of thing, search for bijection. By the way, I believe I owe credit for those functions to one of my math texts at some point, but I cannot remember the reference (or perhaps my younger self was smarter than I recall (or my memory has gotten worse than I think...)). Josh HTH, Jorge __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] CovMMest
Dear experts, I'm having problem in using CovMMest (for MM estimates of location and scale) I have installed robustbase, robust. Unfortunately I could not run the command. It says CovMMest is not a function. What is wrong? Thx. -Mag [[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] CovMMest
On Aug 21, 2011, at 9:56 PM, mazlina Abu Bakar wrote: Dear experts, I'm having problem in using CovMMest (for MM estimates of location and scale) I have installed robustbase, robust. Unfortunately I could not run the command. It says CovMMest is not a function. What is wrong? Thx. Who can tell? No code. In particular we cannot tell if you ever loaded (whatever) package has that function. -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Wiki/revision control to management of CRAN package repository
I propose the following humbly, with little know how as to how to implement, and realize it may have been proposed many times. It is just something I had on my mind. Would it be possible/desirable to have the whole CRAN package repository accessible through a public wiki, forge or version control interface (ideally a fusion of the wiki and forge approach)? It appears it would be a first for a software repository. CRAN package repository is becoming a jungle of R code and may do well with currating and editorial effort. This can not/should not be the task of a single person or small group of people. Using a crowd sourced method by implementing a wiki approach to the CRAN package repository would allow for the rapid editing, sorting and improvement of this impressive and precious resource, while also improving the accessibility, visibility and quality of individual packages. It would also bind the For example, such an interface would allow the cleaning up of the repository through the use of tagging of packages, using similar approaches to the wikipedia project (http://en.wikipedia.org/wiki/Wikipedia:Template_messages#Cleanup). Such a tagging approach could be used within existing vcs, if the repository was migrated/mirrored within one of these systems. Packages could be marked using tags for all following actions prior to implementing the action. Actions could be undertaken directly by package users after a delay or discussion. Packages management/editorial effort: -Merging/ -Combining packages that have: -Large overlap in functionality -Are largely interdependant -Are only minor extensions of another package -… -Split/fork -Subdividing behemoth packages into smaller packages with more specific tasks -Categorization -Packages could be sorted by use, improvement of Task View -Tags, keywords could be added to packages for searching -Packages could be placed in a hierarchy, not only by true dependance and reverse dependance, but also by logical dependance/reverse dependance -ie. which package should probably be used with which package, an improvement on the see also help section -Deletion -Marking/tagging -a stub/prototype -broken Package improvements -Improving help files -Adding functions -Adding examples -Requiring, improving or adding references -References to the theory or approach used... -A section could include a list of articles making use of the package, with package users encourage to enter this information -This would allow package author recognition and allow a package impact factor -Adding key words for indexing and searching -Function improvement -Adding compatibility with other packages/formats (including when merging packages) -Speed improvements Discussion -On package improvements, management steps directly attached to the package -Help discussion These actions would be reversible, possibly with veto power from the author of the package. Links: http://www.rforge.net/ http://sourceforge.net/ http://channel9.msdn.com/Forums/Coffeehouse/174561-Coding-Wiki http://code.google.com/p/mcover/ http://www.tigris.org/ This is just an idea I had on my mind. Thank you __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
