Re: [R] PCA and MDS
On Wed, 22 Jun 2005, Guohui Ding wrote: prcomp(stats) Principal Components Analysis princomp(stats) Principal Components Analysis scale(base) Scaling and Centering of Matrix-like Objects The first two are correct for PCA, but scale is not MDS. MDS is available in cmdscale (stats), isoMDS (MASS), sammon (MASS) and elsewhere. I am not familar with R. I want to use PCA (principal components analysis) and MDS (multidimensional scaling). Can someone tell me which R package I should use for PCA and MDS? I appreciate your help in advance. -- 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 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to use expression in label with xYplot
Dear R-List,
I want to use the label function (from Hmisc library) to allow for the
names of my isotopes.
library(Hmisc)
library(lattice)
library(grid)
num - c(78,137,129m)
nom - c(Ge,Cs,Te)
df - data.frame(GE78=seq(nom),CS137=seq(nom),TE129m=seq(nom))
if I use this function to create the labels :
lab - function(i)
as.expression(bquote(italic(phantom(0)^{.(num[i])}*.(nom[i]
label(df$GE78) - lab(1)
label(df$CS137) - lab(2)
label(df$TE129m) - lab(3)
all works fine when I use text and xyplot :
plot(1:10)
text(6,6,labels=label(df$CS137))
xyplot(CS137~TE129m,data=df,xlab=label(df$CS137),ylab=label(df$TE129m))
but xYplot doesn't work fine
xYplot(CS137~TE129m,data=df)
I have the message :
Error in parse(file, n, text, prompt) : parse error
if I change the lab function
lab - function(i)
as.character(paste(italic(phantom(0)^{\,num[i],\}*\,nom[i],\),sep=
))
text and xyplot work fine if I use the parse function and xYplot works fine.
label(df$GE78) - lab(1)
label(df$CS137) - lab(2)
label(df$TE129m) - lab(3)
plot(1:10)
text(6,6,labels=parse(text=label(df$CS137)))
xyplot(CS137~TE129m,data=df,xlab=parse(text=label(df$CS137)),ylab=parse(text
=label(df$TE129m)))
xYplot(CS137~TE129m,data=df)
What is the good way to use expression in the labels of variables with the
label function ?
thanks
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Re: [R] tapply
AndyL == Liaw, Andy [EMAIL PROTECTED] on Tue, 21 Jun 2005 13:30:54 -0400 writes: AndyL Try: (x - factor(1:2, levels=1:5)) AndyL [1] 1 2 AndyL Levels: 1 2 3 4 5 (x - x[, drop=TRUE]) AndyL [1] 1 2 AndyL Levels: 1 2 or (x - factor(1:2, levels=1:5)) (x2 - factor(x)) which also drops the level Martin AndyL Andy From: Weiwei Shi [mailto:[EMAIL PROTECTED] Even before I tried, I already realize it must be true when I read this reply! Great job! thanks, Andy. str(z) `data.frame': 235 obs. of 2 variables: $ CLAIMNUM : Factor w/ 1907 levels 0,1001849,..: 1083 1083 1083 1582 1582 1084 1681 1681 1391 1391 ... $ SIU.SAVED: int 475 3000 3000 0 0 4352 0 0 4500 3000 ... So, I have another general question: how to avoid this when I do the matching? In my case, claimnum does not have to be a factor. I think I can do as.integer on it to de-factor it. But, I want to know how to do it w/ keeping is as factor? btw, what's your way to drop those levels? :) weiwei On 6/21/05, Liaw, Andy [EMAIL PROTECTED] wrote: What does str(z) say? I suspect the second column is a factor, which, after the subsetting, has some empty levels. If so, just drop those levels. Andy From: Weiwei Shi hi i tried all the methods suggested above: ave and rowsum with with function works for my situation. I think the problem might not be due to tapply. My data z comes from z-y[y[[1]] %in% x[[2]], c(1,9)] while z is supposed to have no entries for those non-matched between x and y. however, when I run tapply, and the result also includes those non-matched entries. I use is.na function to remove those entry from z first and then use tapply again, but the result is the same: those NA's and those non-matched results are still there. That's what I mean by it doesn't work. Is there something I missed here so that z implicitly has some trace back to y dataset? thanks, On 6/20/05, Gabor Grothendieck [EMAIL PROTECTED] wrote: On 6/20/05, Weiwei Shi [EMAIL PROTECTED] wrote: hi, i have another question on tapply: i have a dataset z like this: 5540 389100307391 2600 5541 389100307391 2600 5542 389100307391 2600 5543 389100307391 2600 5544 389100307391 2600 5546 381300302513NA 5547 387000307470NA 5548 387000307470NA 5549 387000307470NA 5550 387000307470NA 5551 387000307470NA 5552 387000307470NA I want to sum the column 3 by column 2. I removed NA by calling: tapply(z[[3]], z[[2]], sum, na.rm=T) but it does not work. then, i used z1-z[!is.na(z[[3]],] and repeat still doesn't work. please help. Depending on what you want you may be able to use rowsum: - display only groups that have at least one non-NA with the sum being the sum of the non-NAs: with(na.omit(z), rowsum(V3, V2)) - display all groups with the sum being NA if any member is NA: rowsum(z$V3, z$V2) -- Weiwei Shi, Ph.D Did you always know? No, I did not. But I believed... ---Matrix III __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Notice: This e-mail message, together with any attachments, contains information of Merck Co., Inc. (One Merck Drive, Whitehouse Station, New Jersey, USA 08889), and/or its affiliates (which may be known outside the United States as Merck Frosst, Merck Sharp Dohme or MSD and in Japan, as Banyu) that may be confidential, proprietary copyrighted and/or legally privileged. It is intended solely for the use of the individual or entity named on this message. If you are not the intended recipient, and have received this message in error, please notify us immediately by reply e-mail and then delete it from your system. -- -- Weiwei Shi, Ph.D Did you
[R] predict.coxph fitted values for failure times
I would like to extract predicted failure times from a coxph model in library(survival). However, none of the prediction options (lp, risk, expected, terms) seem to bear any relationship to failure time. Perhaps I am asking the wrong question, but can coxph provide predicted failure times? Thanks, Dan Bebber Department of Plant Sciences University of Oxford ___ How much free photo storage do you get? Store your holiday __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] A question on time-dependent covariates in the Cox model.
I have a dataset with event=death time (from medical examination until death/censoring) dose (given at examination time) Two groups are considered, a non-exposed group (dose=0), an exposed group (dose between 5 and 60). For some reason there is a theory of the dose increasing its effect over time (however it was only given (and measured) once = at the time of examination). I tested a model: coxph(Surv(time,dod)~dose + dose:time) Previously I tested the model in SAS: proc phreg data=test; model time*dod(0)=dose dosetime /rl ties=efron; dosetime=time*dose; run; Without the interaction terms I get the same results for the two models. By including the interaction terms I do not. The model in R gives a negative coefficient for the interaction term which is expected to be positive (and is so in SAS). The LRTs are also completely different. TWO QUESTIONS: 1) Is it reasonable to bring in an interaction term when dose is only measured once? 2) If yes, can anyone give a hint on explaining the difference between the models in R and SAS? Thanx in advance, marianne _ Nyhet! Hotmail direkt i Mobilen! http://mobile.msn.com/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] predict.coxph fitted values for failure times
On Wed, 22 Jun 2005, Dan Bebber wrote: I would like to extract predicted failure times from a coxph model in library(survival). However, none of the prediction options (lp, risk, expected, terms) seem to bear any relationship to failure time. Perhaps I am asking the wrong question, but can coxph provide predicted failure times? Strictly no, as a Cox proportional hazards model does not model the baseline hazard. However, there is a coxph method for survfit() which allows you to predict the survival distribution for the fitted or new data, presumably using Breslow's estimator of the baseline hazard. There's an example on the help page and more in MASS (the book and its scripts). Unfortunately the help page is misleading, documenting only the km method as if it were the generic (and using \synopsis to avoid this being flagged). -- 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 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] mu^2(1-mu)^2 variance function for GLM
Dear Professor Firth, David Firth said the following on 2005-06-16 17:22: I do not have a ready stock of other examples, but I do have my own version of a family function for this, reproduced below. It differs from yours (apart from being a regular family function rather than using a modified quasi) in the definition of deviance residuals. These necessarily involve an arbitrary constant (see McCullagh and Nelder, 1989, p330); in my function that arbitrariness is in the choice eps - 0.0005. I don't think the deviance contributions as you specified in your code below will have the right derivative (with respect to mu) for observations where y=0 or y=1. I'm sorry for the late reply. You're right -- my definition of the deviance residuals isn't correct. Your code, on the other hand, seems to do the right thing. Many thanks for this note and the provided `wedderburn' function. Henric __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] sweep() and recycling
I think that as the proponents do not agree, we need to leave this as is.
BTW, R-devel is the place to discuss patched to R, rather than R-help.
On Tue, 21 Jun 2005, Heather Turner wrote:
Agreed my examples may be trivial and I'm sure there are more efficient ways
to do the same thing, but I disagree that my examples are against the spirit
of sweep (). In the first example a vector is swept out from the rows, with
one value for odd rows and one value for even rows. In the second example an
array of values is swept out across the third dimension. In the third example
an array of values is swept out from the full array.
The first example is a natural use of recycling. E.g.
sweep(matrix(1:100, 50, 2), 1, c(1, 1000), +, give.warning = TRUE)
is a quicker way of writing
sweep(matrix(1:100, 50, 2), 1, rep(c(1, 1000), 25), +, give.warning = TRUE)
but your code would give a warning in the first case, even though the intent
and the result are exactly the same as in the second case.
As you say, it is only a warning, that can be ignored. However the warning
should at least reflect the warning condition used, i.e. warn that the length
of STATS does not equal the extent of MARGIN, rather warning that STATS does
not recycle exactly.
Heather
Robin Hankin [EMAIL PROTECTED] 06/21/05 02:47pm
Hi
On Jun 21, 2005, at 02:33 pm, Heather Turner wrote:
I think the warning condition in Robin's patch is too harsh - the
following examples seem reasonable to me, but all produce warnings
sweep(array(1:24, dim = c(4,3,2)), 1, 1:2, give.warning = TRUE)
sweep(array(1:24, dim = c(4,3,2)), 1, 1:12, give.warning = TRUE)
sweep(array(1:24, dim = c(4,3,2)), 1, 1:24, give.warning = TRUE)
The examples above do give warnings (as intended) but I think all three
cases above
are inimical to the spirit of sweep(): nothing is being swept out.
So a warning is appropriate, IMO.
In any case, one can always suppress (or ignore!) a warning if one knows
what one is doing. YMMV, but if I wanted to do the above operations I
would
replace
sweep(array(0, dim = c(4,3,2)), c(1,3), 1:12, + , give.warning =
FALSE)
with
aperm(array(1:12,c(4,2,3)),c(1,3,2))
best wishes
rksh
I have written an alternative (given below) which does not give
warnings in the above cases, but does warn in the following case
sweep(array(1:24, dim = c(4,3,2)), 1:2, 1:3)
, , 1
[,1] [,2] [,3]
[1,]036
[2,]039
[3,]069
[4,]369
, , 2
[,1] [,2] [,3]
[1,] 12 15 18
[2,] 12 15 21
[3,] 12 18 21
[4,] 15 18 21
Warning message:
STATS does not recycle exactly across MARGIN
The code could be easily modified to warn in other cases, e.g. when
length of STATS is a divisor of the corresponding array extent (as in
the first example above, with length(STATS) = 2).
The code also includes Gabor's suggestion.
Heather
sweep - function (x, MARGIN, STATS, FUN = -, warn =
getOption(warn), ...)
{
FUN - match.fun(FUN)
dims - dim(x)
perm - c(MARGIN, (1:length(dims))[-MARGIN])
if (warn = 0) {
s - length(STATS)
cumDim - c(1, cumprod(dims[perm]))
if (s max(cumDim))
warning(length of STATS greater than length of array,
call. = FALSE)
else {
upper - min(ifelse(cumDim s, cumDim, max(cumDim)))
lower - max(ifelse(cumDim s, cumDim, min(cumDim)))
if (any(upper %% s != 0, s %% lower != 0))
warning(STATS does not recycle exactly across MARGIN,
call. = FALSE)
}
}
FUN(x, aperm(array(STATS, dims[perm]), order(perm)), ...)
}
Gabor Grothendieck [EMAIL PROTECTED] 06/21/05 01:25pm
\
Perhaps the signature should be:
sweep(...other args go here..., warn=getOption(warn))
so that the name and value of the argument are consistent with
the R warn option.
On 6/21/05, Robin Hankin [EMAIL PROTECTED] wrote:
On Jun 20, 2005, at 04:58 pm, Prof Brian Ripley wrote:
The issue here is that the equivalent command array(1:5, c(6,6)) (to
matrix(1:5,6,6)) gives no warning, and sweep uses array().
I am not sure either should: fractional recycling was normally
allowed
in S3 (S4 tightened up a bit).
Perhaps someone who thinks sweep() should warn could contribute a
tested patch?
OK, modified R code and Rd file below (is this the best way to do
this?)
sweep -
function (x, MARGIN, STATS, FUN = -, give.warning = FALSE, ...)
{
FUN - match.fun(FUN)
dims - dim(x)
if(give.warning length(STATS)1 any(dims[MARGIN] !=
dim(as.array(STATS{
warning(array extents do not recycle exactly)
}
perm - c(MARGIN, (1:length(dims))[-MARGIN])
FUN(x, aperm(array(STATS, dims[perm]), order(perm)), ...)
}
\name{sweep}
\alias{sweep}
\title{Sweep out Array Summaries}
\description{
Return an array obtained from an input array
Re: [R] A question on time-dependent covariates in the Cox model.
Marianne dk [EMAIL PROTECTED] writes: I have a dataset with event=death time (from medical examination until death/censoring) dose (given at examination time) Two groups are considered, a non-exposed group (dose=0), an exposed group (dose between 5 and 60). For some reason there is a theory of the dose increasing its effect over time (however it was only given (and measured) once = at the time of examination). I tested a model: coxph(Surv(time,dod)~dose + dose:time) Previously I tested the model in SAS: proc phreg data=test; model time*dod(0)=dose dosetime /rl ties=efron; dosetime=time*dose; run; Without the interaction terms I get the same results for the two models. By including the interaction terms I do not. The model in R gives a negative coefficient for the interaction term which is expected to be positive (and is so in SAS). The LRTs are also completely different. TWO QUESTIONS: 1) Is it reasonable to bring in an interaction term when dose is only measured once? 2) If yes, can anyone give a hint on explaining the difference between the models in R and SAS? I don't know what SAS does, maybe it second-guesses your intentions, but R will definitely get it completely wrong. If you use time as a covariate, the same time (of death/censoring) will be applied at all death times. Pretty obviously, long observation times tend to be associated with low mortalities! With interactions you get, er, similarly incorrect effects. To do coxph with time-dependent variables, you need to split data into little time segments, according to the death time of every death, inserting a new variable (ntime, say) which is the time of the endpoint of the interval. -- 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Subsetting across a frame for plotting
I have a huge frame holding holding model results for a number of locations and time series: str(tonedata) `data.frame': 434 obs. of 339 variables: $ VALUE : int 101 104 105 106 111 118 119 121 122 123 ... $ COUNT : int 2443 184 1539 1016 132 1208 1580 654 864 560 ... $ AREA: num 6.11e+08 4.60e+07 3.85e+08 2.54e+08 3.30e+07 ... $ D1_1958 : num 470 446 452 457 407 ... $ D2_1958 : num 480 455 461 467 416 ... $ D3_1958 : num 493 469 475 480 429 ... $ D4_1958 : num 542 517 522 526 475 ... $ D5_1958 : num 585 560 565 568 517 ... I would like to be able to take all values, except the three first (value, count, area) and be able to plot them. I have managed to make a subset that looks like what I want, by doing tonedata[11,4:339], but that data set is not a vector that can be plottet, it is treated like a set of single values. I tried to use as.vector on the set, bot to no help. I am probably overlooking somehing quite simple, (not to mention not really understanding R's data model..) so help would be appreciated. -- Morten Sickel Norwegian Radiation Protection Authority __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] legend
I color some area grey with polygon() (with a red border) and then I
want to have the dashed red border in the legend as well. How do I
manage it?
And I want to mix (latex) expressions with text in my legend.
Just execute my lines below and you know want I mean. Or pass by at
http://de.wikipedia.org/wiki/Bild:GBM.png to see the picture online.
Thomas
bm - function(n=500, from=0, to=1) {
x=seq(from=from,to=to,length=n)
BM-c(0,cumsum(rnorm(n-1,mean=0,sd=sqrt(to/n
cbind(x,BM)
}
gbm - function(bm,S0=1,sigma=0.1,mu=1) {
gbm=S0
for (t in 2:length(bm[,1])) {
gbm[t]=S0*exp((mu-sigma^2/2)*bm[t,1]+sigma*bm[t,2])
}
cbind(bm[,1],gbm)
}
set.seed(9826064)
cs=c(dark green, steelblue, red, yellow)
#png(filename = GBM.png, width=1600, height=1200, pointsize = 12)
par(bg=lightgrey)
x=seq(from=0,to=1,length=500)
plot(x=x, y=exp(0.7*x), type=n, xlab=Zeit, ylab=, ylim=c(1,3.5))
polygon(x=c(x,rev(x)),
y=c(exp(0.7*x)+0.4*sqrt(x),rev(exp(0.7*x)-0.4*sqrt(x))), col=grey,
border=cs[3], lty=dashed)
lines(x=x,y=exp(0.7*x), type=l, lwd=3, col=cs[1])
lines(gbm(bm(),S0=1,mu=0.7,sigma=0.4), lwd=3, col=cs[2])
lines(gbm(bm(),S0=1,mu=0.7,sigma=0.2), lwd=3, col=cs[3])
lines(gbm(bm(),S0=1,mu=0.7,sigma=0.1), lwd=3, col=cs[4])
title(main=Geometrische Brownsche Bewegung,cex.main=2.5)
legend(x=0,y=3.5,legend=c(exp(0.7x),mu=0.7, sigma=0.4,mu=0.7,
sigma=0.2,mu=0.7, sigma=0.1,Standardabweichung für
sigma=0.2),lwd=c(4,4,4,4,12),col=c(cs,grey),bg=transparent,cex=1.15)
#dev.off()
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[R] nlme leading minor error
Dear all I am struggling with nlme and error message. Even going through Pinheiro, Bates nlme book did not gave me a clue how to avoid this. fit - nlme(ce ~ fi1 / ((1+exp(fi2-fi3*tepl))^(1/fi4)), data = temp1na.gr, start = c(fi1=30, fi2=-100, fi3=-.05, fi4=40), fixed = fi1+fi2+fi3+fi4~1, random = pdDiag(fi2+fi4~1), groups = ~spol.f) gives Error in chol((value + t(value))/2) : the leading minor of order 1 is not positive definite Is this error due to lack of experimental points? Here you have one typical part of my data. It is for level spol.f = 3/11. teplce 800 28.87 800 29.35 825 29 850 28.73 875 26.83 900 24.07 I have 1-5 points for each level (2 levels with 5 points, 1 level with 4 points, several levels with 2 and 3 points and few with only one point. Fitting this model to each level separately led to several sets of coeficients fi1-fi4 and the separate fits were quite OK. Please give me a hint what can be the cause for this error message and how I shall organize my data to avoid this. (Lack of experimental points is also an answer as I can do some subsequent measurement. R 2.1.0, W 2000, nlme package Best regards Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Subsetting across a frame for plotting
You might save yourself some headaches by turning it into a matrix instead, since all the columns are either integer or numeric: tonedata - data.matrix(tonedata) Data frames are really lists, so even when you get a one-row subset, it's still a one-row data frame. You can use unlist() to turn that into a vector. Andy From: Morten Sickel I have a huge frame holding holding model results for a number of locations and time series: str(tonedata) `data.frame': 434 obs. of 339 variables: $ VALUE : int 101 104 105 106 111 118 119 121 122 123 ... $ COUNT : int 2443 184 1539 1016 132 1208 1580 654 864 560 ... $ AREA: num 6.11e+08 4.60e+07 3.85e+08 2.54e+08 3.30e+07 ... $ D1_1958 : num 470 446 452 457 407 ... $ D2_1958 : num 480 455 461 467 416 ... $ D3_1958 : num 493 469 475 480 429 ... $ D4_1958 : num 542 517 522 526 475 ... $ D5_1958 : num 585 560 565 568 517 ... I would like to be able to take all values, except the three first (value, count, area) and be able to plot them. I have managed to make a subset that looks like what I want, by doing tonedata[11,4:339], but that data set is not a vector that can be plottet, it is treated like a set of single values. I tried to use as.vector on the set, bot to no help. I am probably overlooking somehing quite simple, (not to mention not really understanding R's data model..) so help would be appreciated. -- Morten Sickel Norwegian Radiation Protection Authority __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] the svDialogs package
Hello Philippe, Thanks for the explanation! did you change the guidlgopen as well? Now I get : file - guiDlgOpen(title= Open case/control file,defaultFile=,defaultDir=,multi=FALSE, filters = c(All files (*.*), *.*)) gdata - read.table(file,as.is=T,header=T) Error in file(file, r) : unable to open connection In addition: Warning message: cannot open file 'C:/ProgramFiles/R/Work/dcdc2.txt' and the reason is that file [1] C:/ProgramFiles/R/Work/dcdc2.txt so the space between Program and Files has disappeared and the path is wrong correct path would be C:/Program Files/R/Work/dcdc2.txt Marco - Original Message - From: Philippe Grosjean [EMAIL PROTECTED] To: Marco Zucchelli [EMAIL PROTECTED] Cc: R-help r-help@stat.math.ethz.ch Sent: Friday, June 10, 2005 7:36 AM Subject: Re: [R] the svDialogs package Hello Marco, For the first error, the message is clear: not implemented yet!. Several dialog boxes are not done yet, but the functions already exist, mainly as placeholders for future development. Regarding the second, there was a bug in the function (corrected in SciViews 0.8-6 that I have just uploaded to CRAN), and also a misunderstanding of the its first argument: list. This argument should be a charactger vector containing the list of items... but not a list! So, the correct code is: m_list - 1:10 res - guiDlgList(m_list) # Need SciViews 0.8-6! res Note that guiDlgXXX() functions return results _invisibly_. So, you need to assign its result to a variable, or use something like: (guiDlgList(m_list)) to see the result printed at the console. Best, Philippe ..°})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Pentagone (3D08) ( ( ( ( (Academie Universitaire Wallonie-Bruxelles ) ) ) ) ) 8, av du Champ de Mars, 7000 Mons, Belgium ( ( ( ( ( ) ) ) ) ) phone: + 32.65.37.34.97, fax: + 32.65.37.30.54 ( ( ( ( (email: [EMAIL PROTECTED] ) ) ) ) ) ( ( ( ( (web: http://www.umh.ac.be/~econum ) ) ) ) ) http://www.sciviews.org ( ( ( ( ( .. Marco Zucchelli wrote: Hi Philippe and R community, I am trying to use some functions from the svDialogs package but I get some werid errors I do not understand: library(svDialogs) m_list - as.list(1:10) guiDlgDoubleList(m_list, m_list) Error in guiDlgDoubleList(m_list, m_list) : Not yet implemented! guiDlgList(m_list) Error in guiDlgList(m_list) : couldn't find function guiSetFonts.tcltk Am I doing anything wrong ?? Do I need some other package? Marco [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to use expression in label with xYplot
PANTERA Laurent wrote:
Dear R-List,
I want to use the label function (from Hmisc library) to allow for the
names of my isotopes.
library(Hmisc)
library(lattice)
library(grid)
num - c(78,137,129m)
nom - c(Ge,Cs,Te)
df - data.frame(GE78=seq(nom),CS137=seq(nom),TE129m=seq(nom))
if I use this function to create the labels :
lab - function(i)
as.expression(bquote(italic(phantom(0)^{.(num[i])}*.(nom[i]
label(df$GE78) - lab(1)
label(df$CS137) - lab(2)
label(df$TE129m) - lab(3)
all works fine when I use text and xyplot :
plot(1:10)
text(6,6,labels=label(df$CS137))
xyplot(CS137~TE129m,data=df,xlab=label(df$CS137),ylab=label(df$TE129m))
but xYplot doesn't work fine
xYplot(CS137~TE129m,data=df)
I have the message :
Error in parse(file, n, text, prompt) : parse error
if I change the lab function
lab - function(i)
as.character(paste(italic(phantom(0)^{\,num[i],\}*\,nom[i],\),sep=
))
text and xyplot work fine if I use the parse function and xYplot works fine.
label(df$GE78) - lab(1)
label(df$CS137) - lab(2)
label(df$TE129m) - lab(3)
plot(1:10)
text(6,6,labels=parse(text=label(df$CS137)))
xyplot(CS137~TE129m,data=df,xlab=parse(text=label(df$CS137)),ylab=parse(text
=label(df$TE129m)))
xYplot(CS137~TE129m,data=df)
What is the good way to use expression in the labels of variables with the
label function ?
thanks
In Hmisc, labels are used for many contexts in which expressions aren't
allowed, so as of this moment labels must be ordinary character strings.
I typically pass expressions to xYplot using xlab or ylab directly
without using the label.
Frank
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Department of Biostatistics Vanderbilt University
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[R] A polar.plot BUG in plotrix 1.3.3 ?
Hi,
I just updated to R-2.1.1 and updated packages acordingly
However, after the update, routines that use polar.plot
did not function as correctly.
In plotrix 1.3.3 the polar.plot function does scale label.pos
to radians prior to calling radial.plot
Hence, the command
polar.plot(c(5,10,5,0),c(-10,0,10,20),rp.type='P',
labels=c(A,N,V,S),label.pos=c(0,90,180,270))
produces absurd compass lines but the correct results are obtained if
label.pos is scaled with pi/180 :
polar.plot(c(5,10,5,0),c(-10,0,10,20),rp.type='P',
labels=c(A,N,V,S),label.pos=c(0,90,180,270)*pi/180)
I have attatched the polar.plot function from the two different versions
of the package.
It seems that if not missing then label.pos is not changed at all in
version 1.3.3
So, a feature or a bug?
Sincerely,
--
--
Halldor Bjornsson ([EMAIL PROTECTED])
Vedurstofa Islands (Icelandic Met. Office)
Bustadavegur 9, IS-150, Reykjavik, Iceland
--
In version 1.3.3
polar.plot-function(lengths,polar.pos,labels,label.pos,rp.type=r,...) {
npos-length(lengths)
# if no positions are given, add the average distance between
positions so that
# the first and last line don't overlap
if(missing(polar.pos)) radial.pos-seq(0,(2-2/(npos+1))*pi,length=npos)
else radial.pos-pi*polar.pos/180
if(missing(labels)) {
labels-as.character(seq(0,340,by=20))
label.pos-seq(0,1.9*pi,length=18)
}
if(missing(label.pos)) label.pos-pi*label.pos/180
radial.plot(lengths,radial.pos,range(radial.pos),labels,label.pos,
rp.type=rp.type,...)
}
In version 1.2
polar.plot-function(lengths,polar.pos,labels,label.pos,rp.type=r,...) {
npos-length(lengths)
# if no positions are given, add the average distance between
positions so that
# the first and last line don't overlap
if(missing(polar.pos)) polar.pos-seq(0,360-360/(npos+1),length=npos)
if(missing(labels)) {
label.pos-seq(0,340,by=20)
labels-as.character(label.pos)
label.range-c(0,pi*340/180)
}
if(missing(label.pos)) label.pos-polar.pos
polar.range-range(polar.pos)
newrange-c(pi*polar.range[1]/180,pi*(2-(360-polar.range[2])/180))
# rescale to radians
radial.pos-rescale(c(polar.pos,polar.range),newrange)[1:npos]
nlabels-length(labels)
label.pos-rescale(c(label.pos,0,360),c(0,2*pi))[1:nlabels]
radial.plot(lengths,radial.pos,newrange,labels,label.pos,rp.type=rp.type,...)
}
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Re: [R] the svDialogs package
Hello again, I am using the guidlglist I noticed (on windows) that if you scroll the list by using the up and down arrows of the gui by clicking with the mouse, if you click twice fast the gui disappears and the value on top of the list is selected. It is meant to be like that ? Marco - Original Message - From: Philippe Grosjean [EMAIL PROTECTED] To: Marco Zucchelli [EMAIL PROTECTED] Cc: R-help r-help@stat.math.ethz.ch Sent: Friday, June 10, 2005 7:36 AM Subject: Re: [R] the svDialogs package Hello Marco, For the first error, the message is clear: not implemented yet!. Several dialog boxes are not done yet, but the functions already exist, mainly as placeholders for future development. Regarding the second, there was a bug in the function (corrected in SciViews 0.8-6 that I have just uploaded to CRAN), and also a misunderstanding of the its first argument: list. This argument should be a charactger vector containing the list of items... but not a list! So, the correct code is: m_list - 1:10 res - guiDlgList(m_list) # Need SciViews 0.8-6! res Note that guiDlgXXX() functions return results _invisibly_. So, you need to assign its result to a variable, or use something like: (guiDlgList(m_list)) to see the result printed at the console. Best, Philippe ..°})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Pentagone (3D08) ( ( ( ( (Academie Universitaire Wallonie-Bruxelles ) ) ) ) ) 8, av du Champ de Mars, 7000 Mons, Belgium ( ( ( ( ( ) ) ) ) ) phone: + 32.65.37.34.97, fax: + 32.65.37.30.54 ( ( ( ( (email: [EMAIL PROTECTED] ) ) ) ) ) ( ( ( ( (web: http://www.umh.ac.be/~econum ) ) ) ) ) http://www.sciviews.org ( ( ( ( ( .. Marco Zucchelli wrote: Hi Philippe and R community, I am trying to use some functions from the svDialogs package but I get some werid errors I do not understand: library(svDialogs) m_list - as.list(1:10) guiDlgDoubleList(m_list, m_list) Error in guiDlgDoubleList(m_list, m_list) : Not yet implemented! guiDlgList(m_list) Error in guiDlgList(m_list) : couldn't find function guiSetFonts.tcltk Am I doing anything wrong ?? Do I need some other package? Marco [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Snow package -- Results
First, many thanks to Simon Urbanek for his help. A simulation case (50x34x5x3 250 replications) each based on behavior of 2500 consumers takes - 2h20mins on a laptop (2 MHz, 1Go, WinXp) - 1h on a G5 bi-proc (2.5 Mhz, 2Go) A loop was used for the laptop, the same loop was transposed as function for the Mac Best regards Naji __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] A question on time-dependent covariates in the Cox model.
This is a question about time-varying effects rather than time-varying covariates, even if the SAS method tests for the former by using the latter. SAS evaluates the line dosetime=time*dose; for all observations at each event time as it estimates the model, such that you are not using future information. It has the effect of testing for a linear change in the magnitude of the effect of dose over time. I believe Paul Allison's survival book recommends this as a quick and dirty test for constancy of effect. Had you put that line in a datastep prior to PHREG, rather than in PHREG, you'd get a completely different (and uninformative) result (probably the same as R is giving you), because each observation's total survival time would be used to create a single value for the interaction term. You could manually replicate SAS's behaviour in R if you wanted, but every observation would have to start a new time interval whenever any other observation has an event, as Peter explained below. You might also want to look at Aalen's additive survival model for non-linear changes in effect over time: http://www.med.uio.no/imb/stat/addreg/ hope that helps, Jacob Etches On 2005/06/22, at 06:34, Peter Dalgaard wrote: Marianne dk [EMAIL PROTECTED] writes: I have a dataset with event=death time (from medical examination until death/censoring) dose (given at examination time) Two groups are considered, a non-exposed group (dose=0), an exposed group (dose between 5 and 60). For some reason there is a theory of the dose increasing its effect over time (however it was only given (and measured) once = at the time of examination). I tested a model: coxph(Surv(time,dod)~dose + dose:time) Previously I tested the model in SAS: proc phreg data=test; model time*dod(0)=dose dosetime /rl ties=efron; dosetime=time*dose; run; Without the interaction terms I get the same results for the two models. By including the interaction terms I do not. The model in R gives a negative coefficient for the interaction term which is expected to be positive (and is so in SAS). The LRTs are also completely different. TWO QUESTIONS: 1) Is it reasonable to bring in an interaction term when dose is only measured once? 2) If yes, can anyone give a hint on explaining the difference between the models in R and SAS? I don't know what SAS does, maybe it second-guesses your intentions, but R will definitely get it completely wrong. If you use time as a covariate, the same time (of death/censoring) will be applied at all death times. Pretty obviously, long observation times tend to be associated with low mortalities! With interactions you get, er, similarly incorrect effects. To do coxph with time-dependent variables, you need to split data into little time segments, according to the death time of every death, inserting a new variable (ntime, say) which is the time of the endpoint of the interval. -- 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] nlme leading minor error
On 6/22/05, Petr Pikal [EMAIL PROTECTED] wrote: Dear all I am struggling with nlme and error message. Even going through Pinheiro, Bates nlme book did not gave me a clue how to avoid this. fit - nlme(ce ~ fi1 / ((1+exp(fi2-fi3*tepl))^(1/fi4)), data = temp1na.gr, start = c(fi1=30, fi2=-100, fi3=-.05, fi4=40), fixed = fi1+fi2+fi3+fi4~1, random = pdDiag(fi2+fi4~1), groups = ~spol.f) gives Error in chol((value + t(value))/2) : the leading minor of order 1 is not positive definite Is this error due to lack of experimental points? Here you have one typical part of my data. It is for level spol.f = 3/11. teplce 800 28.87 800 29.35 825 29 850 28.73 875 26.83 900 24.07 I have 1-5 points for each level (2 levels with 5 points, 1 level with 4 points, several levels with 2 and 3 points and few with only one point. Fitting this model to each level separately led to several sets of coeficients fi1-fi4 and the separate fits were quite OK. Please give me a hint what can be the cause for this error message and how I shall organize my data to avoid this. (Lack of experimental points is also an answer as I can do some subsequent measurement. The first thing to do is to plot the data for each level of spol.f and see if it is reasonable that you would be able to estimate four parameters from such a curve. Then try setting verbose = TRUE, control = list(msVerbose = TRUE) in your call to nlme to see how the parameters are being changed during the iterations. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] A polar.plot BUG in plotrix 1.3.3 ?
Halldor Björnsson wrote: Hi, I just updated to R-2.1.1 and updated packages acordingly However, after the update, routines that use polar.plot did not function as correctly. In plotrix 1.3.3 the polar.plot function does scale label.pos to radians prior to calling radial.plot Hence, the command polar.plot(c(5,10,5,0),c(-10,0,10,20),rp.type='P', labels=c(A,N,V,S),label.pos=c(0,90,180,270)) produces absurd compass lines but the correct results are obtained if label.pos is scaled with pi/180 : polar.plot(c(5,10,5,0),c(-10,0,10,20),rp.type='P', labels=c(A,N,V,S),label.pos=c(0,90,180,270)*pi/180) I have attatched the polar.plot function from the two different versions of the package. It seems that if not missing then label.pos is not changed at all in version 1.3.3 So, a feature or a bug? Sincerely, Obviously this is a question for the author and maintainer of the package (in CC). Uwe Ligges __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Subsetting across a frame for plotting
Fra: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] På vegne av Liaw, Andy You might save yourself some headaches by turning it into a matrix instead, since all the columns are either integer or numeric: tonedata - data.matrix(tonedata) Data frames are really lists, so even when you get a one-row subset, it's still a one-row data frame. You can use unlist() to turn that into a vector. Andy Great, Andy, thanks! unlist was what I was looking for. Morten __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] r programming help
Dear list, Is there anyway i can make the following formula short by r-programming? CYCLE.n-c(NA, WET[1]*DRY[1], WET[1]*DRY[2]+WET[2]*DRY[1], WET[1]*DRY[3]+WET[2]*DRY[2]+WET[3]*DRY[1], WET[1]*DRY[4]+WET[2]*DRY[3]+WET[3]*DRY[2]+WET[4]*DRY[1], WET[1]*DRY[5]+WET[2]*DRY[4]+WET[3]*DRY[3]+WET[4]*DRY[2]+WET[5]*DRY[1], WET[1]*DRY[6]+WET[2]*DRY[5]+WET[3]*DRY[4]+WET[4]*DRY[3]+WET[5]*DRY[2]+WET[6]*DRY[1], WET[1]*DRY[7]+WET[2]*DRY[6]+WET[3]*DRY[5]+WET[4]*DRY[4]+WET[5]*DRY[3]+WET[6]*DRY[2]+WET[7]*DRY[1], WET[1]*DRY[8]+WET[2]*DRY[7]+WET[3]*DRY[6]+WET[4]*DRY[5]+WET[5]*DRY[4]+WET[6]*DRY[3]+WET[7]*DRY[2]+WET[8]*DRY[1], WET[1]*DRY[9]+WET[2]*DRY[8]+WET[3]*DRY[7]+WET[4]*DRY[6]+WET[5]*DRY[5]+WET[6]*DRY[4]+WET[7]*DRY[3]+WET[8]*DRY[2]+WET[9]*DRY[1], WET[1]*DRY[10]+WET[2]*DRY[9]+WET[3]*DRY[8]+WET[4]*DRY[7]+WET[5]*DRY[6]+WET[6]*DRY[5]+WET[7]*DRY[4]+WET[8]*DRY[3]+WET[9]*DRY[2]+WET[10]*DRY[1], WET[1]*DRY[11]+WET[2]*DRY[10]+WET[3]*DRY[9]+WET[4]*DRY[8]+WET[5]*DRY[7]+WET[6]*DRY[6]+WET[7]*DRY[5]+WET[8]*DRY[4]+WET[9]*DRY[3]+WET[10]*DRY[2]+WET[11]*DRY[1], WET[1]*DRY[12]+WET[2]*DRY[11]+WET[3]*DRY[10]+WET[4]*DRY[9]+WET[5]*DRY[8]+WET[6]*DRY[7]+WET[7]*DRY[6]+WET[8]*DRY[5]+WET[9]*DRY[4]+WET[10]*DRY[3]+WET[11]*DRY[2]+WET[12]*DRY[1], WET[1]*DRY[13]+WET[2]*DRY[12]+WET[3]*DRY[11]+WET[4]*DRY[10]+WET[5]*DRY[9]+WET[6]*DRY[8]+WET[7]*DRY[7]+WET[8]*DRY[6]+WET[9]*DRY[5]+WET[10]*DRY[4]+WET[11]*DRY[3]+WET[12]*DRY[2]+WET[13]*DRY[1], WET[1]*DRY[14]+WET[2]*DRY[13]+WET[3]*DRY[12]+WET[4]*DRY[11]+WET[5]*DRY[10]+WET[6]*DRY[9]+WET[7]*DRY[8]+WET[8]*DRY[7]+WET[9]*DRY[6]+WET[10]*DRY[5]+WET[11]*DRY[4]+WET[12]*DRY[3]+WET[13]*DRY[2]+WET[14]*DRY[1], WET[1]*DRY[15]+WET[2]*DRY[14]+WET[3]*DRY[13]+WET[4]*DRY[12]+WET[5]*DRY[11]+WET[6]*DRY[10]+WET[7]*DRY[9]+WET[8]*DRY[8]+WET[9]*DRY[7]+WET[10]*DRY[6]+WET[11]*DRY[5]+WET[12]*DRY[4]+WET[13]*DRY[3]+WET[14]*DRY[2]+WET[15]*DRY[1], WET[1]*DRY[16]+WET[2]*DRY[15]+WET[3]*DRY[14]+WET[4]*DRY[13]+WET[5]*DRY[12]+WET[6]*DRY[11]+WET[7]*DRY[10]+WET[8]*DRY[9]+WET[9]*DRY[8]+WET[10]*DRY[7]+WET[11]*DRY[6]+WET[12]*DRY[5]+WET[13]*DRY[4]+WET[14]*DRY[3]+WET[15]*DRY[2]+WET[16]*DRY[1], WET[1]*DRY[17]+WET[2]*DRY[16]+WET[3]*DRY[15]+WET[4]*DRY[14]+WET[5]*DRY[13]+WET[6]*DRY[12]+WET[7]*DRY[11]+WET[8]*DRY[10]+WET[9]*DRY[9]+WET[10]*DRY[8]+WET[11]*DRY[7]+WET[12]*DRY[6]+WET[13]*DRY[5]+WET[14]*DRY[4]+WET[15]*DRY[3]+WET[16]*DRY[2]+WET[17]*DRY[1], WET[1]*DRY[18]+WET[2]*DRY[17]+WET[3]*DRY[16]+WET[4]*DRY[15]+WET[5]*DRY[14]+WET[6]*DRY[13]+WET[7]*DRY[12]+WET[8]*DRY[11]+WET[9]*DRY[10]+WET[10]*DRY[9]+WET[11]*DRY[8]+WET[12]*DRY[7]+WET[13]*DRY[6]+WET[14]*DRY[5]+WET[15]*DRY[4]+WET[16]*DRY[3]+WET[17]*DRY[2]+WET[18]*DRY[1], WET[1]*DRY[19]+WET[2]*DRY[18]+WET[3]*DRY[17]+WET[4]*DRY[16]+WET[5]*DRY[15]+WET[6]*DRY[15]+WET[7]*DRY[13]+WET[8]*DRY[12]+WET[9]*DRY[11]+WET[10]*DRY[10]+WET[11]*DRY[9]+WET[12]*DRY[8]+WET[13]*DRY[7]+WET[14]*DRY[6]+WET[15]*DRY[5]+WET[16]*DRY[4]+WET[17]*DRY[3]+WET[18]*DRY[2]+WET[19]*DRY[1], WET[1]*DRY[20]+WET[2]*DRY[19]+WET[3]*DRY[18]+WET[4]*DRY[17]+WET[5]*DRY[16]+WET[6]*DRY[15]+WET[7]*DRY[14]+WET[8]*DRY[13]+WET[9]*DRY[12]+WET[10]*DRY[11]+WET[11]*DRY[10]+WET[12]*DRY[9]+WET[13]*DRY[8]+WET[14]*DRY[7]+WET[15]*DRY[6]+WET[16]*DRY[5]+WET[17]*DRY[4]+WET[18]*DRY[3]+WET[19]*DRY[2]+WET[20]*DRY[1], WET[1]*DRY[21]+WET[2]*DRY[20]+WET[3]*DRY[19]+WET[4]*DRY[18]+WET[5]*DRY[17]+WET[6]*DRY[16]+WET[7]*DRY[15]+WET[8]*DRY[14]+WET[9]*DRY[13]+WET[10]*DRY[12]+WET[11]*DRY[11]+WET[12]*DRY[10]+WET[13]*DRY[9]+WET[14]*DRY[8]+WET[15]*DRY[7]+WET[16]*DRY[6]+WET[17]*DRY[5]+WET[18]*DRY[4]+WET[19]*DRY[3]+WET[20]*DRY[2]+WET[21]*DRY[1], WET[1]*DRY[22]+WET[2]*DRY[21]+WET[3]*DRY[20]+WET[4]*DRY[19]+WET[5]*DRY[18]+WET[6]*DRY[17]+WET[7]*DRY[16]+WET[8]*DRY[15]+WET[9]*DRY[14]+WET[10]*DRY[13]+WET[11]*DRY[12]+WET[12]*DRY[11]+WET[13]*DRY[10]+WET[14]*DRY[9]+WET[15]*DRY[8]+WET[16]*DRY[7]+WET[17]*DRY[6]+WET[18]*DRY[5]+WET[19]*DRY[4]+WET[20]*DRY[3]+WET[21]*DRY[2]+WET[22]*DRY[1], WET[1]*DRY[23]+WET[2]*DRY[22]+WET[3]*DRY[21]+WET[4]*DRY[20]+WET[5]*DRY[19]+WET[6]*DRY[18]+WET[7]*DRY[17]+WET[8]*DRY[16]+WET[9]*DRY[15]+WET[10]*DRY[14]+WET[11]*DRY[13]+WET[12]*DRY[12]+WET[13]*DRY[11]+WET[14]*DRY[10]+WET[15]*DRY[9]+WET[16]*DRY[8]+WET[17]*DRY[7]+WET[18]*DRY[6]+WET[19]*DRY[5]+WET[20]*DRY[4]+WET[21]*DRY[3]+WET[22]*DRY[2]+WET[23]*DRY[1], WET[1]*DRY[24]+WET[2]*DRY[23]+WET[3]*DRY[22]+WET[4]*DRY[21]+WET[5]*DRY[20]+WET[6]*DRY[19]+WET[7]*DRY[18]+WET[8]*DRY[17]+WET[9]*DRY[16]+WET[10]*DRY[15]+WET[11]*DRY[14]+WET[12]*DRY[13]+WET[13]*DRY[12]+WET[14]*DRY[11]+WET[15]*DRY[10]+WET[16]*DRY[9]+WET[17]*DRY[8]+WET[18]*DRY[7]+WET[19]*DRY[6]+WET[20]*DRY[5]+WET[21]*DRY[4]+WET[22]*DRY[3]+WET[23]*DRY[2]+WET[24]*DRY[1], WET[1]*DRY[25]+WET[2]*DRY[24]+WET[3]*DRY[23]+WET[4]*DRY[22]+WET[5]*DRY[21]+WET[6]*DRY[20]+WET[7]*DRY[19]+WET[8]*DRY[18]+WET[9]*DRY[17]+WET[10]*DRY[16]+WET[11]*DRY[15]+WET[12]*DRY[14]+WET[13]*DRY[13]+WET[14]*DRY[12]+WET[15]*DRY[11]+WET[16]*DRY[10]+WET[17]*DRY[9]+WET[18]*DRY[8]+WET[19]*DRY[7]+WET[20]*DRY[6]+WET[21]*DRY[5]+WET[22]*DRY[4]+WET[23]*DRY[3]+WET[24]*DRY[2]+WET[25]*DRY[1],
[R] Extract Coeff, Std Error, etc from gnls output
Dear list members; Is there any trick to extract the coefficients along with std errors, t-values and p-values for each beta from a gnls fit model (similar to the results obtained using summary(lm)$coeff for linear models)? Thanks for any hint cm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Contour Plots
Hello all. I'm confused a bit about contour plots. After reading the help at ?contour, it seems as though the contour plot is for 3D plots (x,y, and z). My data is in the form of grid coordinates (x,y), and I want to see a contour plot of the data so that I can tell where most observations lie. My question is simple but still evasive. Say my data is called places. One column is X and another is Y. Or, places$X places$Y Coordinate 1 32 50 Coordinate 2 15 33 Coordinate 3 28 20 etc How do I get R to do the contour plot with no third z coordinate? Thanks. -- Do all you can with what you have in the time you have in the place you are! -Nkosi Johnson, 12-year old African hero __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Problem trying to use boot and lme together
At 23:09 21/06/05, Prof Brian Ripley wrote: On Tue, 21 Jun 2005, Douglas Bates wrote: On 6/21/05, Søren Højsgaard [EMAIL PROTECTED] wrote: Thanks everyone for your help, more comments at the foot The problem with simulate.lme is that it only returns logL for a given model fitted to a simulated data set - not the simulated data set itself (which one might have expected a function with that name to do...). It would be nice with that functionality... Søren You could add it. You just need to create a matrix that will be large enough to hold all the simulated data sets and fill a column (or row if you prefer but column is probably better because of the way that matrices are stored) during each iteration of the simulation and remember to include that matrix in the returned object. Note: you don't need to store it: you can do the analysis at that point and return the statistics you want, rather than just logL. I did say `see simulate.lme', not `use simulate.lme'. I know nlme is no longer being developed, but if it were I would be suggesting/contributing a modification that allowed the user to specify an `extraction' function from the fit -- quite a few pieces of bootstrap code work that way. -- 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 This has been most informative for me. Given the rather stern comments in Pinheiro and Bates that most things do not have the reference distribution you might naively think I now realise that simulate.lme is more important than its rather cursory treatment in the book. As suggested I have looked at the code but although I can see broadly what each section does I lack the skill to modify it myself. I will have to wait for someone more gifted. If there is to be a successor edition to Pinheiro and Bates perhaps I could suggest that this topic merits a bit more discussion? Michael Dewey [EMAIL PROTECTED] http://www.aghmed.fsnet.co.uk/home.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Contour Plots
I guess what you want is contours of the density over the x-y plane. There are a few choices that I know of: kde2d in MASS (part of the `VR' bundle) bkde2d in KernSmooth sm.density in sm locfit in locfit Andy From: Bernard L. Dillard Hello all. I'm confused a bit about contour plots. After reading the help at ?contour, it seems as though the contour plot is for 3D plots (x,y, and z). My data is in the form of grid coordinates (x,y), and I want to see a contour plot of the data so that I can tell where most observations lie. My question is simple but still evasive. Say my data is called places. One column is X and another is Y. Or, places$X places$Y Coordinate 1 32 50 Coordinate 2 15 33 Coordinate 3 28 20 etc How do I get R to do the contour plot with no third z coordinate? Thanks. -- Do all you can with what you have in the time you have in the place you are! -Nkosi Johnson, 12-year old African hero __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Kouros Owzar is ooo.
I will be out of the office starting 06/21/2005 and will not return until 06/29/2005. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] r programming help
Mohammad Ehsanul Karim [EMAIL PROTECTED] writes: Dear list, Is there anyway i can make the following formula short by r-programming? CYCLE.n-c(NA, WET[1]*DRY[1], WET[1]*DRY[2]+WET[2]*DRY[1], WET[1]*DRY[3]+WET[2]*DRY[2]+WET[3]*DRY[1], As far as I can see: z - toeplitz(DRY) z[upper.tri(z)] - 0 c(NA, z %*% WET) or convolve() with suitable options, padding, and/or cutting (but beware, there could be devils in the details). convolve(WET,DRY, type=o) gives you about twice what you need, I believe. -- 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] (slightly off topic, but...) More of a stat design question...
With such a wide range of backgrounds here, I thought I'd toss this out here to get ideas. I've lucked into some clinical trial data where schizophrenic patients were randomly assigned to start on one of three drugs, then were followed naturalistically over a year (or more, depending on when they enrolled). They were assessed with a standard battery of instruments at baseline, and at regular intervals. Apart from the initial random assignment, treatment decisions were left up to patients and their doctors. Thus, one of the main outcomes of the trial is the all-cause time to discontinuation which is believed to be the patient-centric tipping point where the risks and costs of using the drug outweight the benefit, so the patient either switches or just stops. The outcome variable, therefore, is a censored time variable. I have the number of days, and a flag to identify if that number is actual discontinuation or just the end of observation. Now, the a priori analysis is done and gone (Cox regression of which drug had longest time to d/c), and a more clinical question has come up: using these data, can we build a model to predict, for each patient, the drug which is likely to be best for that individual. Sounds like a bayesian opportunity to me, a separate model for each drug based on an analysis of the patients randomized to each drug, then apply each model to a holdout sample, and see if patients matched to therapy did better than patients who were not. If I were just predicting a single continuous measure, or a dichotomous outcome, I'd have no problem. The question is, given that the outcome is a censored time variable, which approaches would get me closest to the posterior probability of success or expected number of days, give or take Y in the clinical sense, not the frequentist sense (if you repeated this study K times,...)? Just interested in hearing some thoughts on this. --- David L. Van Brunt, Ph.D. mailto:[EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] predict.coxph fitted values for failure times
On Wed, 22 Jun 2005, Dan Bebber wrote: I would like to extract predicted failure times from a coxph model in library(survival). However, none of the prediction options (lp, risk, expected, terms) seem to bear any relationship to failure time. Perhaps I am asking the wrong question, but can coxph provide predicted failure times? It's tricky, because it depends what you mean by predicted. It's typically impossible to estimate the mean survival time for given covariates when there is censoring. You can use survfit() on your Cox model to get predicted survival curves. -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Is it possible to get the first letter of a word?
Hi, I would to get the first letter of a word like: title_cat TitleCat 1 Training I would like T from Training! Thnaks a lot for your help Sabine - Téléchargez le ici ! [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Is it possible to get the first letter of a word?
Navarre Sabine wrote: Hi, I would to get the first letter of a word like: title_cat TitleCat 1 Training I would like T from Training! Thnaks a lot for your help substr(title_cat,1,1) Uwe Ligges Sabine - Téléchargez le ici ! [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Is it possible to get the first letter of a word?
Use substring() or substr(). Andy From: Navarre Sabine Hi, I would to get the first letter of a word like: title_cat TitleCat 1 Training I would like T from Training! Thnaks a lot for your help Sabine - Téléchargez le ici ! [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Is it possible to get the first letter of a word?
?substring Navarre Sabine wrote: Hi, I would to get the first letter of a word like: title_cat TitleCat 1 Training I would like T from Training! Thnaks a lot for your help Sabine - Téléchargez le ici ! [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 452-1424 (M, W, F) fax: (917) 438-0894 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Is it possible to get the first letter of a word?
or What about; strsplit(Training, split=)[[1]][1] [1] T Ken Knoblauch Inserm U371, Cerveau et Vision Department of Cognitive Neurosciences 18 avenue du Doyen Lepine 69500 Bron France tel: +33 (0)4 72 91 34 77 fax: +33 (0)4 72 91 34 61 portable: 06 84 10 64 10 http://www.lyon.inserm.fr/371/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Is it possible to get the first letter of a word?
On Wed, 2005-06-22 at 16:42 +0200, Navarre Sabine wrote: Hi, I would to get the first letter of a word like: title_cat TitleCat 1 Training I would like T from Training! Thnaks a lot for your help Sabine There are multiple approaches, but you need to be careful, since it appears that your object is a factor. Thus you may need to convert to a character vector first: title_cat - factor(Training) substr(as.character(title_cat), 1, 1) [1] T Otherwise: title_cat - Training substr(title_cat, 1, 1) [1] T See ?substr for more information. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Contour Plots
Bernard L. Dillard wrote: Hello all. I'm confused a bit about contour plots. After reading the help at ?contour, it seems as though the contour plot is for 3D plots (x,y, and z). My data is in the form of grid coordinates (x,y), and I want to see a contour plot of the data so that I can tell where most observations lie. My question is simple but still evasive. Say my data is called places. One column is X and another is Y. Or, places$X places$Y Coordinate 1 32 50 Coordinate 2 15 33 Coordinate 3 28 20 etc How do I get R to do the contour plot with no third z coordinate? For just 2 columns, a contour plot does not make much sense, I guess. I think an image() makes much more sense in this case. Your z matrix consists of the entries of the data.frame/matrix given above, the x and y vectors are the column and row numbers. Uwe Ligges Thanks. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Howto crosstable-ing......
I receive the following meteo dataset regularly, containing the average daily temperatures (tMedia) of a certain month for 24 selected meteo-stations (COD_WMO) whose human-readable names are in (NOME). str(tabella) `data.frame': 1038 obs. of 4 variables: $ COD_WMO: int 16045 16045 16045 16045 16045 16045 16045 16045 16045 16045 ... $ NOME : Factor w/ 24 levels ALGHERO,BARI/PALESE MACCHIE,..: 22 22 22 22 22 22 22 22 22 22 ... $ DATE :'POSIXct', format: chr 2005-05-01 2005-05-02 2005-05-03 2005-05-04 ... $ tMedia : num 11.7 18.6 16.9 19.7 15.0 ... Here you are a short list of it: COD_WMO NOME DATE tMedia 505 16191 FALCONARA 2005-06-01 20.95 506 16191 FALCONARA 2005-06-02 20.15 507 16191 FALCONARA 2005-06-03 18.60 506 16191 FALCONARA 2005-06-02 20.15 507 16191 FALCONARA 2005-06-03 18.60 508 16191 FALCONARA 2005-06-04 22.30 509 16191 FALCONARA 2005-06-05 NA 510 16191 FALCONARA 2005-06-06 NA 511 16191 FALCONARA 2005-06-07 18.20 549 16206 GROSSETO 2005-06-01 20.65 550 16206 GROSSETO 2005-06-02 21.95 551 16206 GROSSETO 2005-06-03 22.25 552 16206 GROSSETO 2005-06-04 20.15 553 16206 GROSSETO 2005-06-05 NA 554 16206 GROSSETO 2005-06-06 NA 555 16206 GROSSETO 2005-06-07 22.35 . . I need to rearrange tMedia into a new dataframe whose column names are COD_WMO (or NOME) and the row is DATE. ex. DATEALGHERO BARI/PALESE FALCONARA GROSSETO .. 2005-06-01 16.3 12.8 17.3 14.0 ... 2005-06-02 18.2 8.918.0 17.9 .. ... I read some pieces of R-docs in the internet and run the MASS chapter 2 examples but without finding anything suitable to my purpose. Could you please help me? Ciao Vittorio __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] legend
Thomas Steiner wrote:
I color some area grey with polygon() (with a red border) and then I
want to have the dashed red border in the legend as well. How do I
manage it?
And I want to mix (latex) expressions with text in my legend.
Both points are not that easy to solve, hence I'd like to suggest to
write your own little function that generates the legend.
Starting at the upper left, calculating the stringheight, painting the
(party very special) symbols, and adding the text line by line seems to
be the most easiest solution here (which is not that nice, though.
Uwe Ligges
Just execute my lines below and you know want I mean. Or pass by at
http://de.wikipedia.org/wiki/Bild:GBM.png to see the picture online.
Thomas
bm - function(n=500, from=0, to=1) {
x=seq(from=from,to=to,length=n)
BM-c(0,cumsum(rnorm(n-1,mean=0,sd=sqrt(to/n
cbind(x,BM)
}
gbm - function(bm,S0=1,sigma=0.1,mu=1) {
gbm=S0
for (t in 2:length(bm[,1])) {
gbm[t]=S0*exp((mu-sigma^2/2)*bm[t,1]+sigma*bm[t,2])
}
cbind(bm[,1],gbm)
}
set.seed(9826064)
cs=c(dark green, steelblue, red, yellow)
#png(filename = GBM.png, width=1600, height=1200, pointsize = 12)
par(bg=lightgrey)
x=seq(from=0,to=1,length=500)
plot(x=x, y=exp(0.7*x), type=n, xlab=Zeit, ylab=, ylim=c(1,3.5))
polygon(x=c(x,rev(x)),
y=c(exp(0.7*x)+0.4*sqrt(x),rev(exp(0.7*x)-0.4*sqrt(x))), col=grey,
border=cs[3], lty=dashed)
lines(x=x,y=exp(0.7*x), type=l, lwd=3, col=cs[1])
lines(gbm(bm(),S0=1,mu=0.7,sigma=0.4), lwd=3, col=cs[2])
lines(gbm(bm(),S0=1,mu=0.7,sigma=0.2), lwd=3, col=cs[3])
lines(gbm(bm(),S0=1,mu=0.7,sigma=0.1), lwd=3, col=cs[4])
title(main=Geometrische Brownsche Bewegung,cex.main=2.5)
legend(x=0,y=3.5,legend=c(exp(0.7x),mu=0.7, sigma=0.4,mu=0.7,
sigma=0.2,mu=0.7, sigma=0.1,Standardabweichung für
sigma=0.2),lwd=c(4,4,4,4,12),col=c(cs,grey),bg=transparent,cex=1.15)
#dev.off()
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Re: [R] Howto crosstable-ing......
See ?reshape Uwe Ligges [EMAIL PROTECTED] wrote: I receive the following meteo dataset regularly, containing the average daily temperatures (tMedia) of a certain month for 24 selected meteo-stations (COD_WMO) whose human-readable names are in (NOME). str(tabella) `data.frame': 1038 obs. of 4 variables: $ COD_WMO: int 16045 16045 16045 16045 16045 16045 16045 16045 16045 16045 ... $ NOME : Factor w/ 24 levels ALGHERO,BARI/PALESE MACCHIE,..: 22 22 22 22 22 22 22 22 22 22 ... $ DATE :'POSIXct', format: chr 2005-05-01 2005-05-02 2005-05-03 2005-05-04 ... $ tMedia : num 11.7 18.6 16.9 19.7 15.0 ... Here you are a short list of it: COD_WMO NOME DATE tMedia 505 16191 FALCONARA 2005-06-01 20.95 506 16191 FALCONARA 2005-06-02 20.15 507 16191 FALCONARA 2005-06-03 18.60 506 16191 FALCONARA 2005-06-02 20.15 507 16191 FALCONARA 2005-06-03 18.60 508 16191 FALCONARA 2005-06-04 22.30 509 16191 FALCONARA 2005-06-05 NA 510 16191 FALCONARA 2005-06-06 NA 511 16191 FALCONARA 2005-06-07 18.20 549 16206 GROSSETO 2005-06-01 20.65 550 16206 GROSSETO 2005-06-02 21.95 551 16206 GROSSETO 2005-06-03 22.25 552 16206 GROSSETO 2005-06-04 20.15 553 16206 GROSSETO 2005-06-05 NA 554 16206 GROSSETO 2005-06-06 NA 555 16206 GROSSETO 2005-06-07 22.35 . . I need to rearrange tMedia into a new dataframe whose column names are COD_WMO (or NOME) and the row is DATE. ex. DATEALGHERO BARI/PALESE FALCONARA GROSSETO .. 2005-06-01 16.3 12.8 17.3 14.0 ... 2005-06-02 18.2 8.918.0 17.9 .. ... I read some pieces of R-docs in the internet and run the MASS chapter 2 examples but without finding anything suitable to my purpose. Could you please help me? Ciao Vittorio __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] chisq test and fisher exact test
Hi, I have a text mining project and currently I am working on feature generation/selection part. My plan is selecting a set of words or word combinations which have better discriminant capability than other words in telling the group id's (2 classes in this case) for a dataset which has 2,000,000 documents. One approach is using contrast-set association rule mining while the other is using chisqr or fisher exact test. An example which has 3 contingency tables for 3 words as followed (word coded by number): tab[,,1:3] , , 1 [,1][,2] [1,] 11266 2151526 [2,] 125 31734 , , 2 [,1][,2] [1,] 43571 2119221 [2,]52 31807 , , 3 [,1][,2] [1,] 427 2162365 [2,]5 31854 I have some questions on this: 1. What's the thumb of rule to use chisq test instead of Fisher exact test. I have a vague memory which said for each cell, the count needs to be over 50 if chisq instead of fisher exact test is going to be used. In the case of word 3, I think I should use fisher test. However, running chisq like below is fine: tab[,,3] [,1][,2] [1,] 427 2162365 [2,]5 31854 chisq.test(tab[,,3]) Pearson's Chi-squared test with Yates' continuity correction data: tab[, , 3] X-squared = 0.0963, df = 1, p-value = 0.7564 but running on the whole set of words (including 14240 words) has the following warnings: p.chisq-as.double(lapply(1:N, function(i) chisq.test(tab[,,i])$p.value)) There were 50 or more warnings (use warnings() to see the first 50) warnings() Warning messages: 1: Chi-squared approximation may be incorrect in: chisq.test(tab[, , i]) 2: Chi-squared approximation may be incorrect in: chisq.test(tab[, , i]) 3: Chi-squared approximation may be incorrect in: chisq.test(tab[, , i]) 4: Chi-squared approximation may be incorrect in: chisq.test(tab[, , i]) 2. So, my second question is, is this warning b/c I am against the assumption of using chisq. But why Word 3 is fine? How to trace the warning to see which word caused this warning? 3. My result looks like this (after some mapping treating from number id to word and some words are stemmed here, like ACCID is accident): of[1:50,] map...2. p.fisher 21 ACCID 0.00e+00 30 CD 0.00e+00 67ROCK 0.00e+00 104 CRACK 0.00e+00 111 CHIP 0.00e+00 179 GLASS 0.00e+00 84BACK 4.199878e-291 395 DRIVEABL 5.335989e-287 60 CAP 9.405235e-285 262 WINDSHIELD 2.691641e-254 13 IV 3.905186e-245 110 HZ 2.819713e-210 11CAMP 9.086768e-207 2 SHATTER 5.273994e-202 297ALP 1.678521e-177 162BED 1.822031e-173 249BCD 1.398391e-160 493 RACK 4.178617e-156 59CAUS 7.539031e-147 3.1 question: Should I use two-sided test instead of one-sided for fisher test? I read some material which suggests using two-sided. 3.2 A big question: Even though the result looks very promising since this is case of classiying fraud cases and the words selected by this approach make sense. However, I think p-values here just indicate the strength to reject null hypothesis, not the strength of association between word and class of document. So, what kind of statistics I should use here to evaluate the strength of association? odds ratio? Any suggestions are welcome! Thanks! -- Weiwei Shi, Ph.D Did you always know? No, I did not. But I believed... ---Matrix III __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Howto crosstable-ing......
Looks to me like you want something like reshape()... Andy From: [EMAIL PROTECTED] I receive the following meteo dataset regularly, containing the average daily temperatures (tMedia) of a certain month for 24 selected meteo-stations (COD_WMO) whose human-readable names are in (NOME). str(tabella) `data.frame': 1038 obs. of 4 variables: $ COD_WMO: int 16045 16045 16045 16045 16045 16045 16045 16045 16045 16045 ... $ NOME : Factor w/ 24 levels ALGHERO,BARI/PALESE MACCHIE,..: 22 22 22 22 22 22 22 22 22 22 ... $ DATE :'POSIXct', format: chr 2005-05-01 2005-05-02 2005-05-03 2005-05-04 ... $ tMedia : num 11.7 18.6 16.9 19.7 15.0 ... Here you are a short list of it: COD_WMO NOME DATE tMedia 505 16191 FALCONARA 2005-06-01 20.95 506 16191 FALCONARA 2005-06-02 20.15 507 16191 FALCONARA 2005-06-03 18.60 506 16191 FALCONARA 2005-06-02 20.15 507 16191 FALCONARA 2005-06-03 18.60 508 16191 FALCONARA 2005-06-04 22.30 509 16191 FALCONARA 2005-06-05 NA 510 16191 FALCONARA 2005-06-06 NA 511 16191 FALCONARA 2005-06-07 18.20 549 16206 GROSSETO 2005-06-01 20.65 550 16206 GROSSETO 2005-06-02 21.95 551 16206 GROSSETO 2005-06-03 22.25 552 16206 GROSSETO 2005-06-04 20.15 553 16206 GROSSETO 2005-06-05 NA 554 16206 GROSSETO 2005-06-06 NA 555 16206 GROSSETO 2005-06-07 22.35 . . I need to rearrange tMedia into a new dataframe whose column names are COD_WMO (or NOME) and the row is DATE. ex. DATEALGHERO BARI/PALESE FALCONARA GROSSETO .. 2005-06-01 16.3 12.8 17.3 14.0 ... 2005-06-02 18.2 8.9 18.0 17.9 .. .. . I read some pieces of R-docs in the internet and run the MASS chapter 2 examples but without finding anything suitable to my purpose. Could you please help me? Ciao Vittorio __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] r programming help
Try: DRY-c(2,5,3,7,11) WET-(1:5)*10 print(filter(c(rep(0,length(WET)),DRY),WET)) Time Series: Start = 1 End = 10 Frequency = 1 [1] NA NA 0 20 90 190 360 640 NA NA CYCLE.n-c(NA, WET[1]*DRY[1], WET[1]*DRY[2]+WET[2]*DRY[1], WET[1]*DRY[3]+WET[2]*DRY[2]+WET[3]*DRY[1], WET[1]*DRY[4]+WET[2]*DRY[3]+WET[3]*DRY[2]+WET[4]*DRY[1], WET[1]*DRY[5]+WET[2]*DRY[4]+WET[3]*DRY[3]+WET[4]*DRY[2]+WET[5]*DRY[1]) print(CYCLE.n) [1] NA 20 90 190 360 640 Peter Wolf Mohammad Ehsanul Karim wrote: Dear list, Is there anyway i can make the following formula short by r-programming? CYCLE.n-c(NA, WET[1]*DRY[1], WET[1]*DRY[2]+WET[2]*DRY[1], WET[1]*DRY[3]+WET[2]*DRY[2]+WET[3]*DRY[1], WET[1]*DRY[4]+WET[2]*DRY[3]+WET[3]*DRY[2]+WET[4]*DRY[1], WET[1]*DRY[5]+WET[2]*DRY[4]+WET[3]*DRY[3]+WET[4]*DRY[2]+WET[5]*DRY[1], WET[1]*DRY[6]+WET[2]*DRY[5]+WET[3]*DRY[4]+WET[4]*DRY[3]+WET[5]*DRY[2]+WET[6]*DRY[1], WET[1]*DRY[7]+WET[2]*DRY[6]+WET[3]*DRY[5]+WET[4]*DRY[4]+WET[5]*DRY[3]+WET[6]*DRY[2]+WET[7]*DRY[1], WET[1]*DRY[8]+WET[2]*DRY[7]+WET[3]*DRY[6]+WET[4]*DRY[5]+WET[5]*DRY[4]+WET[6]*DRY[3]+WET[7]*DRY[2]+WET[8]*DRY[1], WET[1]*DRY[9]+WET[2]*DRY[8]+WET[3]*DRY[7]+WET[4]*DRY[6]+WET[5]*DRY[5]+WET[6]*DRY[4]+WET[7]*DRY[3]+WET[8]*DRY[2]+WET[9]*DRY[1], WET[1]*DRY[10]+WET[2]*DRY[9]+WET[3]*DRY[8]+WET[4]*DRY[7]+WET[5]*DRY[6]+WET[6]*DRY[5]+WET[7]*DRY[4]+WET[8]*DRY[3]+WET[9]*DRY[2]+WET[10]*DRY[1], WET[1]*DRY[11]+WET[2]*DRY[10]+WET[3]*DRY[9]+WET[4]*DRY[8]+WET[5]*DRY[7]+WET[6]*DRY[6]+WET[7]*DRY[5]+WET[8]*DRY[4]+WET[9]*DRY[3]+WET[10]*DRY[2]+WET[11]*DRY[1], WET[1]*DRY[12]+WET[2]*DRY[11]+WET[3]*DRY[10]+WET[4]*DRY[9]+WET[5]*DRY[8]+WET[6]*DRY[7]+WET[7]*DRY[6]+WET[8]*DRY[5]+WET[9]*DRY[4]+WET[10]*DRY[3]+WET[11]*DRY[2]+WET[12]*DRY[1], WET[1]*DRY[13]+WET[2]*DRY[12]+WET[3]*DRY[11]+WET[4]*DRY[10]+WET[5]*DRY[9]+WET[6]*DRY[8]+WET[7]*DRY[7]+WET[8]*DRY[6]+WET[9]*DRY[5]+WET[10]*DRY[4]+WET[11]*DRY[3]+WET[12]*DRY[2]+WET[13]*DRY[1], WET[1]*DRY[14]+WET[2]*DRY[13]+WET[3]*DRY[12]+WET[4]*DRY[11]+WET[5]*DRY[10]+WET[6]*DRY[9]+WET[7]*DRY[8]+WET[8]*DRY[7]+WET[9]*DRY[6]+WET[10]*DRY[5]+WET[11]*DRY[4]+WET[12]*DRY[3]+WET[13]*DRY[2]+WET[14]*DRY[1], WET[1]*DRY[15]+WET[2]*DRY[14]+WET[3]*DRY[13]+WET[4]*DRY[12]+WET[5]*DRY[11]+WET[6]*DRY[10]+WET[7]*DRY[9]+WET[8]*DRY[8]+WET[9]*DRY[7]+WET[10]*DRY[6]+WET[11]*DRY[5]+WET[12]*DRY[4]+WET[13]*DRY[3]+WET[14]*DRY[2]+WET[15]*DRY[1], WET[1]*DRY[16]+WET[2]*DRY[15]+WET[3]*DRY[14]+WET[4]*DRY[13]+WET[5]*DRY[12]+WET[6]*DRY[11]+WET[7]*DRY[10]+WET[8]*DRY[9]+WET[9]*DRY[8]+WET[10]*DRY[7]+WET[11]*DRY[6]+WET[12]*DRY[5]+WET[13]*DRY[4]+WET[14]*DRY[3]+WET[15]*DRY[2]+WET[16]*DRY[1], WET[1]*DRY[17]+WET[2]*DRY[16]+WET[3]*DRY[15]+WET[4]*DRY[14]+WET[5]*DRY[13]+WET[6]*DRY[12]+WET[7]*DRY[11]+WET[8]*DRY[10]+WET[9]*DRY[9]+WET[10]*DRY[8]+WET[11]*DRY[7]+WET[12]*DRY[6]+WET[13]*DRY[5]+WET[14]*DRY[4]+WET[15]*DRY[3]+WET[16]*DRY[2]+WET[17]*DRY[1], WET[1]*DRY[18]+WET[2]*DRY[17]+WET[3]*DRY[16]+WET[4]*DRY[15]+WET[5]*DRY[14]+WET[6]*DRY[13]+WET[7]*DRY[12]+WET[8]*DRY[11]+WET[9]*DRY[10]+WET[10]*DRY[9]+WET[11]*DRY[8]+WET[12]*DRY[7]+WET[13]*DRY[6]+WET[14]*DRY[5]+WET[15]*DRY[4]+WET[16]*DRY[3]+WET[17]*DRY[2]+WET[18]*DRY[1], WET[1]*DRY[19]+WET[2]*DRY[18]+WET[3]*DRY[17]+WET[4]*DRY[16]+WET[5]*DRY[15]+WET[6]*DRY[15]+WET[7]*DRY[13]+WET[8]*DRY[12]+WET[9]*DRY[11]+WET[10]*DRY[10]+WET[11]*DRY[9]+WET[12]*DRY[8]+WET[13]*DRY[7]+WET[14]*DRY[6]+WET[15]*DRY[5]+WET[16]*DRY[4]+WET[17]*DRY[3]+WET[18]*DRY[2]+WET[19]*DRY[1], WET[1]*DRY[20]+WET[2]*DRY[19]+WET[3]*DRY[18]+WET[4]*DRY[17]+WET[5]*DRY[16]+WET[6]*DRY[15]+WET[7]*DRY[14]+WET[8]*DRY[13]+WET[9]*DRY[12]+WET[10]*DRY[11]+WET[11]*DRY[10]+WET[12]*DRY[9]+WET[13]*DRY[8]+WET[14]*DRY[7]+WET[15]*DRY[6]+WET[16]*DRY[5]+WET[17]*DRY[4]+WET[18]*DRY[3]+WET[19]*DRY[2]+WET[20]*DRY[1], WET[1]*DRY[21]+WET[2]*DRY[20]+WET[3]*DRY[19]+WET[4]*DRY[18]+WET[5]*DRY[17]+WET[6]*DRY[16]+WET[7]*DRY[15]+WET[8]*DRY[14]+WET[9]*DRY[13]+WET[10]*DRY[12]+WET[11]*DRY[11]+WET[12]*DRY[10]+WET[13]*DRY[9]+WET[14]*DRY[8]+WET[15]*DRY[7]+WET[16]*DRY[6]+WET[17]*DRY[5]+WET[18]*DRY[4]+WET[19]*DRY[3]+WET[20]*DRY[2]+WET[21]*DRY[1], WET[1]*DRY[22]+WET[2]*DRY[21]+WET[3]*DRY[20]+WET[4]*DRY[19]+WET[5]*DRY[18]+WET[6]*DRY[17]+WET[7]*DRY[16]+WET[8]*DRY[15]+WET[9]*DRY[14]+WET[10]*DRY[13]+WET[11]*DRY[12]+WET[12]*DRY[11]+WET[13]*DRY[10]+WET[14]*DRY[9]+WET[15]*DRY[8]+WET[16]*DRY[7]+WET[17]*DRY[6]+WET[18]*DRY[5]+WET[19]*DRY[4]+WET[20]*DRY[3]+WET[21]*DRY[2]+WET[22]*DRY[1], WET[1]*DRY[23]+WET[2]*DRY[22]+WET[3]*DRY[21]+WET[4]*DRY[20]+WET[5]*DRY[19]+WET[6]*DRY[18]+WET[7]*DRY[17]+WET[8]*DRY[16]+WET[9]*DRY[15]+WET[10]*DRY[14]+WET[11]*DRY[13]+WET[12]*DRY[12]+WET[13]*DRY[11]+WET[14]*DRY[10]+WET[15]*DRY[9]+WET[16]*DRY[8]+WET[17]*DRY[7]+WET[18]*DRY[6]+WET[19]*DRY[5]+WET[20]*DRY[4]+WET[21]*DRY[3]+WET[22]*DRY[2]+WET[23]*DRY[1],
Re: [R] Howto crosstable-ing......
On Wed, 2005-06-22 at 17:12 +0200, [EMAIL PROTECTED] wrote: I receive the following meteo dataset regularly, containing the average daily temperatures (tMedia) of a certain month for 24 selected meteo-stations (COD_WMO) whose human-readable names are in (NOME). str(tabella) `data.frame': 1038 obs. of 4 variables: $ COD_WMO: int 16045 16045 16045 16045 16045 16045 16045 16045 16045 16045 ... $ NOME : Factor w/ 24 levels ALGHERO,BARI/PALESE MACCHIE,..: 22 22 22 22 22 22 22 22 22 22 ... $ DATE :'POSIXct', format: chr 2005-05-01 2005-05-02 2005-05-03 2005-05-04 ... $ tMedia : num 11.7 18.6 16.9 19.7 15.0 ... Here you are a short list of it: COD_WMO NOME DATE tMedia 505 16191 FALCONARA 2005-06-01 20.95 506 16191 FALCONARA 2005-06-02 20.15 507 16191 FALCONARA 2005-06-03 18.60 506 16191 FALCONARA 2005-06-02 20.15 507 16191 FALCONARA 2005-06-03 18.60 508 16191 FALCONARA 2005-06-04 22.30 509 16191 FALCONARA 2005-06-05 NA 510 16191 FALCONARA 2005-06-06 NA 511 16191 FALCONARA 2005-06-07 18.20 549 16206 GROSSETO 2005-06-01 20.65 550 16206 GROSSETO 2005-06-02 21.95 551 16206 GROSSETO 2005-06-03 22.25 552 16206 GROSSETO 2005-06-04 20.15 553 16206 GROSSETO 2005-06-05 NA 554 16206 GROSSETO 2005-06-06 NA 555 16206 GROSSETO 2005-06-07 22.35 . . I need to rearrange tMedia into a new dataframe whose column names are COD_WMO (or NOME) and the row is DATE. ex. DATEALGHERO BARI/PALESE FALCONARA GROSSETO .. 2005-06-01 16.3 12.8 17.3 14.0 ... 2005-06-02 18.2 8.918.0 17.9 .. ... I read some pieces of R-docs in the internet and run the MASS chapter 2 examples but without finding anything suitable to my purpose. Could you please help me? Ciao Vittorio I believe that the following will get you there, based upon your example output above: reshape(tabella[, -1], idvar = DATE, timevar = NOME, v.names = tMedia, direction = wide) DATE tMedia.FALCONARA tMedia.GROSSETO 1 2005-06-0120.95 20.65 2 2005-06-0220.15 21.95 3 2005-06-0318.60 22.25 6 2005-06-0422.30 20.15 7 2005-06-05 NA NA 8 2005-06-06 NA NA 9 2005-06-0718.20 22.35 See ?reshape for more information. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] r programming help
In addition to Peter's suggestion of converting to a matrix operation, here is a simple solution using naive R programming. Using the 7th element of CYCLE.n as an example: replace this: WET[1]*DRY[6]+WET[2]*DRY[5]+WET[3]*DRY[4]+WET[4]*DRY[3]+WET[5]*DRY[2]+WET[6]*DRY[1] with this: sum(WET[1:6]*DRY[6:1]) which then suggests a loop. nin - length(WET) CYCLE.n - rep(NA,nin+1) for (i in 1:nin) CYCLE.n[i+1] - sum(WET[1:i]*DRY[i:1]) -Don At 4:03 PM +0200 6/22/05, Peter Dalgaard wrote: Mohammad Ehsanul Karim [EMAIL PROTECTED] writes: Dear list, Is there anyway i can make the following formula short by r-programming? CYCLE.n-c(NA, WET[1]*DRY[1], WET[1]*DRY[2]+WET[2]*DRY[1], WET[1]*DRY[3]+WET[2]*DRY[2]+WET[3]*DRY[1], As far as I can see: z - toeplitz(DRY) z[upper.tri(z)] - 0 c(NA, z %*% WET) or convolve() with suitable options, padding, and/or cutting (but beware, there could be devils in the details). convolve(WET,DRY, type=o) gives you about twice what you need, I believe. -- 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- -- Don MacQueen Environmental Protection Department Lawrence Livermore National Laboratory Livermore, CA, USA __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Extract Coeff, Std Error, etc from gnls output
On 6/22/05, Christian Mora [EMAIL PROTECTED] wrote: Dear list members; Is there any trick to extract the coefficients along with std errors, t-values and p-values for each beta from a gnls fit model (similar to the results obtained using summary(lm)$coeff for linear models)? The best way to get the coefficients is with the extractor function coef. There is no extractor for the t-values and the p-values so you need to look at the result of str(summary(gnlsfit)) to find the name of the component. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] monitoring objects sizes
I have an R script that loops over market contracts. The script runs well for markets with relatively small number of contracts but seg faults when the number of contracts (loop iterations) is large. Is there a way for me to monitor my objects and their sizes from within the R script? How can I get all of the objects and their sizes? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] chisq test and fisher exact test
Weiwei Shi wrote: Hi, I have a text mining project and currently I am working on feature generation/selection part. My plan is selecting a set of words or word combinations which have better discriminant capability than other words in telling the group id's (2 classes in this case) for a dataset which has 2,000,000 documents. One approach is using contrast-set association rule mining while the other is using chisqr or fisher exact test. An example which has 3 contingency tables for 3 words as followed (word coded by number): tab[,,1:3] , , 1 [,1][,2] [1,] 11266 2151526 [2,] 125 31734 , , 2 [,1][,2] [1,] 43571 2119221 [2,]52 31807 , , 3 [,1][,2] [1,] 427 2162365 [2,]5 31854 I have some questions on this: 1. What's the thumb of rule to use chisq test instead of Fisher exact test. I have a vague memory which said for each cell, the count needs to be over 50 if chisq instead of fisher exact test is going to be used. In the case of word 3, I think I should use fisher test. However, running chisq like below is fine: tab[,,3] [,1][,2] [1,] 427 2162365 [2,]5 31854 chisq.test(tab[,,3]) Pearson's Chi-squared test with Yates' continuity correction data: tab[, , 3] X-squared = 0.0963, df = 1, p-value = 0.7564 but running on the whole set of words (including 14240 words) has the following warnings: p.chisq-as.double(lapply(1:N, function(i) chisq.test(tab[,,i])$p.value)) There were 50 or more warnings (use warnings() to see the first 50) warnings() Warning messages: 1: Chi-squared approximation may be incorrect in: chisq.test(tab[, , i]) 2: Chi-squared approximation may be incorrect in: chisq.test(tab[, , i]) 3: Chi-squared approximation may be incorrect in: chisq.test(tab[, , i]) 4: Chi-squared approximation may be incorrect in: chisq.test(tab[, , i]) 2. So, my second question is, is this warning b/c I am against the assumption of using chisq. But why Word 3 is fine? How to trace the warning to see which word caused this warning? 3. My result looks like this (after some mapping treating from number id to word and some words are stemmed here, like ACCID is accident): of[1:50,] map...2. p.fisher 21 ACCID 0.00e+00 30 CD 0.00e+00 67ROCK 0.00e+00 104 CRACK 0.00e+00 111 CHIP 0.00e+00 179 GLASS 0.00e+00 84BACK 4.199878e-291 395 DRIVEABL 5.335989e-287 60 CAP 9.405235e-285 262 WINDSHIELD 2.691641e-254 13 IV 3.905186e-245 110 HZ 2.819713e-210 11CAMP 9.086768e-207 2 SHATTER 5.273994e-202 297ALP 1.678521e-177 162BED 1.822031e-173 249BCD 1.398391e-160 493 RACK 4.178617e-156 59CAUS 7.539031e-147 3.1 question: Should I use two-sided test instead of one-sided for fisher test? I read some material which suggests using two-sided. 3.2 A big question: Even though the result looks very promising since this is case of classiying fraud cases and the words selected by this approach make sense. However, I think p-values here just indicate the strength to reject null hypothesis, not the strength of association between word and class of document. So, what kind of statistics I should use here to evaluate the strength of association? odds ratio? Any suggestions are welcome! Thanks! You can use chisq.test with sim=TRUE, or call it as usual first, see if there is a warning, and then recall with sim=TRUE. Kjetil -- Kjetil Halvorsen. Peace is the most effective weapon of mass construction. -- Mahdi Elmandjra -- No virus found in this outgoing message. Checked by AVG Anti-Virus. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] monitoring objects sizes
Do RSiteSearch(ls.obj) and click on the first hit. Andy From: Omar Lakkis I have an R script that loops over market contracts. The script runs well for markets with relatively small number of contracts but seg faults when the number of contracts (loop iterations) is large. Is there a way for me to monitor my objects and their sizes from within the R script? How can I get all of the objects and their sizes? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] monitoring objects sizes
hel.search(object size) found a function object.size that should do what you want. spencer graves Omar Lakkis wrote: I have an R script that loops over market contracts. The script runs well for markets with relatively small number of contracts but seg faults when the number of contracts (loop iterations) is large. Is there a way for me to monitor my objects and their sizes from within the R script? How can I get all of the objects and their sizes? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] classes with chron slots
I'd like to define a class with a chron slot, but: R require(chron) R setClass(myclass, representation(datetime = chron)) [1] myclass Warning message: undefined slot classes in definition of myclass: datetime(class chron) in: .completeClassSlots(ClassDef, where) How should such a class be defined? Sebastian -- Sebastian P. Luque __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] analyzing suvival data using splines (a.k.a., piecewise log-hazard-ratio models)
I'm looking for software that makes plots such as fig 4 (a)-(e), fig 5 anf fig 7 of Gray, Robert, Flexible Methods for Analyzing Survival Data Using Splines, with Applications to Breast Cancer Prognosis, 1992, J Am Stat Assoc, pp 942-51. In other words, I'm looking for software that takes survival data and a continuous covariate as input and computes a curve giving log hazard ratio (or rate of failure for a specific time) as a function of the continuous covariate, as well as curves giving +-SE for that curve. A plot of this nature can also be found as figure 4 in Paik, et al., A Multigene Assay to Predict Recurrence of Tamoxifen-Treated, Node-Negative Breast Cancer, 2004, New England Journal of Medicine, pp 2817-26. Any help would be greatly appreciated. Thanks. -Ben Ben Wittner Research fellow, MGH Harvard Medical School [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] analyzing suvival data using splines (a.k.a., piecewise log-hazard-ratio models)
Wittner, Ben wrote: I'm looking for software that makes plots such as fig 4 (a)-(e), fig 5 anf fig 7 of Gray, Robert, Flexible Methods for Analyzing Survival Data Using Splines, with Applications to Breast Cancer Prognosis, 1992, J Am Stat Assoc, pp 942-51. In other words, I'm looking for software that takes survival data and a continuous covariate as input and computes a curve giving log hazard ratio (or rate of failure for a specific time) as a function of the continuous covariate, as well as curves giving +-SE for that curve. library(Design) d - datadist(mydata); options(datadist='d') f - cph(Surv(dtime,death) ~ age + sex + rcs(blood.pressure,5)) plot(f, blood.pressure=NA) Frank A plot of this nature can also be found as figure 4 in Paik, et al., A Multigene Assay to Predict Recurrence of Tamoxifen-Treated, Node-Negative Breast Cancer, 2004, New England Journal of Medicine, pp 2817-26. Any help would be greatly appreciated. Thanks. -Ben Ben Wittner Research fellow, MGH Harvard Medical School [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Rsquare from glmmPQL or another GLMM?
Hi, I know that Rsquare in glm or in non-linear models is wrong, but some people like this. How I make to estimate the Rsquare from a model ajusted with glmmPQL or another GLMM? Thanks for all Ronaldo -- A simplicidade é o último degrau da sabedoria. -- Victor Hugo -- | // | \\ [***] | ( õ õ ) [Ronaldo Reis Júnior] | V [UFV/DBA-Entomologia] |/ \ [36570-000 Viçosa - MG ] | /(.''`.)\ [Fone: 31-3899-4007 ] | /(: :' :)\ [EMAIL PROTECTED]] |/ (`. `'` ) \[ICQ#: 5692561 | LinuxUser#: 205366 ] |( `- ) [***] | _/ \_Powered by GNU/Debian Woody/Sarge __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] load history does not work on OS X
I am running R 2.1 on OSX 10.4 using the precompiled binaries/ Raqua.app. loadhistory() nor load history (from the history 'drawer') do NOT work. I can save history files and even see them in a text editor just fine, but cannot load any information (even after loading library(utils)). Google and list-searches have been to no avail. Thanks in advance for any insight ... Steve -- Steven T. Stoddard Program in Ecology and Evolutionary Biology University of Illinois at Urbana-Champaign www.life.uiuc.edu/~sstoddar [EMAIL PROTECTED] (217) 333 - 2235 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] substitute in a named expression
I have a 'named expression' like expr - expression(rep(1,d)) and would like to replace the argument d with say 5 without actually evaluating the expression. So I try substitute(expr, list(d=5)) in which case R simply returns expr which when I 'evaluate' it gives eval(expr) Error in rep.default(1, d) : invalid number of copies in rep() I've looked at ?substitute and ?expression (and other places) for ideas, but - well I guess there are some details which I haven't quite understood. Can anyone point me in the right direction? Thanks Søren __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to sort a dataset by one column?
I understand how to sort a vector, but I could not find how to sort a data frame or matrix by one variable (column). Could you give me some examples? Thanks! Ling __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] string/character to number
I did a very quick search of the archive and couldn't find a readily available answer to this one: I'd like to convert, for example: c(a, b, a, b) to c(1, -1, 1, -1) In the case of the first vector, it may be any length, but will always only have two unique values. It must always be replaced by corresponding values of 1 and -1. Any thoughts? Thanks in advance, Jake __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] string/character to number
Duh! sub() --Jake On Jun 22, 2005, at 3:35 PM, Jake Michaelson wrote: I did a very quick search of the archive and couldn't find a readily available answer to this one: I'd like to convert, for example: c(a, b, a, b) to c(1, -1, 1, -1) In the case of the first vector, it may be any length, but will always only have two unique values. It must always be replaced by corresponding values of 1 and -1. Any thoughts? Thanks in advance, Jake __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to sort a dataset by one column?
On Wed, 2005-06-22 at 14:34 -0700, Ling Jin wrote: I understand how to sort a vector, but I could not find how to sort a data frame or matrix by one variable (column). Could you give me some examples? Thanks! Ling See the examples in ?order HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] string/character to number
This works x-c(a, b, a, b) x[x==a]-1 x[x==b]- -1 as.numeric(x) [1] 1 -1 1 -1 Fra: [EMAIL PROTECTED] på vegne af Jake Michaelson Sendt: on 22-06-2005 23:35 Til: R-help@stat.math.ethz.ch Emne: [R] string/character to number I did a very quick search of the archive and couldn't find a readily available answer to this one: I'd like to convert, for example: c(a, b, a, b) to c(1, -1, 1, -1) In the case of the first vector, it may be any length, but will always only have two unique values. It must always be replaced by corresponding values of 1 and -1. Any thoughts? Thanks in advance, Jake __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to sort a dataset by one column?
See RSiteSearch(sort.data.frame), or more generally, ?order. Andy From: Ling Jin I understand how to sort a vector, but I could not find how to sort a data frame or matrix by one variable (column). Could you give me some examples? Thanks! Ling __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] string/character to number
Note: sub() returns a character vector not a numeric vector. as.numeric()
will convert it.
Slightly slicker and faster is: 2*(z=='a')-1 where z is your vector,
c('a','b','a','b')
Cheers,
Bert
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Jake Michaelson
Sent: Wednesday, June 22, 2005 2:46 PM
To: Jake Michaelson
Cc: R-help@stat.math.ethz.ch
Subject: Re: [R] string/character to number
Duh!
sub()
--Jake
On Jun 22, 2005, at 3:35 PM, Jake Michaelson wrote:
I did a very quick search of the archive and couldn't find a readily
available answer to this one:
I'd like to convert, for example:
c(a, b, a, b)
to
c(1, -1, 1, -1)
In the case of the first vector, it may be any length, but will always
only have two unique values. It must always be replaced by
corresponding values of 1 and -1.
Any thoughts?
Thanks in advance,
Jake
__
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http://www.R-project.org/posting-guide.html
__
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https://stat.ethz.ch/mailman/listinfo/r-help
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__
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Re: [R] string/character to number
To do this in one step, it would be easier to use ifelse(): Chars [1] a b a b ifelse(Chars == a, 1, -1) [1] 1 -1 1 -1 HTH, Marc Schwartz On Wed, 2005-06-22 at 15:46 -0600, Jake Michaelson wrote: Duh! sub() --Jake On Jun 22, 2005, at 3:35 PM, Jake Michaelson wrote: I did a very quick search of the archive and couldn't find a readily available answer to this one: I'd like to convert, for example: c(a, b, a, b) to c(1, -1, 1, -1) In the case of the first vector, it may be any length, but will always only have two unique values. It must always be replaced by corresponding values of 1 and -1. Any thoughts? Thanks in advance, Jake __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] string/character to number
You can do indexing by name: x - c(a, b, a, b) v - c(a=1, b=-1) v[x] a b a b 1 -1 1 -1 Andy From: Jake Michaelson Duh! sub() --Jake On Jun 22, 2005, at 3:35 PM, Jake Michaelson wrote: I did a very quick search of the archive and couldn't find a readily available answer to this one: I'd like to convert, for example: c(a, b, a, b) to c(1, -1, 1, -1) In the case of the first vector, it may be any length, but will always only have two unique values. It must always be replaced by corresponding values of 1 and -1. Any thoughts? Thanks in advance, Jake __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] substitute in a named expression
On 6/22/05, Søren Højsgaard [EMAIL PROTECTED] wrote: I have a 'named expression' like expr - expression(rep(1,d)) and would like to replace the argument d with say 5 without actually evaluating the expression. So I try substitute(expr, list(d=5)) in which case R simply returns expr which when I 'evaluate' it gives eval(expr) Error in rep.default(1, d) : invalid number of copies in rep() I've looked at ?substitute and ?expression (and other places) for ideas, but - well I guess there are some details which I haven't quite understood. Can anyone point me in the right direction? Try this: eval(substitute(substitute(qq, list(d=5)), list(qq = expr[[1]]))) This aspect of R drove me crazy some time ago but Tony Plate finally figured it out and discussed it some time back: http://tolstoy.newcastle.edu.au/R/help/04/03/1247.html There is also a handy utility routine, esub, defined there. The key points are: - substitute won't go inside expressions but it will go inside call objects. In this case your expr is an expression but expr[[1]] is a call object with the desired contents. Note that quote will return a call object so you can avoid the [[1]] if you define expr as cl - quote(rep(1,d)) i.e. cl - quote(rep(1,d)) eval(substitute(substitute(cl, list(d=5)), list(cl = cl))) - substitute autoquotes anything inside it so one must substitute in the first argument to the inner substitute using a second outer substitute. That is, the outer substitute substitutes expr[[1]] (which is evaluated) into the first argument of the inner substitute. - the outer substitute wraps the result of the inner one in a call so we must perform an eval to get what is within the call. This part is explained in ?substitute Sorry if this is complicated but that seems to be how it works. Using the esub function defined in the link above you can simplify it substantially like this: esub(cl, list(d=5)) # or esub(expr[[1]], list(d=5)) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] chisq test and fisher exact test
Is it b/c my question is too long so no one answers it? I should have splitted it. :( On 6/22/05, Kjetil Brinchmann Halvorsen [EMAIL PROTECTED] wrote: Weiwei Shi wrote: Hi, I have a text mining project and currently I am working on feature generation/selection part. My plan is selecting a set of words or word combinations which have better discriminant capability than other words in telling the group id's (2 classes in this case) for a dataset which has 2,000,000 documents. One approach is using contrast-set association rule mining while the other is using chisqr or fisher exact test. An example which has 3 contingency tables for 3 words as followed (word coded by number): tab[,,1:3] , , 1 [,1][,2] [1,] 11266 2151526 [2,] 125 31734 , , 2 [,1][,2] [1,] 43571 2119221 [2,]52 31807 , , 3 [,1][,2] [1,] 427 2162365 [2,]5 31854 I have some questions on this: 1. What's the thumb of rule to use chisq test instead of Fisher exact test. I have a vague memory which said for each cell, the count needs to be over 50 if chisq instead of fisher exact test is going to be used. In the case of word 3, I think I should use fisher test. However, running chisq like below is fine: tab[,,3] [,1][,2] [1,] 427 2162365 [2,]5 31854 chisq.test(tab[,,3]) Pearson's Chi-squared test with Yates' continuity correction data: tab[, , 3] X-squared = 0.0963, df = 1, p-value = 0.7564 but running on the whole set of words (including 14240 words) has the following warnings: p.chisq-as.double(lapply(1:N, function(i) chisq.test(tab[,,i])$p.value)) There were 50 or more warnings (use warnings() to see the first 50) warnings() Warning messages: 1: Chi-squared approximation may be incorrect in: chisq.test(tab[, , i]) 2: Chi-squared approximation may be incorrect in: chisq.test(tab[, , i]) 3: Chi-squared approximation may be incorrect in: chisq.test(tab[, , i]) 4: Chi-squared approximation may be incorrect in: chisq.test(tab[, , i]) 2. So, my second question is, is this warning b/c I am against the assumption of using chisq. But why Word 3 is fine? How to trace the warning to see which word caused this warning? 3. My result looks like this (after some mapping treating from number id to word and some words are stemmed here, like ACCID is accident): of[1:50,] map...2. p.fisher 21 ACCID 0.00e+00 30 CD 0.00e+00 67ROCK 0.00e+00 104 CRACK 0.00e+00 111 CHIP 0.00e+00 179 GLASS 0.00e+00 84BACK 4.199878e-291 395 DRIVEABL 5.335989e-287 60 CAP 9.405235e-285 262 WINDSHIELD 2.691641e-254 13 IV 3.905186e-245 110 HZ 2.819713e-210 11CAMP 9.086768e-207 2 SHATTER 5.273994e-202 297ALP 1.678521e-177 162BED 1.822031e-173 249BCD 1.398391e-160 493 RACK 4.178617e-156 59CAUS 7.539031e-147 3.1 question: Should I use two-sided test instead of one-sided for fisher test? I read some material which suggests using two-sided. 3.2 A big question: Even though the result looks very promising since this is case of classiying fraud cases and the words selected by this approach make sense. However, I think p-values here just indicate the strength to reject null hypothesis, not the strength of association between word and class of document. So, what kind of statistics I should use here to evaluate the strength of association? odds ratio? Any suggestions are welcome! Thanks! You can use chisq.test with sim=TRUE, or call it as usual first, see if there is a warning, and then recall with sim=TRUE. Kjetil -- Kjetil Halvorsen. Peace is the most effective weapon of mass construction. -- Mahdi Elmandjra -- No virus found in this outgoing message. Checked by AVG Anti-Virus. Version: 7.0.323 / Virus Database: 267.7.7/20 - Release Date: 16/06/2005 -- Weiwei Shi, Ph.D Did you always know? No, I did not. But I believed... ---Matrix III __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] substitute in a named expression
Gabor Grothendieck [EMAIL PROTECTED] writes: On 6/22/05, Søren Højsgaard [EMAIL PROTECTED] wrote: I have a 'named expression' like expr - expression(rep(1,d)) and would like to replace the argument d with say 5 without actually evaluating the expression. So I try substitute(expr, list(d=5)) in which case R simply returns expr which when I 'evaluate' it gives eval(expr) Error in rep.default(1, d) : invalid number of copies in rep() I've looked at ?substitute and ?expression (and other places) for ideas, but - well I guess there are some details which I haven't quite understood. Can anyone point me in the right direction? Try this: eval(substitute(substitute(qq, list(d=5)), list(qq = expr[[1]]))) This aspect of R drove me crazy some time ago but Tony Plate finally figured it out and discussed it some time back: http://tolstoy.newcastle.edu.au/R/help/04/03/1247.html There is also a handy utility routine, esub, defined there. The key points are: - substitute won't go inside expressions but it will go inside call objects. In this case your expr is an expression but expr[[1]] is a call object with the desired contents. Note that quote will return a call object so you can avoid the [[1]] if you define expr as cl - quote(rep(1,d)) i.e. cl - quote(rep(1,d)) eval(substitute(substitute(cl, list(d=5)), list(cl = cl))) - substitute autoquotes anything inside it so one must substitute in the first argument to the inner substitute using a second outer substitute. That is, the outer substitute substitutes expr[[1]] (which is evaluated) into the first argument of the inner substitute. - the outer substitute wraps the result of the inner one in a call so we must perform an eval to get what is within the call. This part is explained in ?substitute Sorry if this is complicated but that seems to be how it works. Using the esub function defined in the link above you can simplify it substantially like this: esub(cl, list(d=5)) # or esub(expr[[1]], list(d=5)) Yes, substitute() is a bass-ackward design and the automatic quoting of the first arg is a pain. It would have been much cleaner if standard semantics were used and you'd just quote() the argument when needed. Your explanation of eval(substitute(substitute(qq, list(d=5)), list(qq = expr[[1]]))) is a tad long-winded though. What happens is that the inner unevaluated substitute(qq, list(d=5)) gets the qq replaced by the value of expr[[1]]. In casu it becomes substitute(rep(1,d),list(d=5)) this then needs to be evaluated, yielding rep(1,5) -- 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to read an excel data into R?
Hi all, Does anybody know the easiest way to import excel data into R? I copied and pasted the excel data into a txt file, and tried read.table, but R reported that Error in read.table(data_support.txt, sep = , header = T) : more columns than column names Thanks! Ling __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to read an excel data into R?
Your error message tells me that you have different numbers of fields in different lines. You say you, copied and pasted the excel data into a txt file. I usually copy what I want into a clean sheet then File - Save, then File - Save As - Save as type = CSV (Comma delimited) (*.csv) or Text (Tab delimited) (*.txt). Excel will ask if I'm sure a couple of times, and I say yes. If that's what you've done and still have a problem, then I have other tools: First, I'll assign the file name to something like File. Then, 'readLines(File, n=9)' tells me if the file starts as I think it does. If I've got extra headers, it will tell me that. Then, I do something like the following: n.flds - count.fields(File, sep=\t) plot(n.flds) sd(n.flds) Then I play with the arguments to count.fields until 'sd(n.flds)' is 0. Then I use read.table with arguments as I used to get everything right in 'count.fields'. If I can't get sd(n.flds) to 0, you can try read.table with 'fill=TRUE'. However, when you do that, you need to check to make sure all the columns line up correctly with the shorter lines. Also, this issue has been discussed many times. 'RSiteSearch(read excel)' just produced 1196 hits for me. If the above doesn't work, you might try skimming a few from that list. hope this helps. spencer graves Ling Jin wrote: Hi all, Does anybody know the easiest way to import excel data into R? I copied and pasted the excel data into a txt file, and tried read.table, but R reported that Error in read.table(data_support.txt, sep = , header = T) : more columns than column names Thanks! Ling __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] legend
Hi
Uwe Ligges wrote:
Thomas Steiner wrote:
I color some area grey with polygon() (with a red border) and then I
want to have the dashed red border in the legend as well. How do I
manage it?
And I want to mix (latex) expressions with text in my legend.
Both points are not that easy to solve, hence I'd like to suggest to
write your own little function that generates the legend.
Starting at the upper left, calculating the stringheight, painting the
(party very special) symbols, and adding the text line by line seems to
be the most easiest solution here (which is not that nice, though.
I don't think it's too bad. For example, try replacing the original ...
legend(x=0,y=3.5,legend=c(exp(0.7x),mu=0.7, sigma=0.4,mu=0.7,
sigma=0.2,mu=0.7, sigma=0.1,Standardabweichung für
sigma=0.2),lwd=c(4,4,4,4,12),col=c(cs,grey),bg=transparent,cex=1.15)
... with ...
# Use grid and gridBase so you've got some sensible
# coordinate systems to work within
library(grid)
library(gridBase)
# Align a grid viewport with the plotting region
vps - baseViewports()
pushViewport(vps$inner, vps$figure, vps$plot)
# Define labels and colours
# Labels are mathematical expressions
labels - expression(exp(0.7x),
list(mu == 0.7,sigma == 0.4),
list(mu == 0.7,sigma == 0.2),
list(mu == 0.7, sigma == 0.1),
paste(Standardabweichung für ,sigma == 0.2))
cols - cs
# Draw each legend item on its own line
# Top line 1cm in from top-left corner
for (i in 1:5) {
x - unit(1, cm)
y - unit(1, npc) - unit(1, cm) - unit(i, lines)
if (i 5) {
grid.lines(unit.c(x, unit(2, cm)), y + unit(0.5, lines),
gp=gpar(col=cols[i], lwd=3))
} else {
grid.rect(x, y, width=unit(1, cm),
height=unit(1, lines),
gp=gpar(fill=grey, col=cs[3], lty=dashed),
just=c(left, bottom))
}
grid.text(labels[i], x + unit(1.5, cm), y,
just=c(left, bottom))
}
# clean up
popViewport(3)
... that's a bit of typing, but if you need to do more than one, it
would go inside a function with labels and cols as arguments (and '5'
replaced by 'length(labels)') without too much trouble.
(In this case, you could also pretty easily just do the main plot using
grid and avoid having to use gridBase.)
Paul
Just execute my lines below and you know want I mean. Or pass by at
http://de.wikipedia.org/wiki/Bild:GBM.png to see the picture online.
Thomas
bm - function(n=500, from=0, to=1) {
x=seq(from=from,to=to,length=n)
BM-c(0,cumsum(rnorm(n-1,mean=0,sd=sqrt(to/n
cbind(x,BM)
}
gbm - function(bm,S0=1,sigma=0.1,mu=1) {
gbm=S0
for (t in 2:length(bm[,1])) {
gbm[t]=S0*exp((mu-sigma^2/2)*bm[t,1]+sigma*bm[t,2])
}
cbind(bm[,1],gbm)
}
set.seed(9826064)
cs=c(dark green, steelblue, red, yellow)
#png(filename = GBM.png, width=1600, height=1200, pointsize = 12)
par(bg=lightgrey)
x=seq(from=0,to=1,length=500)
plot(x=x, y=exp(0.7*x), type=n, xlab=Zeit, ylab=, ylim=c(1,3.5))
polygon(x=c(x,rev(x)),
y=c(exp(0.7*x)+0.4*sqrt(x),rev(exp(0.7*x)-0.4*sqrt(x))), col=grey,
border=cs[3], lty=dashed)
lines(x=x,y=exp(0.7*x), type=l, lwd=3, col=cs[1])
lines(gbm(bm(),S0=1,mu=0.7,sigma=0.4), lwd=3, col=cs[2])
lines(gbm(bm(),S0=1,mu=0.7,sigma=0.2), lwd=3, col=cs[3])
lines(gbm(bm(),S0=1,mu=0.7,sigma=0.1), lwd=3, col=cs[4])
title(main=Geometrische Brownsche Bewegung,cex.main=2.5)
legend(x=0,y=3.5,legend=c(exp(0.7x),mu=0.7, sigma=0.4,mu=0.7,
sigma=0.2,mu=0.7, sigma=0.1,Standardabweichung für
sigma=0.2),lwd=c(4,4,4,4,12),col=c(cs,grey),bg=transparent,cex=1.15)
#dev.off()
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--
Dr Paul Murrell
Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
[EMAIL PROTECTED]
http://www.stat.auckland.ac.nz/~paul/
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Re: [R] How to read an excel data into R?
Ling, You might take a look at the function read.xls() in gdata library. HTH. On 6/22/05, Ling Jin [EMAIL PROTECTED] wrote: Hi all, Does anybody know the easiest way to import excel data into R? I copied and pasted the excel data into a txt file, and tried read.table, but R reported that Error in read.table(data_support.txt, sep = , header = T) : more columns than column names Thanks! Ling __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- WenSui Liu, MS MA Senior Decision Support Analyst Division of Health Policy and Clinical Effectiveness Cincinnati Children Hospital Medical Center __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Large number of covariates in survival regressions
I am trying to run a survival time regression (no time dependence so exponential distribution) with a fairly large but not huge number of right hand side variables (about 60). r seems to not be able to estimate coefficients or standard errors for many of the covariates but can for some of the covariates, so I don't think this is a missing values problem. Stata does not have the same problem (although it is slow). Any thoughts? Thomas Davidoff Assistant Professor Haas School of Business UC Berkeley Berkeley, CA 94720 phone: (510) 643-1425 fax:(510) 643-7357 [EMAIL PROTECTED] http://faculty.haas.berkeley.edu/davidoff [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] mac osx, g95 package port problem
Hi all, I have a working package for linux, including fortran 95 code compiled with g95, that I need to port to OS X. The package works on Linux and seems to load on the Mac, but when I try to run a function that calls C or Fortran I'm told that the symbol is not loaded. I'm developing via a shell account on an OS X system, I don't have access to a desktop. The set up is: R 2.1.0 Patched (2005-05-12). Darwin Kernel Version 8.1.0 G95 (GCC 4.0.0 20050124 (experimental) (g95!) 06/20/05) Here is the Makefile: F90_FILES=\ dbest_dbase_class.f90 \ ... r_estimate.f90 \ FORTRAN_FILES=\ dgletc.f \ ... mecdf.f C_FILES=init.c CFLAGS=-g -fPIC %.o: %.f90 g95 -c -g $ %.o: %.f g95 -c -g $ %.o: %.c gcc -c -g $ bpkg.so: $(F90_FILES:%.f90=%.o) $(C_FILES:%.c=%.o) $(FORTRAN_FILES:%.f=%.o) g95 -L/Library/Frameworks -o $@ $^ -L/Library/Frameworks/R.framewo\ rk/Resources -end makefile Here is the relevant output of R CMD CHECK: * checking for working latex ...sh: line 1: latex: command not found NO * using log directory '/Users/jbremson/dev/bpkg.Rcheck' * using R version 2.1.0, 2005-05-12 * checking for file 'bpkg/DESCRIPTION' ... OK * this is package 'bpkg' version '1.0-1' * checking if this is a source package ... OK * Installing *source* package 'bpkg' ... ** libs g95 -c -g dbest_dbase_class.f90 g95 -c -g cm_class.f90 ... gcc -c -g init.c init.c: In function 'R_g95_init': init.c:22: warning: incompatible implicit declaration of built-in function 'strdup' g95 -c -g dgletc.f g95 -c -g dglfgb.f g95 -c -g dglfg.f g95 -c -g dglfgb.f g95 -c -g dglfg.f g95 -c -g dmdc.f g95 -c -g mecdf.f g95 -shared -L/Library/Frameworks -o bpkg.so dbest_dbase_class.o cm_class.o bgw_cla\ ss.o cm_mle_class.o pcm_dglg_o1.o cm_main.o r_estimate.o init.o dgletc.o dglfgb.o dg\ lfg.o dmdc.o mecdf.o -L/Library/Frameworks/R.framework/Resources g95: unrecognized option '-shared' ** R ** data ** help Building/Updating help pages for package 'bpkg' Formats: text html latex example estimate.model text html latex example ** building package indices ... * DONE (bpkg) * checking package directory ... OK * checking for portable file names ... OK * checking for sufficient/correct file permissions ... OK * checking DESCRIPTION meta-information ... OK * checking package dependencies ... OK * checking index information ... WARNING Empty file 'INDEX'. See the information on INDEX files and package subdirectories in section 'Creating R packages' of the 'Writing R Extensions' manual. * checking package subdirectories ... OK * checking R files for syntax errors ... OK * checking R files for library.dynam ... OK * checking S3 generic/method consistency ... WARNING Error: package/namespace load failed for 'bpkg' Call sequence: 2: stop(gettextf(package/namespace load failed for '%s', libraryPkgName(package)),\ call. = FALSE, domain = NA) 1: library(package, lib.loc = lib.loc, character.only = TRUE, verbose = FALSE) Execution halted See section 'Generic functions and methods' of the 'Writing R Extensions' manual. See section 'Generic functions and methods' of the 'Writing R Extensions' manual. * checking replacement functions ... WARNING Error: package/namespace load failed for 'bpkg' Call sequence: 2: stop(gettextf(package/namespace load failed for '%s', libraryPkgName(package)),\ call. = FALSE, domain = NA) 1: library(package, lib.loc = lib.loc, character.only = TRUE, verbose = FALSE) Execution halted In R, the argument of a replacement function which corresponds to the right hand side must be named 'value'. * checking foreign function calls ... WARNING Error: package/namespace load failed for 'bpkg' Call sequence: 2: stop(gettextf(package/namespace load failed for '%s', libraryPkgName(package)),\ call. = FALSE, domain = NA) 1: library(package, lib.loc = lib.loc, character.only = TRUE, verbose = FALSE) Execution halted See section 'System and foreign language interfaces' of the 'Writing R Extensions' manual. * checking Rd files ... OK * checking for missing documentation entries ... ERROR Error: package/namespace load failed for 'bpkg' -EOF Obviously, there is a package/namespace load filed. I looked at the mentioned R documentation and didn't find anything I thought relevant. Here is the NAMESPACE file for good measure. The package is called bpkg. --- useDynLib(bpkg) export(dummy,estimate.model,em2) -EOF--- I have checked the bpkg.so file with nm and the symbols for all functions are present. How can I get R to see the symbols for the package? Regards, Joel Bremson Graduate Student UC Davis [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to read an excel data into R?
Ling, If any column has text with spaces between words, this will lead to the more columns ... problem. Delete the spaces and try again. e.g., if the Excel file is Var1Var2Var3 text1 2 more text 3 4 yet more5 6 and more7 8 blahblah9 10 On a Mac, this will lead to the error message Error in scan(file = file, what = what, sep = sep, quote = quote, dec = dec, : line 1 did not have 4 elements (which I believe is the equivalent message to what you are getting on a PC) But, if your remove the blanks in column 1, this reads as x - read.table(test.txt,header=T) x Var1 Var2 Var3 1 text12 2 moretext34 3 yetmore56 4 andmore78 5 blahblah9 10 with no error message. Alternatively, for small files, if using a PC try copying the Excel spreadsheet to your clipboard and x - read.table(file(clipboard), header = TRUE) or, if using a Mac x - read.table(pipe(pbpaste), header = TRUE) Bill At 8:38 PM -0400 6/22/05, Wensui Liu wrote: Ling, You might take a look at the function read.xls() in gdata library. HTH. On 6/22/05, Ling Jin [EMAIL PROTECTED] wrote: Hi all, Does anybody know the easiest way to import excel data into R? I copied and pasted the excel data into a txt file, and tried read.table, but R reported that Error in read.table(data_support.txt, sep = , header = T) : more columns than column names Thanks! Ling -- William Revelle http://pmc.psych.northwestern.edu/revelle.html Professor http://personality-project.org/personality.html Department of Psychology http://www.wcas.northwestern.edu/psych/ Northwestern University http://www.northwestern.edu/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to save changed options in Rcmdr
Dear All, I want to change the default options of Rcmdr; it seemed to work when I made changes and click the Exit and Restart R Commander. However, next time I open Rcmdr, it automatically restored to the default options. Is there a way to change Rcmdr's options permanently? Thanks! Shige __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
