Re: [R] help for code in jump diffusion
i tried to run your code, this is how/why you got NaN:
mu-0.2
sig-0.2
S0-100
j-0.2
dt-1/252
int-0.1
i-0
is.nan
function (x) .Primitive(is.nan)
k-rnorm(1,0,1)
k
[1] 0.3214954
theta-ifelse((k(int*dt)),1,0)
theta
[1] 0
m-rnorm(1)
m
[1] -0.4525731
gam-qnorm(m,0,1)
Warning message:
In qnorm(p, mean, sd, lower.tail, log.p) : NaNs produced
gam
[1] NaN
S0-abs(S0*((1+mu*dt+sig*sqrt(dt)+ gam)- j*theta))
S0
[1] NaN
if(!is.nan (S0 = 0))
+ warning(S0 must be positive)
Warning message:
S0 must be positive
cat(NaN,\n)
NaN
cat(S0,\n)
NaN
On Sat, Nov 22, 2008 at 2:37 PM, subbudas [EMAIL PROTECTED] wrote:
hello everyone ,
i have written some code in R for jump diffusion model.
the code generates answer as
NaN
There were 50 or more warnings (use warnings() to see the first 50)
my code is
mu-0.2
sig-0.2
S0-100
j-0.2
dt-1/252
int-0.1
i-0
while(i=1)
{
is.nan
k-rnorm(1,0,1)
theta-ifelse((k(int*dt)),1,0)
m-rnorm(1)
gam-qnorm(m,0,1)
S0-abs(S0*((1+mu*dt+sig*sqrt(dt)+ gam)- j*theta))
if(!is.nan (S0 = 0))
warning(S0 must be positive)
cat(NaN,\n)
cat(S0,\n)
i-i+(1/252)
}
the problem i am facing is i am not able to find out the reason for this NaN
output.
please help
thanks in advance.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] help for code in jump diffusion
On Fri, 2008-11-21 at 22:37 -0800, subbudas wrote:
hello everyone ,
i have written some code in R for jump diffusion model.
the code generates answer as
NaN
There were 50 or more warnings (use warnings() to see the first 50)
my code is
mu-0.2
sig-0.2
S0-100
j-0.2
dt-1/252
int-0.1
i-0
while(i=1)
{
is.nan
k-rnorm(1,0,1)
theta-ifelse((k(int*dt)),1,0)
m-rnorm(1)
gam-qnorm(m,0,1)
S0-abs(S0*((1+mu*dt+sig*sqrt(dt)+ gam)- j*theta))
if(!is.nan (S0 = 0))
warning(S0 must be positive)
cat(NaN,\n)
cat(S0,\n)
i-i+(1/252)
}
the problem i am facing is i am not able to find out the reason for this NaN
output.
please help
thanks in advance.
Hi
I see two problems in your script:
1- m-rnorm(1) will produce a random number with distribution normal
mean =0 and sd=1 so m can 1 or 0
In this cases gam-qnorm(m,0,1) is NAN because m is not a probability
2- I think if(!is.nan (S0 = 0)) ... is wrong
Try this script :
mu-0.2
sig-0.2
S0-100
j-0.2
dt-1/252
int-0.1
i-0
while(i=1){
k-rnorm(1,0,1)
theta-ifelse((k(int*dt)),1,0)
# m-rnorm(1)
m-runif(1,0,1)
gam-qnorm(m,0,1)
S0-abs(S0*((1+mu*dt+sig*sqrt(dt)+ gam)- j*theta))
if(!is.nan(S0)(S0 = 0)){
warning(S0 must be positive)
cat(NaN,\n)
}
cat(S0,\n)
i-i+(1/252)
}
--
Bernardo Rangel Tura, M.D,MPH,Ph.D
National Institute of Cardiology
Brazil
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Hi, I am residing in Taipei, Taiwan. My windows has English R (2.7.2) installed. I noticed that the CRAN mirror for Taipei is Taiwan (Taipeh), Is Taipeh a Deutsch or English name? Thanks, cruz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] summary statistics into table/data base, many factors to analyse
On Fri, Nov 21, 2008 at 5:50 AM, Gerit Offermann [EMAIL PROTECTED] wrote: Dear list, thanks to your help I managed to find means of analysing my data. However, the whole data set contains 264 variables. Of which some are factors, others are not. The factors tend to be grouped, e.g. data$f1304 to data$f1484 and data$f3204 to data$5408. But there are other types of variables in the data set as well, e.g. data$f1504. Not every spot is taken, i.e data$f1345 to data$1399 might not exist in the data set. We can compute on the names like this (using the builtin anscombe data set to get just columns y1, x1, x2, x3, x4). Try this: # display anscombe data set anscombe # names.x are names that start with x names.x - grep(^x, names(anscombe), value = TRUE) anscombe[, c(y1, names.x)] The solution summaryBy works for cross analysis, of which there is a handful. So I am not worried here. The solution from Jorge is fine. However, I am trying to get my head around how to efficiently reduce my data set to the dependet variable and the factors such that the solution is applicable. Having to type each variable into my.reduced.data - cbind(my.data$f1001, my.data$1002, my.data$1003... is an obvious option, but does not seem to be the most efficient one. Are there better ways to go about? Thanks, Gerit -- Sensationsangebot nur bis 30.11: GMX FreeDSL - Telefonanschluss + DSL für nur 16,37 Euro/mtl.!* http://dsl.gmx.de/?ac=OM.AD.PD003K11308T4569a __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error with lapply [addirional clarification needed]
I need one more clarification here : Here I did : fn - function(i) return(list(i, i^2)) ss = sapply(1:4, fn) Here the object ss should be a matrix object : is.matrix(ss) However I feel it lacks some the matrix object properties. For example the syntax min(ss[1,]) generates an error : Error in min(ss[1, ]) : invalid 'type' (list) of argument. What should be the way out? Am I missing something ? Regards, baptiste auguie-2 wrote: Hi, you are feeding lapply i as an optional argument, which is passed to fn() and causes an error. Just use lapply(1:4, fn), or better yet, sapply, fn - function(i) return(i^2) sapply(1:4, fn) [1] 1 4 9 16 Hope this helps, baptiste On 20 Nov 2008, at 16:31, megh wrote: I have written following codes, with intention to get a list with values 1,2,9,16 : fn - function(i) return(i^2) lapply(1:4, fn, i) However I got following error : Error in FUN(1:4[[1L]], ...) : unused argument(s) (1) Can anyone please tell me what will be the correct code here? Regards, -- View this message in context: http://www.nabble.com/Error-with-lapply-tp20605066p20605066.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. _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Error-with-lapply-tp20605066p20634821.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 with lapply [addirional clarification needed]
megh wrote: I need one more clarification here : Here I did : fn - function(i) return(list(i, i^2)) ss = sapply(1:4, fn) Here the object ss should be a matrix object : is.matrix(ss) However I feel it lacks some the matrix object properties. For example the syntax min(ss[1,]) generates an error : Error in min(ss[1, ]) : invalid 'type' (list) of argument. What should be the way out? Am I missing something ? ss is a dim'ed list object, because fn returns a list. This appears to be a feature, although rarely used. One point is that you can do fn - function(i) return(list(as.character(i), i^2)) ss - sapply(1:4, fn) ss [,1] [,2] [,3] [,4] [1,] 1 2 3 4 [2,] 14916 ss[,1] [[1]] [1] 1 [[2]] [1] 1 I.e., the rows can be of different mode. The easiest way out is just not to do that, i.e. return a vectore c(i, i^2) instead. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error with lapply [addirional clarification needed]
Try str(ss) to see what it really looks like. You probably want: fn - function(i) c(i, i^2) On Sat, Nov 22, 2008 at 4:54 AM, megh [EMAIL PROTECTED] wrote: I need one more clarification here : Here I did : fn - function(i) return(list(i, i^2)) ss = sapply(1:4, fn) Here the object ss should be a matrix object : is.matrix(ss) However I feel it lacks some the matrix object properties. For example the syntax min(ss[1,]) generates an error : Error in min(ss[1, ]) : invalid 'type' (list) of argument. What should be the way out? Am I missing something ? Regards, baptiste auguie-2 wrote: Hi, you are feeding lapply i as an optional argument, which is passed to fn() and causes an error. Just use lapply(1:4, fn), or better yet, sapply, fn - function(i) return(i^2) sapply(1:4, fn) [1] 1 4 9 16 Hope this helps, baptiste On 20 Nov 2008, at 16:31, megh wrote: I have written following codes, with intention to get a list with values 1,2,9,16 : fn - function(i) return(i^2) lapply(1:4, fn, i) However I got following error : Error in FUN(1:4[[1L]], ...) : unused argument(s) (1) Can anyone please tell me what will be the correct code here? Regards, -- View this message in context: http://www.nabble.com/Error-with-lapply-tp20605066p20605066.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. _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Error-with-lapply-tp20605066p20634821.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.
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I want to draw following plot, given here http://www.2shared.com/file/4327128/830b82c4/pic.html for the following data : dat - cbind(rnorm(100), sample(c(1:4), 1000, T)) colnames(dat) - c(data,level) Here x-axis should be on data and y-axis is for level and z-axis should be to display the frequency of data for a particular level Is there any R code(s) to doing that? -- View this message in context: http://www.nabble.com/How-to-draw-following-plot-in-R--tp20635099p20635099.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.
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I have question: how can I put the value on the bar chart. This my code: barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber) This's my data: Year Hr Family Tuberize 1 2007 20 A 0.26 2 2007 20 B 6.08 3 2007 20 C 0.00 4 2007 20 D 0.27 5 2008 20 A 1.18 6 2008 20 B 9.17 7 2008 20 C 0.00 8 2008 20 D 2.13 9 2007 14 A 35.83 10 2007 14 B 38.98 11 2007 14 C 7.90 12 2007 14 D 18.75 13 2008 14 A 43.70 14 2008 14 B 51.84 15 2008 14 C 27.11 16 2008 14 D 24.11 17 2008 11 A 85.67 18 2008 11 B 93.73 19 2008 11 C 81.45 20 2008 11 D 90.25 21 2007 8 A 63.04 22 2007 8 B 66.29 23 2007 8 C 70.85 24 2007 8 D 83.48 25 2008 8 A 83.33 26 2008 8 B 92.31 27 2008 8 C 82.21 28 2008 8 D 94.03 Thank you Piya [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
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[R] Need some help in R programming code
Dear R guru,
I am Saikat Sarkar working as a researcher of Economics in Tampere
University, Finland. I am trying to estimate some Garch related tests with
Bayesian analysis by R programme.
I am not good in R but trying to survive.
Anyway I have the coding but not working properly. I have tried to find the
problem but failed. I am writing to all R gurus to help me out.
Could you please look at the problem and help me if you can.
Thanking you
saikat
Could you please look at the problem below:
R message is like below:
---
Error in garch.gjr.d0(param, y, X, Z, iterate = FALSE)$res :
$ operator is invalid for atomic vectors
plot(Ch.mic,Ch.mic.rep,xlim=c(0,6),ylim=c(0,6))
---
Coding
res-garch.gjr.d0(a$estimate,y,X,X,iterate=FALSE)$res
sig2-garch.gjr.d0(a$estimate,y,X,X,iterate=FALSE)$sig2
test.mic - function(param,y,X,Z)
{
n - length(y)
res - garch.gjr.d0(param,y,X,Z,iterate=FALSE)$res
s2 - garch.gjr.d0(param,y,X,Z,iterate=FALSE)$sig2
s-sqrt(s2)
r - res
u - r[-1]
r2- (r[-n]^2*s2[-n])/s[-1]
r3- (r[-n]^3*s2[-n]*s[-n])/s[-1]
U - cbind(r[-n],r2,r3)
U - cbind(1,U)
U - U/matrix(s[-1],nr=nrow(U),nc=ncol(U))
Fstat-summary(lm(u~U-1))$fstatistic[1]
Fstat
}
sim.gjr - function(par,y,X=0,Z=0)
{
n - length(y)
e-numeric(n)
yrep-numeric(n)
yrep[1] - y[1]
n.dummies.mean - dim(X)[2]
if(is.matrix(X)) e[1] - e[1]-sum(X[1,]*par[9:(8+n.dummies.mean)])
delta - par[-(1:(8+n.dummies.mean))]
s2 - var(y)
for(i in 2:n){
s2 -
par[5]+par[6]*e[i-1]^2+par[7]*ifelse(e[i-1]0,1,0)*e[i-1]^2+par[8]*s2+sum(X[i,]*delta)
e[i] - rnorm(1,0,sqrt(s2))
yrep[i]-par[1]+par[2]*yrep[i-1]+par[3]*s2+par[4]*yrep[i-1]*s2+e[i]
}
if(is.matrix(X)) yrep - yrep + X%*%par[9:(8+n.dummies.mean)]
yrep
}
yrep - sim.gjr(a$estimate,y,X,Z)
n.test - 500
Ch.mic - Ch.mic.rep - numeric(n.test)
for(i in 1:n.test){
simulation - sample(1:1,1)
par - sims.matrix[simulation,]
Ch.mic[i] - test.mic(par,y,X,X)
yrep - sim.gjr(par,y,X,X)
Ch.mic.rep[i] - test.mic(par,yrep,X,X)
}
plot(Ch.mic,Ch.mic.rep,xlim=c(0,6),ylim=c(0,6))
lines(c(0,6),c(0,6))
--
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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] lsoda warning too much accuracy requested
A short addendum, resulting from an off-list discussion:
The reason why Colleen's code failed was raising a negative base to a
fractional exponent in the third state equation for certain sets of
parameters, esp. fractional values of beta.
Old versions of odesolve broke down, and recent versions of deSolve (or
the deprecated odesolve) simply returned NaN for the third state. In
order to track such problems down, it is helpful to call the system
separately, e.g.:
model(0, start, parms)
[[1]]
HBA N
0.3702103 1.2229436 NaN
and then isolate the problem in the respective state equation.
Thomas P.
Thomas Petzoldt wrote:
Hi Colleen,
this error was not uncommon and is usually a sign of a numerically
problematic or wrongly implemented model. Please use package deSolve,
the successor of odesolve, that is more robust and has also a bunch
alternative solvers for difficult cases.
I tested your code with deSolve (on R 2.8.0, Windows) and it runs
without problems.
Thomas Petzoldt
BTW: your system worked also with odesolve, so my question: which
versions (R, odesolve) and operating system are you using?
Colleen Carlson wrote:
Dear list -
Does anyone have any ideas / comments about why I am receiving the
following
warning when I run lsoda:
1: lsoda-- at t (=r1), too much accuracy requested in: lsoda(start,
times,
model, parms)
2: for precision of machine.. see tolsf (=r2) in: lsoda(start,
times,
model, parms)
I have tried changing both rtol and atol but without success. I saw the
thread in the R-archive of 11 June 2004 but this has not helped me.
I have built the model in stages and the problem only occurs when the
exponent beta in the third DE is anything other than 0 or 1. If beta
= 0 or
1 then the solver gives me perfectly justifiable results. Just changing
beta to 0.9 or similar causes the problem.
I am still new to R so I am unsure if it is my programming or my
understanding of the way lsoda works.
Any comments or input would be welcome.
Many thanks
Colleen
___
My code is:
library(odesolve)
SI - 80
model - function(t, x, parms) {
H - x[1]
BA - x[2]
N - x[3]
with(as.list(parms), {
dHdt - (b/c)*(((a**c)*((H)**(1-c))-H))
dBAdt - -(BA*b)*(c0+(c1*SI)-log(BA))/(log(1-((H/a)**c)))
dNdt - N*alpha*(((log(1-((H/a)**c)))/b)**beta) - (gamma*BA)
list(c(dHdt, dBAdt, dNdt))
})
}
times - seq(0, 40, 1)
parms - c(a=(SI*1.258621)-1.32759, b=0.1, c=0.4, c0=4.6012, c1=0.013597,
alpha=0.0005, beta=0.5, gamma=0.01)
start - c(H=0.1, BA=0.1, N=600)
out - as.data.frame(lsoda(start, times, model, parms))
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] [Rd] Re Bessel functions of complex argument
I thought I'd share a few workaround routes I've considered (my attempt at using Amos' Fortran routines failed miserably -- if anyone is interested i can explain what I tried), - Ryacas seems to provide a very simple way to evaluate bessel functions with complex argument, yacas(N(BesselJ(2, 1+I))) Complex(0.415798869439e-1,0.247397641513); Not sure yet how fast this is, and how to get data in/out in R format but it looks promising. - Similarly, I get the idea that Sage could provide an interface to Maxima or Pari and return the desired value. There doesn't seem to be much of a documentation centred on the use of Sage with R, though. - Octave has implemented the Amos code (so has Matlab). This should make it possible to call Octave and get the result back in R (Roctave package I believe), or better yet, get inspiration to properly use the underlying Fortran code. Baptiste On 9 Nov 2008, at 12:22, baptiste auguie wrote: Dear all, I'm writing a code that requires Bessel functions with complex argument. Searching the list, I found the continuation of a thread I initiated a few months ago: http://tolstoy.newcastle.edu.au/R/e4/devel/08/03/0746.html As I understand, the most promising option would be to use the fortran or C implementation of Amos, http://portal.acm.org/citation.cfm?id=214331dl=GUIDEcoll=GUIDECFID=9717168CFTOKEN=75957024 Sadly, my limited programming skills don't include any knowledge of fortran, and I have never managed to link external code to R functions (i'm missing a simple example based tutorial, R-ext overwhelms me with information). A few questions for anyone interested: - Does the aforementioned fortran code lends itself to an easy interfacing with R? (in other words, do I stand a chance in less than 10 years?) - Is anyone willing to help me with this? The package is hosted on R- forge and i could take care of the documentation, etc. - Is there a package I've missed that already implements this? (one can dream) Many thanks, baptiste _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ [[alternative HTML version deleted]] ATT1.txt _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 draw following plot in R?
2008/11/22 RON70 [EMAIL PROTECTED]: I want to draw following plot, given here http://www.2shared.com/file/4327128/830b82c4/pic.html for the following data : dat - cbind(rnorm(100), sample(c(1:4), 1000, T)) colnames(dat) - c(data,level) Here x-axis should be on data and y-axis is for level and z-axis should be to display the frequency of data for a particular level Is there any R code(s) to doing that? I'm pretty sure there's no code for doing it in character graphics like in your image link! Something similar can be found in the R graph gallery: http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=116 Or if you want an interactive spinny 3d-histogram, get the rgl package and look at demo(hist3d). See: http://rgl.neoscientists.org/gallery.shtml Those links should get you started. If you really do want ascii character graphics...well, I think you're on your own Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Bug in Kendall for n4?
SM == Stavros Macrakis [EMAIL PROTECTED] on Fri, 21 Nov 2008 14:44:37 -0500 writes: library(Kendall) Kendall(1:3,1:3) SM WARNING: Error exit, tauk2. IFAULT = 12 tau = 1, SM 2-sided pvalue =1 SM I believe Kendall tau is well-defined for this case and SM the reported value is correct; isn't it a bug to give a SM warning? (And if, e.g., the pvalue is not well-defined SM in this case, wouldn't it be better to return NA or NaN SM or something?) Also, shouldn't the error code be given SM in plain English -- or at least the meaning of IFAULT = SM 12 documented on the help page? The real question is *WHY* there needs to be a separate package 'Kendall' when R itself does everything you want and does not show any problems ? cor() cov() cor.test() all have a method = kendall and seem to work alright, even for n=2 cor(1:3, c(3,1,2), method=kendall) [1] -0.333 cov(1:3, c(3,1,2), method=kendall) [1] -2 cor(1:3, 1:3, method=kendall) [1] 1 cor.test(1:3, 1:3, method=kendall) Kendall's rank correlation tau data: 1:3 and 1:3 T = 3, p-value = 0. alternative hypothesis: true tau is not equal to 0 sample estimates: tau 1 Questions about the 'Kendall' package should typically first go to its author ... But those on cor(), cor.test() etc do belong here. Best regards, Martin Maechler, ETH Zurich SM A somewhat less clear case is Kendall(1:2,1:2), which SM gives the same error. Though the usual formula for SM Kendall tau has a zero in the denominator in this case, SM I'd think the correct generalization is 1 if the two SM elements are in the same order, and -1 if they are not SM (the only possibilities). But perhaps I don't fully SM understand the interpretation of this statistic. SM -s SM __ SM R-help@r-project.org mailing list SM https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do SM read the posting guide SM http://www.R-project.org/posting-guide.html and provide SM 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] Need some help in R programming code
You need to also provide the data that your code is using since the
error message indicates that the problem is probably in the way that
the object 'a' is defined and there is no indication of what it looks
like. You should either provide the output of str(a), or the output
of 'dput(a)' so we have an idea of what it is. The error message
seems to indicate the the $ is not legal for 'a' as it is currently
defined.
On Sat, Nov 22, 2008 at 4:55 AM, saikat sarkar [EMAIL PROTECTED] wrote:
Dear R guru,
I am Saikat Sarkar working as a researcher of Economics in Tampere
University, Finland. I am trying to estimate some Garch related tests with
Bayesian analysis by R programme.
I am not good in R but trying to survive.
Anyway I have the coding but not working properly. I have tried to find the
problem but failed. I am writing to all R gurus to help me out.
Could you please look at the problem and help me if you can.
Thanking you
saikat
Could you please look at the problem below:
R message is like below:
---
Error in garch.gjr.d0(param, y, X, Z, iterate = FALSE)$res :
$ operator is invalid for atomic vectors
plot(Ch.mic,Ch.mic.rep,xlim=c(0,6),ylim=c(0,6))
---
Coding
res-garch.gjr.d0(a$estimate,y,X,X,iterate=FALSE)$res
sig2-garch.gjr.d0(a$estimate,y,X,X,iterate=FALSE)$sig2
test.mic - function(param,y,X,Z)
{
n - length(y)
res - garch.gjr.d0(param,y,X,Z,iterate=FALSE)$res
s2 - garch.gjr.d0(param,y,X,Z,iterate=FALSE)$sig2
s-sqrt(s2)
r - res
u - r[-1]
r2- (r[-n]^2*s2[-n])/s[-1]
r3- (r[-n]^3*s2[-n]*s[-n])/s[-1]
U - cbind(r[-n],r2,r3)
U - cbind(1,U)
U - U/matrix(s[-1],nr=nrow(U),nc=ncol(U))
Fstat-summary(lm(u~U-1))$fstatistic[1]
Fstat
}
sim.gjr - function(par,y,X=0,Z=0)
{
n - length(y)
e-numeric(n)
yrep-numeric(n)
yrep[1] - y[1]
n.dummies.mean - dim(X)[2]
if(is.matrix(X)) e[1] - e[1]-sum(X[1,]*par[9:(8+n.dummies.mean)])
delta - par[-(1:(8+n.dummies.mean))]
s2 - var(y)
for(i in 2:n){
s2 -
par[5]+par[6]*e[i-1]^2+par[7]*ifelse(e[i-1]0,1,0)*e[i-1]^2+par[8]*s2+sum(X[i,]*delta)
e[i] - rnorm(1,0,sqrt(s2))
yrep[i]-par[1]+par[2]*yrep[i-1]+par[3]*s2+par[4]*yrep[i-1]*s2+e[i]
}
if(is.matrix(X)) yrep - yrep + X%*%par[9:(8+n.dummies.mean)]
yrep
}
yrep - sim.gjr(a$estimate,y,X,Z)
n.test - 500
Ch.mic - Ch.mic.rep - numeric(n.test)
for(i in 1:n.test){
simulation - sample(1:1,1)
par - sims.matrix[simulation,]
Ch.mic[i] - test.mic(par,y,X,X)
yrep - sim.gjr(par,y,X,X)
Ch.mic.rep[i] - test.mic(par,yrep,X,X)
}
plot(Ch.mic,Ch.mic.rep,xlim=c(0,6),ylim=c(0,6))
lines(c(0,6),c(0,6))
--
View this message in context:
http://www.nabble.com/Need-some-help-in-R-programming-code-tp20634187p20634187.html
Sent from the R help mailing list archive at Nabble.com.
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+1 513 646 9390
What is the problem that you are trying to solve?
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[R] Nested Clade Analysis
Hi Wondering if anyone knows of a package that does Nested Clade Analysis? Thanks -- View this message in context: http://www.nabble.com/Nested-Clade-Analysis-tp20637180p20637180.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] Basic question on concatenating factors
Here is a function I have used. I got it from the list, and
unfortunately don't remember who to credit:
c.Factor -
function (x, y)
{
newlevels = union(levels(x), levels(y))
m = match(levels(y), newlevels)
ans = c(unclass(x), m[unclass(y)])
levels(ans) = newlevels
class(ans) = factor
ans
}
On Fri, Nov 21, 2008 at 12:29 PM, Alain Guillet
[EMAIL PROTECTED] wrote:
Hi,
I have a solution to concatenate two factors in one but I don't believe
it is the best one: factor(c(as.character(f1),as.character(f2)))
[1] a a b b b a
Levels: a b
You can always add a level by assigning a new vector at the level vector:
levels(f1) - c(a,b,c)
f1
[1] a a b
Levels: a b c
udi cohen wrote:
Hi all,
I hope it's not too trivial for the list - I'm trying to concatenate
two factor arrays, and obtain the following:
f1-factor(c(a,a,b))
f1
[1] a a b
Levels: a b
f2-factor(c(b,b,a))
f2
[1] b b a
Levels: a b
c(f1,f2)
[1] 1 1 2 2 2 1
Instead of getting:
[1] a a b b b a
Levels: a b
a related question is: how do I add a level which does not exists yet
in a factored vector, so I'll be able to add later these values,
without getting:
In `[-.factor`(`*tmp*`, 2, value = c) :
invalid factor level, NAs generated
Thanks,
EC
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--
Alain Guillet
Statistician and Computer Scientist
SMCS - Institut de statistique - Université catholique de Louvain
Bureau d.126
Voie du Roman Pays, 20
B-1348 Louvain-la-Neuve
Belgium
tel: +32 10 47 30 50
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--
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What is the problem that you are trying to solve?
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Re: [R] rgl lighting question
On 21/11/2008 2:30 PM, Rajarshi Guha wrote: Hi, I'm using rgl to generate a 3D surface plot and I'm struggling to get the lighting correct. Currently the surface gets plotted, but is very 'shiny'. On rotating the view, I get to see parts of the surface - but overall I don't see much detail because of the spotlight like lighting. I've played around with the specular, ambient and diffuse but I can't bring out the details of the surface. Could anybody point me to some examples of how to make a plain matte surface, which isn't obscured by specular reflections? This gives the regular shiny surface: library(rgl) example(surface3d) This gives one with no specular reflections, because the material doesn't do that: open3d() surface3d(x, y, z, color=col, back=lines, specular=black) And here's another way to get no specular reflections. This time there's no light to reflect that way: open3d() rgl.pop(lights) light3d(specular=black) surface3d(x, y, z, color=col, back=lines) I suspect you missed the rgl.pop() call. If you just call light3d or rgl.light() you'll add an additional light, you don't change the existing one. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] different .Rprofile files for different projects
On Thu, 2008-11-20 at 19:43 -0800, Andrew J. Rominger wrote: Dear list, First off, let me offer my apologies, I know this is a very basic question. After amassing a large number of objects (from multiple projects) in one working directory, I'd like to be able to start using different directories, just for the sake of organization alone. But I have no idea how to do this. I am using a mac, running R 2.5. I'm not very familiar with the MacOS X version of R, but students of mine were using it recently and under the Misc menu there is an entry to change the working directory, which provides a standard Mac directory selection dialogue box IIRC. Alternatively, start R as per usual and then change into the correct directory using the setwd() function. You should probably update your R - 2.5.0 is 18 months old now, 2.8.0 is current. G Searching the FAQ online I find: 12.1 How can I have a per session .Rprofile? You can by writing a .Rprofile file in your favorite session directory... So I think my specific question is how do I write a .Rprofile? I know it should be obvious, but is there a command to be called in the R console during a given session, or do I write a stub file outside of R (in a session directory?), open this using Preferences and then modify it in an R session? If the latter, how do I go about writing a stub and making R recognize it as a working directory, and what is a session directory/how can I find where the default session directory is located? Again, my apologies for being naive, and thanks very much for any help-- Andy Rominger __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Basic question on concatenating factors
You can try this also: unlist(list(f1, f2)) On Fri, Nov 21, 2008 at 3:15 PM, udi cohen [EMAIL PROTECTED] wrote: Hi all, I hope it's not too trivial for the list - I'm trying to concatenate two factor arrays, and obtain the following: f1-factor(c(a,a,b)) f1 [1] a a b Levels: a b f2-factor(c(b,b,a)) f2 [1] b b a Levels: a b c(f1,f2) [1] 1 1 2 2 2 1 Instead of getting: [1] a a b b b a Levels: a b a related question is: how do I add a level which does not exists yet in a factored vector, so I'll be able to add later these values, without getting: In `[-.factor`(`*tmp*`, 2, value = c) : invalid factor level, NAs generated Thanks, EC __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] declaring constants in an Sweave / LaTeX document
On 22/11/2008 2:01 AM, Kyle Matoba wrote:
List,
I would like to set a variable to hold, say, the size of my plots in a
Sweave document. i.e. something like the following in my '.Rnw' file:
==
smallPlotSize = 4
fig1, echo=false, results=hide, height=smallPlotSize, width=smallPlotSize,
fig=true=
dat - read.table(
http://www.stanford.edu/~xing/statfinbook/_BookData/Chap05/q_us_gdp.txt;,
skip=2, header=T)
GDP - ts(data=dat$VALUE, start=c(1947,1), frequency=4)
acf(GDP, type=correlation, main= ACF of US GDP)
@
==
I don't think that is possible, but you can do things like
\SweaveOpts{height=3.5, width=7}
anywhere in your document to set the default size for subsequent plots.
You can also set LaTeX macros to change the displayed size, e.g.
\newcommand{\Fullwidth}{\setkeys{Gin}{width=\textwidth}}
\newcommand{\Smallwidth}{\setkeys{Gin}{width=4in}}
Here you're talking to LaTeX, not to R, so those are setting options to
the \includegraphics{} calls. Unfortunately, you can't put \SweaveOpts
in the macro to also set the R width and height, because Sweave handles
all the \SweaveOpts calls before calling LaTeX.
Duncan Murdoch
This way I can tweak things if I have to include a bunch of graphics of the
same size in a file. I am not great with LaTeX and still very new to using
Sweave, so I was wondering what the preferred method of doing something like
this is between LaTeX and Sweave. I have seen some people doing some tricky
stuff from within R such that when their code is preprocessed it is compiled
as LaTeX source, but this seems like an ugly hack for something so simple.
Please forgive me if this has been addressed elsewhere; I have found
documentation on Sweave to be rather sparse.
Best,
Kyle
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Re: [R] How to add the value on the barchart
Try something like the code below. Unfortunately I cannot remember who wrote the code. === my.values=10:15 x - barplot(my.values, ylim=c(0,11)) text(x, my.values, wibble, pos=3) # always does what you want # whereas: text(x, 0.6+my.values, wibble) # doesn't look very nice = --- On Fri, 11/21/08, Piya Kittipadakul [EMAIL PROTECTED] wrote: From: Piya Kittipadakul [EMAIL PROTECTED] Subject: [R] How to add the value on the barchart To: r-help@r-project.org Received: Friday, November 21, 2008, 7:14 PM I have question: how can I put the value on the bar chart. This my code: barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber) This's my data: Year Hr Family Tuberize 1 2007 20 A 0.26 2 2007 20 B 6.08 3 2007 20 C 0.00 4 2007 20 D 0.27 5 2008 20 A 1.18 6 2008 20 B 9.17 7 2008 20 C 0.00 8 2008 20 D 2.13 9 2007 14 A 35.83 10 2007 14 B 38.98 11 2007 14 C 7.90 12 2007 14 D 18.75 13 2008 14 A 43.70 14 2008 14 B 51.84 15 2008 14 C 27.11 16 2008 14 D 24.11 17 2008 11 A 85.67 18 2008 11 B 93.73 19 2008 11 C 81.45 20 2008 11 D 90.25 21 2007 8 A 63.04 22 2007 8 B 66.29 23 2007 8 C 70.85 24 2007 8 D 83.48 25 2008 8 A 83.33 26 2008 8 B 92.31 27 2008 8 C 82.21 28 2008 8 D 94.03 Thank you Piya [[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. __ Yahoo! Canada Toolbar: Search from anywhere on the web, and bookmark your favourite sites. Download it now at http://ca.toolbar.yahoo.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] rgl lighting question
Thanks a lot for the pointer to rgl.pop() - that works (as does looking at the examples!) On Nov 22, 2008, at 10:28 AM, Duncan Murdoch wrote: On 21/11/2008 2:30 PM, Rajarshi Guha wrote: Hi, I'm using rgl to generate a 3D surface plot and I'm struggling to get the lighting correct. Currently the surface gets plotted, but is very 'shiny'. On rotating the view, I get to see parts of the surface - but overall I don't see much detail because of the spotlight like lighting. I've played around with the specular, ambient and diffuse but I can't bring out the details of the surface. Could anybody point me to some examples of how to make a plain matte surface, which isn't obscured by specular reflections? This gives the regular shiny surface: library(rgl) example(surface3d) This gives one with no specular reflections, because the material doesn't do that: open3d() surface3d(x, y, z, color=col, back=lines, specular=black) And here's another way to get no specular reflections. This time there's no light to reflect that way: open3d() rgl.pop(lights) light3d(specular=black) surface3d(x, y, z, color=col, back=lines) I suspect you missed the rgl.pop() call. If you just call light3d or rgl.light() you'll add an additional light, you don't change the existing one. Duncan Murdoch --- Rajarshi Guha [EMAIL PROTECTED] GPG Fingerprint: D070 5427 CC5B 7938 929C DD13 66A1 922C 51E7 9E84 --- Paper or plastic? Not 'Not paper AND not plastic!!' -- Augustus DeMorgan in a grocery store __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 add the value on the barchart
Following the example in
https://stat.ethz.ch/pipermail/r-help/2006-January/086985.html
You might want to try something like:
barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber,
+ panel = function(y,x,...){
+ panel.barchart(x,y,...)
+ panel.text(x,y,label = tuber$Tuberize, pos=3)
+ }
+ )
It still needs some work to increase the ylim so the values are not
obscured at the upper ranges of the panels, but this should point you
in the right direction.
--
David Winsemius
On Nov 21, 2008, at 7:14 PM, Piya Kittipadakul wrote:
I have question: how can I put the value on the bar chart.
This my code:
barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber)
This's my data:
Year Hr Family Tuberize
1 2007 20 A 0.26
2 2007 20 B 6.08
3 2007 20 C 0.00
4 2007 20 D 0.27
5 2008 20 A 1.18
6 2008 20 B 9.17
7 2008 20 C 0.00
8 2008 20 D 2.13
9 2007 14 A35.83
10 2007 14 B38.98
11 2007 14 C 7.90
12 2007 14 D18.75
13 2008 14 A43.70
14 2008 14 B51.84
15 2008 14 C27.11
16 2008 14 D24.11
17 2008 11 A85.67
18 2008 11 B93.73
19 2008 11 C81.45
20 2008 11 D90.25
21 2007 8 A63.04
22 2007 8 B66.29
23 2007 8 C70.85
24 2007 8 D83.48
25 2008 8 A83.33
26 2008 8 B92.31
27 2008 8 C82.21
28 2008 8 D94.03
Thank you
Piya
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Re: [R] How to add the value on the barchart
On second look I see that although values appear in every panel, they
are the same in every panel as well. Some sort of use of the
subscripting facility probably needs to be employed.
--
David Winsemius
On Nov 22, 2008, at 1:05 PM, David Winsemius wrote:
Following the example in
https://stat.ethz.ch/pipermail/r-help/2006-January/086985.html
You might want to try something like:
barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber,
+ panel = function(y,x,...){
+ panel.barchart(x,y,...)
+ panel.text(x,y,label = tuber$Tuberize, pos=3)
+ }
+ )
It still needs some work to increase the ylim so the values are not
obscured at the upper ranges of the panels, but this should point
you in the right direction.
--
David Winsemius
On Nov 21, 2008, at 7:14 PM, Piya Kittipadakul wrote:
I have question: how can I put the value on the bar chart.
This my code:
barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber)
This's my data:
Year Hr Family Tuberize
1 2007 20 A 0.26
2 2007 20 B 6.08
3 2007 20 C 0.00
4 2007 20 D 0.27
5 2008 20 A 1.18
6 2008 20 B 9.17
7 2008 20 C 0.00
8 2008 20 D 2.13
9 2007 14 A35.83
10 2007 14 B38.98
11 2007 14 C 7.90
12 2007 14 D18.75
13 2008 14 A43.70
14 2008 14 B51.84
15 2008 14 C27.11
16 2008 14 D24.11
17 2008 11 A85.67
18 2008 11 B93.73
19 2008 11 C81.45
20 2008 11 D90.25
21 2007 8 A63.04
22 2007 8 B66.29
23 2007 8 C70.85
24 2007 8 D83.48
25 2008 8 A83.33
26 2008 8 B92.31
27 2008 8 C82.21
28 2008 8 D94.03
Thank you
Piya
[[alternative HTML version deleted]]
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Re: [R] Fitting a sine wave using solver
To follow Dieter's comment,
You can in fact fit a data to a sine in Excel using LINEST. I've done
it. I don't recommend it :-) .
What I did was create columns containing sin(x) and cos(x) , roughly
speaking, and fit using
LINEST([y=values],{sines, cosines},...)
Ya need the cosines or something similar to get rid of phase ambiguity.
But using either FFTs or glm() should do fine.
And stay away from Excel! :-(
Carl
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Re: [R] How to add the value on the barchart
No, it only needed closer attention to the example:
This gets you to the point where you need to fix the y scale settings
but the values are properly cycled through.
barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber,
panel = function(y,x,...){
panel.barchart(x,y,...)
panel.text(x,y,label = y, pos=3)
}
)
And this fixes the lack of space problem, albeit with not a
particularly general solution:
barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber,
prepanel = function(x,y){ list(ylim = c(0,120) )},
panel = function(y,x,...){
panel.barchart(x,y,...)
panel.text(x,y,label = y, pos=3)
}
)
As we say in medical circles, ... Thank you for this interesting
consult.
--
David Winsemius, MD
On Nov 22, 2008, at 2:16 PM, David Winsemius wrote:
On second look I see that although values appear in every panel,
they are the same in every panel as well. Some sort of use of the
subscripting facility probably needs to be employed.
--
David Winsemius
On Nov 22, 2008, at 1:05 PM, David Winsemius wrote:
Following the example in
https://stat.ethz.ch/pipermail/r-help/2006-January/086985.html
You might want to try something like:
barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber,
+ panel = function(y,x,...){
+ panel.barchart(x,y,...)
+ panel.text(x,y,label = tuber$Tuberize, pos=3)
+ }
+ )
It still needs some work to increase the ylim so the values are not
obscured at the upper ranges of the panels, but this should point
you in the right direction.
--
David Winsemius
On Nov 21, 2008, at 7:14 PM, Piya Kittipadakul wrote:
I have question: how can I put the value on the bar chart.
This my code:
barchart(Tuberize~Family|factor(Year)*factor(Hr),data=tuber)
This's my data:
Year Hr Family Tuberize
1 2007 20 A 0.26
2 2007 20 B 6.08
3 2007 20 C 0.00
4 2007 20 D 0.27
5 2008 20 A 1.18
6 2008 20 B 9.17
7 2008 20 C 0.00
8 2008 20 D 2.13
9 2007 14 A35.83
10 2007 14 B38.98
11 2007 14 C 7.90
12 2007 14 D18.75
13 2008 14 A43.70
14 2008 14 B51.84
15 2008 14 C27.11
16 2008 14 D24.11
17 2008 11 A85.67
18 2008 11 B93.73
19 2008 11 C81.45
20 2008 11 D90.25
21 2007 8 A63.04
22 2007 8 B66.29
23 2007 8 C70.85
24 2007 8 D83.48
25 2008 8 A83.33
26 2008 8 B92.31
27 2008 8 C82.21
28 2008 8 D94.03
Thank you
Piya
[[alternative HTML version deleted]]
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Re: [R] Bug in Kendall for n4?
On Sat, Nov 22, 2008 at 9:04 AM, Martin Maechler [EMAIL PROTECTED] wrote: SM I believe Kendall tau is well-defined for this case... The real question is *WHY* there needs to be a separate package 'Kendall' when R itself does everything you want and does not show any problems? Thanks for pointing me to cor(...,method=kendall), which I did not know about; I used the Kendall CRAN package out of pure ignorance. In my defense, I think it is excusable ignorance, as Search on the R Project home page finds the Kendall package (which only mentions cor as a See Also). I only more recently discovered the advantages of help.search. By the way, is Kendall well-defined when the arguments are not permutations of each other? cor seems to return results even in this case: a-factor(c(Alice,Bob,Chris)) b-a[1:2] c-a[2:3] cor(a,b,method=kendall) = 1 apparently interpreting b as c(1,2) and c as c(1,2) based on alphabetical order (even though it is an UNordered factor), which seems to make the value depend on the subjects' names, which I'd think was wrong for a rank-order statistic. Thanks again, -s __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] applying an operation for several dataframes
Dear R community,
I am trying to apply a simple operation to several dataframes (e.g.
nrow) and cannot get the looping to work. My objective is to get an output
that indicates me the number of rows for every dataframe.
c
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1 1 11 21 31 41 51 61 71 81 91
2 2 12 22 32 42 52 62 72 82 92
3 3 13 23 33 43 53 63 73 83 93
4 4 14 24 34 44 54 64 74 84 94
5 5 15 25 35 45 55 65 75 85 95
cc
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1 1 11 21 31 41 51 61 71 81 91
2 2 12 22 32 42 52 62 72 82 92
3 3 13 23 33 43 53 63 73 83 93
4 4 14 24 34 44 54 64 74 84 94
5 5 15 25 35 45 55 65 75 85 95
6 6 16 26 36 46 56 66 76 86 96
7 7 17 27 37 47 57 67 77 87 97
8 8 18 28 38 48 58 68 78 88 98
9 9 19 29 39 49 59 69 79 89 99
10 10 20 30 40 50 60 70 80 90 100
names-c(c,cc)
for(i in names){nrow(i)}
Thank you!
Georg.
*
Georg Ehret
[EMAIL PROTECTED]
Geneva University Hospital
Switzerland
[[alternative HTML version deleted]]
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Re: [R] applying an operation for several dataframes
?get
names - c('c', 'cc')
for (i in names){
print(nrow(get(i)))
}
On Sat, Nov 22, 2008 at 4:06 PM, Georg Ehret [EMAIL PROTECTED] wrote:
Dear R community,
I am trying to apply a simple operation to several dataframes (e.g.
nrow) and cannot get the looping to work. My objective is to get an output
that indicates me the number of rows for every dataframe.
c
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1 1 11 21 31 41 51 61 71 81 91
2 2 12 22 32 42 52 62 72 82 92
3 3 13 23 33 43 53 63 73 83 93
4 4 14 24 34 44 54 64 74 84 94
5 5 15 25 35 45 55 65 75 85 95
cc
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1 1 11 21 31 41 51 61 71 81 91
2 2 12 22 32 42 52 62 72 82 92
3 3 13 23 33 43 53 63 73 83 93
4 4 14 24 34 44 54 64 74 84 94
5 5 15 25 35 45 55 65 75 85 95
6 6 16 26 36 46 56 66 76 86 96
7 7 17 27 37 47 57 67 77 87 97
8 8 18 28 38 48 58 68 78 88 98
9 9 19 29 39 49 59 69 79 89 99
10 10 20 30 40 50 60 70 80 90 100
names-c(c,cc)
for(i in names){nrow(i)}
Thank you!
Georg.
*
Georg Ehret
[EMAIL PROTECTED]
Geneva University Hospital
Switzerland
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] syntax and package for generalized linear mixed models
On Nov 21, 2008, at 11:28 AM, Douglas Bates wrote: As tempting as it may be to want to have several dials and knobs on statistical models to tune their behavior, we still need to be careful to specify a mathematical model that is consistent. -- Douglas Bates Yet another fortune candidate. -- David Winsemius [[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] unique columns in a matrix and number of each
I need help determining the unique columns of a matrix and the numbers of each unique column. For example, let's say I have a matrix with 6 columns, 2 of these are filled with the value of 1 and the other 4 are filled with the value of 0. I would then like to have a command that tells me what the unique columns are (so columns with 1s and columns with 0s) and the number of each type of column (so 2 columns of 1s and 4 columns of 0s) that occur in the matrix. Thank you for any help anyone can provide!! I have been very impressed with the help that I have received so far!! 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.
[R] Where is Hong Kong in maps？
It seems that there is no Hong Kong in maps' world cities. world.cities[substr(world.cities$name,1,3)==Hon,] namecountry.etcpop latlong capital 14623 Honami Japan 26040 33.61 130.68 0 14624 Honaz Turkey 8073 37.75 29.27 0 14625 Honda Colombia 28055 5.19 -74.75 0 14626 Hondaidhoo Maldives 83 6.77 73.08 0 14627 Hondarribia Spain 16159 43.37 -1.80 0 14628 Hondo Japan 40728 32.46 130.19 0 14629 Hondo Valle Dominican Republic 3695 18.72 -71.70 0 14630Honefoss Norway 13587 60.19 10.23 0 14631HongDenmark 3792 55.52 11.30 0 14632Hong GaiVietnam 149427 20.97 107.08 0 14633 Hongan China 65273 31.28 114.61 0 14634HongchonKorea South 33721 37.69 127.89 0 14635Honggang China 150194 46.47 124.87 0 14636 Honghu China 178779 29.82 113.47 0 14637 Hongjiang China 59377 27.12 109.99 0 14638HongsongKorea South 38208 36.60 126.66 0 14639 HongwonKorea North 71617 40.02 127.97 0 14640 HoniaraSolomon Islands 57410 -9.43 159.91 1 14641 Honiton UK 11516 50.81 -3.20 0 14642 Honjo Japan 46939 39.44 140.09 0 14643 Honjo Japan 61961 36.24 139.18 0 14644 HonkajokiFinland 1950 61.83 21.53 0 14645 Honningsvag Norway 2521 70.98 25.98 0 14646HonoluluUSA 386345 21.32 -157.80 0 The info of the maps package: Package: maps Title: Draw Geographical Maps Version: 2.0-40 Date: 2008-04-17 I wonder maybe I have made some mistakes using maps since I am a beginner in R. -- View this message in context: http://www.nabble.com/Where-is-Hong-Kong-in-maps%EF%BC%9F-tp20635934p20635934.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] What's the BEST way in R to adapt this vector?
Goal: Suppose you have a vector that is a discrete variable with values ranging from 1 to 3, and length of 10. We'll use this as the example: y - c(1,2,3,1,2,3,1,2,3,1) ...and suppose you want your new vector (y.new) to be equal in length to the possible discrete values (3) times the length (10), and formatted in such a way that if y[1] == 1, then y.new[1:3] == c(1,0,0), and if y[2] == 2, then y.new[4:6] == c(0,1,0). For example, the final goal should be: y.new - c(1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,0,0,1,1,0,0) Note: I know how to do this with loops, but that's not taking advantage of R's capabilities with vectors and, I suspect, matrices. So far, my best guess would be to start as follows: y1 - ifelse(y == 1, 1, 0) y2 - ifelse(y == 2, 1, 0) y3 - ifelse(y == 3, 1, 0) From here, maybe put these into a 10x3 matrix, and read them out by row into y.new? Is that even the most efficient way? If it is, I'm sure I can get them into a matrix, but how do I read them out correctly? Thanks for any input. -- View this message in context: http://www.nabble.com/What%27s-the-BEST-way-in-R-to-adapt-this-vector--tp20638991p20638991.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] What's the BEST way in R to adapt this vector?
On Sat, Nov 22, 2008 at 12:00 PM, zerfetzen [EMAIL PROTECTED] wrote: Goal: Suppose you have a vector that is a discrete variable with values ranging from 1 to 3, and length of 10. We'll use this as the example: y - c(1,2,3,1,2,3,1,2,3,1) ...and suppose you want your new vector (y.new) to be equal in length to the possible discrete values (3) times the length (10), and formatted in such a way that if y[1] == 1, then y.new[1:3] == c(1,0,0), and if y[2] == 2, then y.new[4:6] == c(0,1,0). For example, the final goal should be: y.new - c(1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,0,0,1,1,0,0) Note: I know how to do this with loops, but that's not taking advantage of R's capabilities with vectors and, I suspect, matrices. How about: as.vector(diag(3)[, y]) Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] What's the BEST way in R to adapt this vector?
Try this: outer(y, sort(unique(y)), ==)+0 On Sat, Nov 22, 2008 at 3:37 PM, zerfetzen [EMAIL PROTECTED] wrote: Goal: Suppose you have a vector that is a discrete variable with values ranging from 1 to 3, and length of 10. We'll use this as the example: y - c(1,2,3,1,2,3,1,2,3,1) ...and suppose you want your new vector (y.new) to be equal in length to the possible discrete values (3) times the length (10), and formatted in such a way that if y[1] == 1, then y.new[1:3] == c(1,0,0), and if y[2] == 2, then y.new[4:6] == c(0,1,0). For example, the final goal should be: y.new - c(1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,0,0,1,1,0,0) Note: I know how to do this with loops, but that's not taking advantage of R's capabilities with vectors and, I suspect, matrices. So far, here's my best: y - c(1,2,3,1,2,3,1,2,3,1) y.k - length(unique(y)) #Num. of Categories of y y.n - NROW(y) y.nk - y.n * y.k #Length of new vector y.1 - ifelse(y == 1,1,0) y.2 - ifelse(y == 2,1,0) y.3 - ifelse(y == 3,1,0) z - cbind(y.1, y.2, y.3) z.trans - t(z) y.new - c(1:y.nk); y.new[1:y.nk] - NA y.new - as.vector(z.trans) rm(y.k, y.n, y.nk, y.1, y.2, y.3, z, z.trans) y y.new Is there a better a way? I'm still pretty new. Thanks. -- View this message in context: http://www.nabble.com/What%27s-the-BEST-way-in-R-to-adapt-this-vector--tp20638991p20638991.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] What's the BEST way in R to adapt this vector?
On Sat, 22 Nov 2008 10:00:18 -0800 (PST) zerfetzen [EMAIL PROTECTED] wrote: Goal: Suppose you have a vector that is a discrete variable with values ranging from 1 to 3, and length of 10. We'll use this as the example: y - c(1,2,3,1,2,3,1,2,3,1) ...and suppose you want your new vector (y.new) to be equal in length to the possible discrete values (3) times the length (10), and formatted in such a way that if y[1] == 1, then y.new[1:3] == c(1,0,0), and if y[2] == 2, then y.new[4:6] == c(0,1,0). For example, the final goal should be: y.new - c(1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,0,0,1,1,0,0) Note: I know how to do this with loops, but that's not taking advantage of R's capabilities with vectors and, I suspect, matrices. So far, my best guess would be to start as follows: [] My guess would be to use: R y - c(1,2,3,1,2,3,1,2,3,1) R y.new - as.numeric(as.vector(outer(sort(unique(y)), y, ==))) R y.new [1] 1 0 0 0 1 0 0 0 1 1 0 0 0 1 0 0 0 1 1 0 0 0 1 0 0 0 1 1 0 0 [...] From here, maybe put these into a 10x3 matrix, and read them out by row into y.new? Better put them into a 3x10 matrix and then turn the matrix into a vector by as.vector(). Essentially, a matrix is a vector with an attribute giving the dimensions of a matrix and as.vector() removes that attribute. Furthermore, in R (just as in Fortran), matrices are stored in column major format; thus, as soon as the dim attribute is removed, the values are in the correct order. HTH. Cheers, Berwin === Full address = Berwin A TurlachTel.: +65 6516 4416 (secr) Dept of Statistics and Applied Probability+65 6516 6650 (self) Faculty of Science FAX : +65 6872 3919 National University of Singapore 6 Science Drive 2, Blk S16, Level 7 e-mail: [EMAIL PROTECTED] Singapore 117546http://www.stat.nus.edu.sg/~statba __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Basic question on concatenating factors
On Sat, Nov 22, 2008 at 10:20 AM, jim holtman [EMAIL PROTECTED] wrote:
c.Factor -
function (x, y)
{
newlevels = union(levels(x), levels(y))
m = match(levels(y), newlevels)
ans = c(unclass(x), m[unclass(y)])
levels(ans) = newlevels
class(ans) = factor
ans
}
This algorithm depends crucially on union preserving the order of the
elements of its arguments. As far as I can tell, the spec of union
does not require this. If union were to (for example) sort its
arguments then merge them (generally a more efficient algorithm), this
function would no longer work.
Fortunately, the fix is simple. Instead of union, use:
newlevels - c(levels(x),setdiff(levels(y),levels(x))
which is guaranteed to preserve the order of levels(x).
-s
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Re: [R] Basic question on concatenating factors
You are right. union used 'unique(c(x,y))' and I am not sure if
'unique' preserves the order, but the help page seems to indicate that
an element is omitted if it is identical to any previous element ;
this might mean that the order is preserved.
On Sat, Nov 22, 2008 at 11:43 PM, Stavros Macrakis
[EMAIL PROTECTED] wrote:
On Sat, Nov 22, 2008 at 10:20 AM, jim holtman [EMAIL PROTECTED] wrote:
c.Factor -
function (x, y)
{
newlevels = union(levels(x), levels(y))
m = match(levels(y), newlevels)
ans = c(unclass(x), m[unclass(y)])
levels(ans) = newlevels
class(ans) = factor
ans
}
This algorithm depends crucially on union preserving the order of the
elements of its arguments. As far as I can tell, the spec of union
does not require this. If union were to (for example) sort its
arguments then merge them (generally a more efficient algorithm), this
function would no longer work.
Fortunately, the fix is simple. Instead of union, use:
newlevels - c(levels(x),setdiff(levels(y),levels(x))
which is guaranteed to preserve the order of levels(x).
-s
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] unique columns in a matrix and number of each
One way is to 'paste' together the values in a column and then use 'table' to count them. 'duplicated' can probably do the same thing with the MARGIN option to find the duplicated one. You still them have to find the original ones. On Sat, Nov 22, 2008 at 3:42 PM, Salas, Andria Kay [EMAIL PROTECTED] wrote: I need help determining the unique columns of a matrix and the numbers of each unique column. For example, let's say I have a matrix with 6 columns, 2 of these are filled with the value of 1 and the other 4 are filled with the value of 0. I would then like to have a command that tells me what the unique columns are (so columns with 1s and columns with 0s) and the number of each type of column (so 2 columns of 1s and 4 columns of 0s) that occur in the matrix. Thank you for any help anyone can provide!! I have been very impressed with the help that I have received so far!! 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Basic question on concatenating factors
On Sun, 23 Nov 2008, jim holtman wrote:
You are right. union used 'unique(c(x,y))' and I am not sure if
'unique' preserves the order, but the help page seems to indicate that
an element is omitted if it is identical to any previous element ;
this might mean that the order is preserved.
It says
'unique' returns a vector, data frame or array like 'x' but with
duplicate elements/rows removed.
Although it is a generic function, it is hard to see how that can be
interpreted to allow the order to be changed.
The claim that union would be more efficiently implemented via sorting is
made with no evidence: do look up a basic computer science textbook for
this kind of thing, as well as how R actually does it. (Also 'efficient'
was not defined: both speed and memory usage are potentially measures of
efficiency.) But for example
x - rnorm(1e7)
system.time(unique(x))
user system elapsed
2.258 0.261 2.523
system.time(sort(x))
user system elapsed
4.102 0.112 4.231
system.time(sort(x, method=quick))
user system elapsed
1.928 0.109 2.047
will indicate that unique() is comparable in speed to sorting.
On Sat, Nov 22, 2008 at 11:43 PM, Stavros Macrakis
[EMAIL PROTECTED] wrote:
On Sat, Nov 22, 2008 at 10:20 AM, jim holtman [EMAIL PROTECTED] wrote:
c.Factor -
function (x, y)
{
newlevels = union(levels(x), levels(y))
m = match(levels(y), newlevels)
ans = c(unclass(x), m[unclass(y)])
levels(ans) = newlevels
class(ans) = factor
ans
}
This algorithm depends crucially on union preserving the order of the
elements of its arguments. As far as I can tell, the spec of union
does not require this. If union were to (for example) sort its
arguments then merge them (generally a more efficient algorithm), this
function would no longer work.
Fortunately, the fix is simple. Instead of union, use:
newlevels - c(levels(x),setdiff(levels(y),levels(x))
which is guaranteed to preserve the order of levels(x).
-s
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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