Re: [R] ED50 from logistic model with interactions
On Wed, 02 May 2007 11:37:22 +1000 Kate Stark [EMAIL PROTECTED] wrote: [...] My model is: fit - glm(Mature ~ Season * Size - 1, family = binomial, data=dat) where Mature is a binary response, 0 for immature, 1 for mature. There are 3 Seasons. I would use: fit - glm(Mature ~ Season/Size - 1, family=binomial, data=dat) With this parameterisation you get the three intercepts and the three slopes directly (together with there standard errors from summary()). Makes life simpler for your calculations. In Faraway(2006) he has an example using the delta method to calculate the StdErr, but again without any interactions. I can apply this for the first Season, as there is just one intercept and one slope coefficient, but for the other 2 Seasons, the slope is a combination of the Size coefficient and the Size*Season coefficient, [...] Not with the above parameterisation, so life is easier. I don't have my copy of Faraway (2006) handy at the moment, so I cannot vouch that you can use the method the describes now directly. But I expect you can. :) I could divide the data and do 3 different logistic regressions, one for each season, but while the Mat50 (i.e. mean Size at 50% maturity) is the same as that calculated by the separate lines regression, Im not sure how this may change the StdErr? As far as I can tell, there should be no difference. Compare the estimates and their standard errors that you obtain from the 3 different logistic fits with the estimates and standard errors from the parameterisation that I suggested. They should be the same. Hope this helps. Cheers, Berwin __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Simulation using parts of density function
Thanks for your code! It is not exactly what I really want - but it is my
fault, because my description was wrong...
It is not sim but rhater exp(rgamma(...)) that should not exceed 500. So
I tried to modify your code but it doesn't really work. sim.test returns just
1 value and not 999.
My modified code:
sim.test - NULL
for(i in 1:999)
{sim-NULL
remain - rpois(1,2000)
x - remain
while(remain0){
sim0-replicate(10*remain,
exp(rgamma(1, scale = 0.5, shape = 12))
)
sim-c(sim,sim0[sim0=500])
remain-(x - length(sim))
}
sim-sim[1:x]}
sim.test - rbind(sim.test, c(value=sum(sim)))
Thanks for any help,
Brigitte
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Gesendet: Dienstag, 1. Mai 2007 20:18
An: Thür Brigitte
Cc: r-help@stat.math.ethz.ch
Betreff: RE: [R] Simulation using parts of density function
On 01-May-07 17:03:46, Thür Brigitte wrote:
Hi
My simulation with the followin R code works perfectly:
sim - replicate(999, sum(exp(rgamma(rpois(1,2000),
scale = 0.5, shape = 12
But now I do not want to have values in object sim exceeding
5'000'000, that means that I am just using the beginning of
densitiy function gamma x 15.4. Is there a possibility to
modify my code in an easy way?
Thanks for any help!
Regards, Brigitte
A somewhat extreme problem!
The easiest way to modify the code is as below -- certiainly easier
than writing a special function to draw random samples from the
truncated gamma distribution.
A bit of experimentation shows that, from your code above, about
10% of the results are = 500. So:
sim-NULL
remain - 999
while(remain0){
sim0-replicate(10*remain,
sum(exp(rgamma(rpois(1,2000), scale = 0.5, shape = 12)))
)
sim-c(sim,sim0[sim0=500])
remain-(999 - length(sim))
}
sim-sim[1:999]
Results of a run:
sum(sim500)
[1] 0
max(sim)
[1] 4999696
length(sim)
[1] 999
It may be on the slow side (though not hugely -- on a quite slow
machine the above run was completed in 2min 5sec, while the
999-replicate in your original took 15sec. So about 8 times as long.
Most of this, of course, is taken up with the first round.
Hoping this helps,
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 01-May-07 Time: 19:18:01
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[R] Log-likelihood function
I've computed a loglinear model on a categorical dataset. I would like to test whether an interaction can be dropped by comparing the log-likelihoods from two models(the model with the interaction vs. the model without). Since R does not immediately print the log-likelihood when I use the glm function, I used SAS initially. After searching for an extracting function, I found one in R. But, the log-likelihood given by SAS is different from the one given by R. I'm not sure if the logLik function in R is giving me something I don't want. Or if I'm misinterpreting the SAS output. Can anyone help? -- View this message in context: http://www.nabble.com/Log-likelihood-function-tf3678882.html#a10280717 Sent from the R help mailing list archive at Nabble.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 and provide commented, minimal, self-contained, reproducible code.
[R] Get the difference of values to their own median value
Hello, I've got a matrix (mail end) with the colnames x, y, z. In this matrix are different measurements. x and y are risign coordinates. With the following line I got the median value of z for all x AND y witch are the same (not every measurment in my list hast the same number of x and y values. Sometimes lines are missing. MEDIAN - na.omit( aggregate(INPUT[,3], by=list(INPUT[,2],INPUT[,1] ), FUN=median ) ) To see the failin of my measurments I want to get the difference between every measurment and the detected median values. The aim would be to have a list with different colums (for every measurment) with the difference between measurment 1,2,3, ..., x and the median. Has anybody an idea? I though I could split the measurment list. The beginning of every measurment can be found with START - grep( 0 0.0 0., INPUT) But not every measurment has the same length and I don't have always the same number of measurments. I hope you can help me. Thank's a lot. Felix ### ## My R Code ## ### INPUT - readLines(dat.dat) INPUT - gsub(^ , , INPUT) INPUT - t( sapply( strsplit(INPUT, split= ), as.numeric ) ) colnames(INPUT) - c(x, y, z ) # ## dat.dat ## # 29 4.5 1.505713 29 4.6 1.580402 29 4.7 1.656875 29 4.8 1.735054 30 0 0 30 0.1 0.00096108 30 0.2 0.00323831 29 4.5 1.495148 29 4.6 1.568961 29 4.7 1.644467 30 0 0 30 0.1 0.00093699 30 0.2 0.00319411 30 0.3 0.00676619 __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Simulation using parts of density function
Please do not send everything twice: you are using R-help in both the To:
and Cc: fields.
I disagree with Ted: it _is_ much easier to create a generator for this
purpose.
Consider
rtgamma - function(n, ..., tr = log(500))
{
p - pgamma(tr, ...)
qgamma(p*runif(n), ...)
}
as inversion (especially at C level) is plenty fast enough.
On Wed, 2 May 2007, Thür Brigitte wrote:
Thanks for your code! It is not exactly what I really want - but it is my
fault, because my description was wrong...
It is not sim but rhater exp(rgamma(...)) that should not exceed 500. So I tried to
modify your code but it doesn't really work. sim.test returns just 1 value and not 999.
My modified code:
sim.test - NULL
for(i in 1:999)
{sim-NULL
remain - rpois(1,2000)
x - remain
while(remain0){
sim0-replicate(10*remain,
exp(rgamma(1, scale = 0.5, shape = 12))
)
sim-c(sim,sim0[sim0=500])
remain-(x - length(sim))
}
sim-sim[1:x]}
sim.test - rbind(sim.test, c(value=sum(sim)))
Thanks for any help,
Brigitte
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Gesendet: Dienstag, 1. Mai 2007 20:18
An: Thür Brigitte
Cc: r-help@stat.math.ethz.ch
Betreff: RE: [R] Simulation using parts of density function
On 01-May-07 17:03:46, Thür Brigitte wrote:
Hi
My simulation with the followin R code works perfectly:
sim - replicate(999, sum(exp(rgamma(rpois(1,2000),
scale = 0.5, shape = 12
But now I do not want to have values in object sim exceeding
5'000'000, that means that I am just using the beginning of
densitiy function gamma x 15.4. Is there a possibility to
modify my code in an easy way?
Thanks for any help!
Regards, Brigitte
A somewhat extreme problem!
The easiest way to modify the code is as below -- certiainly easier
than writing a special function to draw random samples from the
truncated gamma distribution.
A bit of experimentation shows that, from your code above, about
10% of the results are = 500. So:
sim-NULL
remain - 999
while(remain0){
sim0-replicate(10*remain,
sum(exp(rgamma(rpois(1,2000), scale = 0.5, shape = 12)))
)
sim-c(sim,sim0[sim0=500])
remain-(999 - length(sim))
}
sim-sim[1:999]
Results of a run:
sum(sim500)
[1] 0
max(sim)
[1] 4999696
length(sim)
[1] 999
It may be on the slow side (though not hugely -- on a quite slow
machine the above run was completed in 2min 5sec, while the
999-replicate in your original took 15sec. So about 8 times as long.
Most of this, of course, is taken up with the first round.
Hoping this helps,
Ted.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] missing package
l'm trying to find the survdiff package/function but it seems i cant get it how do l instal it if its not there thanks - [[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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Log-likelihood function
I think you need to learn about deviances, which R does print. Log-likelihoods are only defined up to additive constants. In this case the conventional constant differs if you view this as a Poisson or as a product-multinomial log-linear model, and R gives you the log-likelihood for a Poisson log-linear model (assuming you specified family=poisson). However, deviances and differences in log-likelihoods do not depend on which. More details and worked examples can be found in MASS (the book, see the FAQ), including other ways to fit log-linear models in R. On Tue, 1 May 2007, someone ashamed of his real name wrote: I've computed a loglinear model on a categorical dataset. I would like to test whether an interaction can be dropped by comparing the log-likelihoods from two models(the model with the interaction vs. the model without). Since R does not immediately print the log-likelihood when I use the glm function, I used SAS initially. After searching for an extracting function, I found one in R. But, the log-likelihood given by SAS is different from the one given by R. I'm not sure if the logLik function in R is giving me something I don't want. Or if I'm misinterpreting the SAS output. Can anyone help? -- 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] ? R 2.5.0 alpha bug
The version 2.5.0 has left Alpha status long time ago and its final version has been released so please try the new version. Inman, Brant A. M.D. wrote: This email is intended to highlight 2 problems that I encountered running R 2.5.0 alpha on a Windows XP machine. #1 - Open script error If I click the Open folder icon on the toolbar, R opens my script files perfectly. However, when I select File Open Script MyFileLocation, I get a fatal error that causes R to close immediately. This error was reproduced on 3 consecutive occasions but has been intermittent thereafter. One of these fatal errors resulted in a typical error reporting box being generated which I sent off. I was not able to verify if this error has been reported and corrected in subsequent versions of 2.5. #2 - Bug reporting link on CRAN website broken I tried to report the bug listed above on the CRAN website but when I clicked on the bug reporting link on the left-hand side panel of the main site (http://bugs.r-project.org/cgi-bin/R) , I get an error page with the following message: The system encountered a fatal error cannot open config file /home/sfe/r-bugs/jitterbug/R : No such file or directory The last error code was: No such file or directory uid/gid=30/8 This has been submitted to r-devel. Brant Inman __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Simulation using parts of density function
On 02-May-07 07:45:48, Prof Brian Ripley wrote:
Please do not send everything twice: you are using R-help in both the
To:
and Cc: fields.
I disagree with Ted: it _is_ much easier to create a generator for this
purpose.
Consider
rtgamma - function(n, ..., tr = log(500))
{
p - pgamma(tr, ...)
qgamma(p*runif(n), ...)
}
as inversion (especially at C level) is plenty fast enough.
Of course ... !!
Just to explain Brian's solution above:
Since pgamma(rgamma(...),...) is uniformly distributed on (0,1),
if rgamma is truncated to (0,tr) them pgamma(rgamma) will be
truncated to (0,pgamma(tr)), and hence uniformly distributed
on this range.
Best wishes,
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 02-May-07 Time: 09:16:08
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[R] hello
hello,
I have a problem with a R program I don't understand my errors
my program looks like this
for(i in LE) {
+for(j in LEC[[i]]) {
+for(k in LR) {
+ donRep[[k]] - subset(don2, Id_Essai == 1006961
Id_Cara == j Id_Rep == k Id_Geno != 65125, select = Val_O)
+ M[[j]] - matrix(rep(1, 3*length(donRep[[1]][,1])),
nrow =length(donRep[[1]][,1]), ncol = 3)
+for(k in LR) {
+ M[[j]][,k] - as.numeric(as.character(donRep[[k]][,1]))
+}
+ }
+print(M[[j]])
+if (subset(donParCara, Id_Cara == j , select = Ana_C) == AV)
+ {print(cor(M[[j]], method = pearson))}
else{print(cor(M[[j]], method = spearman))}
+ }
+ }
Erreur dans M[[j]][, k] - as.numeric(as.character(donRep[[k]][, 1])) :
le nombre d'objets à remplacer n'est pas multiple de la taille du
remplacement
and sometimes it run good but sometimes there are errors I don't understand
why. Can you help me please?
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Re: [R] Log-likelihood function
You're right. I do need to learn more. I never learned null/residual deviance. I know the deviance is equivalent to an anova decompostion. But I've never dealt with it seperated like this. I understand deviance as the difference between two model's log-likelihood difference between them and the most complex. I want to compare two models that are not the most complex. That is why I wanted the log-likelihood. I am using the poisson distribution because my response is count data, so the link is the log. If the deviance in R is computed by comparing the fitted model against the most saturated (which would make sense). Then yes, I can use that. I just picked the log-likelihood because I'm comparing two models. And that's the best way. But, it's equivalent if R compares the fitted to the most complex. I assumed the deviance print out tested the fitted model against the least complex. This tests whether the current model parameters can be dropped (that's what I thought the null deviance meant). I'm not sure what the residaul deviance means though. My main concern is why the likelihood functions differed between SAS and R. If anyone has encountered this or understands why, I would appreciate some help. Prof Brian Ripley wrote: I think you need to learn about deviances, which R does print. Log-likelihoods are only defined up to additive constants. In this case the conventional constant differs if you view this as a Poisson or as a product-multinomial log-linear model, and R gives you the log-likelihood for a Poisson log-linear model (assuming you specified family=poisson). However, deviances and differences in log-likelihoods do not depend on which. More details and worked examples can be found in MASS (the book, see the FAQ), including other ways to fit log-linear models in R. On Tue, 1 May 2007, someone ashamed of his real name wrote: I've computed a loglinear model on a categorical dataset. I would like to test whether an interaction can be dropped by comparing the log-likelihoods from two models(the model with the interaction vs. the model without). Since R does not immediately print the log-likelihood when I use the glm function, I used SAS initially. After searching for an extracting function, I found one in R. But, the log-likelihood given by SAS is different from the one given by R. I'm not sure if the logLik function in R is giving me something I don't want. Or if I'm misinterpreting the SAS output. Can anyone help? -- 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 and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Log-likelihood-function-tf3678882.html#a10282152 Sent from the R help mailing list archive at Nabble.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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] missing package
--- raymond chiruka [EMAIL PROTECTED] wrote: l'm trying to find the survdiff package/function but it seems i cant get it how do l instal it if its not there thanks I think you want the survival package. __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] missing values
hello,
I need your help for this example
for(k in LR) {
+ donGeno[[k]] - as.numeric(levels(factor(subset(don2, Id_Essai == 1006961
Id_Cara == LC[1] Id_Rep == k, select = Id_Geno)[,1])))
+ print(donGeno[[k]])}
[1] 65125 65126 65127 65128 65129 65130 65131 65132 65133 65134 65135 65136
65137 65138 65139 65140 65141 65142 65143 65144 65171
[1] 65126 65127 65128 65129 65130 65131 65132 65133 65134 65135 65136 65137
65138 65139 65140 65141 65142 65143 65144 65171
[1] 65125 65126 65127 65128 65129 65130 65131 65132 65133 65134 65135 65136
65137 65138 65139 65140 65141 65142 65143 65144 65171
there are a missing value for the vector donGeno[[2]] in fact there aren't the
value 65125 and I wanna cut this value in the others vectors and I tried to do
this as follow
C - vector()
for(k in LR) {
C[k] - length(donGeno[[k]])
}
print(C)
na=match(rep(0,length(C)-sum(match(C,C[1],nomatch=0))),match(C,C[1],nomatch=0))
#print(na)
if(na==length(C)){
pos=match(0,match(donGeno[[na-1]],donGeno[[na]],nomatch=0))
for(k in 1:(na-1)) {
donGeno[[k]] - donGeno[[k]][1:(na-1)]
}
}
else{
pos=match(0,match(donGeno[[na+1]],donGeno[[na]],nomatch=0))
for(k in 1:(.))
}
but I wonder if there's better from this script?
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Re: [R] missing package
help.search(survdiff) tells you it is in package survival. Most probably this package is not missing, nevertheless if it is, you can install it in a standard way using install.packages() or from the menu (Windows). Petr raymond chiruka napsal(a): l'm trying to find the survdiff package/function but it seems i cant get it how do l instal it if its not there thanks - [[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 and provide commented, minimal, self-contained, reproducible code. -- Petr Klasterecky Dept. of Probability and Statistics Charles University in Prague Czech Republic __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] length of the object is not a multiple of... [was: Re: hello]
elyakhlifi mustapha napsal(a):
hello,
I have a problem with a R program I don't understand my errors
my program looks like this
for(i in LE) {
+for(j in LEC[[i]]) {
+for(k in LR) {
+ donRep[[k]] - subset(don2, Id_Essai == 1006961
Id_Cara == j Id_Rep == k Id_Geno != 65125, select = Val_O)
+ M[[j]] - matrix(rep(1, 3*length(donRep[[1]][,1])),
nrow =length(donRep[[1]][,1]), ncol = 3)
+for(k in LR) {
+ M[[j]][,k] - as.numeric(as.character(donRep[[k]][,1]))
+}
+ }
+print(M[[j]])
+if (subset(donParCara, Id_Cara == j , select = Ana_C) ==
AV)
+ {print(cor(M[[j]], method = pearson))}
else{print(cor(M[[j]], method = spearman))}
+ }
+ }
Erreur dans M[[j]][, k] - as.numeric(as.character(donRep[[k]][, 1])) :
le nombre d'objets ŕ remplacer n'est pas multiple de la taille du
remplacement
and sometimes it run good but sometimes there are errors I don't understand
why. Can you help me please?
No we can't since we cannot reproduce your program. The error message
says you are trying to assign a longer/shorter vector into a matrix. Add
some print statements into your code and check how M[[j]] and
donRep[[k]][,1] look like in each step of the loop.
And next time, read the posting guide and give a meaningfull subject and
a reproducible example.
Petr
--
Petr Klasterecky
Dept. of Probability and Statistics
Charles University in Prague
Czech Republic
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Re: [R] to draw a smooth arc
(Ted Harding) wrote: This thread prompts me to ask about something I've been pondering for a while, as to whether there's an implementation somewhere ticked away in the R resources. So far, people have been responding to the original query in terms of increasing the numbers of points, and joining these by lines. However, if you're using PostScript output, you can draw really smooth curves by exploiting PS's curveto operator. This draws a cubic-curve segment in the following way: ... Anyway. The Question: is there a general function for the above kind of smooth curve-drawing? Hi Ted, My experience with this some years ago was in anti-aliasing visual stimuli presented on a PC monitor. The intended line was calculated pixel by pixel and the each pixel that the line touched was assigned a value based on a sawtooth function. The peak of the function was always exactly on the intended line and the luminance of the pixel was a linear function of the distance of its center along a perpendicular to the line. We used the width of a pixel as the lateral extent of the function with adequate success. The lines appeared smooth until the viewer could resolve the individual pixels. Wider lines used a trapezoidal distribution of luminance with the same slope on the limbs. I must note that we did not have to create our stimuli in real time, and this method might be too slow. However, I am fairly certain that if the function knew about the characteristics of the output device, one could base a curveto function on this. Jim __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Polar graph of time and tide
Alan E. Davis wrote: I have been trying to visualize times of lowest tides, month by month. I have tide predictions with times either in unix time or a text format, and heights in feet or meters. I had been able to derive the clock times of each prediction. I would now like to graph this data with points showing heights as r and times as theta, from to 2355. There is a seasonal component: I am interested in displaying times of lowest tides in particular. I am sure this is so simple as to burden those on the list; I however have spent two evenings trying to figure out how to use polar.plot, and I'm not sure that's the best way to do this. May I request some advice? The docs with polar.plot are not complete, I fear. Thank you, begging for your indulgence, Hi Alan, Earl Glynn's advice is spot-on if you are trying to map tides onto the diurnal cycle. However, I get the impression that you want an annual cycle. If this is the case, it is probably best to go to the underlying function, radial.plot. Here is an example with some imaginary tides. lowtide-sin(seq(1:12)+sin(seq(1,24,by=2)/10)) lowtide-rescale(lowtide,c(0,2355)) month.abbr-c(jan,Feb,Mar,Apr,May,Jun, Jul,Aug,Sep,Oct,Nov,Dec) radial.plot(lowtide,labels=month.abbr,rp.type=s) This plots time of lowest tide in a month as a symbol at a radial distance proportional to the 24 hour time of day. As you can see from running the example, the months of the year are displayed around the circumference of the plot. One annoyance is that the hours are displayed as 0-2500. As pretty handles that, you would have to set show.radial.grid to FALSE and manually draw the desired grid (a pain, but not impossible). You could also place the symbols at the position in the month of the lowest tide by specifying radial.pos instead of letting the function spread them out evenly. If you want to overlay a number of years, x should be a matrix and point.col can be chosen so that each year has a different color. I would like to know of any deficiencies in the docs, as while I think I know what I'm talking about, it is you out there who have to understand it. Jim __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Log-likelihood function
Alternatively generate the log-likelihood using the sum(dpois(y, fitted(model), log = TRUE)) Regards Ross Darnell Doxastic wrote: You're right. I do need to learn more. I never learned null/residual deviance. I know the deviance is equivalent to an anova decompostion. But I've never dealt with it seperated like this. I understand deviance as the difference between two model's log-likelihood difference between them and the most complex. I want to compare two models that are not the most complex. That is why I wanted the log-likelihood. I am using the poisson distribution because my response is count data, so the link is the log. If the deviance in R is computed by comparing the fitted model against the most saturated (which would make sense). Then yes, I can use that. I just picked the log-likelihood because I'm comparing two models. And that's the best way. But, it's equivalent if R compares the fitted to the most complex. I assumed the deviance print out tested the fitted model against the least complex. This tests whether the current model parameters can be dropped (that's what I thought the null deviance meant). I'm not sure what the residaul deviance means though. My main concern is why the likelihood functions differed between SAS and R. If anyone has encountered this or understands why, I would appreciate some help. Prof Brian Ripley wrote: I think you need to learn about deviances, which R does print. Log-likelihoods are only defined up to additive constants. In this case the conventional constant differs if you view this as a Poisson or as a product-multinomial log-linear model, and R gives you the log-likelihood for a Poisson log-linear model (assuming you specified family=poisson). However, deviances and differences in log-likelihoods do not depend on which. More details and worked examples can be found in MASS (the book, see the FAQ), including other ways to fit log-linear models in R. On Tue, 1 May 2007, someone ashamed of his real name wrote: I've computed a loglinear model on a categorical dataset. I would like to test whether an interaction can be dropped by comparing the log-likelihoods from two models(the model with the interaction vs. the model without). Since R does not immediately print the log-likelihood when I use the glm function, I used SAS initially. After searching for an extracting function, I found one in R. But, the log-likelihood given by SAS is different from the one given by R. I'm not sure if the logLik function in R is giving me something I don't want. Or if I'm misinterpreting the SAS output. Can anyone help? -- 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 and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Log-likelihood-function-tf3678882.html#a10283755 Sent from the R help mailing list archive at Nabble.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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Concepts question: environment, frame, search path
On 01/05/2007 11:34 AM, Prof Brian Ripley wrote: On Tue, 1 May 2007, Duncan Murdoch wrote: On 01/05/2007 12:29 AM, Graham Wideman wrote: [...] Refman p122: Environments consist of a frame, or collection of named objects, and a pointer to an enclosing environment. Is the or here explaining parenthetically that a frame is a collection of named objects, or is separating too alternative structures for an environment? The former. If the former, does this imply that a frame can contain arbitrary variables? Yes, but a frame isn't an R object, it's a concept that appears in descriptions, e.g. part of an environment, or the local variables created during function evaluation, etc. And pointer? Is that a type of thing in R? No, there are no pointers in R. There are a couple of tricks to fake them (e.g. environment objects aren't copied when assigned, you just get a new reference to the same environment; this allows you to construct something like a pointer by wrapping an object in an environment), but I don't recommend using these routinely. Nevertheless, the statement is true. R is implemented using pointers. Yes, definitely. I misread the original posting, and didn't notice that the pointer question was connected to the previous quote. Duncan Murdoch __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] Degrees of freedom in repeated measures glmmPQL
Hello, I've just carried out my first good-looking model using glmmPQL, and the output makes perfect sense in terms of how it fits with our hypothesis and the graphical representation of the data. However, please could you clarify whether my degrees of freedom are appropriate? I had 106 subjects, each of them was observed about 9 times, creating 882 data points. The subjects were in 3 treatment groups, so I have told the model to include subject as a random factor nested within treatment. There are two other variables and I'm interested in their two-way interactions with Treatment. I'm happy with the model structure, and the output generally looks right, but... In the 'DF' column of the output table, it has 882 as the degrees of freedom for all the variables (except Treatment itself, which has 0 degrees of freedom). At the bottom of the output, it says Groups: Subjects = 106, Treatment = 3. Should I be worried or is this what to expect?! I was expecting it to be more like an ANOVA table, where the error degrees of freedom should reflect the number of subjects, not all the data points. I can't see the usual differentiation between the numerater and denominator/error degrees of freedom, so am I right in thinking the DF column shows the error degrees of freedom? Or do glmms not work like this? Thank you very much in advance, Charlotte __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Log-likelihood function
Thanks. I used this and it gave me the same result as the logLik function. The reason I ask is the SAS output gives me a loglik = 1089. R gives me -298.09583. Both for my reduced model. For the saturated (or complex) model, SAS gives me an loglik = 1143. R gives me -298.1993. The problem is these give two very different pictures about whether I can drop the interaction. However, I think the residual deviance in the R output is equal to G^2. So, I can just take the difference between those two. If I do this, I get a difference with an interpretation similar to that of what comes from SAS. So I think I'll just go with that. But who knows if I'm right (not me)? Thanks! Ross Darnell wrote: Alternatively generate the log-likelihood using the sum(dpois(y, fitted(model), log = TRUE)) Regards Ross Darnell Doxastic wrote: You're right. I do need to learn more. I never learned null/residual deviance. I know the deviance is equivalent to an anova decompostion. But I've never dealt with it seperated like this. I understand deviance as the difference between two model's log-likelihood difference between them and the most complex. I want to compare two models that are not the most complex. That is why I wanted the log-likelihood. I am using the poisson distribution because my response is count data, so the link is the log. If the deviance in R is computed by comparing the fitted model against the most saturated (which would make sense). Then yes, I can use that. I just picked the log-likelihood because I'm comparing two models. And that's the best way. But, it's equivalent if R compares the fitted to the most complex. I assumed the deviance print out tested the fitted model against the least complex. This tests whether the current model parameters can be dropped (that's what I thought the null deviance meant). I'm not sure what the residaul deviance means though. My main concern is why the likelihood functions differed between SAS and R. If anyone has encountered this or understands why, I would appreciate some help. Prof Brian Ripley wrote: I think you need to learn about deviances, which R does print. Log-likelihoods are only defined up to additive constants. In this case the conventional constant differs if you view this as a Poisson or as a product-multinomial log-linear model, and R gives you the log-likelihood for a Poisson log-linear model (assuming you specified family=poisson). However, deviances and differences in log-likelihoods do not depend on which. More details and worked examples can be found in MASS (the book, see the FAQ), including other ways to fit log-linear models in R. On Tue, 1 May 2007, someone ashamed of his real name wrote: I've computed a loglinear model on a categorical dataset. I would like to test whether an interaction can be dropped by comparing the log-likelihoods from two models(the model with the interaction vs. the model without). Since R does not immediately print the log-likelihood when I use the glm function, I used SAS initially. After searching for an extracting function, I found one in R. But, the log-likelihood given by SAS is different from the one given by R. I'm not sure if the logLik function in R is giving me something I don't want. Or if I'm misinterpreting the SAS output. Can anyone help? -- 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 and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Log-likelihood-function-tf3678882.html#a10284217 Sent from the R help mailing list archive at Nabble.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 and provide commented, minimal, self-contained, reproducible code.
[R] I need help
hello,
I need help because I don't understand the syntaxe else how can I write it
for example I writed a script to cut missings values and I have errors
if(na==length(C)){
+ pos=match(0,match(donGeno[[na-1]],donGeno[[na]],nomatch=0))
+ for(k in 1:(na-1)) {
+ if(pos==1) {donGeno[[k]] -
donGeno[[k]][2:C[k]]}
+if(pos==C[k]){donGeno[[k]] - donGeno[[k]][1:(C[k]-1)]}
+else{donGeno[[k]] -
c(donGeno[[k]][1:(pos-1)],donGeno[[k]][(pos+1):C(k)])}
+ }
+}
if(na==1){
+ pos=match(0,match(donGeno[[na+1]],donGeno[[na]],nomatch=0))
+ for(k in 2:length(C)){
+ if(pos==1) {donGeno[[k]] - donGeno[[k]][2:C[k]]}
+ if(pos==C[k]){donGeno[[k]] - donGeno[[k]][1:(C[k]-1)]}
+ else{donGeno[[k]] - c(donGeno[[k]][1:(pos-1)],donGeno[[k]][(pos+1):C(k)])}
+ }
+ }
else{for(k in 1:(na-1)){
Erreur : erreur de syntaxe dans else
if(pos==1) {donGeno[[k]] - donGeno[[k]][2:C[k]]}
if(pos==C[k]){donGeno[[k]] - donGeno[[k]][1:(C[k]-1)]}
else{donGeno[[k]] - c(donGeno[[k]][1:(pos-1)],donGeno[[k]][(pos+1):C(k)])}
Erreur : erreur de syntaxe dans else
}
Erreur : erreur de syntaxe dans }
for(k in 1:(na-1)){
+if(pos==1) {donGeno[[k]] - donGeno[[k]][2:C[k]]}
+ if(pos==C[k]){donGeno[[k]] - donGeno[[k]][1:(C[k]-1)]}
+ else{donGeno[[k]] - c(donGeno[[k]][1:(pos-1)],donGeno[[k]][(pos+1):C(k)])}
+ }
Erreur dans C(k) : objet non interprétable comme un facteur
}
Erreur : erreur de syntaxe dans }
Have you got some ideas?
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] I need help
On 02/05/2007 8:11 AM, elyakhlifi mustapha wrote:
hello,
I need help because I don't understand the syntaxe else how can I write it
for example I writed a script to cut missings values and I have errors
if(na==length(C)){
+ pos=match(0,match(donGeno[[na-1]],donGeno[[na]],nomatch=0))
+ for(k in 1:(na-1)) {
+ if(pos==1) {donGeno[[k]] -
donGeno[[k]][2:C[k]]}
+if(pos==C[k]){donGeno[[k]] - donGeno[[k]][1:(C[k]-1)]}
+else{donGeno[[k]] -
c(donGeno[[k]][1:(pos-1)],donGeno[[k]][(pos+1):C(k)])}
+ }
+}
if(na==1){
+ pos=match(0,match(donGeno[[na+1]],donGeno[[na]],nomatch=0))
+ for(k in 2:length(C)){
+ if(pos==1) {donGeno[[k]] - donGeno[[k]][2:C[k]]}
+ if(pos==C[k]){donGeno[[k]] - donGeno[[k]][1:(C[k]-1)]}
+ else{donGeno[[k]] - c(donGeno[[k]][1:(pos-1)],donGeno[[k]][(pos+1):C(k)])}
+ }
+ }
else{for(k in 1:(na-1)){
Erreur : erreur de syntaxe dans else
R will parse a complete statement if you enter one. Your if() was
complete before the else was entered, so the else got orphaned.
The usual convention to avoid this is to put the else on the same line
as the brace that closes the if, e.g.
if(na==1){
...
} else {
...
}
Duncan Murdoch
if(pos==1) {donGeno[[k]] - donGeno[[k]][2:C[k]]}
if(pos==C[k]){donGeno[[k]] - donGeno[[k]][1:(C[k]-1)]}
else{donGeno[[k]] - c(donGeno[[k]][1:(pos-1)],donGeno[[k]][(pos+1):C(k)])}
Erreur : erreur de syntaxe dans else
}
Erreur : erreur de syntaxe dans }
for(k in 1:(na-1)){
+if(pos==1) {donGeno[[k]] - donGeno[[k]][2:C[k]]}
+ if(pos==C[k]){donGeno[[k]] - donGeno[[k]][1:(C[k]-1)]}
+ else{donGeno[[k]] -
c(donGeno[[k]][1:(pos-1)],donGeno[[k]][(pos+1):C(k)])}
+ }
Erreur dans C(k) : objet non interprétable comme un facteur
}
Erreur : erreur de syntaxe dans }
Have you got some ideas?
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__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Odp: I need help
Hi
[EMAIL PROTECTED] napsal dne 02.05.2007 14:11:51:
hello,
I need help because I don't understand the syntaxe else how can I
write it
for example I writed a script to cut missings values and I have errors
if(na==length(C)){
+ pos=match(0,match(donGeno[[na-1]],donGeno[[na]],nomatch=0))
+ for(k in 1:(na-1)) {
+ if(pos==1) {donGeno[[k]]
-
donGeno[[k]][2:C[k]]}
+if(pos==C[k]){donGeno[[k]] - donGeno[[k]][1:(C[k]-1)]}
+else{donGeno[[k]] -
c(donGeno[[k]][1:(pos-1)],donGeno[[k]][(pos+1):C(k)])}
+ }
+}
if(na==1){
+ pos=match(0,match(donGeno[[na+1]],donGeno[[na]],nomatch=0))
+ for(k in 2:length(C)){
+ if(pos==1) {donGeno[[k]] - donGeno[[k]][2:C[k]]}
+ if(pos==C[k]){donGeno[[k]] - donGeno[[k]][1:(C[k]-1)]}
+ else{donGeno[[k]] -
c(donGeno[[k]][1:(pos-1)],donGeno[[k]][(pos+1):C(k)])}
+ }
+ }
else{for(k in 1:(na-1)){
Erreur : erreur de syntaxe dans else
if(pos==1) {donGeno[[k]] - donGeno[[k]][2:C[k]]}
if(pos==C[k]){donGeno[[k]] - donGeno[[k]][1:(C[k]-1)]}
else{donGeno[[k]] -
c(donGeno[[k]][1:(pos-1)],donGeno[[k]][(pos+1):C(k)])}
Erreur : erreur de syntaxe dans else
}
Erreur : erreur de syntaxe dans }
for(k in 1:(na-1)){
+if(pos==1) {donGeno[[k]] - donGeno[[k]][2:C[k]]}
+ if(pos==C[k]){donGeno[[k]] - donGeno[[k]][1:(C[k]-1)]}
+ else{donGeno[[k]] -
c(donGeno[[k]][1:(pos-1)],donGeno[[k]][(pos+1):C(k)])}
+ }
Erreur dans C(k) : objet non interprétable comme un facteur
}
Erreur : erreur de syntaxe dans }
Have you got some ideas?
What about to try to provide some reproducible example as suggested in
posting guide. I believe your messy code can be evaluated in much more
neat and concise way without so many ifs and fors. Maybe you can uncover
some by yourself what trying to write a simple reproducible example. I am
reluctant to decipher what you want to achieve but maybe you want retain
only common values of several sets. So e.g. from match help page
## The intersection of two sets :
intersect - function(x, y) y[match(x, y, nomatch = 0)]
x-sample(1:100, 50)
y-1:50
x2-sample(1:100,50)
intersect(x,x2)
[1] 39 87 66 7 64 79 62 98 6 95 96 35 74 36 3 50 58 97 52 33 61 88 47
17 32 11 76 25
intersect(y,intersect(x,x2))
[1] 3 6 7 11 17 25 32 33 35 36 39 47 50
Regarding the error message
if (1==1) print(25) else print(30)
[1] 25
if (1==2) print(25) else print(30)
[1] 30
if (1==1) print(25)
[1] 25
else print(30)
Error: syntax error, unexpected ELSE in else
From help page
In particular, you ***should not have a newline between } and else to
avoid a syntax error*** in entering a if ... else construct at the
keyboard or via source. For that reason, one
(somewhat extreme) attitude of defensive programming is to always use
braces, e.g., for if clauses
Regards
Petr
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Get the difference of values to their own median value
Use ave. Also its easier to use read.table to read it in and then convert it to a matrix if that's what you want. Input - 29 4.5 1.505713 29 4.6 1.580402 29 4.7 1.656875 29 4.8 1.735054 30 0 0 30 0.1 0.00096108 30 0.2 0.00323831 29 4.5 1.495148 29 4.6 1.568961 29 4.7 1.644467 30 0 0 30 0.1 0.00093699 30 0.2 0.00319411 30 0.3 0.00676619 cn - c(x, y, z) # replace next line with this: # INPUT - as.matrix(read.table(dat.dat, col.names = cn)) INPUT - as.matrix(read.table(textConnection(Input), col.names = cn)) INPUT[,3] - ave(INPUT[,3], INPUT[,1], INPUT[,2], FUN = median) On 5/2/07, Felix Wave [EMAIL PROTECTED] wrote: Hello, I've got a matrix (mail end) with the colnames x, y, z. In this matrix are different measurements. x and y are risign coordinates. With the following line I got the median value of z for all x AND y witch are the same (not every measurment in my list hast the same number of x and y values. Sometimes lines are missing. MEDIAN - na.omit( aggregate(INPUT[,3], by=list(INPUT[,2],INPUT[,1] ), FUN=median ) ) To see the failin of my measurments I want to get the difference between every measurment and the detected median values. The aim would be to have a list with different colums (for every measurment) with the difference between measurment 1,2,3, ..., x and the median. Has anybody an idea? I though I could split the measurment list. The beginning of every measurment can be found with START - grep( 0 0.0 0., INPUT) But not every measurment has the same length and I don't have always the same number of measurments. I hope you can help me. Thank's a lot. Felix ### ## My R Code ## ### INPUT - readLines(dat.dat) INPUT - gsub(^ , , INPUT) INPUT - t( sapply( strsplit(INPUT, split= ), as.numeric ) ) colnames(INPUT) - c(x, y, z ) # ## dat.dat ## # 29 4.5 1.505713 29 4.6 1.580402 29 4.7 1.656875 29 4.8 1.735054 30 0 0 30 0.1 0.00096108 30 0.2 0.00323831 29 4.5 1.495148 29 4.6 1.568961 29 4.7 1.644467 30 0 0 30 0.1 0.00093699 30 0.2 0.00319411 30 0.3 0.00676619 __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] repeated measures and recurrent events data
Hello, I need a data set on repeated measures and recurrent events data which had been registered on the same subject. This is very important to my thesis work. Thanks a lot, Luis Guillermo Díaz Monroy Profesor Asociado Universidad Nacional de Colombia __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] ED50 from logistic model with interactions
Hi Kate, try looking at the package 'drc' on CRAN and in particular look at the example in the help page for the dataset 'daphnids' (?daphnids). You can obtain arbitrary ED values with approximate standard errors using the function 'ED'. Christian __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] how to concatinate the elements of some text vectors cat() or print() ?
I have some comment text taken from a SAS data file.
It is stored in two vectors and is difficult to read.
I would like to simply concatentate the individual
entries and end up with a character vector that give
me one line of text per comment.
I cannot see how to do this, yet it must be very easy.
I have played around with cat() and print with no
success. Would someone kindly point out where I
am going wrong?
Thanks
Simple Example:
aa - LETTERS[1:5]
bb - letters[1:5]
cat(aa[1], bb[1])# works for individuals
cat(aa,bb)#(concatinates entire vectors)
# Using sink I might get it to work if I could figure
out how to escape a
# new line command. encodeString does not seem
appropriate here.
harry - c(rep(NA,5))
for (i in 1:5 ) {
cat (aa[i],bb[i])
}
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[R] NAs introduced by coercion in dist()
I work with Windows and use R version 2.4.1. I am JUST starting to learn this program... I get this warning message 'NAs introduced by coercion' while trying to build a distance matrix (to be analyzed with NMDS later) from a 336 x 100 data matrix. The original matrix has lots of zeros and no missing values, but I don't think this should matter. I searched this forum and people have suggested that the warning should be ignored but when I try to print the distance matrix I only get the row numbers (the matrix seems to be 'empty') and I'm not being able to judge whether the matrix worked or not. To get the distance matrix I wrote: dist.PxMx - dist (PeaksMatrix, method='euclidean', diag=FALSE, upper=FALSE) I tried including the p argument (included in the help for dist()) and leaving it out, but that didn't seem to change anything. I think that's required for one distance measure though, not for euclidean dist. Should I really ignore this warning? If so, why am I not being able to see the distance matrix? -- View this message in context: http://www.nabble.com/NAs-introduced-by-coercion-in-dist%28%29-tf3680727.html#a10286358 Sent from the R help mailing list archive at Nabble.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 and provide commented, minimal, self-contained, reproducible code.
[R] KS test pvalue estimation using mctest (library truncgof)
Hi, I'm trying to evaluate a Monte Carlo p-value (using truncgof package) on a left truncated sample. From an empirical sample I've estimated a generalized pareto distribution parameters (xi, beta, threshold) (I've used fExtremes pkg). I'm in doubt on what of the following command is the most appropriate: Let: x-sample t-threshold xt-x[xt] xihat-gpdFit(x, threshold=t, type = pwm)$par.ests[1] betahat-gpdFit(x, threshold=t, type = pwm)$par.ests[2] (1) ks.test(xt,pgpd,list(xi=xihat,beta=betahat),H=t,estfun = as.list(gpdFit(x, 0)$par.ests), tol = 1e-02) (2) ks.test(xt,pgpd,list(xi=xihat,beta=betahat),H=t,estfun = as.list(gpdFit(x, t)$par.ests), tol = 1e-02) (3) ks.test(xt,pgpd,list(xi=xihat,beta=betahat,mu=t),estfun = as.list(gpdFit(x, t)$par.ests), tol = 1e-02) Someone have ever faced this problem? I need to understand the role of threshold in the Monte Carlo sampling from the GPD. In the 1st case I've obtained high value of statistics and p-value, in the second same value of statistic and very low p-value, in the 3rd low statistic and p-value always equal to 1. Thank you very much in advance Regards Claudia __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] Few contributed binaries for Mac OS X?
Dear all, After upgrading to R 2.5.0 on my Intel-based Mac I find that there are only a few binary packages available. I've installed most of the packages I need from source, but am having trouble compiling RGtk2. Will Mac binary packages be available for R 2.5.0? Thanks, Ista Zahn http://izahn.homedns.org __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Degrees of freedom in repeated measures glmmPQL
Apologies, I made a mistake with my maths. The degrees of freedom look correct, assuming they are the denominator and that glmms work this way. I just under-estimated the number of data points I had. Sorry. Charlotte On 02/05/07, Charlotte Burn [EMAIL PROTECTED] wrote: Hello, I've just carried out my first good-looking model using glmmPQL, and the output makes perfect sense in terms of how it fits with our hypothesis and the graphical representation of the data. However, please could you clarify whether my degrees of freedom are appropriate? I had 106 subjects, each of them was observed about 9 times, creating 882 data points. The subjects were in 3 treatment groups, so I have told the model to include subject as a random factor nested within treatment. There are two other variables and I'm interested in their two-way interactions with Treatment. I'm happy with the model structure, and the output generally looks right, but... In the 'DF' column of the output table, it has 882 as the degrees of freedom for all the variables (except Treatment itself, which has 0 degrees of freedom). At the bottom of the output, it says Groups: Subjects = 106, Treatment = 3. Should I be worried or is this what to expect?! I was expecting it to be more like an ANOVA table, where the error degrees of freedom should reflect the number of subjects, not all the data points. I can't see the usual differentiation between the numerater and denominator/error degrees of freedom, so am I right in thinking the DF column shows the error degrees of freedom? Or do glmms not work like this? Thank you very much in advance, Charlotte -- -- Dr Charlotte C. Burn Department of Animal Welfare and Behaviour School of Clinical Veterinary Science University of Bristol Langford House Bristol BS40 5DU Tel: 0117 9219134 http://seis.bristol.ac.uk/~frccb/charlotteburn.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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Polar graph of time and tide
On 5/2/07, Jim Lemon [EMAIL PROTECTED] wrote: Alan E. Davis wrote: I have been trying to visualize times of lowest tides, month by month. I have tide predictions with times either in unix time or a text format, and heights in feet or meters. I had been able to derive the clock times of each prediction. I would now like to graph this data with points showing heights as r and times as theta, from to 2355. There is a seasonal component: I am interested in displaying times of lowest tides in particular. I am sure this is so simple as to burden those on the list; I however have spent two evenings trying to figure out how to use polar.plot, and I'm not sure that's the best way to do this. May I request some advice? The docs with polar.plot are not complete, I fear. Thank you, begging for your indulgence, Hi Alan, Earl Glynn's advice is spot-on if you are trying to map tides onto the diurnal cycle. However, I get the impression that you want an annual cycle. If this is the case, it is probably best to go to the underlying function, radial.plot. Here is an example with some imaginary tides. lowtide-sin(seq(1:12)+sin(seq(1,24,by=2)/10)) lowtide-rescale(lowtide,c(0,2355)) month.abbr-c(jan,Feb,Mar,Apr,May,Jun, Jul,Aug,Sep,Oct,Nov,Dec) radial.plot(lowtide,labels=month.abbr,rp.type=s) Note that month.abb is built into R so one could eliminate the month.abbr- line and write: radial.plot(lowtide, labels = month.abb, rp.type = s) __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to concatinate the elements of some text vectors cat() or print() ?
Is paste() what you're looking for?
Andy
From: John Kane
I have some comment text taken from a SAS data file.
It is stored in two vectors and is difficult to read.
I would like to simply concatentate the individual
entries and end up with a character vector that give
me one line of text per comment.
I cannot see how to do this, yet it must be very easy.
I have played around with cat() and print with no
success. Would someone kindly point out where I
am going wrong?
Thanks
Simple Example:
aa - LETTERS[1:5]
bb - letters[1:5]
cat(aa[1], bb[1])# works for individuals
cat(aa,bb)#(concatinates entire vectors)
# Using sink I might get it to work if I could figure
out how to escape a
# new line command. encodeString does not seem
appropriate here.
harry - c(rep(NA,5))
for (i in 1:5 ) {
cat (aa[i],bb[i])
}
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--
Notice: This e-mail message, together with any attachments,...{{dropped}}
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Re: [R] ED50 from logistic model with interactions
Kate Stark wrote:
Hi,
I was wondering if someone could please help me. I am doing a logistic
regression to compare size at maturity between 3 seasons. My model is:
fit - glm(Mature ~ Season * Size - 1, family = binomial, data=dat)
where Mature is a binary response, 0 for immature, 1 for mature. There
are 3 Seasons.
The Season * Size interaction is significant. I would like to compare the
size at 50% maturity between Seasons, which I have calculated as:
Mat50_S1 - -fit$coef[1]/fit$coef[4]
Mat50_S2 - -fit$coef[2]/(fit$coef[4] + fit$coef[5])
Mat50_S3 - -fit$coef[3]/(fit$coef[4] + fit$coef[6])
But I am not sure how to calculate the standard error around each of
these estimates. The p.dose function from the MASS package does this
automatically, but it doesn’t seem to allow interaction terms.
In Faraway(2006) he has an example using the delta method to calculate
the StdErr, but again without any interactions. I can apply this for the
first Season, as there is just one intercept and one slope coefficient,
but for the other 2 Seasons, the slope is a combination of the Size
coefficient and the Size*Season coefficient, and I am not sure how to use
the covariance matrix in the delta calculation.
I could divide the data and do 3 different logistic regressions, one for
each season, but while the Mat50 (i.e. mean Size at 50% maturity) is the
same as that calculated by the separate lines regression, Im not sure how
this may change the StdErr?
Regards,
Kate
Hi,
Maybe you can re-parameterize the logistic model in such a way that the
size at 50% maturity for each season are explicit parameters in the
model, so the support function is maximized for those arguments and you
get the standard errors directly. For example, the following code
estimate the size at 50% maturity when the only predictor variable is
size. I guess you can use it to generalize to the case of an additional
factor such as season. I wrote this code as a translation from ADMB to R
so it is rather detailed (it uses nlm for maximization). count is real
data from a squid population and size (l) is in cm. Note in prop.ini,
prop.est, and prop.fit the reparameterization that introduces the size
at 50% maturity directly (along with the size at 95% maturity).
Rubén Roa-Ureta
l-sort(c(8:33,8:33))
mat-rep.int(x=c(0,1),times=26)
imat-rep.int(x=c(1,0),times=26)
count-c(2,0,3,0,3,0,6,0,7,0,9,0,6,0,3,0,6,0,2,0,4,0,1,0,4,1,2,1,2,0,1,2,1,1,2,0,1,2,2,2,0,0,1,1,0,0,0,0,0,0,0,1)
l.plot-8:33
contot-vector(mode=numeric,length=26)
prop.obs-vector(mode=numeric,length=26)
for (i in 1:26) {
countot[i]-count[i*2-1]+count[i*2]
ifelse(countot[i]0,prop.obs[i]-count[i*2]/countot[i],NA)
}
l_50=25
l_95=35
prop.ini-1/(1+exp((log(1/19))*((l.plot-l_50)/(l_95-l_50
plot(l.plot,prop.obs)
lines(l.plot,prop.ini)
fn-function(p){
prop.est-1/(1+exp(log(1/19)*(l-p[1])/(p[2]-p[1])));
iprop.est-1-prop.est;
negloglik--sum(count*(mat*log(prop.est)+imat*log(iprop.est)));
}
prop.lik-nlm(fn,p=c(25.8,33.9),hessian=TRUE)
prop.lik
L_50-prop.lik$estimate[1]
L_95-prop.lik$estimate[2]
prop.covmat-solve(prop.lik$hessian)
seL_50-sqrt(prop.covmat[1,1])
seL_95-sqrt(prop.covmat[2,2])
covL_50_95-prop.covmat[1,2]
prop.fit-1/(1+exp((log(1/19))*((l_plot-L_50)/(L_95-L_50
plot(l.plot,prop.obs,pch=19,xlab=Length (cm),ylab=Proportion Mature)
lines(l.plot,prop.fit)
text(x=12.5,y=0.9,expression(paste(p(l)=,frac(1,1+e^frac(ln(1/19)(l-l[50]),(l[95]-l[50]))
text(x=12.5,y=0.7,expression(paste(hat(l)[50],=25.8)))
text(x=12.5,y=0.55,expression(paste(hat(l)[95],=34.0)))
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Re: [R] NAs introduced by coercion in dist()
slomascolo wrote: I work with Windows and use R version 2.4.1. I am JUST starting to learn this program... I get this warning message 'NAs introduced by coercion' while trying to build a distance matrix (to be analyzed with NMDS later) from a 336 x 100 data matrix. The original matrix has lots of zeros and no missing values, but I don't think this should matter. I searched this forum and people have suggested that the warning should be ignored but when I try to print the distance matrix I only get the row numbers (the matrix seems to be 'empty') and I'm not being able to judge whether the matrix worked or not. To get the distance matrix I wrote: dist.PxMx - dist (PeaksMatrix, method='euclidean', diag=FALSE, upper=FALSE) I tried including the p argument (included in the help for dist()) and leaving it out, but that didn't seem to change anything. I think that's required for one distance measure though, not for euclidean dist. Should I really ignore this warning? If so, why am I not being able to see the distance matrix? That error message is often generated by things like as.numeric(X) [1] NA Warning message: NAs introduced by coercion so the immediate suspicion is that your data are not what you think they are. The output of str(PeaksMatrix) should be illuminating. -- 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] ROracle issues
You may have a confusion between 64 bit and 32 bit versions. (I'm not certain; I haven't studied your transcript in complete detail, but rather just noticed a couple of things.) There are a lot of references in your transcript to 64 bit libraries, e.g. gcc -I/usr/lib64 ... but I think that $ORACLE_HOME/lib is a reference to Oracle's 32 bit libraries. ROracle offers some INSTALL flags to specify 32 bit or 64 bit, so you might try figuring out which one you want, and specifying it. There is also the R-sig-db mailing list, which might be a better place to ask. -Don At 11:15 AM -0600 4/30/07, Matt Anthony wrote: Hi all - I am trying to install ROracle for linux machines ... I have read the INSTALL documentation and followed the directions for setting the paths as follows: export PATH=/app/oracle/product/10.2.0/db_1/bin:$PATH [EMAIL PROTECTED] ~]$ export LD_LIBRARY_PATH=/usr/lib:/usr/lib64:/lib:/app/oracle/product/10.1.0/db_1 /lib32:/app/oracle/product/10.1.0/db_1/lib:/app/oracle/product/10.1.0/db _1/lib64 As directed by the INSTALL document. Executing the install by running sudo R CMD INSTALL ROracle_0.5-8.tar.gz gives me the following: * Installing *source* package 'ROracle' ... checking for gcc... gcc checking for C compiler default output... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ANSI C... none needed checking how to run the C preprocessor... gcc -E configure: creating ./config.status config.status: creating src/Makevars config.status: creating src/Makefile ** libs ** arch - R CMD COMPILE RS-DBI.c make[1]: Entering directory `/tmp/R.INSTALL.e14574/ROracle/src' gcc -I/usr/lib64/R/include -I/usr/lib64/R/include -DRS_ORA_SQLGLS_WORKAROUND -I/usr/local/include-fpic -O2 -g -std=gnu99 -c RS-DBI.c -o RS-DBI.o make[1]: Leaving directory `/tmp/R.INSTALL.e14574/ROracle/src' proc CODE=ANSI_C MODE=ORACLE INCLUDE=/usr/lib64/R/include \ PARSE=NONE LINES=false PREFETCH=1 RS-Oracle.pc proc: error while loading shared libraries: libclntsh.so.10.1: cannot open shared object file: No such file or directory make: *** [RS-Oracle.c] Error 127 ERROR: compilation failed for package 'ROracle' ** Removing '/usr/lib64/R/library/ROracle' Now, to check for the libclntsh.so.10.1 file, did the following: CD $ORACLE_HOME/lib Ls In which the file libclntsh.so.10.1 actually does exist, even though the error message seems to imply it doesn't. I have searched the archives but found no record of a solution to this problem, and the documentation provided with the package does not address this issue as far as I can tell. Can someone please explain/point me in the right direction? Thank you Matt Matt Anthony | Senior Statistician| 303.327.1761 | [EMAIL PROTECTED] 10155 Westmoor Drive | Westminster, CO 80021 | FAX 303.327.1650 [[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 and provide commented, minimal, self-contained, reproducible code. -- -- 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 and provide commented, minimal, self-contained, reproducible code.
[R] RE : add arrows to barchart with groups
Hello and thank you for this first response
I'll rephrase what I mean by more general: in a case where I have only 2
levels in groups, but in which I have more than 2 levels for variable s, and in
a case where the second bar is not necessarily higher than the first.
Thanks in advance.
Best regards,
David Gouache
Arvalis - Institut du Végétal
-Message d'origine-
De : Deepayan Sarkar [mailto:[EMAIL PROTECTED]
Envoyé : vendredi 27 avril 2007 19:52
À : GOUACHE David
Cc : r-help@stat.math.ethz.ch
Objet : Re: [R] add arrows to barchart with groups
On 4/27/07, GOUACHE David [EMAIL PROTECTED] wrote:
Hello Rhelpers,
I am trying to represent the following data (hereafter named donnees) in a
barchart with a grouping variable :
sitetraitement dates res
1 NT 17/10/2005 normal 76.2
1 T 17/10/2005 normal 103.2
1 NT 23/11/2005 tardif 81.6
1 T 23/11/2005 tardif 98
2 NT 15/10/2005 normal 72.71
2 T 15/10/2005 normal 94.47
2 NT 15/11/2005 tardif 79.65
2 T 15/11/2005 tardif 94.7
barchart(res~s|site,groups=traitement,data=donnees)
What I'd like to do is for each site represent with an arrow the difference
in value of variable res between normal and tardif values of variable s.
I've found one way of doing it:
trellis.focus(panel,1,1)
xx-trellis.panelArgs()$x
yy-trellis.panelArgs()$y
panel.arrows(as.numeric(xx)[c(1,3)]-0.1,yy[c(1,3)],as.numeric(xx)[c(1,3)]-0.1,yy[c(2,4)],lwd=2,code=3)
panel.text(as.numeric(xx)[c(1,3)]-0.35,c(87,87),paste(yy[c(2,4)]-yy[c(1,3)],\nq/ha),font=2)
trellis.focus(panel,2,1)
xx-trellis.panelArgs()$x
yy-trellis.panelArgs()$y
panel.arrows(as.numeric(xx)[c(1,3)]-0.1,yy[c(1,3)],as.numeric(xx)[c(1,3)]-0.1,yy[c(2,4)],lwd=2,code=3)
panel.text(as.numeric(xx)[c(1,3)]-0.35,c(87,87),paste(yy[c(2,4)]-yy[c(1,3)],\nq/ha),font=2)
trellis.unfocus()
But I would prefer doing this within a custom panel function so I can apply
it more generally, and I haven't been able to figure out how...
Could anyone give me a hand?
The obvious analog (just copy/pasting your code) is:
my.panel - function(x, y, ...)
{
panel.barchart(x, y, ...)
xx - x
yy - y
panel.arrows(as.numeric(xx)[c(1,3)]-0.1, yy[c(1,3)],
as.numeric(xx)[c(1,3)]-0.1, yy[c(2,4)],
lwd = 2, code = 3)
panel.text(as.numeric(xx)[c(1,3)] - 0.35, c(87,87),
paste(yy[c(2,4)] - yy[c(1,3)], \nq/ha),
font=2)
}
and this seems to work:
barchart(res~s|site,groups=traitement,data=donnees,
panel = my.panel)
barchart(res~s|site,groups=traitement,data=donnees,
panel = my.panel,
origin = 0)
I'm not sure what else you are looking for and what you mean by more
general. For example, it's not clear what you want to happen If you
have more than 2 levels in 'groups', or if the second bar is not
always higher than the first.
-Deepayan
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Re: [R] NAs introduced by coercion in dist()
It was suggested that the 'NAs introduced by coercion' message might be warning me that my data are not what they should be. I checked this using str(PeaksMatrix), as suggested, and the data seem to be what I thought they were: 'data.frame': 335 obs. of 127 variables: $ Code : Factor w/ 335 levels A1MR,A1MU,..: 1 2 3 4 5 6 7 8 9 10 ... $ P3.70 : num 0 0 0 0 0 0 0 0 0 0 ... $ P3.97 : num 0 0 0 0 0 0 0 0 0 0 ... $ P4.29 : num 0 0 0 0 0 0 0 0 0 0 ... $ P4.90 : num 0 0 0 0 0 0 0 0 0 0 ... $ P6.30 : num 0 0 0 0 0 0 0 0 0 0 ... $ P6.45 : num 7.73 0 0 0 0 0 4.03 0 0 0 ... $ P6.55 : num 0 0 0 0 0 0 0 0 0 0 ... ... I do have 335 observations, 127 variables that are named P3.70, 3.97, P4.29, etc.. This was a relief, but I still don't know whether the distance matrix is what it should be. I tried 'str(dist.PxMx)', which is the name of my distance matrix, but I get something that has not much meaning to me, an unexperienced R user: Class 'dist' atomic [1:55945] 329.6 194.9 130.1 70.7 116.9 ... ..- attr(*, Size)= int 335 ..- attr(*, Labels)= chr [1:335] 1 2 3 4 ... ..- attr(*, Diag)= logi FALSE ..- attr(*, Upper)= logi FALSE ..- attr(*, method)= chr euclidean ..- attr(*, call)= language dist(x = PeaksMatrix, method = euclidean, diag = FALSE, upper = FALSE, p = 2) Any more suggestions, please? Silvia Lomascolo wrote: I work with Windows and use R version 2.4.1. I am JUST starting to learn this program... I get this warning message 'NAs introduced by coercion' while trying to build a distance matrix (to be analyzed with NMDS later) from a 336 x 100 data matrix. The original matrix has lots of zeros and no missing values, but I don't think this should matter. I searched this forum and people have suggested that the warning should be ignored but when I try to print the distance matrix I only get the row numbers (the matrix seems to be 'empty') and I'm not being able to judge whether the matrix worked or not. To get the distance matrix I wrote: dist.PxMx - dist (PeaksMatrix, method='euclidean', diag=FALSE, upper=FALSE) I tried including the p argument (included in the help for dist()) and leaving it out, but that didn't seem to change anything. I think that's required for one distance measure though, not for euclidean dist. Should I really ignore this warning? If so, why am I not being able to see the distance matrix? -- View this message in context: http://www.nabble.com/NAs-introduced-by-coercion-in-dist%28%29-tf3680727.html#a10286882 Sent from the R help mailing list archive at Nabble.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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Log-likelihood function
At 07:30 AM 5/2/2007, Doxastic wrote: Thanks. I used this and it gave me the same result as the logLik function. The reason I ask is the SAS output gives me a loglik = 1089. R gives me -298.09583. Both for my reduced model. For the saturated (or complex) model, SAS gives me an loglik = 1143. R gives me -298.1993. The problem is these give two very different pictures about whether I can drop the interaction. However, I think the residual deviance in the R output is equal to G^2. So, I can just take the difference between those two. If I do this, I get a difference with an interpretation similar to that of what comes from SAS. So I think I'll just go with that. But who knows if I'm right (not me)? Some comments: 1. Use summary() on your glm() object to get a fuller display of post-fit statistics, including the starting (null) and residual deviances. 2. The deviance is - 2 L, where L = ln(likelihood). 3. To test two nested models for the difference in covariates, subtract the two residual deviances and two d.f. and perform a chi-square test. This can be done nicely by anova() on the two glm() objects. 4. Check the coefficients in your SAS and R models and make sure you are performing the same fit in both cases. Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: [EMAIL PROTECTED] Least Cost Formulations, Ltd.URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239Fax: 757-467-2947 Vere scire est per causas scire __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] stringification magic in subset?
dear R wizards: I am trying to replace subset() with my own version that first checks that each name in the select statement has a corresponding name in the data set. preferably, it would have the same syntax and semantics as subset() otherwise. alas, subset works in interesting ways: subset(d, select=col1) works just like subset(d, select=col1) somehow, subset manages to stringify its argument in such cases. Is it possible to duplicate the subset method one-for-one? a minor question---where do I find the source definition such as that of subset(), which is defined in subset function (x, ...) UseMethod(subset) environment: namespace:base any help, as usual, appreciated. regards, /ivo __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] upgrade to 2.5
Hi I am using R version 2.4.1. How can I upgrade to version 2.5 without having to install all the packages again? Thanks Jason Dr. Iasonas Lamprianou Department of Education The University of Manchester Oxford Road, Manchester M13 9PL, UK Tel. 0044 161 275 3485 [EMAIL PROTECTED] ___ now. [[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 and provide commented, minimal, self-contained, reproducible code.
[R] A Question about knnFinder
Hello, Thanks to the author for writing knnFinder. I seem to have problem with a data set found here (http://www.stat.purdue.edu/~sguha/random/ mydata.csv) This is the R-Code ma-read.csv(~/mydata.csv) ma-as.matrix(ma) rownames(ma)-NULL colnames(ma)-NULL #ma is a 390x2 matrix library(knnFinder) ne-nn(ma,p=5) However, ne$nn.idx is the zero matrix and ne$nn.dist is the matrix whose entries are all 1000. Could someone point out where I'm going wrong? Thanks Saptarshi Saptarshi Guha | [EMAIL PROTECTED] | http://www.stat.purdue.edu/~sguha PGP.sig Description: This is a digitally signed message part __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] Query about RODBC to access MySQL from Windows
Hi I am trying to use RODBC in R installed on Windows to access MySQL database (on a linux box). I set up a DSN and specified this DSN in R as follows library(RODBC); channel - odbcConnect(mysqldsn); RODB Connection 5 Details: case=nochange PORT=3306 Although this seems to connect properly, running any command yields NO results. i.e. sqlQuery(channel, show tables) yields 0 rows when there are close to 500 tables in the database. Ditto with any other query. It does not cause an error, but it returns 0 rows. The USER DSN mysqldsn is set up as follows :- host : zion.xxx.xxx.xxx default database : default_db port : 3306 username : uname password : pwd Running use default_db; show tables; command from the command prompt on the db server returns 500 rows. I find this problem while running any query. Running select * from tname limit 100 returns 0 rows whereas tname has around a million records. In the past, I have used MySQL clients for Windows to access the database without encountering any such problem I even tried setting up the mysqldsn DSN as a system DSN instead of a user DSN. I would like to know a) whether this is a permissions issue at some level b) whether there is any solution for this problem in R Thanks Lalitha __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] stringification magic in subset?
ivo welch said the following on 5/2/2007 8:13 AM: dear R wizards: I am trying to replace subset() with my own version that first checks that each name in the select statement has a corresponding name in the data set. preferably, it would have the same syntax and semantics as subset() otherwise. alas, subset works in interesting ways: subset(d, select=col1) works just like subset(d, select=col1) somehow, subset manages to stringify its argument in such cases. Is it possible to duplicate the subset method one-for-one? a minor question---where do I find the source definition such as that of subset(), which is defined in subset function (x, ...) UseMethod(subset) environment: namespace:base any help, as usual, appreciated. regards, /ivo __ 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 and provide commented, minimal, self-contained, reproducible code. Try: methods(subset) which will point you to subset.data.frame. The latter code will answer your questions. HTH, --sundar __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] NAs introduced by coercion in dist()
Silvia Lomascolo wrote: It was suggested that the 'NAs introduced by coercion' message might be warning me that my data are not what they should be. I checked this using str(PeaksMatrix), as suggested, and the data seem to be what I thought they were: 'data.frame': 335 obs. of 127 variables: $ Code : Factor w/ 335 levels A1MR,A1MU,..: 1 2 3 4 5 6 7 8 9 10 ... $ P3.70 : num 0 0 0 0 0 0 0 0 0 0 ... $ P3.97 : num 0 0 0 0 0 0 0 0 0 0 ... $ P4.29 : num 0 0 0 0 0 0 0 0 0 0 ... $ P4.90 : num 0 0 0 0 0 0 0 0 0 0 ... $ P6.30 : num 0 0 0 0 0 0 0 0 0 0 ... $ P6.45 : num 7.73 0 0 0 0 0 4.03 0 0 0 ... $ P6.55 : num 0 0 0 0 0 0 0 0 0 0 ... ... I do have 335 observations, 127 variables that are named P3.70, 3.97, P4.29, etc.. This was a relief, but I still don't know whether the distance matrix is what it should be. I tried 'str(dist.PxMx)', which is the name of my distance matrix, but I get something that has not much meaning to me, an unexperienced R user: Class 'dist' atomic [1:55945] 329.6 194.9 130.1 70.7 116.9 ... ..- attr(*, Size)= int 335 ..- attr(*, Labels)= chr [1:335] 1 2 3 4 ... ..- attr(*, Diag)= logi FALSE ..- attr(*, Upper)= logi FALSE ..- attr(*, method)= chr euclidean ..- attr(*, call)= language dist(x = PeaksMatrix, method = euclidean, diag = FALSE, upper = FALSE, p = 2) Any more suggestions, please? Actually, you seem to have 126 variables plus a factor called Code, which has non-numeric levels. I think you probably want to lose that one before calculating distances. Silvia Lomascolo wrote: I work with Windows and use R version 2.4.1. I am JUST starting to learn this program... I get this warning message 'NAs introduced by coercion' while trying to build a distance matrix (to be analyzed with NMDS later) from a 336 x 100 data matrix. The original matrix has lots of zeros and no missing values, but I don't think this should matter. I searched this forum and people have suggested that the warning should be ignored but when I try to print the distance matrix I only get the row numbers (the matrix seems to be 'empty') and I'm not being able to judge whether the matrix worked or not. To get the distance matrix I wrote: dist.PxMx - dist (PeaksMatrix, method='euclidean', diag=FALSE, upper=FALSE) I tried including the p argument (included in the help for dist()) and leaving it out, but that didn't seem to change anything. I think that's required for one distance measure though, not for euclidean dist. Should I really ignore this warning? If so, why am I not being able to see the distance matrix? -- 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 and provide commented, minimal, self-contained, reproducible code.
[R] reference in article
Hi all R positive, does anyone know how to refer R in article? thanks tomas __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Read every second line from ASCII file
Dear all,
I just realized that I forgotten to write some kind of final email for
this thread and to thank you for your help.
It seems that the recommeneded procedure in such circumstances has three
steps:
1) readLines()
2) select the desired lines
3) strsplit()
Thanks Ferdinand, Jim, and Paul!
Roland
Roland Rau wrote:
Dear all,
I have an ASCII file where records are separated by a blank. I would
like to read those data; however, only the data in rows 1, 3, 5, 7, ...
are important; the other lines (2,4,6,8,) contain no useful
information for me.
So far I used awk/gawk to do it:
gawk '{if ((FNR % 2) != 0) {print $0}}' infile.txt outfile.txt
What is the recommended way to accomplish this in R?
Simply reading the whole file, and deleting all the even-numbered lines
is not straightforward since these lines have different length (whereas
lines 1,3,5,7,... have the same length).
I 'RSiteSearched' for read every second line from a file but this
search did not yield the desired result.
Also trying out the arguments nrows and skip from read.table() did not
help.
Maybe someone knows an easy way to do it from within R? -- of course not
using system(gawk ) :-)
If not, it does not matter too much since I get the job done easily with
awk.
Thanks,
Roland
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] choose.dir
Hi all, I have written a R-script under Windows using choose.dir. Now, I have seen that this function is missing at MacOS. Does anybody know an alternative? Antje __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sorting in barplot
Thank you for your answer, But I think it doesn't work, because the function bargraph.CI (I use it because barplot doesn't have a standard error option) use two internally function(one for mean and the other for standard error) which use something like tapply (which sort the factors by alphabet). If I decompose the two function and than reorder one time with mean and one other time with standard error, so it doesn't match on the graph. I try also so: bargraph.CI(x.factor = REGION, response = MOY_FST, data = Fst2,ylab = Average Fst by pair of populations,reorder(Fst2$REGION,Fst2$MOY_FST,mean)) with tihs Fst2 table: REGION MOY_FST 1 AR 0.039 2 AR 0.040 3 AR 0.041 4 AR 0.041 5 AR 0.044 6 AR 0.051 7 AR 0.055 8 AR 0.058 9 AR 0.069 10 AR 0.076 11 HT-S 0.080 12 HT-S 0.084 13 HT-S 0.090 14 HT-S 0.091 15 HT-S 0.094 16 HT-S 0.096 17 GE 0.079 18 GE 0.086 19 GE 0.095 20 GE 0.104 21 GE 0.107 22 GE 0.118 23 GE 0.119 24 GE 0.137 25 GE 0.139 26 GE 0.152 27 GE 0.178 But it doesn't work well. RM, best regards. Marc Schwartz [EMAIL PROTECTED] mcast.net A [EMAIL PROTECTED] 01.05.2007 19:49 cc r-help@stat.math.ethz.ch Objet Veuillez répondre Re: [R] sorting in barplot à [EMAIL PROTECTED] cast.net On Tue, 2007-05-01 at 19:33 +0200, [EMAIL PROTECTED] wrote: Hello, I'm trying to sort my bargraph.CI plot (function like barplot in the SCIPLOT package) by the mean of the response variable. Does somebody have a trick for it. Thank you. Romain Mayor, PHD student. If it is built on top of barplot(), then by default, the factor levels of your response variable will determine the order of the bars in the plot. See ?reorder.factor for more details relative to defining the order based upon the mean of the variable. There is an example there of using the median. 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] reference in article
citation() -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Tomas Mikoviny Sent: Wednesday, May 02, 2007 11:44 AM To: r-help@stat.math.ethz.ch Subject: [R] reference in article Hi all R positive, does anyone know how to refer R in article? thanks tomas __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] reference in article
try citation() Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: Tomas Mikoviny [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Wednesday, May 02, 2007 5:44 PM Subject: [R] reference in article Hi all R positive, does anyone know how to refer R in article? thanks tomas __ 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 and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] Query about finding correlations
Hi I have a dataframe which has 3 columns of numeric data A,B,C each of which has been obtained independent of the other. We are trying to find out, which of A or B cause C i.e. We are hypothesising that C is the effect and either A or B, not both is the cause. i.e. A causes C and this cause-effect relationship explains B. The data for A contains more noise than that for B. We are working with around 1000 points. I would greatly appreciate any inputs on the best statistcal approach to tackle this problem. I am thinking that we can find correlation coefficients between A and C, and between B and C, but I am not sure this answers the question. Also we do not know whether the correlation between them is linear or non linear. Thanks Lalitha __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] stringification magic in subset?
Beautiful. thank you. /ivo Try: methods(subset) which will point you to subset.data.frame. The latter code will answer your questions. HTH, --sundar __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] reference in article
On Wed, 2 May 2007, Tomas Mikoviny wrote: Hi all R positive, does anyone know how to refer R in article? Every time you start R it says (in part) R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. -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 and provide commented, minimal, self-contained, reproducible code.
[R] choose.dir
Hi all, I have written a R-script under Windows using choose.dir. Now, I have seen that this function is missing at MacOS. Does anybody know an alternative? Antje - Kennt man wirklich jeden über 3 Ecken? Die Antworten gibt's bei Yahoo! Clever. [[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 and provide commented, minimal, self-contained, reproducible code.
[R] Reenviar: repeated measures and recurrent events data
Luis Guillermo Díaz Monroy Profesor Asociado Universidad Nacional de Colombia---BeginMessage--- Hello, I need a data set on repeated measures and recurrent events data which had been registered on the same subject. This is very important to my thesis work. Thanks a lot, Luis Guillermo Díaz Monroy Profesor Asociado Universidad Nacional de Colombia __ 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 and provide commented, minimal, self-contained, reproducible code. ---End Message--- __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Read every second line from ASCII file
I typically wouldn't use strsplit but would reread it using read.table
and textConnection as in:
http://www.mail-archive.com/r-help@stat.math.ethz.ch/msg84752.html
On 5/2/07, Roland Rau [EMAIL PROTECTED] wrote:
Dear all,
I just realized that I forgotten to write some kind of final email for
this thread and to thank you for your help.
It seems that the recommeneded procedure in such circumstances has three
steps:
1) readLines()
2) select the desired lines
3) strsplit()
Thanks Ferdinand, Jim, and Paul!
Roland
Roland Rau wrote:
Dear all,
I have an ASCII file where records are separated by a blank. I would
like to read those data; however, only the data in rows 1, 3, 5, 7, ...
are important; the other lines (2,4,6,8,) contain no useful
information for me.
So far I used awk/gawk to do it:
gawk '{if ((FNR % 2) != 0) {print $0}}' infile.txt outfile.txt
What is the recommended way to accomplish this in R?
Simply reading the whole file, and deleting all the even-numbered lines
is not straightforward since these lines have different length (whereas
lines 1,3,5,7,... have the same length).
I 'RSiteSearched' for read every second line from a file but this
search did not yield the desired result.
Also trying out the arguments nrows and skip from read.table() did not
help.
Maybe someone knows an easy way to do it from within R? -- of course not
using system(gawk ) :-)
If not, it does not matter too much since I get the job done easily with
awk.
Thanks,
Roland
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] choose.dir
I don't have a Mac but perhaps this works: library(tcltk) tclvalue(tkchooseDirectory()) On 5/2/07, Antje [EMAIL PROTECTED] wrote: Hi all, I have written a R-script under Windows using choose.dir. Now, I have seen that this function is missing at MacOS. Does anybody know an alternative? Antje __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sorting in barplot
Romain, Try this before calling bargraph.CI(): Fst2$REGION - reorder(Fst2$REGION, Fst2$MOY_FST, mean) This will reorder the factor levels for REGION, based upon the mean value of MOY_FST for each region. Thus, your REGIONS should then be sorted by increasing mean values in the plot. For example, note the factor levels for REGION at first: str(Fst2) 'data.frame': 27 obs. of 2 variables: $ REGION : Factor w/ 3 levels AR,GE,HT-S: 1 1 1 1 1 1 1 1 1 1 ... $ MOY_FST: num 0.039 0.04 0.041 0.041 0.044 0.051 0.055 0.058 0.069 0.076 ... Now, note them after using reorder.factor(): Fst2$REGION - reorder(Fst2$REGION, Fst2$MOY_FST, mean) str(Fst2) 'data.frame': 27 obs. of 2 variables: $ REGION : Factor w/ 3 levels AR,HT-S,GE: 1 1 1 1 1 1 1 1 1 1 ... $ MOY_FST: num 0.039 0.04 0.041 0.041 0.044 0.051 0.055 0.058 0.069 0.076 ... This is consistent with the means for each REGION: tapply(Fst2$MOY_FST, Fst2$REGION, mean) AR HT-S GE 0.0514 0.08916667 0.11945455 Unless the author of the CRAN package that you are using has internally altered the factor levels for the variable, you should be OK. Another alternative, if needed, would be to use the barplot2() function in the gplots package, or simply use barplot() and then add the CI's using arrows() or segments(). There is an example of doing this in the help for both functions. HTH, Marc On Wed, 2007-05-02 at 17:36 +0200, [EMAIL PROTECTED] wrote: Thank you for your answer, But I think it doesn't work, because the function bargraph.CI (I use it because barplot doesn't have a standard error option) use two internally function(one for mean and the other for standard error) which use something like tapply (which sort the factors by alphabet). If I decompose the two function and than reorder one time with mean and one other time with standard error, so it doesn't match on the graph. I try also so: bargraph.CI(x.factor = REGION, response = MOY_FST, data = Fst2,ylab = Average Fst by pair of populations,reorder(Fst2$REGION,Fst2$MOY_FST,mean)) with tihs Fst2 table: REGION MOY_FST 1 AR 0.039 2 AR 0.040 3 AR 0.041 4 AR 0.041 5 AR 0.044 6 AR 0.051 7 AR 0.055 8 AR 0.058 9 AR 0.069 10 AR 0.076 11 HT-S 0.080 12 HT-S 0.084 13 HT-S 0.090 14 HT-S 0.091 15 HT-S 0.094 16 HT-S 0.096 17 GE 0.079 18 GE 0.086 19 GE 0.095 20 GE 0.104 21 GE 0.107 22 GE 0.118 23 GE 0.119 24 GE 0.137 25 GE 0.139 26 GE 0.152 27 GE 0.178 But it doesn't work well. RM, best regards. Marc Schwartz [EMAIL PROTECTED] mcast.net A [EMAIL PROTECTED] 01.05.2007 19:49 cc r-help@stat.math.ethz.ch Objet Veuillez répondre Re: [R] sorting in barplot à [EMAIL PROTECTED] cast.net On Tue, 2007-05-01 at 19:33 +0200, [EMAIL PROTECTED] wrote: Hello, I'm trying to sort my bargraph.CI plot (function like barplot in the SCIPLOT package) by the mean of the response variable. Does somebody have a trick for it. Thank you. Romain Mayor, PHD student. If it is built on top of barplot(), then by default, the factor levels of your response variable will determine the order of the bars in the plot. See ?reorder.factor for more details relative to defining the order based upon the mean of the variable. There is an example there of using the median. 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Query about finding correlations
Hi This is not a homework assignment :) Me and my manager are trying to understand the problem better. In the meanwhile, we thought we would post the problem on this forum to seek some input from statisticians who possibly do this kind of analyses everyday and hence are possibly more proficient with R and/or any recommended methodologies. Lalitha On 5/2/07, Stefan Grosse [EMAIL PROTECTED] wrote: How about making your homeworks yourselfes? lalitha viswanath wrote: Hi I have a dataframe which has 3 columns of numeric data A,B,C each of which has been obtained independent of the other. We are trying to find out, which of A or B cause C i.e. We are hypothesising that C is the effect and either A or B, not both is the cause. i.e. A causes C and this cause-effect relationship explains B. The data for A contains more noise than that for B. We are working with around 1000 points. I would greatly appreciate any inputs on the best statistcal approach to tackle this problem. I am thinking that we can find correlation coefficients between A and C, and between B and C, but I am not sure this answers the question. Also we do not know whether the correlation between them is linear or non linear. Thanks Lalitha __ 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 and provide commented, minimal, self-contained, reproducible code. [[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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Query about finding correlations
I am sorry if I misinterpreted this but this questions looks/looked very like a very basic one about statistical inference... (have a look at the Manuals/contributed documentation e.g. Faraway) Lalitha Viswanath wrote: Hi This is not a homework assignment :) Me and my manager are trying to understand the problem better. In the meanwhile, we thought we would post the problem on this forum to seek some input from statisticians who possibly do this kind of analyses everyday and hence are possibly more proficient with R and/or any recommended methodologies. Lalitha On 5/2/07, *Stefan Grosse* [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] wrote: How about making your homeworks yourselfes? lalitha viswanath wrote: Hi I have a dataframe which has 3 columns of numeric data A,B,C each of which has been obtained independent of the other. We are trying to find out, which of A or B cause C i.e. We are hypothesising that C is the effect and either A or B, not both is the cause. i.e. A causes C and this cause-effect relationship explains B. The data for A contains more noise than that for B. We are working with around 1000 points. I would greatly appreciate any inputs on the best statistcal approach to tackle this problem. I am thinking that we can find correlation coefficients between A and C, and between B and C, but I am not sure this answers the question. Also we do not know whether the correlation between them is linear or non linear. Thanks Lalitha __ R-help@stat.math.ethz.ch mailto: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 http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. No virus found in this incoming message. Checked by AVG Free Edition. Version: 7.5.467 / Virus Database: 269.6.2/784 - Release Date: 01.05.2007 14:57 __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] reference in article
Tomas Mikoviny wrote:
Hi all R positive,
does anyone know how to refer R in article?
thanks
tomas
Like this (well, almost! This is from R 2.4.1 - 2007 on the most recent
version)
citation()
To cite R in publications use:
R Development Core Team (2006). R: A language and environment for
statistical computing. R Foundation for Statistical Computing,
Vienna, Austria. ISBN 3-900051-07-0, URL http://www.R-project.org.
A BibTeX entry for LaTeX users is
@Manual{,
title = {R: A Language and Environment for Statistical Computing},
author = {{R Development Core Team}},
organization = {R Foundation for Statistical Computing},
address = {Vienna, Austria},
year = {2006},
note = {{ISBN} 3-900051-07-0},
url = {http://www.R-project.org},
}
We have invested a lot of time and effort in creating R, please cite it
when using it for data analysis. See also ‘citation(pkgname)’ for
citing R packages.
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and provide commented, minimal, self-contained, reproducible code.
--
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
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[R] Multiple scatterplots
Hi, I have to plot three Ziph distributions for three languages where the x value represents the rank of a given word and the y value represents the relative frequency of this word in the corpus. Is there some way so that I can plot all three distributions on a single scatterplot, preferably with different colours :) I tried to find something in the R manual but there are no such examples :( Thank you! __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Query about finding correlations
How about making your homeworks yourselfes? lalitha viswanath wrote: Hi I have a dataframe which has 3 columns of numeric data A,B,C each of which has been obtained independent of the other. We are trying to find out, which of A or B cause C i.e. We are hypothesising that C is the effect and either A or B, not both is the cause. i.e. A causes C and this cause-effect relationship explains B. The data for A contains more noise than that for B. We are working with around 1000 points. I would greatly appreciate any inputs on the best statistcal approach to tackle this problem. I am thinking that we can find correlation coefficients between A and C, and between B and C, but I am not sure this answers the question. Also we do not know whether the correlation between them is linear or non linear. Thanks Lalitha __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] Multiple scatterplots
Hi, I have to plot three Ziph distributions for three languages where the x value represents the rank of a given word and the y value represents the relative frequency of this word in the corpus. Is there some way so that I can plot all three distributions on a single scatterplot, preferably with different colours :) I tried to find something in the R manual but there are no such examples :( Thank you! __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] upgrade to 2.5
Iasonas Lamprianou said the following on 5/2/2007 8:25 AM: Hi I am using R version 2.4.1. How can I upgrade to version 2.5 without having to install all the packages again? Thanks Jason You may find the following link relevant. http://finzi.psych.upenn.edu/R/Rhelp02a/archive/75359.html HTH, --sundar __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] choose.dir
Thank you. For Windows it works, I'll check whether it also works for Mac :) Gabor Grothendieck schrieb: I don't have a Mac but perhaps this works: library(tcltk) tclvalue(tkchooseDirectory()) On 5/2/07, Antje [EMAIL PROTECTED] wrote: Hi all, I have written a R-script under Windows using choose.dir. Now, I have seen that this function is missing at MacOS. Does anybody know an alternative? Antje __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Query about finding correlations
Lalitha Viswanath wrote: We are trying to find out, which of A or B cause C i.e. We are hypothesising that C is the effect and either A or B, not both is the cause. (...) I would greatly appreciate any inputs on the best statistcal approach to tackle this problem. I am thinking that we can find correlation coefficients between A and C, and between B and C, but I am not sure this answers the question. Also we do not know whether the correlation between them is linear or non linear. If the causation (not the correlation) is not linear, then the correlation (which is linear, always) may not be the best indicator. Take, as an extreme case, this: A - (-50:50) + 100 * rnorm(101) B - abs((-50):50) + 10 * rnorm(101) C - A^2 / 50 + rnorm(101) cor(A, C) cor(B, C) A is obviously the cause of C, but B (in some cases) is better correlated to C than A to C. Alberto Monteiro __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] upgrade to 2.5
On 5/2/07, Sundar Dorai-Raj [EMAIL PROTECTED] wrote: Iasonas Lamprianou said the following on 5/2/2007 8:25 AM: Hi I am using R version 2.4.1. How can I upgrade to version 2.5 without having to install all the packages again? Thanks Jason You may find the following link relevant. http://finzi.psych.upenn.edu/R/Rhelp02a/archive/75359.html if you use Windows XP. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Query about finding correlations
Hi Thanks for your input. I stand corrected. The causation is not linear. We wish to find out which is the cause and under what circumstances. i.e. at what points along a scale for C, is A the cause and when does B become the cause, if at all. As a crude analyses, we assumed that the above is not the case (i.e. either A causes C all the time or B causes C all the time) and obtained correlation coefficients using lmFit , however as you rightly mentioned, it is not of much help to us. We are trying to find out whether ages of proteins(A) or their rates of evolution(B) influences parameter C. There is an obvious correlation between A and B which needs to fulfil the hypothesis as well. I am checking out wald.test(eba), HypothesisTesting(fBasics), O8.Tests, O6.LinearModels(limma) amongst others presently. Thanks Lalitha On 5/2/07, Alberto Monteiro [EMAIL PROTECTED] wrote: Lalitha Viswanath wrote: We are trying to find out, which of A or B cause C i.e. We are hypothesising that C is the effect and either A or B, not both is the cause. (...) I would greatly appreciate any inputs on the best statistcal approach to tackle this problem. I am thinking that we can find correlation coefficients between A and C, and between B and C, but I am not sure this answers the question. Also we do not know whether the correlation between them is linear or non linear. If the causation (not the correlation) is not linear, then the correlation (which is linear, always) may not be the best indicator. Take, as an extreme case, this: A - (-50:50) + 100 * rnorm(101) B - abs((-50):50) + 10 * rnorm(101) C - A^2 / 50 + rnorm(101) cor(A, C) cor(B, C) A is obviously the cause of C, but B (in some cases) is better correlated to C than A to C. Alberto Monteiro [[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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Multiple scatterplots
Your title and your posting do not say the same thing. Assuming you want all three distributions on one scatter plot does this help? aa - 1:10 bb - 11:2 cc - bb^2 dd - c(3,4,7,9,11,32,11,14,5,9) plot(aa,cc, col=red) points(aa,bb, col=blue) points(aa,dd, col=green) Also in plotting it is a good idea to look at all the variations etc that you can get with par() Type ?par --- Kostadin Cholakov [EMAIL PROTECTED] wrote: Hi, I have to plot three Ziph distributions for three languages where the x value represents the rank of a given word and the y value represents the relative frequency of this word in the corpus. Is there some way so that I can plot all three distributions on a single scatterplot, preferably with different colours :) I tried to find something in the R manual but there are no such examples :( Thank you! __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Free Webinar: Vendor Neutral Intro to Data Mining for AbsoluteBeginners, May 23, 2007
Lisa: Can we expect to see R used [exclusively, I would hope] during your demonstration? Learning how data mining models work: the inputs, the outputs, and the nature of the predictive mechanism only makes sense for me if I can follow/retrace your steps on my systems. Thank you. Brian J. Koch Data Manager Decision Development Inc -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Lisa Solomon Sent: Tuesday, May 01, 2007 1:46 PM To: r-help Subject: [R] Free Webinar: Vendor Neutral Intro to Data Mining for AbsoluteBeginners, May 23, 2007 ONLINE VENDOR NEUTRAL INTRO TO DATA MINING FOR ABSOLUTE BEGINNERS (no charge) A non-technical data mining introduction for absolute beginners May 23, 2007, 10AM - 11AM PST Future Sessions (June 14, Sept 7) To register for the webinar --- 1. Go to https://salford.webex.com/salford/onstage/g.php?d=928318845t=a 2. Click Enroll. 3. On the registration form, enter your information and then click Submit. Once you have registered, you will receive a confirmation email message with instructions on how to join the event, as well as audio and system requirements. Please read this confirmation email carefully! This one-hour webinar is a perfect place to start if you are new to data mining and have little-to-no background in statistics or machine learning. In one hour, we will discuss: **Data basics: what kind of data is required for data mining and predictive analytics; In what format must the data be; what steps are necessary to prepare data appropriately **What kinds of questions can we answer with data mining **How data mining models work: the inputs, the outputs, and the nature of the predictive mechanism **Evaluation criteria: how predictive models can be assessed and their value measured **Specific background knowledge to prepare you to begin a data mining project. Please do not hesitate to contact me if you have any questions. Sincerely, Lisa Solomon [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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] How to install previous packages after upgrading to R 2.5.0?
Hello, I have just upgraded from R-2.4.1 to R-2.5.0 for Windows. I had installed a large number of add-on packages under 2.4.1. Is there an easy way to install (or load, if that's the easier way) those packages under 2.5.0, without having to install each package by hand? Thanks, Jeff __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] R GUI in Ubuntu Feisty?
Hello Andy, I had the same problem in Feisty and now I fixed it. I updated to Java 6 through the repositories from Feisty. Then I choosed java 6 with sudo update-alternatives --config java. After that I started R with sudo R. If you had already installed JGR, but it doesn't run you have to write: JGR(update=TRUE) This is shown when you run library(JGR) I wish you luck, greetings Ralph Andy Weller wrote: Dear all, After an update from Ubuntu Edgy to Feisty, I seem to have lost package JGR()!? I have updated my sources.list to point to the Feisty repos at http://cran.ch.r-project.org/ and re-installed JGR() via: $ sudo rm -rf /usr/local/lib/R/site-packages/* $ sudo R CMD javareconf $ sudo R install.packages(JGR,dep=TRUE) library(JGR) JGR() However, I get the problem: export: 41: graphics,: bad variable name and JGR() doesn't want to start, R just hangs. I have r-base-dev and sun-java5-jdk installed. Does anyone have any clues as to how to get this working? Thanks in advance, Andy __ 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 and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/R-GUI-in-Ubuntu-Feisty--tf3670710.html#a10295067 Sent from the R help mailing list archive at Nabble.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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] upgrade to 2.5
At 01:41 PM 5/2/2007, you wrote: On 5/2/07, Sundar Dorai-Raj [EMAIL PROTECTED] wrote: Iasonas Lamprianou said the following on 5/2/2007 8:25 AM: Hi I am using R version 2.4.1. How can I upgrade to version 2.5 without having to install all the packages again? Thanks Jason You may find the following link relevant. http://finzi.psych.upenn.edu/R/Rhelp02a/archive/75359.html if you use Windows XP. This link was useful to me, as I am new to R. (Win2000, R-2.5.0) What I have been doing is using a file compare utility (Beyond Compare in my case) to move files in the old library directory to the new one, if the files are missing in the new one. Then I perform an update.packages command. This procedure appears to work without problem. It would seem much preferable to have all packages saved in an installation-independent directory, instead of a library directory under R's installation directory. Then, of course, no update would be necessary. I can't find how this option is settable in R, other than a direct argument to library() or install.package(). How does one shift the R default libraries location to a particular directory? Thanks. Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: [EMAIL PROTECTED] Least Cost Formulations, Ltd.URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239Fax: 757-467-2947 Vere scire est per causas scire __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to install previous packages after upgrading to R 2.5.0?
On 02/05/2007 5:12 PM, Jeffrey Wood wrote: Hello, I have just upgraded from R-2.4.1 to R-2.5.0 for Windows. I had installed a large number of add-on packages under 2.4.1. Is there an easy way to install (or load, if that's the easier way) those packages under 2.5.0, without having to install each package by hand? I'm not sure if this counts as installing by hand, but (assuming you have a network connection) you can use the menu system to see the list of CRAN packages, and install the ones you want from there. If you have some of your own non-CRAN packages, Install package from local .zip file allows multiple files to be selected and installed. The only combination that's not so easy is installing from source .tar.gz files, but a shell command to loop over *.tar.gz could probably be written. Duncan Murdoch __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] upgrade to 2.5
Robert A LaBudde said the following on 5/2/2007 2:39 PM: At 01:41 PM 5/2/2007, you wrote: On 5/2/07, Sundar Dorai-Raj [EMAIL PROTECTED] wrote: Iasonas Lamprianou said the following on 5/2/2007 8:25 AM: Hi I am using R version 2.4.1. How can I upgrade to version 2.5 without having to install all the packages again? Thanks Jason You may find the following link relevant. http://finzi.psych.upenn.edu/R/Rhelp02a/archive/75359.html if you use Windows XP. This link was useful to me, as I am new to R. (Win2000, R-2.5.0) What I have been doing is using a file compare utility (Beyond Compare in my case) to move files in the old library directory to the new one, if the files are missing in the new one. Then I perform an update.packages command. This procedure appears to work without problem. It would seem much preferable to have all packages saved in an installation-independent directory, instead of a library directory under R's installation directory. Then, of course, no update would be necessary. I can't find how this option is settable in R, other than a direct argument to library() or install.package(). How does one shift the R default libraries location to a particular directory? Set an environment variable R_LIBS. See the R Installation and Administration manual (Section 6). HTH, --sundar __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to install previous packages after upgrading to R 2.5.0?
Rename the library subdirectory in your new installation to something like library.new. Copy the library subdirectory in the old distribution to the new installation. Copy the contents of the renamed library.new to the the directory copied from the old distribution. You should then start your new R and update packages from the menu. Check and ensure that you have not enabled automatic loading of packages in /etc/Rprofile.site (eg to run Sciviews or similar) For further information see the Windows FAQ. Regards John Frain On 02/05/07, Jeffrey Wood [EMAIL PROTECTED] wrote: Hello, I have just upgraded from R-2.4.1 to R-2.5.0 for Windows. I had installed a large number of add-on packages under 2.4.1. Is there an easy way to install (or load, if that's the easier way) those packages under 2.5.0, without having to install each package by hand? Thanks, Jeff __ 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 and provide commented, minimal, self-contained, reproducible code. -- John C Frain Trinity College Dublin Dublin 2 Ireland www.tcd.ie/Economics/staff/frainj/home.html mailto:[EMAIL PROTECTED] mailto:[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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Multiple scatterplots
Please note: in R you can specify (some of the) graphics parameters as the appropriate length vectors. So your plot example below can also be done as, for example: plot( rep.int(aa,3),c(cc,bb,dd),col=rep(c(red,blue,green),e=length(aa))) However, this doesn't seem to fit the posted request, where maybe something like a trellis plot of the different distributions is what is wanted?? -- but I may well misunderstand. Bert Gunter Genentech Nonclinical Statistics -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of John Kane Sent: Wednesday, May 02, 2007 11:06 AM To: Kostadin Cholakov; r-help@stat.math.ethz.ch Subject: Re: [R] Multiple scatterplots Your title and your posting do not say the same thing. Assuming you want all three distributions on one scatter plot does this help? aa - 1:10 bb - 11:2 cc - bb^2 dd - c(3,4,7,9,11,32,11,14,5,9) plot(aa,cc, col=red) points(aa,bb, col=blue) points(aa,dd, col=green) Also in plotting it is a good idea to look at all the variations etc that you can get with par() Type ?par --- Kostadin Cholakov [EMAIL PROTECTED] wrote: Hi, I have to plot three Ziph distributions for three languages where the x value represents the rank of a given word and the y value represents the relative frequency of this word in the corpus. Is there some way so that I can plot all three distributions on a single scatterplot, preferably with different colours :) I tried to find something in the R manual but there are no such examples :( Thank you! __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code. __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] Is R's fast fourier transform function different from fft2 in Matlab?
Hi All, I found mvfft in R and fft2 in Matlab give different result and can't figure out why. My example is: In R: matrix(c(1,4,2,20), nrow=2) [,1] [,2] [1,]12 [2,]4 20 mvfft(matrix(c(1,4,2,20), nrow=2)) [,1] [,2] [1,] 5+0i 22+0i [2,] -3+0i -18+0i In Matlab: fft2([1,2;4,20]) ans= 27 -17 -21 15 Does any function in R generate teh same result as what from Matlab? Thanks, Li [[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 and provide commented, minimal, self-contained, reproducible code.
[R] how to reproduce the same sampled units?
Hi all,
Is it possible to generate the same sample number of times in R? In
SAS, using the option seed it is possible to reproduce exactly the
same sample. Is there any such feature in R which I can use?
For further clarity,
for (i in 1:2)
{
samp = sample(1:1000,100,replace = FALSE)
print(samp)
}
For the above simulation, is it possible to generate the same sampled
units twice?
Any suggestion would be gratefully acknowledged.
Thanks,
Santanu
Santanu Pramanik
JPSM,1218J Lefrak Hall
University of Maryland,College Park
Phone no.-301-314-9916
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to reproduce the same sampled units?
On 02-May-07 23:25:18, Santanu Pramanik wrote:
Hi all,
Is it possible to generate the same sample number of times in R?
In SAS, using the option seed it is possible to reproduce exactly
the same sample. Is there any such feature in R which I can use?
For further clarity,
for (i in 1:2)
{
samp = sample(1:1000,100,replace = FALSE)
print(samp)
}
For the above simulation, is it possible to generate the same sampled
units twice?
Any suggestion would be gratefully acknowledged.
Thanks,
Santanu
Have a look at set.seed():
?set.seed
You will of course have to invoke it before you do your first run.
You cannot find out what the seed was afterwards!
For example (your example somewhat reduced):
for (i in 1:2)
{
samp = sample(1:1000,10,replace = FALSE)
print(samp)
}
# [1] 984 587 26 920 304 247 434 392 650 279
# [1] 178 619 458 208 389 629 988 263 598 308
for (i in 1:2)
{
set.seed(512534)
samp = sample(1:1000,10,replace = FALSE)
print(samp)
}
# [1] 271 106 570 621 257 663 399 454 983 805
# [1] 271 106 570 621 257 663 399 454 983 805
Best wishes,
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 03-May-07 Time: 00:56:43
-- XFMail --
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Re: [R] upgrade to 2.5
On 5/2/07, Robert A LaBudde [EMAIL PROTECTED] wrote: At 01:41 PM 5/2/2007, you wrote: On 5/2/07, Sundar Dorai-Raj [EMAIL PROTECTED] wrote: Iasonas Lamprianou said the following on 5/2/2007 8:25 AM: Hi I am using R version 2.4.1. How can I upgrade to version 2.5 without having to install all the packages again? Thanks Jason You may find the following link relevant. http://finzi.psych.upenn.edu/R/Rhelp02a/archive/75359.html if you use Windows XP. This link was useful to me, as I am new to R. (Win2000, R-2.5.0) What I have been doing is using a file compare utility (Beyond Compare in my case) to move files in the old library directory to the new one, if the files are missing in the new one. Then I perform an update.packages command. This procedure appears to work without problem. It would seem much preferable to have all packages saved in an installation-independent directory, instead of a library directory under R's installation directory. Then, of course, no update would be necessary. You can do that but there is the limitation that the old version of R will become unusable with the common library as newer versions of packages, unable to work with the older version of R, get installed. If each version of R has its own library then you won't likely have that problem. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to install previous packages after upgrading to R 2.5.0?
On 5/2/07, Jeffrey Wood [EMAIL PROTECTED] wrote: Hello, I have just upgraded from R-2.4.1 to R-2.5.0 for Windows. I had installed a large number of add-on packages under 2.4.1. Is there an easy way to install (or load, if that's the easier way) those packages under 2.5.0, without having to install each package by hand? This was just discussed in another thread but batchfiles on CRAN (which is a collection of Windows XP batch files, not an R package) contains movedir.bat and copydir.bat that will move or copy your packages from one version of R to another. See the home page: http://code.google.com/p/batchfiles/ which gives links to the README where its explained and to the CRAN download area from where you can download it. movedir.bat is very fast (only a few seconds) but your packages will then only reside in the new version of R. copydir.bat is slower but makes copies rather than moving the packages. Neither will overwrite anything so they are both pretty safe to use. __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is R's fast fourier transform function different from fft2 in Matlab?
Li Li said the following on 5/2/2007 4:06 PM: Hi All, I found mvfft in R and fft2 in Matlab give different result and can't figure out why. My example is: In R: matrix(c(1,4,2,20), nrow=2) [,1] [,2] [1,]12 [2,]4 20 mvfft(matrix(c(1,4,2,20), nrow=2)) [,1] [,2] [1,] 5+0i 22+0i [2,] -3+0i -18+0i In Matlab: fft2([1,2;4,20]) ans= 27 -17 -21 15 Does any function in R generate teh same result as what from Matlab? Thanks, Li I don't know Matlab or any of its functions, but the following produces the same output. z - matrix(c(1, 4, 2, 20), nrow = 2) Re(fft(z)) And from ?fft: When 'z' contains an array, 'fft' computes and returns the multivariate (spatial) transform. HTH, --sundar __ 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 and provide commented, minimal, self-contained, reproducible code.
[R] about using read.table
Hi,Dear R users, I have a file text nommed chif which contains 16 lines and 4 columns in the disc dur. I have a difficulty to read this file in R console I have used the following command chif - read.table(c:/chif.txt, header=T, sep= ) I have obtained from R console:undefined file! Can you please help me.Thank you in advance. [[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 and provide commented, minimal, self-contained, reproducible code.
[R] Building package: What does this message about rcompgen imply?
Hi all
I'm updating a package for submission to CRAN, and I am getting
an error message I never have seen before, and can't find out
what it implies or means (and hence, what I need to do to fix
it).
After running R CMD check, I get this:
snip
* checking foreign function calls ... OK
* checking R code for possible problems ... OK
* checking Rd files ... OK
* checking Rd cross-references ... WARNING
Error in .find.package(package, lib.loc) :
there is no package called 'rcompgen'
Execution halted
* checking for missing documentation entries ... OK
* checking for code/documentation mismatches ... OK
* checking Rd \usage sections ... OK
snip
I am informed: There is no package called 'rcompgen'. True;
I never knew there was one, so I certainly never asked for
it to be loaded. (I can search all the files in the package directory
and the string rcompgen never appears either, and nor does the
string find.package.) So somewhere, this package must be deemed
necessary to install. (I have checked all dependencies too: none of
those packages need rcompgen either.)
I can get this package from CRAN and install it of course, but I
don't think I need it. (Not that I fully understand what it is for
or what it does...). I'm sure it implies there is a problem in
my package, but i don't know what problem that is.
So can anyone help me: Why does R think I need package rcompgen?
What error/mistake in my Rd files would invoke this? Where should
I be looking, and what sort of problem might I be looking for?
Thanks as always.
P.
--
Dr Peter Dunn | dunn at usq.edu.au
Faculty of Sciences, USQ; http://www.sci.usq.edu.au/staff/dunn
Aust. Centre for Sustainable Catchments: www.usq.edu.au/acsc
This email (including any attached files) is confidential an...{{dropped}}
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] how to reproduce the same sampled units?
?set.seed
On Wed, 02 May 2007 18:25:18 -0500, Santanu Pramanik
[EMAIL PROTECTED] wrote:
Hi all,
Is it possible to generate the same sample number of times in R? In
SAS, using the option seed it is possible to reproduce exactly the
same sample. Is there any such feature in R which I can use?
For further clarity,
for (i in 1:2)
{
samp = sample(1:1000,100,replace = FALSE)
print(samp)
}
For the above simulation, is it possible to generate the same sampled
units twice?
Any suggestion would be gratefully acknowledged.
Thanks,
Santanu
Santanu Pramanik
JPSM,1218J Lefrak Hall
University of Maryland,College Park
Phone no.-301-314-9916
[[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
and provide commented, minimal, self-contained, reproducible code.
--
Kenneth Roy Cabrera Torres
Cel 315 504 9339
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Building package: What does this message about rcompgen imply?
We don't have any of the basic details the posting guide asks for! If this is R 2.5.0, rcompgen is a recommended package and ought to be installed. If it is not, then .check_Rd_refs cannot cross-check references against all the standard and recommended packages, and that looks like the cause of the error. If it is an earler version of R, the advice in the posting guide to update applies. CRAN will be checking on 2.5.0 and later. On Thu, 3 May 2007, Peter Dunn wrote: Hi all I'm updating a package for submission to CRAN, and I am getting an error message I never have seen before, and can't find out what it implies or means (and hence, what I need to do to fix it). After running R CMD check, I get this: snip * checking foreign function calls ... OK * checking R code for possible problems ... OK * checking Rd files ... OK * checking Rd cross-references ... WARNING Error in .find.package(package, lib.loc) : there is no package called 'rcompgen' Execution halted * checking for missing documentation entries ... OK * checking for code/documentation mismatches ... OK * checking Rd \usage sections ... OK snip I am informed: There is no package called 'rcompgen'. True; I never knew there was one, so I certainly never asked for it to be loaded. (I can search all the files in the package directory and the string rcompgen never appears either, and nor does the string find.package.) So somewhere, this package must be deemed necessary to install. (I have checked all dependencies too: none of those packages need rcompgen either.) I can get this package from CRAN and install it of course, but I don't think I need it. (Not that I fully understand what it is for or what it does...). I'm sure it implies there is a problem in my package, but i don't know what problem that is. So can anyone help me: Why does R think I need package rcompgen? What error/mistake in my Rd files would invoke this? Where should I be looking, and what sort of problem might I be looking for? Thanks as always. P. -- 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Building package: What does this message about rcompgen imply?
Peter, On 3 May 2007 at 10:43, Peter Dunn wrote: | After running R CMD check, I get this: | | snip | * checking foreign function calls ... OK | * checking R code for possible problems ... OK | * checking Rd files ... OK | * checking Rd cross-references ... WARNING | Error in .find.package(package, lib.loc) : | there is no package called 'rcompgen' | Execution halted Are you by chance on Debian (or Ubuntu)? I couldn't make r-recommended 'Depends:' on r-cran-rcompgen because the later was taking some time to come throught the 'new package' queue that has to be manually inspected, checked, approved, ... by the Debian ftpmaster. It is now in unstable. In any event, with R 2.5.0, and with or without Debian, install.packages(rcompgen) from within R (possibly running as root) should install it for you. Subsequent Debian R packages will have r-recommended depends on r-cran-rcompgen, and r-cran-codetools (which is delayed a bit more as I made a mistake in the very first version I uploaded). Regards, Dirk -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Building package: What does this message about rcompgen imply?
Dirk
On 3 May 2007 at 10:43, Peter Dunn wrote:
| After running R CMD check, I get this:
|
| snip
| * checking foreign function calls ... OK
| * checking R code for possible problems ... OK
| * checking Rd files ... OK
| * checking Rd cross-references ... WARNING
| Error in .find.package(package, lib.loc) :
| there is no package called 'rcompgen'
| Execution halted
Are you by chance on Debian (or Ubuntu)?
Bingo.
The short answer seems to be that I do need rcompgen to be
installed, and my Debian system didn't do that. Easy fixed though.
Thanks.
P.
version
_
platform i486-pc-linux-gnu
arch i486
os linux-gnu
system i486, linux-gnu
status
major 2
minor 5.0
year 2007
month 04
day23
svn rev41293
language R
version.string R version 2.5.0 (2007-04-23)
--
Dr Peter Dunn | dunn at usq.edu.au
Faculty of Sciences, USQ; http://www.sci.usq.edu.au/staff/dunn
Aust. Centre for Sustainable Catchments: www.usq.edu.au/acsc
This email (including any attached files) is confidential an...{{dropped}}
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and provide commented, minimal, self-contained, reproducible code.
[R] Survival statistics--displaying multiple plots
Hello all! I am once again analyzing patient survival data with chronic liver disease. The severity of the liver disease is given by a number which is continuously variable. I have referred to this number as meld--model for end stage liver disease--which is the result of a mathematical calculation on underlying laboratory values. So, for example, I can generate a Kaplan-Meier plot of patients undergoing a TIPS procedure with the following: plot(survfit(Surv(days,status==1),subset(tips,meld10)) where tips is my data set, days is the number of days alive, and meld is the meld score. What I would like to do is display the survival graphs of patients with meld10, 10meld20, and meld20. I am unsure about how to go about this. Any suggestions would be appreciated. Greg __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Building package: What does this message about rcompgen imply?
Peter, On 3 May 2007 at 12:23, Peter Dunn wrote: | Are you by chance on Debian (or Ubuntu)? | | Bingo. | | The short answer seems to be that I do need rcompgen to be | installed, and my Debian system didn't do that. Easy fixed though. Yes, sorry. I realized relatively late that we needed the two new packages, and did not get them into the queue in time for the R 2.5.0 release, despite an upload made during my vacation :) As this tends to take two to three weeks with the number of packages going into Debian, I ended up missing this by a few days. Sorry for the trouble, and glad you got it working now. Dirk -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is R's fast fourier transform function different from fft2 in Matlab?
Thanks for both replies. Then I found the ifft2 from Matlab gives different result from fft( , inverse=T) from R. An example: in R: temp - matrix(c(1,4,2, 20), nrow=2) fft(temp) [,1] [,2] [1,] 27+0i -17+0i [2,] -21+0i 15+0i fft(temp,inverse=T) [,1] [,2] [1,] 27+0i -17+0i [2,] -21+0i 15+0i In Matlab: A = [1,2;4,20]; fft2(A) Ans = 27-17 -21 15 ifft2(A) Ans= 6.7500-4.2500 -5.2500 3.7500 I also tried mvfft with inverse but can't get same result with ifft2. Does any function work? Thanks, Li On 5/2/07, Sundar Dorai-Raj [EMAIL PROTECTED] wrote: I don't know Matlab or any of its functions, but the following produces the same output. z - matrix(c(1, 4, 2, 20), nrow = 2) Re(fft(z)) And from ?fft: When 'z' contains an array, 'fft' computes and returns the multivariate (spatial) transform. HTH, --sundar [[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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Survival statistics--displaying multiple plots
I should clarify. I can generate plots for each category individually but not for all three on the same chart. Greg -Original Message- From: Gregory Pierce [mailto:[EMAIL PROTECTED] Sent: Wednesday, May 02, 2007 10:21 PM To: 'r-help@stat.math.ethz.ch' Subject: Survival statistics--displaying multiple plots Hello all! I am once again analyzing patient survival data with chronic liver disease. The severity of the liver disease is given by a number which is continuously variable. I have referred to this number as meld--model for end stage liver disease--which is the result of a mathematical calculation on underlying laboratory values. So, for example, I can generate a Kaplan-Meier plot of patients undergoing a TIPS procedure with the following: plot(survfit(Surv(days,status==1),subset(tips,meld10)) where tips is my data set, days is the number of days alive, and meld is the meld score. What I would like to do is display the survival graphs of patients with meld10, 10meld20, and meld20. I am unsure about how to go about this. Any suggestions would be appreciated. Greg __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is R's fast fourier transform function different from fft2 in Matlab?
Discrete Fourier transforms can be normalized in different ways. Some apply the whole normalization to the forward transform, some to the reverse transform, some apply the square root to each, and some don't normalize at all (in which case the reverse of the forward transform will need scaling). The latter apparently the case with R, according to your values. Note that the R and the MatLab answers agree to within a scale factor for each row. At 10:53 PM 5/2/2007, Li-Li wrote: Thanks for both replies. Then I found the ifft2 from Matlab gives different result from fft( , inverse=T) from R. An example: in R: temp - matrix(c(1,4,2, 20), nrow=2) fft(temp) [,1] [,2] [1,] 27+0i -17+0i [2,] -21+0i 15+0i fft(temp,inverse=T) [,1] [,2] [1,] 27+0i -17+0i [2,] -21+0i 15+0i In Matlab: A = [1,2;4,20]; fft2(A) Ans = 27-17 -21 15 ifft2(A) Ans= 6.7500-4.2500 -5.2500 3.7500 I also tried mvfft with inverse but can't get same result with ifft2. Does any function work? Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: [EMAIL PROTECTED] Least Cost Formulations, Ltd.URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239Fax: 757-467-2947 Vere scire est per causas scire __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Survival statistics--displaying multiple plots
? layout() ? par() E.g., layout(matrix(c(1,2,3),1,3,byrow=TRUE) #3 plots side-by-side Then use plot() three times to generate each of your graphs. At 11:14 PM 5/2/2007, Greg wrote: I should clarify. I can generate plots for each category individually but not for all three on the same chart. Greg -Original Message- From: Gregory Pierce [mailto:[EMAIL PROTECTED] Sent: Wednesday, May 02, 2007 10:21 PM To: 'r-help@stat.math.ethz.ch' Subject: Survival statistics--displaying multiple plots Hello all! I am once again analyzing patient survival data with chronic liver disease. The severity of the liver disease is given by a number which is continuously variable. I have referred to this number as meld--model for end stage liver disease--which is the result of a mathematical calculation on underlying laboratory values. So, for example, I can generate a Kaplan-Meier plot of patients undergoing a TIPS procedure with the following: plot(survfit(Surv(days,status==1),subset(tips,meld10)) where tips is my data set, days is the number of days alive, and meld is the meld score. What I would like to do is display the survival graphs of patients with meld10, 10meld20, and meld20. I am unsure about how to go about this. Any suggestions would be appreciated. Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: [EMAIL PROTECTED] Least Cost Formulations, Ltd.URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239Fax: 757-467-2947 Vere scire est per causas scire __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is R's fast fourier transform function different from fft2 in Matlab?
Li Li said the following on 5/2/2007 7:53 PM: Thanks for both replies. Then I found the ifft2 from Matlab gives different result from fft( , inverse=T) from R. An example: in R: temp - matrix(c(1,4,2, 20), nrow=2) fft(temp) [,1] [,2] [1,] 27+0i -17+0i [2,] -21+0i 15+0i fft(temp,inverse=T) [,1] [,2] [1,] 27+0i -17+0i [2,] -21+0i 15+0i In Matlab: A = [1,2;4,20]; fft2(A) Ans = 27-17 -21 15 ifft2(A) Ans= 6.7500-4.2500 -5.2500 3.7500 I also tried mvfft with inverse but can't get same result with ifft2. Does any function work? This is easily explained if you read ?fft and the description of the 'inverse' argument in the Value section. Please do read the help pages as the posting guide suggests. Re(fft(temp, inverse = TRUE)/4) --sundar __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Survival statistics--displaying multiple plots
Have a look at ?lines ?points ?plot - the option add, add=TRUE You will have to specify proper limits on both axes, otherwise you may only see parts of the graphs. Some functions in the survival library also allow stratified analyses in which case the plots account for different strata. Petr Gregory Pierce napsal(a): I should clarify. I can generate plots for each category individually but not for all three on the same chart. Greg -Original Message- From: Gregory Pierce [mailto:[EMAIL PROTECTED] Sent: Wednesday, May 02, 2007 10:21 PM To: 'r-help@stat.math.ethz.ch' Subject: Survival statistics--displaying multiple plots Hello all! I am once again analyzing patient survival data with chronic liver disease. The severity of the liver disease is given by a number which is continuously variable. I have referred to this number as meld--model for end stage liver disease--which is the result of a mathematical calculation on underlying laboratory values. So, for example, I can generate a Kaplan-Meier plot of patients undergoing a TIPS procedure with the following: plot(survfit(Surv(days,status==1),subset(tips,meld10)) where tips is my data set, days is the number of days alive, and meld is the meld score. What I would like to do is display the survival graphs of patients with meld10, 10meld20, and meld20. I am unsure about how to go about this. Any suggestions would be appreciated. Greg __ 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 and provide commented, minimal, self-contained, reproducible code. -- Petr Klasterecky Dept. of Probability and Statistics Charles University in Prague Czech Republic __ 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 and provide commented, minimal, self-contained, reproducible code.
