Re: [R] ggplot2 / reshape / Question on manipulating data
On 7/12/07, Pete Kazmier [EMAIL PROTECTED] wrote: I'm an R newbie but recently discovered the ggplot2 and reshape packages which seem incredibly useful and much easier to use for a beginner. Using the data from the IMDB, I'm trying to see how the average movie rating varies by year. Here is what my data looks like: ratings - read.delim(groomed.list, header = TRUE, sep = |, comment.char = ) ratings - subset(ratings, VoteCount 100) head(ratings) Title Histogram VoteCount VoteMean Year 1!Huff (2004) (TV) 16 299 8.4 2004 8 'Allo 'Allo! (1982) 000125 829 8.6 1982 50 .hack//SIGN (2002) 001113 150 7.0 2002 561-800-Missing (2003) 000103 118 5.4 2003 66 Greatest Artists (2000) (mini) 00..16 110 7.8 2000 77 00 Scariest Movie (2004) (mini) 00..000115 256 8.6 2004 Have you tried using the movies dataset included in ggplot? Or is there some data that you want that is not in that dataset. The above data is not aggregated. So after playing around with basic R functionality, I stumbled across the 'aggregate' function and was able to see the information in the manner I desired (average movie rating by year). byYear - aggregate(ratings$VoteMean, list(Year = ratings$Year), mean) plot(byYear) Having just discovered gglot2, I wanted to create the same graph but augment it with a color attribute based on the total number of votes in a year. So first I tried to see if I could reproduce the above: library(ggplot2) qplot(Year, x, byYear) This did not work as expected because the x-axis contained labels for each and every year making it impossible to read whereas the plot created with basic R had nice x-axis labels. How do I get 'qplot' to treat the x-axis in a similar manner to 'plot'? The problem is probably that Year is a factor - and factors are labelled on every level (even if they overlap - which is a bug). There's no terribly easy way to fix this, but the following will work: qplot(as.numeric(as.character(Year)), x, data=byYear) After playing around further, I was able to get 'qplot' to work in a manner similar to 'plot' with regards to the x-axis labels by using 'melt' and 'cast'. The 'qplot' now behaves correctly: mratings - melt(ratings, id = c(Title, Year), measure = c(VoteCount, VoteMean)) byYear2 - cast(mratings, Year ~ variable, mean, subset = variable == VoteMean) qplot(Year, VoteMean, data = byYear2) How do 'byYear' and 'byYear2' differ? I am trying to use 'typeof' but both seem to be lists. However, they are clearly different in some way because 'qplot' graphs them differently. Try using str - it's much more helpful, and you should see the different quickly. Finally, I'd like to use a color attribute to 'qplot' to augment each point with a color based on the total number of votes for the year. Using attributes with 'qplot' seems simple, but I'm having a hard time grooming my data appropriately. I believe this requires aggregation by summing the VoteCount column. Is there a way to cast the data using different aggregation functions for various columns? In my Not easily, unfortunately. However, you could do: cast(mratings, Year ~ variable, c(mean, sum)), subset = variable %in% c(VoteMean, VoteCount)) which will give you a mean and sum for both. case, I want the mean of the VoteMean column, and the sum of the VoteCount column. Then I want to produce a graph showing the average movie rating per year but with each point colored to reflect the total number of votes for that year. Any pointers? Using the built in movies data: mm - melt(movies, id=1:2, m=c(rating, votes)) msum - cast(mm, year ~ variable, c(mean, sum)) qplot(year, rating_mean, data=msum, colour=votes_sum) qplot(year, rating_mean, data=msum, colour=votes_sum, geom=line) Hadley __ 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] ggplot doesnt work in loops?
Dear list members
I am still a newbie so might be asking a stupid question, but I can't get
ggplot to work in a loop (or a while statement for that matter).
# to take a minimal example -
mydata$varc = c(1,2,3)
for (i in 1:1){
jpeg(test3.jpg)
plot(mydata$varc)
#ggplot(mydata, aes(x=mydata$varc)) + geom_bar()
dev.off()
}
this produces an empty jpeg, whereas the content of the loop produces the
jpeg correctly.
a standard plot() does work inside the loop.
Any ideas? This is with R 2.4.0 and ggplot2
thanks in advance
Steve Powell
proMENTE social research
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Re: [R] ggplot doesnt work in loops?
Hi Steve,
You need to explicitly print the ggplot object:
ggplot(mydata, aes(x=mydata$varc)) + geom_bar()
(this is a R-faq for lattice plots, and ggplot works the same way)
In the latest version of ggplot (0.5.4) you can construct the plot
before hand and modify the aesthetics in each instance of the loop:
p - ggplot(mydata) + geom_bar()
mydata$varc = c(1,2,3)
for (i in 1:1){
jpeg(test3.jpg)
p + aes(x = mydata$varc)
dev.off()
}
(not that this will actually work because you're not using i inside
your loop anywhere)
(and to be strictly correct you should probably use list(x =
as.name(names(mydata)[i])) instead of the aes call - but I haven't
written any documentation for this yet)
Hadley
On 7/12/07, Steve Powell [EMAIL PROTECTED] wrote:
Dear list members
I am still a newbie so might be asking a stupid question, but I can't get
ggplot to work in a loop (or a while statement for that matter).
# to take a minimal example -
mydata$varc = c(1,2,3)
for (i in 1:1){
jpeg(test3.jpg)
plot(mydata$varc)
#ggplot(mydata, aes(x=mydata$varc)) + geom_bar()
dev.off()
}
this produces an empty jpeg, whereas the content of the loop produces the
jpeg correctly.
a standard plot() does work inside the loop.
Any ideas? This is with R 2.4.0 and ggplot2
thanks in advance
Steve Powell
proMENTE social research
__
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[R] How to get weekly Covariance with R
Hi: I am trying to migrate from Systat to R but I am facing my first challenge. While I easily can get weekly co-variance for my data, I can't seem to acomplish this with R ( I can't figure out how is done) If interested in looking a sample of my data, please check the data below. In Systat from Week 28 I get a covariance of 1055 fish. Thanks in advance Fish Passage Week number Weekly Covariance0 26 00 27 00 27 0 27 0 27 0 27 0 27 0 27 0 28 10550 28 71 28 132 28 223 28 224 28 218 28 228 29 224 29 488 29 525 29 80 29 417 29 82 29 413 30 373 30 914 30 213 30 651 30 521 30 979 30 177 31 604 31 824 31 1190 31 926 31 1257 31 1071 31 1709 32 1166 32 704 32 1424 32 2586 32 1163 32 647 32 Felipe D. Carrillo Fishery Biologist US Fish Wildlife Service Red Bluff, California 96080 - [[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 get the p-values from an lm function ?
Hi, dear R-users, I am computing a liner regression by rating category using the 'by' function as stated below: tmp - by(projet, rating, function(x) lm(defaults ~ CGDP+CSAVE+SP500, data = x)) I would like to get not only the coefficients but also their p-values. I can't find the command in the help pages to get them. Does anyone have a suggestion ? Thank you, Benoit. [[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] how to get the p-values from an lm function ?
On 7/12/07, Benoit Chemineau [EMAIL PROTECTED] wrote: Hi, dear R-users, I am computing a liner regression by rating category using the 'by' function as stated below: tmp - by(projet, rating, function(x) lm(defaults ~ CGDP+CSAVE+SP500, data = x)) I would like to get not only the coefficients but also their p-values. I can't find the command in the help pages to get them. Does anyone have a suggestion ? Hi Benoit, A general approach to find p-values: m - lm(wt ~ mpg, data=mtcars) First figure out how to display them on screen: m # nope coef(m) # nope summary(m) # got it # Then use str to look at the components str(summary(m)) # And pick out the one want summary(m)$coef coef(summary(m)) # slighty better style, but won't work in general # In general, you may also need to try str(print(summary(m))) # as sometimes the print method calculates the data you're looking for Hadley __ 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] aov() question
On 7/11/07, Leigh E Alexander [EMAIL PROTECTED] wrote: i have a 3 x 4 x 2 repeated measures design. All of the IVs are within subjects.I do also have missing values (unequal N), as I have to remove any incorrect trials for each subject. Hum, you mean there are empty cells in table(sid, tran, block, half)? That is probably the reason for the error message. It is not clear to me that 'aov' can handle empty cells in such a design (maybe 'lmer' can) Christophe Here is the code I entered and the error message: a-aov(log(rt)~(tran*block*half) + Error (sid/ (tran*block*half)), data=mydata2) Warning message: Error() model is singular in: aov(log(rt) ~ (tran * block * half) + Error(sid/(tran * block * I then do summary(a) and am able to get an output, but I am not sure whether or not I can trust that output since I got the error message. Any body have any thoughts/solutions for this? Also, are there any benefits of you aov() vs. use some of the linear model functions or vice versa? Thanks for any help you can offer!! ~Leigh Alexander __ 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. -- Christophe Pallier (http://www.pallier.org) [[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] how to get the p-values from an lm function ?
try the following:
tmp - by(projet, rating, function (x) Thursday, 12.July.2007{
fit - lm(defaults ~ CGDP + CSAVE + SP500, data = x)
summary(fit)$coefficients
})
I hope it helps.
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: Benoit Chemineau [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Thursday, July 12, 2007 10:51 AM
Subject: [R] how to get the p-values from an lm function ?
Hi, dear R-users,
I am computing a liner regression by rating category using the 'by'
function
as stated below:
tmp - by(projet, rating, function(x) lm(defaults ~
CGDP+CSAVE+SP500, data =
x))
I would like to get not only the coefficients but also their
p-values. I
can't find the command in the help pages to get them.
Does anyone have a suggestion ?
Thank you,
Benoit.
[[alternative HTML version deleted]]
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http://www.R-project.org/posting-guide.html
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Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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[R] problems with memory in Mac
Dear friends, I am having some doubts about the amount of memory that is being used by R in my Mac (MacBook Pro, 2Gig). Is there a way to increase the amount of memory used? When I type: mem.limits() the result is: nsize vsize NANA and I can't change it, tough my computing in R isn't using all the memory at it's disposal. Best regards, Carlos -- Carlos GUERRA Gabinete de Sistemas de Informacao Geografica Escola Superior Agraria de Ponte de Lima Mosteiro de Refoios do Lima 4990-706 Ponte de Lima Tlm: +351 91 2407109 Tlf: +351 258 909779 Reclaim your Inbox...!!! http://www.mozilla.org/products/thunderbird/ __ 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] type III ANOVA for a nested linear model
The aliasing problem arises in alias(), which is what Anova uses to detect
aliasing. It is simply the fact that anova more or less blithely ignores the
NA's that makes anova behave apparently more 'sensibly' than Anova.
But like Carsten, I found this difficult to understand. Unordered factors are
supposed to be arbitrary. I can understand why A:C behaves differently from A:D
(where D-factor(rep(1:3,6)). A:C generates a 27-level factor with only 18
levels populated; A:D has only nine levels. But it is not so simple.
I was goig to suggest using AC-factor(A:C) instead of A %in% C as a fix, as
this generates a 9-level factor with all levels populated - just as A:D does.
But
alias(lm(resp~A+AC))
generates an alias, where
alias(lm(resp~A+A:D))
does not.
That seemed to me entirely bizarre. table(A,AC) and table(A,A:D) show identical
balance and nesting. Something, somewhere, is expanding the two differently.
But what?
model.matrix is the 'culprit', I think. The two model matrices differ;
model.matrix(resp~A+A:D) has 9 columns, and model.matrix(resp~A+AC) has eleven.
So now I know what has happened. What I don't understand is why, except that
'that is what model.matrix does with this factor level numbering' (refactoring
either AC or A:D via as.numeric finally generates identical - and aliased -
behaviour)) . I think I am going to invoke the law of unintended consequences
and go find a cold compress.
Steve E
Peter Dalgaard [EMAIL PROTECTED] 11/07/2007 16:03:13
A term C %in% A (or A/C) is not a _specification_ that C is nested in
A, it is a _directive_ to include the terms A and C:A. Now, C:A involves
a term for each combination of A and C, of which many are empty if C is
strictly coarser than A. This may well be what is confusing Anova().
In fact, with this (c(1:3,6:11)) coding of C, A:C is completely
equivalent to C, but if you look at summary(lm()) you will see a lot
of NA coefficients in the A:C case. If you use resp ~ A*B+C, then you
still get a couple of missing coefficients in the C terms because of
collinearity with the A terms. (Notice that this is one case where the
order inside the model formula will matter; C+A*B is not the same.)
***
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Re: [R] Subsetting problem
Hi Massimo, Professor Ripley has given you your answer. It may help you further to know that factor levels aren't automatically dropped when you subset a data set; you have to do it manually. Some time ago I scrounged the following command from Andy Liaw's randomForest package: it removes all empty factor levels in a subsetted data set. I subset a great deal, and find it extremely useful. MyDat[] - lapply(MyDat, function(x) if (is.factor(x)) x[, drop=T] else x) ## Liaw's code for doing this Regards, Mark. On Thu, 12 Jul 2007, Cressoni, Massimo (NIH/NHLBI) [F] wrote: I need to perform the Exact Wilcoxon Mann-Whitney on a subset of my database. Assuming that IPPO is my data frame and IPPOBIS is the subset my variable still have 3 different levels and the function wilcox_test (package coin) does not accept it. I do not know how to overcome this problem. ippo - c(rep(A,10),rep(B,10),rep(C,10)) ippo2 - c(rnorm(10,0,1),rnorm(10,10,10),rnorm(10,10,10)) IPPO - data.frame(ippo,ippo2) IPPOBIS - IPPO[IPPO$ippo == A | IPPO$ippo == B,] wilcox_test(ippo2 ~ ippo,data=IPPOBIS,distribution=exact()) Error in check(itp) : 'object' does not represent a two sample problem levels(IPPOBIS$ippo) [1] A B C Massimo Cressoni __ 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. -- 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/Subsetting-problem-tf4066094.html#a11557188 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] how to get the p-values from an lm function ?
On Thu, 12 Jul 2007, hadley wickham wrote: On 7/12/07, Benoit Chemineau [EMAIL PROTECTED] wrote: Hi, dear R-users, I am computing a liner regression by rating category using the 'by' function as stated below: tmp - by(projet, rating, function(x) lm(defaults ~ CGDP+CSAVE+SP500, data = x)) I would like to get not only the coefficients but also their p-values. I can't find the command in the help pages to get them. Does anyone have a suggestion ? Hi Benoit, A general approach to find p-values: m - lm(wt ~ mpg, data=mtcars) First figure out how to display them on screen: m # nope coef(m) # nope summary(m) # got it # Then use str to look at the components str(summary(m)) # And pick out the one want summary(m)$coef coef(summary(m)) # slighty better style, but won't work in general If x$coef works, coef(x) will almost certainly work at least as well. But note that in most cases it is x$coefficients and so x$coef is liable to partially match erroneously. # In general, you may also need to try str(print(summary(m))) # as sometimes the print method calculates the data you're looking for But a print method should always return its input, so str(summary(m)) str(print(summary(m))) should be the same. Reading the help pages would be a very good idea, as they usually not only tell you the names of the components of the result but also what they mean. -- 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] problems with memory in Mac
2007/7/12, Carlos Guerra [EMAIL PROTECTED]: Dear friends, I am having some doubts about the amount of memory that is being used by R in my Mac (MacBook Pro, 2Gig). Is there a way to increase the amount of memory used? When I type: mem.limits() the result is: nsize vsize NANA and I can't change it, tough my computing in R isn't using all the memory at it's disposal. Best regards, Carlos -- Carlos GUERRA Gabinete de Sistemas de Informacao Geografica Escola Superior Agraria de Ponte de Lima Mosteiro de Refoios do Lima 4990-706 Ponte de Lima Tlm: +351 91 2407109 Tlf: +351 258 909779 Reclaim your Inbox...!!! http://www.mozilla.org/products/thunderbird/ __ 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. read the help ?memory.limit - Rod. __ 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 get the p-values from an lm function ?
On 7/12/07, Prof Brian Ripley [EMAIL PROTECTED] wrote: On Thu, 12 Jul 2007, hadley wickham wrote: On 7/12/07, Benoit Chemineau [EMAIL PROTECTED] wrote: Hi, dear R-users, I am computing a liner regression by rating category using the 'by' function as stated below: tmp - by(projet, rating, function(x) lm(defaults ~ CGDP+CSAVE+SP500, data = x)) I would like to get not only the coefficients but also their p-values. I can't find the command in the help pages to get them. Does anyone have a suggestion ? Hi Benoit, A general approach to find p-values: m - lm(wt ~ mpg, data=mtcars) First figure out how to display them on screen: m # nope coef(m) # nope summary(m) # got it # Then use str to look at the components str(summary(m)) # And pick out the one want summary(m)$coef coef(summary(m)) # slighty better style, but won't work in general If x$coef works, coef(x) will almost certainly work at least as well. But note that in most cases it is x$coefficients and so x$coef is liable to partially match erroneously. I meant in general that x$y, does not correspond to y(x) - I realised after I wrote it that I was unclear. # In general, you may also need to try str(print(summary(m))) # as sometimes the print method calculates the data you're looking for But a print method should always return its input, so str(summary(m)) str(print(summary(m))) Oh yes, I was getting confused with print functions which compute values and print them but do not return them. And that comment has made me realise many of my print methods don't return x. - something to fix. Hadley __ 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] dose-response on a grid
I have the following problem. I have measured a dose response curve (binary response, continuous dose) on a grid of x,y positions. I would like to produce a grey-level plot that shows the LD50 at each (x,y) position. I am thinking that I have to do something like fit-glm(resp ~ x*y + dose, family = binomial) Corrections welcome. But from here I don't know how to get LD50, and certainly not at each x,y, position. Thanks very much for any help. Bill __ 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] RWeka control parameters classifiers interface
Hi, many thanks for the answer. It ist true, that for example m1 - SMO(Species ~ ., data = iris, control = Weka_control( K = weka.classifiers.functions.supportVector.PolyKernel)) m2 - SMO(Species ~ ., data = iris, control = Weka_control( K = weka.classifiers.functions.supportVector.RBFKernel)) deliver different results but m3 - SMO(Species ~ ., data = iris, control = Weka_control( K = weka.classifiers.functions.supportVector.PolyKernel,E=2)) m4 - SMO(Species ~ ., data = iris, control = Weka_control( K = weka.classifiers.functions.supportVector.RBFKernel),G=0.2) m3 does not differ from m1 (from the point of view of the setup, irrespective of the data!) m4 does not differ from m2 (from the point of view of the setup, irrespective of the data!) which can be seen, when looking at the results: m1 # (Linear Kernel, okay ) m2 # (RBF Kernel, okay) m3 # still uses a linear kernel, but should be a x,y^2 kernel m4 # G is ignored, resulting in m2 Thx Bjoern - original Nachricht Betreff: Re: [R] RWeka control parameters classifiers interface Gesendet: Mi 11 Jul 2007 14:42:10 CEST Von: Achim Zeileis[EMAIL PROTECTED] On Wed, 11 Jul 2007 [EMAIL PROTECTED] wrote: The problem is, that the functions result=classifier(formula, data, subset, na.action, control = Weka_control(mycontrol)) do not seem to be manipulated by the mycontrol- arguments Yes, they are...not all parameter changes have always an effect on the specified learner. Perhaps this should be resepected via the handlers- argument , but the documentation in this regard is rather sparse. Handlers are not needed here. Re: sparse docs. In case you have not seen that paper already, there is a technical report on the ideas behind RWeka: http://epub.wu-wien.ac.at/dyn/openURL?id=oai:epub.wu-wien.ac.at:epub-wu-01_b a6 Re: SMO. Compare m1 - SMO(Species ~ ., data = iris) m2 - SMO(Species ~ ., data = iris, control = Weka_control( K = weka.classifiers.functions.supportVector.RBFKernel)) which yield different results so the Weka_control() works. The same happens if you register the mySMO() interface yourself. I'm not sure why the E = ... argument has no influence on the SMO, please check the Weka docs for this particular learner. Best, Z --- original Nachricht Ende __ 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] time-varying recursive filter - vectorized
A question about vectorized operations (avoiding loops, for speed)... I need to run a simple recursive (autoregressive) filter with a time-varying coefficient. It is just a one-step recursive filter, so it would be an exponential decay if the filter was constant. I just want to do this, where 'x' is the data and 'w' is the weight to apply to the previous time step: x - c(1, 1, 0, 2, 0, 0) w - c(NA, 0.1, 0.5, 0.4, 0.3, 0.2) y - x for (i in seq_along(x)[-1]) y[i] - y[i] + w[i] * y[i-1] print(y) [1] 1. 1.1000 0.5500 2.2200 0.6660 0.1332 But, since loops are slow, I would like a vectorized method, like filter(, method=recursive). Any ideas? -- Felix Andrews / 安福立 PhD candidate, The Fenner School of Environment and Society The Australian National University (Building 48A), ACT 0200 Beijing Bag, Locked Bag 40, Kingston ACT 2604 http://www.neurofractal.org/felix/ voice:+86_1051404394 (in China) mobile:+86_13522529265 (in China) mobile:+61_410400963 (in Australia) xmpp:[EMAIL PROTECTED] 3358 543D AAC6 22C2 D336 80D9 360B 72DD 3E4C F5D8 __ 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] matrix of scatterplots
Hi, I would like to use the function pairs() to plot a matrix of scatterplots. For each scatterplot, the data are plotted in circles, can I add some argument to change the circles into dots? Could anyone give me some advice?Many thanks -- View this message in context: http://www.nabble.com/matrix-of-scatterplots-tf4067527.html#a11558049 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] How to activate the R commands in SciViews
Hello everybody, I have a problem similar to that reported by Felipe. I installed R 2.5.0, Rcmdr from CRAN and SciViews-R 0.8-9 (with all the required and optional components). When accessing the R Commander menu from within SciViews, the links cannot be clicked. When pointing at them, the mouse transforms in a hand (as it normally does, similar to Internet hyperlinks). However, the links cannot be activated. I had the impression that in some way SciViews makes use of Internet Explorer in its GUI, and that the problem was somewhere there. But I could not get to the bottom of it. Is there any way to make SciViews correctly use Rcmdr functionality? Thanks in advance, Liviu On 6/20/07, Felipe Carrillo [EMAIL PROTECTED] wrote: Please help, I have SciViews(svGUI) and Rcmdr but when the SciViews Console opens the R commander menu don't work. Any ideas anybody? -- Liviu __ 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] Please Help
Hello, I got this email address from http://tolstoy.newcastle.edu.au/R/e2/help/06/10/2516.html, I got started to use R recently, Can I ask you a question ? this is what I am using: platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status major 2 minor 4.0 year 2006 month 10 day03 svn rev39566 language R version.string R version 2.4.0 (2006-10-03) I wanna to call R in shell( bash ) , write all R commands in the shell script and make it a cron job to execute automatically. do you know how to do this ? Looking forward to hearing from you, thanks a million. Tanya Li -- Regards Xiaohui 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.
Re: [R] matrix of scatterplots
m - matrix( rnorm(300), nc=3 ) pairs(m, pch=20) or pairs(m, pch=.) See help(par) for more details. livia wrote: Hi, I would like to use the function pairs() to plot a matrix of scatterplots. For each scatterplot, the data are plotted in circles, can I add some argument to change the circles into dots? Could anyone give me some advice?Many thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to activate the R commands in SciViews
As explained on the web page from where you downloaded SciViews 0.8-9, this version is not compatible with R 2.5.0. Best, Philippe Grosjean ..∞})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Belgium ( ( ( ( ( .. Liviu Andronic wrote: Hello everybody, I have a problem similar to that reported by Felipe. I installed R 2.5.0, Rcmdr from CRAN and SciViews-R 0.8-9 (with all the required and optional components). When accessing the R Commander menu from within SciViews, the links cannot be clicked. When pointing at them, the mouse transforms in a hand (as it normally does, similar to Internet hyperlinks). However, the links cannot be activated. I had the impression that in some way SciViews makes use of Internet Explorer in its GUI, and that the problem was somewhere there. But I could not get to the bottom of it. Is there any way to make SciViews correctly use Rcmdr functionality? Thanks in advance, Liviu On 6/20/07, Felipe Carrillo [EMAIL PROTECTED] wrote: Please help, I have SciViews(svGUI) and Rcmdr but when the SciViews Console opens the R commander menu don't work. Any ideas anybody? __ 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] lead
Hi,
is there any function in R that shifts elements of a vector to the
opposite direction of what Lag() of the Hmisc package does? (something
like, Lag(x, shift = -1) )
Thanks
Zava
This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}
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Re: [R] elementary statistics with R (rkward?)
--- Donatas G. [EMAIL PROTECTED] wrote: Hi, I am trying to learn some basic statistics stuff but I cannot find any elementary statistics exercises using R language. Using RKward would be even better... I need that in analysing sociological data, obtained through questionnairres - findind corelations between variables, relations between different types of data, etc. Could anyone recommend simple tutorials/exercises, available on www for me to work on? I realize it would be much simple to do this introductory stuff with spss, that everyone around me is using here in Lithuania, but I'd really like to learn to do it with R instead... -- Donatas G. http://www.math.ilstu.edu/dhkim/Rstuff/Rtutor.html is not a bad place to start. John Verzani's notes on Simple R http://www.math.csi.cuny.edu/Statistics/R/simpleR/ may also help but his book is better. Peter Dalgarrd's book Introductory Statistics with R is also very good. __ 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] dose-response on a grid
Hi Bill, have a look at the following artificial example: ## Loading the package 'drc' (on CRAN) library(drc) ## Generating dataset with four dose-response curves finneyx4 - rbind(finney71, finney71, finney71, finney71) ## Generating artificial points (x,y) ## different pairs for each of the 4 curves in the above dataset finneyx4$x - rep(1, 24) finneyx4$y - rep(1:4, c(6, 6, 6, 6)) ## Fitting the two-parameter log-logistic model (logistic regression) m1 - drm(affected/total ~ dose, as.factor(x):as.factor(y), weights = total, data = finneyx4, fct = LL.2(), type = binomial) ## Calculating ED50/LD50 for each location (they are all the same for this dataset) ED(m1, 50) You could try the same approach for your data! 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.
Re: [R] Please Help
This is the R-help mailing list. See help(BATCH). You will need to write the required R commands in a separate script, say script.R and then execute it as R --no-save script.R logfile You may need to augment the code above to include directory paths etc. There are other useful documentations at http://www.r-project.org/ Regards, Adai Tanya Li wrote: Hello, I got this email address from http://tolstoy.newcastle.edu.au/R/e2/help/06/10/2516.html, I got started to use R recently, Can I ask you a question ? this is what I am using: platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status major 2 minor 4.0 year 2006 month 10 day03 svn rev39566 language R version.string R version 2.4.0 (2006-10-03) I wanna to call R in shell( bash ) , write all R commands in the shell script and make it a cron job to execute automatically. do you know how to do this ? Looking forward to hearing from you, thanks a million. Tanya Li __ 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] [Fwd: Re: How to activate the R commands in SciViews]
Well.. plans are there from a long time to rewrite SciViews completely and make it platform independent (to work on Linux/Unix and MacOS X, as well as Windows). I have done some work in this direction when time permitted, but I am pretty busy with other work. During the holidays, I will continue to work in this direction. I will try to package a first running version of SciViews compatible with latest R and Rcmdr for Windows and Linux for next September. At least, I *have* to do so, because I *need* it for one of my teachings which will be done in a different University (where all machines are running with Linux)! So, my advice would be to check the web site again in September/October for some news. Unfortunately, I cannot promise more. Sorry for the long delays. I do my best, and the second programmer (Eric Lecoutre) has left the project a long time ago... so, I am alone on this! Best, Philippe Grosjean ..∞})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Belgium ( ( ( ( ( .. Liviu Andronic wrote: Thank you for the pointer. On the SciViews official site (http://www.sciviews.org/SciViews-R/index.html) I could not find any indications on this incompatibility. At any rate, I would have two questions related to the future of SciViews-R. Are there any developments planned in some near future that would make SciViews compatible with recent versions of R and Rcmdr? At the Unviversity of Social Sciences of Toulouse, for example, the most recent version of R is installed. And for statistics introductory classes, SciViews would be an important advantage. R should be downgraded to which version for the two to be compatible? Secondly, is there any chance that SciViews become available on Linux, again in some not so distant future? That is, would the tcltk2 package be ported to other OS's and would SciViews be subsequently enhanced to build on such systems? Last time I checked the net, no interesting information was available as to such developments. Maybe there is some beta version of tcltk2 that I don't know of.. Regards, Liviu On 7/12/07, Philippe Grosjean [EMAIL PROTECTED] wrote: As explained on the web page from where you downloaded SciViews 0.8-9, this version is not compatible with R 2.5.0. Best, Philippe Grosjean ..∞})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Belgium ( ( ( ( ( .. Liviu Andronic wrote: Hello everybody, I have a problem similar to that reported by Felipe. I installed R 2.5.0, Rcmdr from CRAN and SciViews-R 0.8-9 (with all the required and optional components). When accessing the R Commander menu from within SciViews, the links cannot be clicked. When pointing at them, the mouse transforms in a hand (as it normally does, similar to Internet hyperlinks). However, the links cannot be activated. I had the impression that in some way SciViews makes use of Internet Explorer in its GUI, and that the problem was somewhere there. But I could not get to the bottom of it. Is there any way to make SciViews correctly use Rcmdr functionality? Thanks in advance, Liviu On 6/20/07, Felipe Carrillo [EMAIL PROTECTED] wrote: Please help, I have SciViews(svGUI) and Rcmdr but when the SciViews Console opens the R commander menu don't work. Any ideas anybody? -- ..∞})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Belgium ( ( ( ( ( .. __ 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] matrix of scatterplots
Thank you very much for your help. Adaikalavan Ramasamy wrote: m - matrix( rnorm(300), nc=3 ) pairs(m, pch=20) or pairs(m, pch=.) See help(par) for more details. livia wrote: Hi, I would like to use the function pairs() to plot a matrix of scatterplots. For each scatterplot, the data are plotted in circles, can I add some argument to change the circles into dots? Could anyone give me some advice?Many thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/matrix-of-scatterplots-tf4067527.html#a11558687 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] How to activate the R commands in SciViews
Thank you for the pointer. On the SciViews official site (http://www.sciviews.org/SciViews-R/index.html) I could not find any indications on this incompatibility. At any rate, I would have two questions related to the future of SciViews-R. Are there any developments planned in some near future that would make SciViews compatible with recent versions of R and Rcmdr? At the Unviversity of Social Sciences of Toulouse, for example, the most recent version of R is installed. And for statistics introductory classes, SciViews would be an important advantage. R should be downgraded to which version for the two to be compatible? Secondly, is there any chance that SciViews become available on Linux, again in some not so distant future? That is, would the tcltk2 package be ported to other OS's and would SciViews be subsequently enhanced to build on such systems? Last time I checked the net, no interesting information was available as to such developments. Maybe there is some beta version of tcltk2 that I don't know of.. Regards, Liviu On 7/12/07, Philippe Grosjean [EMAIL PROTECTED] wrote: As explained on the web page from where you downloaded SciViews 0.8-9, this version is not compatible with R 2.5.0. Best, Philippe Grosjean ..∞})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Belgium ( ( ( ( ( .. __ 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] eMail results out of R
Hi everyone, I did my homework and read the posting guideline :-) I want to eMail the results of a computing automatically. So I get the results (the parameters of a garch process) and I want to eMail them to another person. How can I do that? Thx __ Thomas Schwander MVV Energie Konzern-Risikocontrolling Telefon 0621 - 290-3115 Telefax 0621 - 290-3664 E-Mail: [EMAIL PROTECTED] . Internet: www.mvv.de MVV Energie AG . Augustaanlage 67 . 68159 Mannheim Handelsregister-Nr. HRB 1780, Amtsgericht Mannheim Vorsitzender des Aufsichtsrates: Oberbürgermeister Gerhard Widder Vorstand: Dr. Rudolf Schulten (Vorsitzender) . Dr. Werner Dub . Hans-Jürgen Farrenkopf [[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] lead
How about
revLag - function(x, shift=1) rev( Lag(rev(x), shift) )
x - 1:5
revLag(x, shift=2)
As a matter of fact, here is a generalized version of Lag to include
negative shifts.
myLag - function (x, shift = 1){
xLen - length(x)
ret - as.vector(character(xLen), mode = storage.mode(x))
attrib - attributes(x)
if (!is.null(attrib$label))
atr$label - paste(attrib$label, lagged, shift, observations)
if (shift == 0) return(x)
if( xLen = abs(shift) ) return(ret)
if (shift 0) x - rev(x)
retrange = 1:abs(shift)
ret[-retrange] - x[1:(xLen - abs(shift))]
if (shift 0) ret - rev(ret)
attributes(ret) - attrib
return(ret)
}
and some test examples:
myLag(1:5, shift=2)
[1] NA NA 1 2 3
myLag(letters[1:4], shift=2)
[1] a b
myLag(factor(letters[1:4]), shift=2)
[1] NA NA ab
Levels: a b c d
myLag(1:5, shift=-2)
[1] 3 4 5 NA NA
myLag(letters[1:4], shift=-2)
[1] c d
myLag(factor(letters[1:4]), shift=-2)
[1] cdNA NA
Levels: a b c d
Regards, Adai
Aydemir, Zava (FID) wrote:
Hi,
is there any function in R that shifts elements of a vector to the
opposite direction of what Lag() of the Hmisc package does? (something
like, Lag(x, shift = -1) )
Thanks
Zava
This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}
__
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.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] XAML
Hi I wonder if anyone has had any thoughts about rendering R graphical output into XAML? Gordon M. Morrison HSBC Bank plc 8 Canada Square London E14 5HQ - SAVE PAPER - THINK BEFORE YOU PRINT! This transmission has been issued by a member of the HSBC Group HSBC for the information of the addressee only and should not be reproduced and/or distributed to any other person. Each page attached hereto must be read in conjunction with any disclaimer which forms part of it. Unless otherwise stated, this transmission is neither an offer nor the solicitation of an offer to sell or purchase any investment. Its contents are based on information obtained from sources believed to be reliable but HSBC makes no representation and accepts no responsibility or liability as to its completeness or accuracy. __ 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] type III ANOVA for a nested linear model
Dear Carsten In this test, factor B would be representing to a factor of block or repetition according to as the levels of A, B, and C are in the data. Factor C this nested in A, then the model should include: B, A and C nested in A, the difference it is the error. Model: B 1 A 2 C(A)6 Error (2+6)*1 = 8 Total mydata-read.table(mydata.txt,header=T) mydata[,1]- as.factor(mydata[,1]) mydata[,2]- as.factor(mydata[,2]) mydata[,3]- as.factor(mydata[,3]) model - aov(resp ~ B + A + C/A, mydata) summary(model) Df Sum Sq Mean Sq F valuePr(F) B1 915.21 915.21 89.6476 1.274e-05 *** A2 33.12 16.56 1.6223 0.25621 C6 199.50 33.25 3.2570 0.06316 . Residuals8 81.67 10.21 Best regards, Felipe de Mendiburu Statistician -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Carsten Jaeger Sent: Tuesday, July 10, 2007 6:15 AM To: R help list Subject: [R] type III ANOVA for a nested linear model Hello, is it possible to obtain type III sums of squares for a nested model as in the following: lmod - lm(resp ~ A * B + (C %in% A), mydata)) I have tried library(car) Anova(lmod, type=III) but this gives me an error (and I also understand from the documentation of Anova as well as from a previous request (http://finzi.psych.upenn.edu/R/Rhelp02a/archive/64477.html) that it is not possible to specify nested models with car's Anova). anova(lmod) works, of course. My data (given below) is balanced so I expect the results to be similar for both type I and type III sums of squares. But are they *exactly* the same? The editor of the journal which I'm sending my manuscript to requests what he calls conventional type III tests and I'm not sure if can convince him to accept my type I analysis. R mydata A B C resp 1 1 1 1 34.12 2 1 1 2 32.45 3 1 1 3 44.55 4 1 2 1 20.88 5 1 2 2 22.32 6 1 2 3 27.71 7 2 1 6 38.20 8 2 1 7 31.62 9 2 1 8 38.71 102 2 6 18.93 112 2 7 20.57 122 2 8 31.55 133 1 9 40.81 143 1 10 42.23 153 1 11 41.26 163 2 9 28.41 173 2 10 24.07 183 2 11 21.16 Thanks a lot, Carsten __ 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] p-value from survreg
The question was how to get the p-value from the fit below, as an S object sr-survreg(s~groups, dist=gaussian) Coefficients: (Intercept) groups -0.02138485 0.03868351 Scale= 0.01789372 Loglik(model)= 31.1 Loglik(intercept only)= 25.4 Chisq= 11.39 on 1 degrees of freedom, p= 0.00074 n= 16 In general, good places to start are names(sr) help(survreg.object) ssr - summary(sr) names(ssr) As someone else pointed out, it's also easy to look at the print.survreg function and see how the value was created -- one of the things I love about S. Unfortunately, doing the above myself showed that I have let the documentation page for survreg.object get seriously out of date -- quite embarassing as that is logically the first place to start. As to the print function creating things on the fly: there is an area where there is no good answer. Does one make the return object from a fit such that it contains only minimal data, or add in all of the other computations that can be derived from these? The Chambers and Hastie book Statistical Models in S, which was the starting point for model objects, leaned towards the former, and this still influences many functions. Often the summary function will fill in these derived values, the std and t-tests for the individual coefficients for instance. Terry T. __ 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] Fitting a Gamma Curve
Hi there, I hope someone can help me before I tear all my hair out. I have a set transition intensities and when plotted the curve looks like a gamma density. I want to fit a gamma density curve to these intensities. It is just a curve fitting problem but whats causing the trouble is that I need to use least squares minimization to calculate the parameters for the gamma curve. How do I do this??? The curve will be a truncated gamma function so it will have 3 paramaters a, b, c. I tried to do the following nls(lograte ~ log(c) + a*log(b) + (a-1)*log(age)+b*(age)-lgamma(a), start=list(a=1,b=1,c=1)), where lograte, and age are my data. a,b the gamma parameters and c the parameter we need because we are fitting a truncated distribution. I also tried defining fn = function(p) sum((log(y)-log(dgamma(x,p[1],p[2])*p[3]))^2) a residual sum of squares and using nlm to minimise this and find paramaters but this doesnt work either. Can anyone help me ?? Please :) -- View this message in context: http://www.nabble.com/Fitting-a-Gamma-Curve-tf4068543.html#a11561404 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] lead
The lag.zoo method of lag in the zoo package supports positive, negative
and multiple lags and has an na.pad= argument. (zoo also has a
lag.zooreg method, not shown, for zooreg objects):
library(zoo)
z - zoo(11:15)
z
1 2 3 4 5
11 12 13 14 15
lag(z, na.pad = TRUE)
1 2 3 4 5
12 13 14 15 NA
lag(z, 1, na.pad = TRUE) # same
1 2 3 4 5
12 13 14 15 NA
# negative lag
lag(z, -1, na.pad = TRUE)
1 2 3 4 5
NA 11 12 13 14
# mulitple lags
lag(z, 1:3, na.pad = TRUE)
lag1 lag2 lag3
1 12 13 14
2 13 14 15
3 14 15 NA
4 15 NA NA
lag(z, -(1:3), na.pad = TRUE)
lag-1 lag-2 lag-3
211NANA
31211NA
4131211
5141312
vignette(zoo) # more info on zoo
On 7/12/07, Aydemir, Zava (FID) [EMAIL PROTECTED] wrote:
Hi,
is there any function in R that shifts elements of a vector to the
opposite direction of what Lag() of the Hmisc package does? (something
like, Lag(x, shift = -1) )
Thanks
Zava
This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}
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[R] Package for .632 (and .632+) bootstrap and the cross-validation of ROC Parameters
Hi users, I need to calculate .632 (and .632+) bootstrap and the cross-validation of area under curve (AUC) to compare my models. Is there any package for the same. I know about 'ipred' and using it i can calculate misclassification errors. Please help. It's urgent. -- View this message in context: http://www.nabble.com/Package-for-.632-%28and-.632%2B%29-bootstrap-and-the-cross-validation-of-ROC-Parameters-tf4068544.html#a11561405 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] p-value from survreg
On 7/12/07, Terry Therneau [EMAIL PROTECTED] wrote: The question was how to get the p-value from the fit below, as an S object sr-survreg(s~groups, dist=gaussian) Coefficients: (Intercept) groups -0.02138485 0.03868351 Scale= 0.01789372 Loglik(model)= 31.1 Loglik(intercept only)= 25.4 Chisq= 11.39 on 1 degrees of freedom, p= 0.00074 n= 16 In general, good places to start are names(sr) help(survreg.object) ssr - summary(sr) names(ssr) As someone else pointed out, it's also easy to look at the print.survreg function and see how the value was created -- one of the things I love about S. Unfortunately, doing the above myself showed that I have let the documentation page for survreg.object get seriously out of date -- quite embarassing as that is logically the first place to start. As to the print function creating things on the fly: there is an area where there is no good answer. Does one make the return object from a fit such that it contains only minimal data, or add in all of the other computations that can be derived from these? The Chambers and Hastie book Statistical Models in S, which was the starting point for model objects, leaned towards the former, and this still influences many functions. Often the summary function will fill in these derived values, the std and t-tests for the individual coefficients for instance. I think this is where it's nice to have a separate function that does the filling in - then you can have the best of both worlds. That's the role that summary often plays. Hadley __ 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] ggplot2 / reshape / Question on manipulating data
hadley wickham [EMAIL PROTECTED] writes: On 7/12/07, Pete Kazmier [EMAIL PROTECTED] wrote: I'm an R newbie but recently discovered the ggplot2 and reshape packages which seem incredibly useful and much easier to use for a beginner. Using the data from the IMDB, I'm trying to see how the average movie rating varies by year. Here is what my data looks like: ratings - read.delim(groomed.list, header = TRUE, sep = |, comment.char = ) ratings - subset(ratings, VoteCount 100) head(ratings) Title Histogram VoteCount VoteMean Year 1!Huff (2004) (TV) 16 299 8.4 2004 8 'Allo 'Allo! (1982) 000125 829 8.6 1982 50 .hack//SIGN (2002) 001113 150 7.0 2002 561-800-Missing (2003) 000103 118 5.4 2003 66 Greatest Artists (2000) (mini) 00..16 110 7.8 2000 77 00 Scariest Movie (2004) (mini) 00..000115 256 8.6 2004 Have you tried using the movies dataset included in ggplot? Or is there some data that you want that is not in that dataset. It's funny that you mention this because I had intended to write this email about a month ago but was delayed due to other reasons. In any case, when I was typing this up last night, I wanted to recreate my steps but I could not find the IMDB movie data I had used originally. I searched everywhere to no avail so I downloaded the data myself and groomed it. Only now do I remember that I had used the movies dataset included in ggplot. How do 'byYear' and 'byYear2' differ? I am trying to use 'typeof' but both seem to be lists. However, they are clearly different in some way because 'qplot' graphs them differently. Try using str - it's much more helpful, and you should see the different quickly. Thanks! This is the function I've been looking for in my quest to learn about internal data types of R. Too bad it has such a terrible name! Using the built in movies data: mm - melt(movies, id=1:2, m=c(rating, votes)) msum - cast(mm, year ~ variable, c(mean, sum)) qplot(year, rating_mean, data=msum, colour=votes_sum) qplot(year, rating_mean, data=msum, colour=votes_sum, geom=line) Great! This is exactly what I was looking to do. By the way, does any of your documentation use the movie dataset as an example? I'm curious what else I can do with the dataset. For example, how can I use ggplot's facets to see the same information by type of movie? I'm unsure of how to manipulate the binary variables into a single variable so that it can be treated as levels. Thanks! Pete __ 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] eMail results out of R
We use a program called Blat (www.blat.net) on Windows to email out
results of overnight runs. If you're on Unix/Linux you can definitely do
a similar thing using one of the hundreds of command line utils.
The R code is similar to below:
sendEmail - function(from, to, subject, body)
{
BLAT - PATH TO BLAT.EXE
MAILSERVER - your mail server here;
command - paste(BLAT, -, -to, dQuote(to), -server,
MAILSERVER, -s, dQuote(subject), -f, dQuote(from))
system(command, input=body)
}
HTH,
John Scillieri
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: Thursday, July 12, 2007 9:53 AM
To: r-help@stat.math.ethz.ch
Subject: [R] eMail results out of R
Hi everyone,
I did my homework and read the posting guideline :-)
I want to eMail the results of a computing automatically. So I get the
results (the parameters of a garch process) and I want to eMail them to
another person. How can I do that?
Thx
This e-mail and any attachments are confidential, may contain legal,
professional or other privileged information, and are intended solely for the
addressee. If you are not the intended recipient, do not use the information
in this e-mail in any way, delete this e-mail and notify the sender. CEG-IP2
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] eMail results out of R
On 7/12/2007 9:52 AM, [EMAIL PROTECTED] wrote: Hi everyone, I did my homework and read the posting guideline :-) I want to eMail the results of a computing automatically. So I get the results (the parameters of a garch process) and I want to eMail them to another person. How can I do that? This will depend on the system you're using. If the command emailit would work from the command line on your system, then system(emailit) should work from within R. Writing that command is the hard part, of course. 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] eMail results out of R
Hi, There is a paper in the April 2007 issue of R News that might be of help here. http://##cran mirror##/doc/Rnews/Rnews_2007-1.pdf Romain Duncan Murdoch wrote: On 7/12/2007 9:52 AM, [EMAIL PROTECTED] wrote: Hi everyone, I did my homework and read the posting guideline :-) I want to eMail the results of a computing automatically. So I get the results (the parameters of a garch process) and I want to eMail them to another person. How can I do that? This will depend on the system you're using. If the command emailit would work from the command line on your system, then system(emailit) should work from within R. Writing that command is the hard part, of course. Duncan Murdoch -- Mango Solutions data analysis that delivers Tel: +44(0) 1249 467 467 Fax: +44(0) 1249 467 468 Mob: +44(0) 7813 526 123 __ 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] eMail results out of R
Here is a small function that I used on Debian. It requires exim4 :
send.mail-function(addr='[EMAIL PROTECTED]',subject='A
message from R',
text=paste(I have finished to work
,Sys.time(),coll=)){
# send an email
# it requires the reconfiguration of exim4
# you have to connect as root and
# then type dpkg-reconfigure exim4config
mail.cmd-paste(mail ,
-s \,subject,\ ,
addr,
EOT \n,
text,\n,
EOT,
sep=,collapse=)
system(mail.cmd,intern=FALSE)
}
Cheers,
Romain Francois wrote:
Hi,
There is a paper in the April 2007 issue of R News that might be of help
here.
http://##cran mirror##/doc/Rnews/Rnews_2007-1.pdf
Romain
Duncan Murdoch wrote:
On 7/12/2007 9:52 AM, [EMAIL PROTECTED] wrote:
Hi everyone,
I did my homework and read the posting guideline :-)
I want to eMail the results of a computing automatically. So I get the
results (the parameters of a garch process) and I want to eMail them to
another person. How can I do that?
This will depend on the system you're using. If the command emailit
would work from the command line on your system, then
system(emailit)
should work from within R. Writing that command is the hard part, of
course.
Duncan Murdoch
--
Stéphane DRAY ([EMAIL PROTECTED] )
Laboratoire BBE-CNRS-UMR-5558, Univ. C. Bernard - Lyon I
43, Bd du 11 Novembre 1918, 69622 Villeurbanne Cedex, France
Tel: 33 4 72 43 27 57 Fax: 33 4 72 43 13 88
http://biomserv.univ-lyon1.fr/~dray/
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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] Difference in linear regression results for Stata and R
Hi I recently imported data from r into Stata. I then ran the linear regression model I've been working on, only to discover that the results are somewhat (though not dramatically different). the standard errors vary more between the two programs than do the coefficients themselves. Any suggestions on what I've done that causes this mismatch? Thanks, Kate -- View this message in context: http://www.nabble.com/Difference-in-linear-regression-results-for-Stata-and-R-tf4069072.html#a11563283 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] Stepwise GLM selection by LRT?
Thank you very much for the prompt reply. Seems like I had not fully understood what the k-parameter to stepAIC is doing. Your suggested approach looks indeed fine to me, actually I do not quite understand why you say that it's only an approximation to the LRT? Best wishes- Lutz On 7/11/07, Ravi Varadhan [EMAIL PROTECTED] wrote: Check out the stepAIC function in MASS package. This is a nice tool, where you can actually implement any penalty even though the function's name has AIC in it because it is the default. Although this doesn't do an LRT test based variable selection, you can sort of approximate it by using a penalty of k = qchisq(1-p, df=1), where p is the p-value for variable selection. This penalty means that a variable enters/exits an existing model, when the absolute value of change in log-likelihood is greater than qchisq(1-p, df=1). For p = 0.1, k = 2.71, and for p=0.05, k = 3.84. Is this whhant you'd like to do? Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Lutz Ph. Breitling Sent: Wednesday, July 11, 2007 3:06 PM To: r-help@stat.math.ethz.ch Subject: [R] Stepwise GLM selection by LRT? Dear List, having searched the help and archives, I have the impression that there is no automatic model selection procedure implemented in R that includes/excludes predictors in logistic regression models based on LRT P-values. Is that true, or is someone aware of an appropriate function somewhere in a custom package? Even if automatic model selection and LRT might not be the most appropriate methods, I actually would like to use these in order to simulate someone else's modeling approach... Many thanks for all comments- Lutz - Lutz Ph. Breitling German Cancer Research Center Heidelberg/Germany __ 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] eMail results out of R
On 12-Jul-07 13:52:56, [EMAIL PROTECTED] wrote: Hi everyone, I did my homework and read the posting guideline :-) I want to eMail the results of a computing automatically. So I get the results (the parameters of a garch process) and I want to eMail them to another person. How can I do that? Thx As well as the answers from John Scillieri and Duncan Murdoch: if it does happen that you're using a Unix/Linux system, then the appropriate variant of the following illustration will work (the 'mail' command should always be present on variants of these systems): In the little project I'm working on in R at the moment, I have variables x1 and x2 (each length-50 numeric vectors). So I can, in R, do: sink(file=myoutput) cbind(x1,x2) sink() system(mail ted -s \Test No 2\ myoutput) which has stored the 2-column output from cbind(x1,x2) in the file myoutput, and then used the 'mail' command to mail its contents to 'ted' with subject 'Test No 2' (I used a multi-word subject to illustrate the quotes the command line will need, to avoid splitting the subject into separate tokens). See ?sink for how to use the sink() command. Then 'ted' duly received an email with contents: === Date: Thu, 12 Jul 2007 17:10:44 +0100 From: Ted Harding [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Test No 2 x1 x2 [1,] 0.37282844 0.002743146 [2,] 0.93293155 -0.108009247 ... [49,] -0.08681427 0.828313288 [50,] -0.23621908 0.385269729 === You can of course encapsulate something like the above sequence of commands into an R function, depending on how you organise the storage of the results you want to email. Hoping this helps, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 12-Jul-07 Time: 17:25:05 -- XFMail -- __ 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] eMail results out of R
On Thu, 12 Jul 2007, Duncan Murdoch wrote: On 7/12/2007 9:52 AM, [EMAIL PROTECTED] wrote: Hi everyone, I did my homework and read the posting guideline :-) But failed to follow it, and not telling us the OS makes this very much harder to answer adequately. I want to eMail the results of a computing automatically. So I get the results (the parameters of a garch process) and I want to eMail them to another person. How can I do that? This will depend on the system you're using. If the command emailit would work from the command line on your system, then system(emailit) should work from within R. Writing that command is the hard part, of course. But bug.report() will give you a good start on command-line oriented systems which have mailx (the POSIX mail client). (mailx is more standard than mail that a couple of others have referred to.) The adventurous could use make.socket and friends to talk to the sendmail daemon on a Unix-alike. Windows is somewhat harder: 'blat' provides a command-line mail client that talks to a SMTP server elsewhere. If you use MS Exchange you will need to find other ways to talk to it (DCOM?) Long ago we talked about have a mailer() function in R, but making one that was close to universal proved to be far too difficult for its utility. -- 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] eMail results out of R
On 12-Jul-07 16:10:46, Stéphane Dray wrote:
Here is a small function that I used on Debian. It requires exim4 :
send.mail-function(addr='[EMAIL PROTECTED]',subject='A
message from R',
text=paste(I have finished to work
,Sys.time(),coll=)){
# send an email
# it requires the reconfiguration of exim4
# you have to connect as root and
# then type dpkg-reconfigure exim4config
I'm a bit puzzled by this. On any Unix/Linux system (unless
something has changed very recently which I haven't heard about),
the 'mail' command simply works, for any user (without having
to become root); does not require exim4 (or any particular version
of any particular mail agent--so long as something has to be set up
so that email can be sent at all), and (for the purpose of using
'mail' from R) does not require exim4 or any other mail agent to
be re-configured. The email will be sent From: the user who
is running R.
In the example I posted just now, I just used 'mail' in R's
system() command without doing anything special. The mail transfer
agent in my case is 'sendmail', but it's a standard configuration
and nothing special has been done.
mail.cmd-paste(mail ,
-s \,subject,\ ,
addr,
EOT \n,
text,\n,
EOT,
sep=,collapse=)
system(mail.cmd,intern=FALSE)
}
Cheers,
Romain Francois wrote:
Hi,
There is a paper in the April 2007 issue of R News that might be of
help
here.
http://##cran mirror##/doc/Rnews/Rnews_2007-1.pdf
Romain
Duncan Murdoch wrote:
On 7/12/2007 9:52 AM, [EMAIL PROTECTED] wrote:
Hi everyone,
I did my homework and read the posting guideline :-)
I want to eMail the results of a computing automatically. So I get
the results (the parameters of a garch process) and I want to eMail
them to another person. How can I do that?
This will depend on the system you're using. If the command
emailit
would work from the command line on your system, then
system(emailit)
should work from within R. Writing that command is the hard part, of
course.
Duncan Murdoch
--
Stéphane DRAY ([EMAIL PROTECTED] )
Laboratoire BBE-CNRS-UMR-5558, Univ. C. Bernard - Lyon I
43, Bd du 11 Novembre 1918, 69622 Villeurbanne Cedex, France
Tel: 33 4 72 43 27 57 Fax: 33 4 72 43 13 88
http://biomserv.univ-lyon1.fr/~dray/
__
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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.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 12-Jul-07 Time: 18:03:20
-- XFMail --
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[R] error problem with glht
Can anyone help me? I'm having problems with the following code where I want to test the null hypothesis that regression slopes are the same among regressions. Here's the code I've written with comments that include the final error I get. ... initial.dir - getwd() library(systemfit) library(multcomp) basdata - read.table(data_into7_test.txt, header=TRUE,sep=) #show basdata basdata #output from above looks like this ... # basin Order lnLcl lnArea #1SFK 6 3.46322000 5.19766000 #2SFK 6 3.51767000 5.11809000 #3SFK 6 3.79962000 5.79118000 #4SFK 6 2.90242000 4.09768000 #5SFK 4 1.92408000 2.31617000 # #21 NFK 6 3.88342000 5.50927000 #22 NFK 6 3.84522000 5.13474000 #23 NFK 6 3.4913 5.28586000 # #linear model fit and analysis of variance fit_area_1 - lm(lnArea~Order*basin,data=basdata) anova_area_1 - anova(fit_area_1) #get coefficients, covariance matrix, terms coef(fit_area_1) vcov(fit_area_1) terms(fit_area_1) #perform multiple comparisons to test null hypothesis that #linear model fits are the same among basins fit1_mc - glht(anova_area_1,linfct=mcp(basin=Tukey)) #the line above gives ... #Error in terms.default(object) : no terms component #Error in factor_contrasts(model) : no model.matrix method for model found! I appreciate any help offered. Cheers, Pete -- Peter Furey Research Scientist Northwest Research Associates / CORA 3380 Mitchell Lane Boulder, CO 80301 Also affiliated with: Cooperative Institute for Research in Environmental Sciences (CIRES) University of Colorado, Boulder __ 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] ggplot doesnt work in loops?
On 7/12/07, hadley wickham [EMAIL PROTECTED] wrote:
Hi Steve,
You need to explicitly print the ggplot object:
ggplot(mydata, aes(x=mydata$varc)) + geom_bar()
(this is a R-faq for lattice plots, and ggplot works the same way)
In the latest version of ggplot (0.5.4) you can construct the plot
before hand and modify the aesthetics in each instance of the loop:
p - ggplot(mydata) + geom_bar()
mydata$varc = c(1,2,3)
for (i in 1:1){
jpeg(test3.jpg)
p + aes(x = mydata$varc)
dev.off()
}
(not that this will actually work because you're not using i inside
your loop anywhere)
(and to be strictly correct you should probably use list(x =
as.name(names(mydata)[i])) instead of the aes call - but I haven't
written any documentation for this yet)
Actually a better solution (will be included in the next version of ggplot) is:
aes_string - function(...) structure(lapply(list(...), as.name),
class=uneval)
p + aes_string(x = names(mydata)[i])
It converts aes(x = x, y=y) to aes(x=x, y=y). The first is easy
to generate programmatically, the second is less to type.
Hadley
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[R] contour and filled contour plots
Hello,
I'm trying to overlap a contour and a filled.contour:
as I read in a previous post this can be done
calling contour() in the plot.axes of filled.contour.
When I do it (see example below) the contour is placed
instead of the color bar, while the filled.contour
is not drawn at all, which is the right way?
Thanks
Andrea
postscript(covtd.xy.eps, width=6.5, height=5.5,
horizontal=FALSE, onefile=FALSE, paper = special)
filled.contour(plot.axes={ contour(seq(1,nlev),seq(1,nlev),real2) },
seq(1,nlev),seq(1,nlev),real2,levels=c(-40,-20,-15,-10,-7.5,-5,-2.5,-1.5,-0.5,0.5,1.5,2.5,5,7.5,10,15,20,40),xlab=Model
levels,ylab=Model levels,
color = colorRampPalette(c(blue, white,
red)),main=expression(paste(Unit: ,10^-6, ,K%.%s^-1)))
dev.off()
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Re: [R] Package for .632 (and .632+) bootstrap and the cross-validation of ROC Parameters
spime wrote: Hi users, I need to calculate .632 (and .632+) bootstrap and the cross-validation of area under curve (AUC) to compare my models. Is there any package for the same. I know about 'ipred' and using it i can calculate misclassification errors. Please help. It's urgent. See the validate* functions in the Design package. Note that some simulations (see http://biostat.mc.vanderbilt.edu/rms) indicate that the advantages of .632 and .632+ over the ordinary bootstrap are highly dependent on the choice of the accuracy measure being validated. The bootstrap variants seem to have advantages mainly if an improper, inefficient, discontinuous scoring rule such as the percent classified correct is used. -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Generating bivariate or multivariate data with known parameter values
David Kaplan-2 wrote: Greetings, I'm interested in generating data from various bivariate or mulitivariate distributions (e.g. gamma, t, etc), where I can specify the parameter values, including the correlations among the variables. I haven't been able to dig anything up on the faq, but I probably missed something. A nudge in the right direction would be appreciated. David -- David Kaplan, Ph.D. Professor Department of Educational Psychology University of Wisconsin - Madison Educational Sciences, Room 1061 1025 W. Johnson Street Madison, WI 53706 email: [EMAIL PROTECTED] Web: http://www.education.wisc.edu/edpsych/facstaff/kaplan/kaplan.htm Phone: 608-262-0836 Fax: 608-262-0843 __ 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. i'm interesting to use multivariate gamma distributed gamma data, in copula packages, you say that it can generate bivariate gamma distribution, i have try this and succes. my question, 1. are the copula can be used for generating multivariate gamma distributed gamma data? 2. when i try to generating bivariate gamma, can i set the correlation? for knowing, i have try many combination of parameters, but the result is taht correlation value is about 0.5. thankyou very much for your responses. dian -- View this message in context: http://www.nabble.com/-R--Generating-bivariate-or-multivariate-data-with-known-parameter-values-tf2405919.html#a11560289 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] Error in dyn.load()
Hi, All; I tried to load the 'genetics' library but it failed with the following message. Can you help me with this? Thanks in advance. library(genetics) Loading required package: gdata Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library '/Library/Frameworks/R.framework/ Versions/2.5/Resources/library/gtools/libs/ppc/gtools.so': dlopen(/Library/Frameworks/R.framework/Versions/2.5/Resources/ library/gtools/libs/ppc/gtools.so, 6): Library not loaded: /Library/ Frameworks/R.framework/Versions/2.4/Resources/lib/libR.dylib Referenced from: /Library/Frameworks/R.framework/Versions/2.5/ Resources/library/gtools/libs/ppc/gtools.so Reason: Incompatible library version: gtools.so requires version 2.4.0 or later, but libR.dylib provides version 2.2.0 Error: package 'gdata' could not be loaded I am using Mac OS X R 2.5.0. The libraries, 'genetics' v1.2.1, 'gtools' v2.3.1, and 'gdata' v2.3.1 are already installed. Tae-Hoon Chung Post-Doctoral Researcher Computational Biology Division, TGEN 445 N 5th St. Phoenix, AZ 85004 USA O: 1-602-343-8724 F: 1-602-343-8840 [[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] multiple plots in a graph
Hi, I have to generate 10 cdfs in a graph. I need to compare the cdf's nature by plotting ten cdfs in a graph. Thus, I need multiple plots in a graph. I would appreciate if you could give some solution to the problem asap. Thanking you, Sincerely, Ajay. -- Ajay Singh Research Scientist, SOM, IIT-Bombay, Powai, MUMBAI-400076, MH (INDIA). __ 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] calculating percent error from 2 vectors
Hello, I believe this is an easy scripting problem, but one I am stumbling on. I have a known vector of 3 colors with nrow=10: known-c(red, blue, red, red, yellow, blue, yellow, blue, blue, yellow) and a model output vector: modelout-c(red, red, red, blue, yellow, blue, blue, red, blue, yellow) I would like to determine the proportion (in)correctly identified for each color. In other words: % correct red= % correct blue= % correct yellow= How would I code this (assuming the actual dataset is more complex)? Any help would be much appreciated. Thank you, Scott __ 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] Interpreting a string as a variable in a column header
This must be a very simple question, but I can't find any information on it elsewhere, sorry. When extracting information from a list using column headers, how do I get R to interpret something as a variable rather than a string? For example: xx$YAL002 works, but this doesn't: gene - YAL002 xx$gene neither do xx$parse(gene) xx$eval(gene) xx$eval(parse(gene)) or a variety of other constructions I have tried. Background: I have a table of information about yeast genes, and I also have a list of yeast genes and their GO terms (xx) from the YEAST package that I downloaded. I want to go through all the genes in my table and look up their GO terms in the list from the YEAST package. They might not contain exactly the same genes (ideally they should, but I'd be surprised if they do) and I don't think they're in the same order, so I do want to use the column names, but there are a lot of them so I'm not typing them all out individually. It might be possible to turn the GO data into something other than a list, but the help page recommends using xx - as.list(YEASTGO) and doesn't make any other suggestions, so I should probably do as I'm told. Thanks for any help or suggestions of where to look, Rosa __ 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] ggplot2 / histogram / y-axis
Is there a way in ggplot to make a histogram with the left-hand y-axis label as frequency, and a right-hand y-axis label as percentage? Thanks! Pete __ 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] Compute rank within factor groups
Hi,
I have a data.frame which is ordered by score, and has a factor column:
Browse[1] wc[c(report,score)]
report score
9 ADEA 0.96
8 ADEA 0.90
11 Asylum_FED9 0.86
3 ADEA 0.75
14 Asylum_FED9 0.60
5 ADEA 0.56
13 Asylum_FED9 0.51
16 Asylum_FED9 0.51
2 ADEA 0.42
7 ADEA 0.31
17 Asylum_FED9 0.27
1 ADEA 0.17
4 ADEA 0.17
6 ADEA 0.12
10ADEA 0.11
12 Asylum_FED9 0.10
15 Asylum_FED9 0.09
18 Asylum_FED9 0.07
Browse[1]
I need to add a column indicating rank within each factor group, which I
currently accomplish like so:
wc$rank - 0
for(report in as.character(unique(wc$report))) {
wc[wc$report==report,]$rank - 1:sum(wc$report==report)
}
I have to wonder whether there's a better way, something that gets rid of
the for() loop using tapply() or by() or similar. But I haven't come up
with anything.
I've tried these:
by(wc, wc$report, FUN=function(pr){pr$rank - 1:nrow(pr)})
by(wc, wc$report, FUN=function(pr){wc[wc$report %in% pr$report,]$rank -
1:nrow(pr)})
But in both cases the effect of the assignment is lost, there's no $rank
column generated for wc.
Any suggestions?
-Ken
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Re: [R] ggplot2 / reshape / Question on manipulating data
On 7/12/07, Pete Kazmier [EMAIL PROTECTED] wrote: hadley wickham [EMAIL PROTECTED] writes: On 7/12/07, Pete Kazmier [EMAIL PROTECTED] wrote: I'm an R newbie but recently discovered the ggplot2 and reshape packages which seem incredibly useful and much easier to use for a beginner. Using the data from the IMDB, I'm trying to see how the average movie rating varies by year. Here is what my data looks like: ratings - read.delim(groomed.list, header = TRUE, sep = |, comment.char = ) ratings - subset(ratings, VoteCount 100) head(ratings) Title Histogram VoteCount VoteMean Year 1!Huff (2004) (TV) 16 299 8.4 2004 8 'Allo 'Allo! (1982) 000125 829 8.6 1982 50 .hack//SIGN (2002) 001113 150 7.0 2002 561-800-Missing (2003) 000103 118 5.4 2003 66 Greatest Artists (2000) (mini) 00..16 110 7.8 2000 77 00 Scariest Movie (2004) (mini) 00..000115 256 8.6 2004 Have you tried using the movies dataset included in ggplot? Or is there some data that you want that is not in that dataset. It's funny that you mention this because I had intended to write this email about a month ago but was delayed due to other reasons. In any case, when I was typing this up last night, I wanted to recreate my steps but I could not find the IMDB movie data I had used originally. I searched everywhere to no avail so I downloaded the data myself and groomed it. Only now do I remember that I had used the movies dataset included in ggplot. How do 'byYear' and 'byYear2' differ? I am trying to use 'typeof' but both seem to be lists. However, they are clearly different in some way because 'qplot' graphs them differently. Try using str - it's much more helpful, and you should see the different quickly. Thanks! This is the function I've been looking for in my quest to learn about internal data types of R. Too bad it has such a terrible name! Using the built in movies data: mm - melt(movies, id=1:2, m=c(rating, votes)) msum - cast(mm, year ~ variable, c(mean, sum)) qplot(year, rating_mean, data=msum, colour=votes_sum) qplot(year, rating_mean, data=msum, colour=votes_sum, geom=line) Great! This is exactly what I was looking to do. By the way, does any of your documentation use the movie dataset as an example? I'm curious what else I can do with the dataset. For example, how can I use ggplot's facets to see the same information by type of movie? I'm unsure of how to manipulate the binary variables into a single variable so that it can be treated as levels. A lot of the examples do use the movies data, but I don't think any of it is particularly revealing. You might want to look at the results for the 2007 infovis visualisation challenge (http://www.apl.jhu.edu/Misc/Visualization/) which uses similar data. Submission isn't complete yet, but you can see my teams entry at http://had.co.nz/infovis-2007/. There are lots of interesting stories to pursue. I think I will update the movies data to include the first genre as another column. That will make it easier to facet by genre Hadley __ 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] Interpreting a string as a variable in a column header
Use xx[[gene]] instead of xx$gene (the $ is a shorthand for [[ with some extra magic to be more convenient, the magic is getting in your way, so go back to the [[ syntax (make sure you double the braces)). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of rosa clements Sent: Thursday, July 12, 2007 11:39 AM To: r-help@stat.math.ethz.ch Subject: [R] Interpreting a string as a variable in a column header This must be a very simple question, but I can't find any information on it elsewhere, sorry. When extracting information from a list using column headers, how do I get R to interpret something as a variable rather than a string? For example: xx$YAL002 works, but this doesn't: gene - YAL002 xx$gene neither do xx$parse(gene) xx$eval(gene) xx$eval(parse(gene)) or a variety of other constructions I have tried. Background: I have a table of information about yeast genes, and I also have a list of yeast genes and their GO terms (xx) from the YEAST package that I downloaded. I want to go through all the genes in my table and look up their GO terms in the list from the YEAST package. They might not contain exactly the same genes (ideally they should, but I'd be surprised if they do) and I don't think they're in the same order, so I do want to use the column names, but there are a lot of them so I'm not typing them all out individually. It might be possible to turn the GO data into something other than a list, but the help page recommends using xx - as.list(YEASTGO) and doesn't make any other suggestions, so I should probably do as I'm told. Thanks for any help or suggestions of where to look, Rosa __ 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] calculating percent error from 2 vectors
Try something like: mytable - table(known, modelout) prop.table( mytable, 1 ) Also look at ?addmargins and the CrossTable function in the gmodels package. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Scott Bearer Sent: Thursday, July 12, 2007 11:32 AM To: r-help@stat.math.ethz.ch Subject: [R] calculating percent error from 2 vectors Hello, I believe this is an easy scripting problem, but one I am stumbling on. I have a known vector of 3 colors with nrow=10: known-c(red, blue, red, red, yellow, blue, yellow, blue, blue, yellow) and a model output vector: modelout-c(red, red, red, blue, yellow, blue, blue, red, blue, yellow) I would like to determine the proportion (in)correctly identified for each color. In other words: % correct red= % correct blue= % correct yellow= How would I code this (assuming the actual dataset is more complex)? Any help would be much appreciated. Thank you, Scott __ 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] Resolved: Error in dyn.load()
Hi, All; I simply upgraded my R from version 2.5.0 to 2.5.1 and the whole problems disappeared. So I am closing the question myself. It could have been the R version problem. Tae-Hoon Chung Post-Doctoral Researcher Computational Biology Division, TGEN 445 N 5th St. Phoenix, AZ 85004 USA O: 1-602-343-8724 F: 1-602-343-8840 [[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] Compute rank within factor groups
Look at ?ave and try something like:
wc$rank - ave( wc$score, wc$report, FUN=rank )
This works even if the dataframe is not pre sorted.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Ken Williams
Sent: Thursday, July 12, 2007 12:09 PM
To: R-help@stat.math.ethz.ch
Subject: [R] Compute rank within factor groups
Hi,
I have a data.frame which is ordered by score, and has a
factor column:
Browse[1] wc[c(report,score)]
report score
9 ADEA 0.96
8 ADEA 0.90
11 Asylum_FED9 0.86
3 ADEA 0.75
14 Asylum_FED9 0.60
5 ADEA 0.56
13 Asylum_FED9 0.51
16 Asylum_FED9 0.51
2 ADEA 0.42
7 ADEA 0.31
17 Asylum_FED9 0.27
1 ADEA 0.17
4 ADEA 0.17
6 ADEA 0.12
10ADEA 0.11
12 Asylum_FED9 0.10
15 Asylum_FED9 0.09
18 Asylum_FED9 0.07
Browse[1]
I need to add a column indicating rank within each factor
group, which I currently accomplish like so:
wc$rank - 0
for(report in as.character(unique(wc$report))) {
wc[wc$report==report,]$rank - 1:sum(wc$report==report)
}
I have to wonder whether there's a better way, something that
gets rid of the for() loop using tapply() or by() or similar.
But I haven't come up with anything.
I've tried these:
by(wc, wc$report, FUN=function(pr){pr$rank - 1:nrow(pr)})
by(wc, wc$report, FUN=function(pr){wc[wc$report %in%
pr$report,]$rank -
1:nrow(pr)})
But in both cases the effect of the assignment is lost,
there's no $rank column generated for wc.
Any suggestions?
-Ken
__
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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.
Re: [R] Stepwise GLM selection by LRT?
On Thu, 12 Jul 2007, Lutz Ph. Breitling wrote: Thank you very much for the prompt reply. Seems like I had not fully understood what the k-parameter to stepAIC is doing. Your suggested approach looks indeed fine to me, actually I do not quite understand why you say that it's only an approximation to the LRT? So this is computing AIC_k = -2L + kp. If you compare models with p and p+q parameters, this is equvalent to comparing 2 log LR with kq and so for q=1 the Wilks' LRT is found for k = qchisq(1-p, df=1) (which is just a squared Normal). However, no one said q would always be one, and stepAIC steps in terms, not individual coefficients. Therein lies one of the approximations (another is in the asympototic distribution theory of the test). Best wishes- Lutz On 7/11/07, Ravi Varadhan [EMAIL PROTECTED] wrote: Check out the stepAIC function in MASS package. This is a nice tool, where you can actually implement any penalty even though the function's name has AIC in it because it is the default. Although this doesn't do an LRT test based variable selection, you can sort of approximate it by using a penalty of k = qchisq(1-p, df=1), where p is the p-value for variable selection. This penalty means that a variable enters/exits an existing model, when the absolute value of change in log-likelihood is greater than qchisq(1-p, df=1). For p = 0.1, k = 2.71, and for p=0.05, k = 3.84. Is this whhant you'd like to do? Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Lutz Ph. Breitling Sent: Wednesday, July 11, 2007 3:06 PM To: r-help@stat.math.ethz.ch Subject: [R] Stepwise GLM selection by LRT? Dear List, having searched the help and archives, I have the impression that there is no automatic model selection procedure implemented in R that includes/excludes predictors in logistic regression models based on LRT P-values. Is that true, or is someone aware of an appropriate function somewhere in a custom package? Even if automatic model selection and LRT might not be the most appropriate methods, I actually would like to use these in order to simulate someone else's modeling approach... Many thanks for all comments- Lutz - Lutz Ph. Breitling German Cancer Research Center Heidelberg/Germany __ 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. -- 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] ggplot2 / histogram / y-axis
On 7/12/07, Pete Kazmier [EMAIL PROTECTED] wrote: Is there a way in ggplot to make a histogram with the left-hand y-axis label as frequency, and a right-hand y-axis label as percentage? Not currently. I did a quick exploration to see if it was feasible to draw another axis on with grid, but it doesn't look like it's possible: p - qplot(rating, data=movies, geom=histogram) # Map aesthetics to data data - p$layers[[1]]$make_aesthetics(p) # Calculate statistic by hand (we'll need this to get the scales right) binned - StatBin$calculate(data=data, p$scales) n - nrow(movies) # Manually recreate the y scale sp - scale_y_continuous() sp$train(binned$count) # rescale the labels labels - formatC(sp$breaks() / n, digits=2) # Have to do without labels because of bug in grid print(p, pretty=FALSE) downViewport(panel_1_1) grid.draw(ggaxis(sp$breaks(), as.list(labels), right, sp$frange())) # Why don't labels line up? - I'm not sure # How could you make space for the extra axis? - Not sure either # How would this worked for a facetted graphic - not well Also how were you expecting the axes/gridlines to line up? Would both axes be labelled nicely (with whole numbers) and the secondary axis wouldn't have gridlines; or would the second axis match the lines of the primary, even though the number wouldn't be so attractive? Hadley __ 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 plots in a graph
Ajay Singh wrote: Hi, I have to generate 10 cdfs in a graph. I need to compare the cdf's nature by plotting ten cdfs in a graph. Thus, I need multiple plots in a graph. I would appreciate if you could give some solution to the problem asap. plot(ecdf(rnorm(10))) plot(ecdf(rnorm(12)), add=TRUE, col.points=red, col.hor=red) etc... Uwe Ligges Thanking you, Sincerely, Ajay. __ 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.null doesn't work
Seems to work, if I unlist the argument at first ;-) Atte Hi, What's wrong here?: v=c(`-`,`+`,1,`^`,`^`,NA,NA,X,9,X,2) i2=16 v[i2] [[1]] NULL is.null(v[i2]) [1] FALSE Is it a bug or have I misunderstood something? Atte Tenkanen University of Turku, Finland __ 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] calculating percent error from 2 vectors
On 12-Jul-07 17:32:03, Scott Bearer wrote:
Hello,
I believe this is an easy scripting problem, but one I am stumbling on.
I have a known vector of 3 colors with nrow=10:
known-c(red, blue, red, red, yellow, blue, yellow,
blue,
blue, yellow)
and a model output vector:
modelout-c(red, red, red, blue, yellow, blue, blue,
red,
blue, yellow)
I would like to determine the proportion (in)correctly identified for
each
color. In other words:
% correct red=
% correct blue=
% correct yellow=
How would I code this (assuming the actual dataset is more complex)?
For your example:
tbl-table(known,modelout)
tbl
modelout
knownblue red yellow
blue 22 0
red12 0
yellow 10 2
dim(tbl)
[1] 3 3
for(i in (1:dim(tbl)[1])){print(sum(tbl[i,-i])/sum(tbl[i,]))}
[1] 0.5
[1] 0.333
[1] 0.333
and you can modify the print command produce a desired format,
e.g. using rownames(tbl)[i] for the successive colour names.
Hoping this helps (as a start),
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 12-Jul-07 Time: 20:15:34
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Re: [R] Compute rank within factor groups
Is this what you are looking for:
x
report score
9 ADEA 0.96
8 ADEA 0.90
11 Asylum_FED9 0.86
3 ADEA 0.75
14 Asylum_FED9 0.60
5 ADEA 0.56
13 Asylum_FED9 0.51
16 Asylum_FED9 0.51
2 ADEA 0.42
7 ADEA 0.31
17 Asylum_FED9 0.27
1 ADEA 0.17
4 ADEA 0.17
6 ADEA 0.12
10ADEA 0.11
12 Asylum_FED9 0.10
15 Asylum_FED9 0.09
18 Asylum_FED9 0.07
x$rank - ave(x$score, x$report, FUN=rank)
x
report score rank
9 ADEA 0.96 10.0
8 ADEA 0.90 9.0
11 Asylum_FED9 0.86 8.0
3 ADEA 0.75 8.0
14 Asylum_FED9 0.60 7.0
5 ADEA 0.56 7.0
13 Asylum_FED9 0.51 5.5
16 Asylum_FED9 0.51 5.5
2 ADEA 0.42 6.0
7 ADEA 0.31 5.0
17 Asylum_FED9 0.27 4.0
1 ADEA 0.17 3.5
4 ADEA 0.17 3.5
6 ADEA 0.12 2.0
10ADEA 0.11 1.0
12 Asylum_FED9 0.10 3.0
15 Asylum_FED9 0.09 2.0
18 Asylum_FED9 0.07 1.0
On 7/12/07, Ken Williams [EMAIL PROTECTED] wrote:
Hi,
I have a data.frame which is ordered by score, and has a factor column:
Browse[1] wc[c(report,score)]
report score
9 ADEA 0.96
8 ADEA 0.90
11 Asylum_FED9 0.86
3 ADEA 0.75
14 Asylum_FED9 0.60
5 ADEA 0.56
13 Asylum_FED9 0.51
16 Asylum_FED9 0.51
2 ADEA 0.42
7 ADEA 0.31
17 Asylum_FED9 0.27
1 ADEA 0.17
4 ADEA 0.17
6 ADEA 0.12
10ADEA 0.11
12 Asylum_FED9 0.10
15 Asylum_FED9 0.09
18 Asylum_FED9 0.07
Browse[1]
I need to add a column indicating rank within each factor group, which I
currently accomplish like so:
wc$rank - 0
for(report in as.character(unique(wc$report))) {
wc[wc$report==report,]$rank - 1:sum(wc$report==report)
}
I have to wonder whether there's a better way, something that gets rid of
the for() loop using tapply() or by() or similar. But I haven't come up
with anything.
I've tried these:
by(wc, wc$report, FUN=function(pr){pr$rank - 1:nrow(pr)})
by(wc, wc$report, FUN=function(pr){wc[wc$report %in% pr$report,]$rank -
1:nrow(pr)})
But in both cases the effect of the assignment is lost, there's no $rank
column generated for wc.
Any suggestions?
-Ken
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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[R] is.null doesn't work
Hi, What's wrong here?: v=c(`-`,`+`,1,`^`,`^`,NA,NA,X,9,X,2) i2=16 v[i2] [[1]] NULL is.null(v[i2]) [1] FALSE Is it a bug or have I misunderstood something? Atte Tenkanen University of Turku, Finland __ 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 estimate treatment-interaction contrasts
Hello, R experts, Sorry for asking this question again again since I really want a help! I have a two-factor experiment data and like to calculate estimates of interation contrasts say factor A has levels of a1, a2, and B has levels of b1, b2, b3, b4, and b5 with 3 replicates. I am not sure the constrast estimate I got is right using the script below: score-c(7.2,6.5,6.9,6.4,6.9,6.1,6.9,5.3,7.2,5.7,5.1,5.9,7.6,6.9,6.8, 7.2,6.6,6.9,6.4,6.0,6.0,6.9,6.9,6.4,7.5,7.7,7.0,8.6,8.8,8.3) A - gl(2, 15, labels=c(a1, a2)) B - rep(gl(5, 3, labels=c(b1, b2, b3, b4, b5)), 2) contrasts(B)-cbind(c(-4,rep(1,4)),c(rep(-3,2),rep(2,3)), + c(rep(-2,3),rep(3,2)),c(rep(-1,4), rep(4,1))) fit1 - aov(score ~ A*B) summary(fit1, split=list(B=1:4), expand.split = TRUE) Df Sum Sq Mean Sq F valuePr(F) A1 3.2013 3.2013 15.1483 0.0009054 *** B4 8.7780 2.1945 10.3841 0.0001019 *** B: C1 1 0.0301 0.0301 0.1424 0.7099296 B: C2 1 2.0335 2.0335 9.6221 0.0056199 ** B: C3 1 1.2469 1.2469 5.9004 0.0246876 * B: C4 1 5.4675 5.4675 25.8715 5.637e-05 *** A:B 4 5.3420 1.3355 6.3194 0.0018616 ** A:B: C11 0.7207 0.7207 3.4105 0.0796342 . A:B: C21 2.6068 2.6068 12.3350 0.0021927 ** A:B: C31 1.9136 1.9136 9.0549 0.0069317 ** A:B: C41 0.1008 0.1008 0.4771 0.4976647 Residuals 20 4.2267 0.2113 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Now I like to get interaction contrast estimate for b1 and b2 vs b3, b4 and b5 cont - c(1, -1)[A] * c(-3, -3, 2, 2, 2)[B] estimat-sum(cont*score) # value of the contrast estimate for A:B C2 estimat [1] -24.1 I am not sure the estimate for A:B C2 contrast (-24.1) is correct because the F value given the output above(12.3350) does not equal to those I calculate below (15.2684): t.stat - sum(cont*score)/se.contrast(fit1, as.matrix(cont)) t.stat^2 Contrast 1 15.2684 Could you please help me calculate the correct the estimate of interaction contrast and corresponding F value? Thanks in advance! Joshua __ 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] ggplot2 / histogram / y-axis
hadley wickham [EMAIL PROTECTED] writes: On 7/12/07, Pete Kazmier [EMAIL PROTECTED] wrote: Is there a way in ggplot to make a histogram with the left-hand y-axis label as frequency, and a right-hand y-axis label as percentage? Not currently. I did a quick exploration to see if it was feasible to draw another axis on with grid, but it doesn't look like it's possible: Thank you for trying. Also how were you expecting the axes/gridlines to line up? Would both axes be labelled nicely (with whole numbers) and the secondary axis wouldn't have gridlines; or would the second axis match the lines of the primary, even though the number wouldn't be so attractive? I hadn't thought that far ahead. Depending on the audience, I render histograms differently, and was curious if I could just put both on a single graph. However, you bring up some interesting questions in terms of the presentation. On another note, and feel free to defer me to the documentation which I'm still in the process of reading, but will I be able to take advantage of some of Tufte's recommendations in terms of the typical histogram and/or scatterplots (pp126-134 in Visual Display of Quantitative Information)? For example, with histograms, he would eliminates the use of coordinate lines in favor of using a white grid to improve the data/ink ratio. Likewise in scatterplots, he uses range-frames and dot-dash-plots. Will I be able to use ggplot for these types of enhancements? Thanks, Pete __ 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] Compute rank within factor groups
Why are you using order instead of rank? If the data is pre sorted then they tend to give the same result (unless there are ties), but if your data is not presorted, then the results will be different. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: Ken Williams [mailto:[EMAIL PROTECTED] Sent: Thursday, July 12, 2007 2:50 PM To: Greg Snow; R-help@stat.math.ethz.ch Subject: Re: [R] Compute rank within factor groups On 7/12/07 3:42 PM, Ken Williams [EMAIL PROTECTED] wrote: I ended up using: wc$rank - ave( wc$score, wc$report, FUN=function(x) order(x, decreasing=TRUE) ) Which gives me the 1-based rank integers I was looking for. Of course, immediately after sending I realized a simpler way: wc$rank - ave( -wc$score, wc$report, FUN=order ) And as a newbie I think I get to be blissfully ignorant of which one is faster. =) -- Ken Williams Research Scientist The Thomson Corporation Eagan, MN __ 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] automatically generating and accessing data frames of varying dimensions
Hi All,
I want to automatically generate a number of data frames, each with an
automatically generated name and an automatically generated number of
rows. The number of rows has been calculated before and is different for
all data frames (e.g. c(4,5,2)). The number of columns is known a priori
and the same for all data frames (e.g. c(3,3,3)). The resulting data
frames could look something like this:
auto.data.1
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
auto.data.2
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
5 0 0 0
auto.data.3
X1 X2 X3
1 0 0 0
2 0 0 0
Later, I want to fill the elements of the data frames with values read
from somewhere else, automatically looping through the previously
generated data frames.
I know that I can automatically generate variables with the right number
of elements with something like this:
auto.length - c(12,15,6)
for(i in 1:3) {
+ nam - paste(auto.data,i, sep=.)
+ assign(nam, 1:auto.length[i])
+ }
auto.data.1
[1] 1 2 3 4 5 6 7 8 9 10 11 12
auto.data.2
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
auto.data.3
[1] 1 2 3 4 5 6
But how do I turn these variables into data frames or give them any
dimensions? Any commands such as 'as.matrix', 'data.frame', or 'dim' do
not seem to work. I also seem not to be able to access the variables
with something like auto.data.i since:
auto.data.i
Error: object auto.data.i not found
Thus, how would I be able to automatically write to the elements of the
data frames later in a loop such as ...
for(i in 1:3) {
+ for(j in 1:nrow(auto.data.i)) { ### this obviously does not work
since 'Error in nrow(auto.data.i) : object auto.data.i not found'
+ for(k in 1:ncol(auto.data.i)) {
+ auto.data.i[j,k] - 'some value'
+ }}}
Thanks a bunch for all your help.
Best, Michael
Michael Drescher
Ontario Forest Research Institute
Ontario Ministry of Natural Resources
1235 Queen St East
Sault Ste Marie, ON, P6A 2E3
Tel: (705) 946-7406
Fax: (705) 946-2030
__
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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] JRI problem on 64bit Linux
Sorry if this isn't the correct list, but I couldn't find any mailing lists on the JRI site. I am able to install JRI on a 32 bin linux machine without any problem, but I am unable to isntall on a 64 bit machine. I have R installed with the correct option, R_HOME is set up and libR.so is in there. However, I get the following error when running ./configure, any help is appreciated checking build system type... x86_64-unknown-linux-gnu checking host system type... x86_64-unknown-linux-gnu checking for gcc... gcc checking for C compiler default output file name... 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 checking for egrep... grep -E checking for ANSI C header files... yes checking for java... /usr/java/jdk1.5.0_11/bin/java checking for javac... /usr/java/jdk1.5.0_11/bin/javac checking for javah... /usr/java/jdk1.5.0_11/bin/javah checking for jar... /usr/java/jdk1.5.0_11/bin/jar checking whether Java interpreter works... yes checking for Java environment... in /usr/java/jdk1.5.0_11 checking for /usr/java/jdk1.5.0_11/include/jni.h... yes checking for /usr/java/jdk1.5.0_11/include/./jni_md.h... no checking for /usr/java/jdk1.5.0_11/include/linux/jni_md.h... yes checking whether JNI programs can be compiled... configure: error: Cannot compile a simple JNI program. See config.log for details. __ 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] Compute rank within factor groups
Ken Williams wrote:
Hi,
I have a data.frame which is ordered by score, and has a factor column:
Browse[1] wc[c(report,score)]
report score
9 ADEA 0.96
8 ADEA 0.90
11 Asylum_FED9 0.86
3 ADEA 0.75
14 Asylum_FED9 0.60
5 ADEA 0.56
13 Asylum_FED9 0.51
16 Asylum_FED9 0.51
2 ADEA 0.42
7 ADEA 0.31
17 Asylum_FED9 0.27
1 ADEA 0.17
4 ADEA 0.17
6 ADEA 0.12
10ADEA 0.11
12 Asylum_FED9 0.10
15 Asylum_FED9 0.09
18 Asylum_FED9 0.07
Browse[1]
I need to add a column indicating rank within each factor group, which I
currently accomplish like so:
wc$rank - 0
for(report in as.character(unique(wc$report))) {
wc[wc$report==report,]$rank - 1:sum(wc$report==report)
}
I have to wonder whether there's a better way, something that gets rid of
the for() loop using tapply() or by() or similar. But I haven't come up
with anything.
I've tried these:
by(wc, wc$report, FUN=function(pr){pr$rank - 1:nrow(pr)})
by(wc, wc$report, FUN=function(pr){wc[wc$report %in% pr$report,]$rank -
1:nrow(pr)})
But in both cases the effect of the assignment is lost, there's no $rank
column generated for wc.
Any suggestions?
There's a little known and somewhat unfortunately named function called
ave() which does just that sort of thing.
ave(wc$score, wc$report, FUN=rank)
[1] 10.0 9.0 8.0 8.0 7.0 7.0 5.5 5.5 6.0 5.0 4.0 3.5 3.5
2.0 1.0
[16] 3.0 2.0 1.0
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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] automatically generating and accessing data frames of varying dimensions
Replace
assign(nam, 1:auto.length[i])
with
assign(nam,data.frame(matrix(1:auto.length[i], ncol=3)))
Verify if the matrix (and hence the data frame) is filled the way you
intended. Otherwise, use the argument byrow=T.
Julian M. Burgos
Fisheries Acoustics Research Lab
School of Aquatic and Fishery Science
University of Washington
Drescher, Michael (MNR) wrote:
Hi All,
I want to automatically generate a number of data frames, each with an
automatically generated name and an automatically generated number of
rows. The number of rows has been calculated before and is different for
all data frames (e.g. c(4,5,2)). The number of columns is known a priori
and the same for all data frames (e.g. c(3,3,3)). The resulting data
frames could look something like this:
auto.data.1
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
auto.data.2
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
5 0 0 0
auto.data.3
X1 X2 X3
1 0 0 0
2 0 0 0
Later, I want to fill the elements of the data frames with values read
from somewhere else, automatically looping through the previously
generated data frames.
I know that I can automatically generate variables with the right number
of elements with something like this:
auto.length - c(12,15,6)
for(i in 1:3) {
+ nam - paste(auto.data,i, sep=.)
+ assign(nam, 1:auto.length[i])
+ }
auto.data.1
[1] 1 2 3 4 5 6 7 8 9 10 11 12
auto.data.2
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
auto.data.3
[1] 1 2 3 4 5 6
But how do I turn these variables into data frames or give them any
dimensions? Any commands such as 'as.matrix', 'data.frame', or 'dim' do
not seem to work. I also seem not to be able to access the variables
with something like auto.data.i since:
auto.data.i
Error: object auto.data.i not found
Thus, how would I be able to automatically write to the elements of the
data frames later in a loop such as ...
for(i in 1:3) {
+ for(j in 1:nrow(auto.data.i)) { ### this obviously does not work
since 'Error in nrow(auto.data.i) : object auto.data.i not found'
+ for(k in 1:ncol(auto.data.i)) {
+ auto.data.i[j,k] - 'some value'
+ }}}
Thanks a bunch for all your help.
Best, Michael
Michael Drescher
Ontario Forest Research Institute
Ontario Ministry of Natural Resources
1235 Queen St East
Sault Ste Marie, ON, P6A 2E3
Tel: (705) 946-7406
Fax: (705) 946-2030
__
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
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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] (no subject)
Hi All,
I want to automatically generate a number of data frames, each with an
automatically generated name and an automatically generated number of
rows. The number of rows has been calculated before and is different for
all data frames (e.g. c(4,5,2)). The number of columns is known a priori
and the same for all data frames (e.g. c(3,3,3)). The resulting data
frames could look something like this:
auto.data.1
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
auto.data.2
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
5 0 0 0
auto.data.3
X1 X2 X3
1 0 0 0
2 0 0 0
Later, I want to fill the elements of the data frames with values read
from somewhere else, automatically looping through the previously
generated data frames.
I know that I can automatically generate variables with the right number
of elements with something like this:
auto.length - c(12,15,6)
for(i in 1:3) {
+ nam - paste(auto.data,i, sep=.)
+ assign(nam, 1:auto.length[i])
+ }
auto.data.1
[1] 1 2 3 4 5 6 7 8 9 10 11 12
auto.data.2
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
auto.data.3
[1] 1 2 3 4 5 6
But how do I turn these variables into data frames or give them any
dimensions? Any commands such as 'as.matrix', 'data.frame', or 'dim' do
not seem to work. I also seem not to be able to access the variables
with something like auto.data.i since:
auto.data.i
Error: object auto.data.i not found
Thus, how would I be able to automatically write to the elements of the
data frames later in a loop such as ...
for(i in 1:3) {
+ for(j in 1:nrow(auto.data.i)) { ### this obviously does not work
since 'Error in nrow(auto.data.i) : object auto.data.i not found'
+ for(k in 1:ncol(auto.data.i)) {
+ auto.data.i[j,k] - 'some value'
+ }}}
Thanks a bunch for all your help.
Best, Michael
Michael Drescher
Ontario Forest Research Institute
Ontario Ministry of Natural Resources
1235 Queen St East
Sault Ste Marie, ON, P6A 2E3
Tel: (705) 946-7406
Fax: (705) 946-2030
__
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.null doesn't work
On 7/12/07, Atte Tenkanen [EMAIL PROTECTED] wrote: Seems to work, if I unlist the argument at first ;-) Atte Hi, What's wrong here?: v=c(`-`,`+`,1,`^`,`^`,NA,NA,X,9,X,2) i2=16 v[i2] [[1]] NULL is.null(v[i2]) [1] FALSE Is it a bug or have I misunderstood something? v[2] is a list with a single element which happens to be NULL. v[[2]], on the other hand, is NULL. A subset of a list, obtained with [, is a list. An element of a list, obtained with [[, is the native type of that element. Atte Tenkanen University of Turku, Finland __ 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] sub-function default arguments
Hi.
I have defined a function, f1, that calls another function, f2. Inside f1
an intermediate variable called nm1 is created; it is a matrix. f2 takes a
matrix argument, and I defined f2 (schematically) as follows:
f2-function(nmArg1=nm1,...){nC-ncol(nmArg1); ... }
so that it expects nm1 as the default value of its argument. f1 is defined
(schematically) as:
f1-function(...){result1-f2(); ... }
When I ran f1 I got the following error message:
Error in ncol(nmArg1) : object nm1 not found.
If I redefine f1 schematically as:
f1-function(...){result1-f2(nmArg1=nm1); ... }
it runs okay. If I have nm1 defined outside of f1 and I run result1-f2()
it also runs okay. So f2 doesn't seem to pick up the default argument value
inside the function f1, even when the default argument is defined inside f1.
Is there a way to have the subfunction default arguments recognized inside
of a function (perhaps a better protocol for using defaults than what I
did?), or do I just have to spell them out explicitly?
Thanks!
-- TMK --
212-460-5430home
917-656-5351cell
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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] sub-function default arguments
On 12/07/2007 7:23 PM, Talbot Katz wrote:
Hi.
I have defined a function, f1, that calls another function, f2. Inside f1
an intermediate variable called nm1 is created; it is a matrix. f2 takes a
matrix argument, and I defined f2 (schematically) as follows:
f2-function(nmArg1=nm1,...){nC-ncol(nmArg1); ... }
so that it expects nm1 as the default value of its argument. f1 is defined
(schematically) as:
f1-function(...){result1-f2(); ... }
When I ran f1 I got the following error message:
Error in ncol(nmArg1) : object nm1 not found.
If I redefine f1 schematically as:
f1-function(...){result1-f2(nmArg1=nm1); ... }
it runs okay. If I have nm1 defined outside of f1 and I run result1-f2()
it also runs okay. So f2 doesn't seem to pick up the default argument value
inside the function f1, even when the default argument is defined inside f1.
Is there a way to have the subfunction default arguments recognized inside
of a function (perhaps a better protocol for using defaults than what I
did?), or do I just have to spell them out explicitly?
Defaults to arguments are evaluated in the evaluation frame of the
function being called, f2 in your case. Since nm1 is meaningless within
f2, you get the error.
If you want nm1 to be meaningful within f2, you could define f2 within
f1, e.g.
f1-function(...){
nm1 - something
f2 - function (...) {}
result1-f2(nmArg1=nm1)
...
}
(which is the best way to do it), or you could explicitly manipulate the
environment of f2 (which is an ugly way), or you could store nm1 in some
place that's visible to both f1 and f2 and use - when you set it from
within f1 (another ugly way, but sometimes less ugly than my second
suggestion).
Duncan Murdoch
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Re: [R] (no subject)
Is this what you want to do:
auto.length - c(12,15,6)
for(i in 1:3) {
+ nam - paste(auto.data,i, sep=.)
+ assign(nam, as.data.frame(matrix(1:auto.length[i], ncol=3)))
+ }
auto.data.1
V1 V2 V3
1 1 5 9
2 2 6 10
3 3 7 11
4 4 8 12
auto.data.2
V1 V2 V3
1 1 6 11
2 2 7 12
3 3 8 13
4 4 9 14
5 5 10 15
# output the data
for(i in 1:3){
+ cat(x - paste('auto.data.', i, sep=''), '\n')
+ print(get(x))
+ }
auto.data.1
V1 V2 V3
1 1 5 9
2 2 6 10
3 3 7 11
4 4 8 12
auto.data.2
V1 V2 V3
1 1 6 11
2 2 7 12
3 3 8 13
4 4 9 14
5 5 10 15
auto.data.3
V1 V2 V3
1 1 3 5
2 2 4 6
On 7/12/07, Drescher, Michael (MNR) [EMAIL PROTECTED] wrote:
Hi All,
I want to automatically generate a number of data frames, each with an
automatically generated name and an automatically generated number of
rows. The number of rows has been calculated before and is different for
all data frames (e.g. c(4,5,2)). The number of columns is known a priori
and the same for all data frames (e.g. c(3,3,3)). The resulting data
frames could look something like this:
auto.data.1
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
auto.data.2
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
5 0 0 0
auto.data.3
X1 X2 X3
1 0 0 0
2 0 0 0
Later, I want to fill the elements of the data frames with values read
from somewhere else, automatically looping through the previously
generated data frames.
I know that I can automatically generate variables with the right number
of elements with something like this:
auto.length - c(12,15,6)
for(i in 1:3) {
+ nam - paste(auto.data,i, sep=.)
+ assign(nam, 1:auto.length[i])
+ }
auto.data.1
[1] 1 2 3 4 5 6 7 8 9 10 11 12
auto.data.2
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
auto.data.3
[1] 1 2 3 4 5 6
But how do I turn these variables into data frames or give them any
dimensions? Any commands such as 'as.matrix', 'data.frame', or 'dim' do
not seem to work. I also seem not to be able to access the variables
with something like auto.data.i since:
auto.data.i
Error: object auto.data.i not found
Thus, how would I be able to automatically write to the elements of the
data frames later in a loop such as ...
for(i in 1:3) {
+ for(j in 1:nrow(auto.data.i)) { ### this obviously does not work
since 'Error in nrow(auto.data.i) : object auto.data.i not found'
+ for(k in 1:ncol(auto.data.i)) {
+ auto.data.i[j,k] - 'some value'
+ }}}
Thanks a bunch for all your help.
Best, Michael
Michael Drescher
Ontario Forest Research Institute
Ontario Ministry of Natural Resources
1235 Queen St East
Sault Ste Marie, ON, P6A 2E3
Tel: (705) 946-7406
Fax: (705) 946-2030
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] is.null doesn't work
'v' appears to be a list:
v=c(`-`,`+`,1,`^`,`^`,NA,NA,X,9,X,2)
i2=16
v[i2]
[[1]]
NULL
str(v)
List of 11
$ :function (e1, e2)
$ :function (e1, e2)
$ : num 1
$ :function (e1, e2)
$ :function (e1, e2)
$ : logi NA
$ : logi NA
$ : chr X
$ : num 9
$ : chr X
$ : num 2
because you used backquotes(`) on the '-'; notice the difference:
str(c(`-`,1))
List of 2
$ :function (e1, e2)
$ : num 1
str(c('-',1))
chr [1:2] - 1
On 7/12/07, Atte Tenkanen [EMAIL PROTECTED] wrote:
Hi,
What's wrong here?:
v=c(`-`,`+`,1,`^`,`^`,NA,NA,X,9,X,2)
i2=16
v[i2]
[[1]]
NULL
is.null(v[i2])
[1] FALSE
Is it a bug or have I misunderstood something?
Atte Tenkanen
University of Turku, Finland
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] automatically generating and accessing data frames of varying dimensions
Its desirable to put the data frames into a list so they can be easily
iterated over in the future. Try this:
auto.length - 2:4 # replace with desired row lengths
f - function(names, nr) {
x - rep(0, nr)
data.frame(X1 = x, X2 = x, X3 = x)
}
auto.names - paste(auto.data, seq_along(auto.length), sep = .)
mapply(f, auto.names, auto.length, SIMPLIFY = FALSE)
On 7/12/07, Drescher, Michael (MNR) [EMAIL PROTECTED] wrote:
Hi All,
I want to automatically generate a number of data frames, each with an
automatically generated name and an automatically generated number of
rows. The number of rows has been calculated before and is different for
all data frames (e.g. c(4,5,2)). The number of columns is known a priori
and the same for all data frames (e.g. c(3,3,3)). The resulting data
frames could look something like this:
auto.data.1
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
auto.data.2
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
5 0 0 0
auto.data.3
X1 X2 X3
1 0 0 0
2 0 0 0
Later, I want to fill the elements of the data frames with values read
from somewhere else, automatically looping through the previously
generated data frames.
I know that I can automatically generate variables with the right number
of elements with something like this:
auto.length - c(12,15,6)
for(i in 1:3) {
+ nam - paste(auto.data,i, sep=.)
+ assign(nam, 1:auto.length[i])
+ }
auto.data.1
[1] 1 2 3 4 5 6 7 8 9 10 11 12
auto.data.2
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
auto.data.3
[1] 1 2 3 4 5 6
But how do I turn these variables into data frames or give them any
dimensions? Any commands such as 'as.matrix', 'data.frame', or 'dim' do
not seem to work. I also seem not to be able to access the variables
with something like auto.data.i since:
auto.data.i
Error: object auto.data.i not found
Thus, how would I be able to automatically write to the elements of the
data frames later in a loop such as ...
for(i in 1:3) {
+ for(j in 1:nrow(auto.data.i)) { ### this obviously does not work
since 'Error in nrow(auto.data.i) : object auto.data.i not found'
+ for(k in 1:ncol(auto.data.i)) {
+ auto.data.i[j,k] - 'some value'
+ }}}
Thanks a bunch for all your help.
Best, Michael
Michael Drescher
Ontario Forest Research Institute
Ontario Ministry of Natural Resources
1235 Queen St East
Sault Ste Marie, ON, P6A 2E3
Tel: (705) 946-7406
Fax: (705) 946-2030
__
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.
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and provide commented, minimal, self-contained, reproducible code.
[R] R and HTTP get 'has file changed'
Is there a way, maybe using Duncan TL's RCurl, to efficiently test whether
an URL such as
http://$CRAN/src/contrib/
has changed? I.e. one way is via a test of a page in that directory as per
(sorry about the long line, and this would be on Linux with links and awk
installed)
strptime(system(links -width 160 -dump
http://cran.r-project.org/src/contrib/ | awk '/PACKAGES.html/ {print $3,$4}\',
intern=TRUE), %d-%b-%Y %H:%M)
[1] 2007-07-12 18:16:00
and one can then compare the POSIXt with a cached value --- but requesting
the header would presumably be more efficient.
Is there are way to request the 'has changed' part of the http 1.1 spe
directly in R?
Thanks, Dirk
--
Hell, there are no rules here - we're trying to accomplish something.
-- Thomas A. Edison
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] R and HTTP get 'has file changed'
Hi Dirk,
Dirk Eddelbuettel [EMAIL PROTECTED] writes:
Is there a way, maybe using Duncan TL's RCurl, to efficiently test whether
an URL such as
http://$CRAN/src/contrib/
has changed? I.e. one way is via a test of a page in that directory as per
(sorry about the long line, and this would be on Linux with links and awk
installed)
strptime(system(links -width 160 -dump
http://cran.r-project.org/src/contrib/ | awk '/PACKAGES.html/ {print
$3,$4}\', intern=TRUE), %d-%b-%Y %H:%M)
[1] 2007-07-12 18:16:00
and one can then compare the POSIXt with a cached value --- but requesting
the header would presumably be more efficient.
Is there are way to request the 'has changed' part of the http 1.1 spe
directly in R?
Here's a way to use RCurl obtain HTTP headers:
h - basicTextGatherer()
junk - getURI(url, writeheader=h$update, header=TRUE, nobody=TRUE)
h - h$value()
If you want to check many URLs, I think you will find the following
much faster as opposed to looping the above:
h - multiTextGatherer(urls)
junk - getURIAsynchronous(urls, write=h, header=TRUE, nobody=TRUE)
yourInfo - sapply(h, function(x) something(x$value()))
I've used this in the pkgDepTools package to retrieve package download
sizes.
Cheers,
+ seth
--
Seth Falcon | Computational Biology | Fred Hutchinson Cancer Research Center
http://bioconductor.org
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] R and HTTP get 'has file changed'
On 12 July 2007 at 19:46, Seth Falcon wrote:
| Hi Dirk,
|
| Dirk Eddelbuettel [EMAIL PROTECTED] writes:
| Is there a way, maybe using Duncan TL's RCurl, to efficiently test whether
| an URL such as
|
| http://$CRAN/src/contrib/
|
| has changed? I.e. one way is via a test of a page in that directory as per
| (sorry about the long line, and this would be on Linux with links and awk
| installed)
|
| strptime(system(links -width 160 -dump
http://cran.r-project.org/src/contrib/ | awk '/PACKAGES.html/ {print $3,$4}\',
intern=TRUE), %d-%b-%Y %H:%M)
| [1] 2007-07-12 18:16:00
|
|
| and one can then compare the POSIXt with a cached value --- but requesting
| the header would presumably be more efficient.
|
| Is there are way to request the 'has changed' part of the http 1.1 spe
| directly in R?
|
| Here's a way to use RCurl obtain HTTP headers:
|
| h - basicTextGatherer()
| junk - getURI(url, writeheader=h$update, header=TRUE, nobody=TRUE)
| h - h$value()
Sweet:
library(RCurl)
h - basicTextGatherer()
junk - getURI(http://cran.r-project.org/src/contrib/PACKAGES.html;,
writeheader=h$update, header=TRUE, nobody=TRUE)
h - h$value()
h
[1] HTTP/1.1 200 OK\r\nDate: Fri, 13 Jul 2007 02:58:03 GMT\r\nServer:
Apache/2.2.3 (Debian)\r\nLast-Modified: Thu, 12 Jul 2007 16:16:08 GMT\r\nETag:
\a7c11e-21f34-4fe68200\\r\nAccept-Ranges: bytes\r\nContent-Length:
139060\r\nContent-Type: text/html\r\n\r\n
So I can just filter the Date and Last-Modified fields from here, without
having to worry the particular header request. Nice!
| If you want to check many URLs, I think you will find the following
I don't. I just want something 'light and easy' as the script (to feed
CRANberries) may get run a few times from crontan and should stop early if
no new data will be there to be processed.
Thanks!
Dirk
--
Hell, there are no rules here - we're trying to accomplish something.
-- Thomas A. Edison
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[R] ICC
Hello all. I'm working on a PhD thesis analyzing reproducibility/repeatability of a nuclear medicine examination. I have two sets of data (50 patients each) and 3 observers. In each case (patient), two variables are evaluated, first with a range of 0..100 and second 1..3. Each case is also evaluated twice by each observer (for additional intraobserver reproducibility). I'm using R package psy to calculate ICC (the other method I'm using is repeatability). My question with ICC is: how do I evaluate if the difference between observers (and patient groups) is statistically significant? I have been searching through the R archives, but I haven't been able to find an answer. Thank you, Luka Lezaic -- Nobody is perfect. My name is nobody. - This mail was scanned by BitDefender For more informations please visit http://www.bitdefender.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] imposing constraints on the covariance matrix of random effects in lme4?
Hello all, I am using lme4 to fit some mixed logistic regressions. I need to impose an identification constraint of the following form: (1sig12) (sig12 sig22) and have not figured out how to do it, i.e., sig11 = 1 but the rest of the parameters are free to vary. Is this possible and, if so, how? I've been looking through the archive and help to no avail, but perhaps I'm just missing something. Thanks for any help, Jay -- JVVerkuilen. PhD [EMAIL PROTECTED] If you've been playing poker for half an hour and you still don't know who the patsy is, you're the patsy. --Warren Buffett __ 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] automatically generating and accessing data frames of varyingdimensions
I'm not sure why you want to do it this way (it would probably help if
we had a more complete picture of what you were really trying to do, but
here are a few possibilities for the questions you ask.
1. generating data frames.
rw - c(4,5,2)
cl - c(3,3,3)
for(i in 1:length(rw))
assign(paste(auto.data, i, sep=.),
as.data.frame(array(0, dim=c(rw[i], cl[i]),
dimnames = list(NULL, paste(X, 1:cl[i],
sep=)
check:
auto.data.1
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
auto.data.2
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
5 0 0 0
auto.data.3
X1 X2 X3
1 0 0 0
2 0 0 0
2. filling them up (... are you sure you want to do it this way?)
The simplest way is probably through an intermediary
for(nam in paste(auto.data, 1:3, sep=.)) { # loop over the names
tmp - get(nam)
for(i in 1:nrow(tmp))
for(j in 1:ncol(tmp))
tmp[i, j] - i+j-i*j # 'some value'
assign(nam, tmp)
rm(tmp)
}
check:
auto.data.1
X1 X2 X3
1 1 1 1
2 1 0 -1
3 1 -1 -3
4 1 -2 -5
auto.data.2
X1 X2 X3
1 1 1 1
2 1 0 -1
3 1 -1 -3
4 1 -2 -5
5 1 -3 -7
auto.data.3
X1 X2 X3
1 1 1 1
2 1 0 -1
It may work, but I have to say, though, I'm almost sure this is a
mistake. There has to be a better way using the facilities that R
provides for avoiding heavy loops like this.
Just a hunch...
Bill Venables
CSIRO Laboratories
PO Box 120, Cleveland, 4163
AUSTRALIA
Office Phone (email preferred): +61 7 3826 7251
Fax (if absolutely necessary): +61 7 3826 7304
Mobile: +61 4 8819 4402
Home Phone: +61 7 3286 7700
mailto:[EMAIL PROTECTED]
http://www.cmis.csiro.au/bill.venables/
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Drescher, Michael
(MNR)
Sent: Friday, 13 July 2007 9:19 AM
To: r-help@stat.math.ethz.ch
Subject: [R] automatically generating and accessing data frames of
varyingdimensions
Hi All,
I want to automatically generate a number of data frames, each with an
automatically generated name and an automatically generated number of
rows. The number of rows has been calculated before and is different for
all data frames (e.g. c(4,5,2)). The number of columns is known a priori
and the same for all data frames (e.g. c(3,3,3)). The resulting data
frames could look something like this:
auto.data.1
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
auto.data.2
X1 X2 X3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
5 0 0 0
auto.data.3
X1 X2 X3
1 0 0 0
2 0 0 0
Later, I want to fill the elements of the data frames with values read
from somewhere else, automatically looping through the previously
generated data frames.
I know that I can automatically generate variables with the right number
of elements with something like this:
auto.length - c(12,15,6)
for(i in 1:3) {
+ nam - paste(auto.data,i, sep=.)
+ assign(nam, 1:auto.length[i])
+ }
auto.data.1
[1] 1 2 3 4 5 6 7 8 9 10 11 12
auto.data.2
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
auto.data.3
[1] 1 2 3 4 5 6
But how do I turn these variables into data frames or give them any
dimensions? Any commands such as 'as.matrix', 'data.frame', or 'dim' do
not seem to work. I also seem not to be able to access the variables
with something like auto.data.i since:
auto.data.i
Error: object auto.data.i not found
Thus, how would I be able to automatically write to the elements of the
data frames later in a loop such as ...
for(i in 1:3) {
+ for(j in 1:nrow(auto.data.i)) { ### this obviously does not work
since 'Error in nrow(auto.data.i) : object auto.data.i not found'
+ for(k in 1:ncol(auto.data.i)) {
+ auto.data.i[j,k] - 'some value'
+ }}}
Thanks a bunch for all your help.
Best, Michael
Michael Drescher
Ontario Forest Research Institute
Ontario Ministry of Natural Resources
1235 Queen St East
Sault Ste Marie, ON, P6A 2E3
Tel: (705) 946-7406
Fax: (705) 946-2030
__
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.
__
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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] legend and x,y cordinate values
Hi, I have two problems in R. 1. I need 10 cdfs on a graph, the graph needs to have legend. Can you let me know how to get legend on the graph? 2. In ecdf plot, I need to know the x and y co-ordinates. I have to get corresponding y coordinate values to x coordinate value so that I could be able to know the particular percentile value to the x-coordinate value. Can you let me know how could I be able the corresponding values of x to the y coordinates? Thanking you, Looking forward to your kind response, Sincerely, Ajay. -- Ajay Singh Research Scientist, SOM, IIT-Bombay, Powai, MUMBAI-400076, MH (INDIA). __ 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] JRI problem on 64bit Linux
Our experience (and that of several others) is that you need Java 1.6 on x86_64 Linux for any of the JNI-using R packages to work correctly. E.g. the CRAN check machine is using 1.6. On Thu, 12 Jul 2007, chuckanut wrote: Sorry if this isn't the correct list, but I couldn't find any mailing lists on the JRI site. Please see the posting guide: that asks you to ask the maintainer (who knows about the problems). I am able to install JRI on a 32 bin linux machine without any problem, but I am unable to isntall on a 64 bit machine. I have R installed with the correct option, R_HOME is set up and libR.so is in there. However, I get the following error when running ./configure, any help is appreciated You need to read config.log: we can't read it for you as it is on your machine. checking build system type... x86_64-unknown-linux-gnu checking host system type... x86_64-unknown-linux-gnu checking for gcc... gcc checking for C compiler default output file name... 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 checking for egrep... grep -E checking for ANSI C header files... yes checking for java... /usr/java/jdk1.5.0_11/bin/java checking for javac... /usr/java/jdk1.5.0_11/bin/javac checking for javah... /usr/java/jdk1.5.0_11/bin/javah checking for jar... /usr/java/jdk1.5.0_11/bin/jar checking whether Java interpreter works... yes checking for Java environment... in /usr/java/jdk1.5.0_11 checking for /usr/java/jdk1.5.0_11/include/jni.h... yes checking for /usr/java/jdk1.5.0_11/include/./jni_md.h... no checking for /usr/java/jdk1.5.0_11/include/linux/jni_md.h... yes checking whether JNI programs can be compiled... configure: error: Cannot compile a simple JNI program. See config.log for details. ^^ -- 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.
