Re: [R] Binary Search
?findInterval On 8/8/07, Matthew Walker [EMAIL PROTECTED] wrote: Hi! R is an amazing piece of software and with so many libraries it can do almost anything... but I was very surprised that a standard binary search function seems not to exist. I can find other much more highly-complex search routines (optimize, uniroot, nlm) in the standard no-extra-packages-loaded version of R, but not this simple alternative. I searched and found an implementation inside the package genetics, but that requires the loading of an awful lot of other code to get to this simple function. Have I missed something? Perhaps I just didn't find what is already in there. Can anyone point me to what I'm after? If not, perhaps the developers might consider making it more readily available? Cheers, Matthew __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Felix Andrews / 安福立 PhD candidate Integrated Catchment Assessment and Management Centre 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.
Re: [R] XL files
Hi Animesh, Can you send an example of an Excel file you need to process (and the result you wish to get)? Regards, Moshe. --- Acharjee, Animesh [EMAIL PROTECTED] wrote: Dear All, I am new to R. I need to read XLs files, parse the data and convert them into txt format for further calculation . If any one have code for that , please send me code or suggestions for the same.Waiting for reply. Thanking you Animesh [[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-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 include bar values in a barplot?
On Wednesday 08 August 2007 00:40:56 Donatas G. wrote: On Tuesday 07 August 2007 22:09:52 Donatas G. wrote: How do I include bar values in a barplot (or other R graphics, where this could be applicable)? To make sure I am clear I am attaching a barplot created with OpenOffice.org which has barplot values written on top of each barplot. After more than two hours search I finally found a solution: http://tolstoy.newcastle.edu.au/R/help/06/05/27286.html Hey, the solution happens to be only partiall... If the values are not real numbers, and have a lot of digits after the dot, the graph might become unreadable... see this vals - c(1,1.1236886,4.77554676,5.3345245,1,1.1236886,4.77554676,5.3345245,5.5345245,5.4345245,1.1236886,4.77554676,5.3345245,1.1236886,4.77554676,5.3345245) names(vals) - LETTERS[1:16] mp - barplot(vals, ylim = c(0, 6)) text(mp, vals, labels = vals, pos = 3) Is there any way to round up those numbers? I tried using options(digits=2) , and it does change the display of a table, but it does not influence the barplot... -- Donatas Glodenis http://dg.lapas.info __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Creating netcdf from table
Hi! I have a table with 3 columns, 2 for long/lat coordinates and 1 for values (radon concentration). I have Tables for every week of the year 2006, always same coordinates but other values, i.e. time series. Now I want to create a NetCDF file for, which is much easier to handle than 52 tables. I tried the following for the first dataset (i.e first week): W01-read.table(RN_weekly/KW01_RN.dat) long-W01$V1 lat-W01$V2 RN01-W01$V3 dim1 - dim.def.ncdf( EW,degrees, as.double(long)) dim2 - dim.def.ncdf( SN,degrees, as.double(lat)) varz - var.def.ncdf(Radon,Bq/m2/h1, list(dim1,dim2), -1, longname=Radon flux rate) nc.rn - create.ncdf(rn_weekly.nc,varz) put.var.ncdf(nc.rn,varz,RN01) close.ncdf(nc.rn) The problem now is that the last step (put.var.ncdf) doesn't work, because it says I am trying to error: you asked to write 111788329 values, but the passed data array only has 10573 entries!. So I think the problem is I need an array with two dimensions (coordinates...) for my values. But do I get this from my tables?? Thanks for any help! Thomas My problem now is that __ Thomas Szegvary Institute of Environmental Geosciences Department of Geosciences University of Basel Bernoullistrasse 30 CH - 4056 Basel Tel. 41-61-267 04 82 Fax. 41-61-267 04 79 Email: [EMAIL PROTECTED] www.radon.unibas.ch www.unibas.ch/environment __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 include bar values in a barplot?
Please see ?format ?round Note that text() is said to expect a character vector, so why did you supply a numeric vector? labels: a character vector or expression specifying the _text_ to be written. An attempt is made to coerce other language objects (names and calls) to expressions, and vectors and other classed objects to character vectors by 'as.character'. If 'labels' is longer than 'x' and 'y', the coordinates are recycled to the length of 'labels'. and try as.character(vals) for yourself. Is there any way to round up those numbers? See library(fortunes); fortune(Yoda) On Wed, 8 Aug 2007, Donatas G. wrote: On Wednesday 08 August 2007 00:40:56 Donatas G. wrote: On Tuesday 07 August 2007 22:09:52 Donatas G. wrote: How do I include bar values in a barplot (or other R graphics, where this could be applicable)? To make sure I am clear I am attaching a barplot created with OpenOffice.org which has barplot values written on top of each barplot. After more than two hours search I finally found a solution: http://tolstoy.newcastle.edu.au/R/help/06/05/27286.html Hey, the solution happens to be only partiall... If the values are not real numbers, and have a lot of digits after the dot, the graph might become unreadable... see this vals - c(1,1.1236886,4.77554676,5.3345245,1,1.1236886,4.77554676,5.3345245,5.5345245,5.4345245,1.1236886,4.77554676,5.3345245,1.1236886,4.77554676,5.3345245) names(vals) - LETTERS[1:16] mp - barplot(vals, ylim = c(0, 6)) text(mp, vals, labels = vals, pos = 3) Is there any way to round up those numbers? I tried using options(digits=2) , and it does change the display of a table, but it does not influence the barplot... Well, it does not affect as.character, nor should it. -- 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.
[R] Reading data from a string
Hi, I'am looking for ways in R to read in data from a character string in my wiki. Currently I see two possibilities: 1.) write my own routine 2.) using read.table(textConnection(mydata)) Of course, I want to avoid writing my own routine and textConnection is a forbidden command by the R-PHP people (and I think they had good reasons for this). Does anybody know other possibilities? Thanks in advance Sigbert Klinke __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Find out the workspace name
R-help, Sometimes there might be several R sessions open at the same time. In Windows no name appears in the R main bar (just R Console) Is it possible to know the name of the workspace. I ussually write it on the script I am working on but I wish to know without having to search in a text file. Thanks in advance __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Odp: Error in as.double.default(x) : (list) object cannot be coerced to 'double'
Hi Far from beeing an expert I can decide from your post that you probably mix R objects matrix x data frame x list. All of them have specific features and you sometimes cannot convert one to another easily. See what your objects are and how they look like by e.g. str(X1) or mode or typeof I suspect that your X are lists and using c(some lists) result again into list. Plot expect x and y values usually as vectors but there are plot methods for other objects (data.frames, functions, ...) Best starting point would be help page for plot. Regards Petr [EMAIL PROTECTED] napsal dne 07.08.2007 15:33:41: Dear experts, I have in all 14 matrices which stands for gene expression divergence and 14 matrices which stands for gene sequence divergence. I have tried joining them by using the concatanation function giving SequenceDivergence - c(X1,X2,X3,X4,X5,X6,X7,X8,X9,X10,X11,X12,X13,X14) ExpressionDivergence - c(Y1,Y2,Y3,Y4,Y5,Y6,Y7,Y8,Y9,Y10,Y11,Y12,Y13,Y14) where X1,X2..X14 are the expression matrices containing r-values and Y1,Y2..Y14 are the ones with patristic distances Now, I want to plot SequenceDivergence vs. Expression Divergence Tried doing that using plot (Sequence Divergence vs. Expression Divergence) But then getting the error Error in as.double.default(x) : (list) object cannot be coerced to 'double' Note: The diagonal values for X1 was neglected as its seems to give some bias in my results. Code to derive X1,X2,X3..and Y1,Y2,Y3 is given as: EDant - read.table(C:/ant.txt,sep=\t) ED1 - as.matrix(EDant) X1 - lapply(1:ncol(ED15), function(a) ED15[-a, a]) # to neglect diagonal values I am attaching couple of matrices for your reference. ANT.txt is X1 after deleting the diagnol values, similarly,ANTEXP.TXT is Y1, apetella.txt is X2 and apetellaexp.txt is Y2 I tried reading R-manual and other sources but have'nt cracked it yet. Can anyone please help me regarding that? Thanks very much Yours sincerely, Urmi Trivedi. - Once upon a time there was 1 GB storage in your inbox. Click here for happy ending.0 1.21133 1.420221 1.499358 1.475149 1.513886 1.608816 1. 675916 1.859038 1.992958 2.420123 2.309414 2.413222 1.21133 0 1.463585 1.542722 1.518513 1.55725 1.65218 1.71928 1.902402 2.036322 2.463487 2.352778 2.456586 1.420221 1.463585 0 1.418619 1.39441 1.433147 1.528077 1.595177 1.778299 1.912219 2.339384 2.228675 2.332483 1.499358 1.542722 1.418619 0 0.931235 0.969972 1.354498 1.421598 1.60472 1.73864 2.165805 2.055096 2.158904 1.475149 1.518513 1.39441 0.931235 0 0.206013 1.330289 1.397389 1.580511 1.714431 2.141596 2.030887 2.134695 1.513886 1.55725 1.433147 0.969972 0.206013 0 1.369026 1.436126 1.619248 1.753168 2.180333 2.069624 2.173432 1.608816 1.65218 1.528077 1.354498 1.330289 1.369026 0 0.604972 1.537696 1.671616 2.098781 1.988072 2.09188 1.675916 1.71928 1.595177 1.421598 1.397389 1.436126 0.604972 0 1.604796 1.738716 2.165881 2.055172 2.15898 1.859038 1.902402 1.778299 1.60472 1.580511 1.619248 1.537696 1. 604796 0 1.09121 1.968427 1.857718 1.961526 1.992958 2.036322 1.912219 1.73864 1.714431 1.753168 1.671616 1. 738716 1.09121 0 2.102347 1.991638 2.095446 2.420123 2.463487 2.339384 2.165805 2.141596 2.180333 2.098781 2.165881 1.968427 2.102347 0 1.128701 1.232509 2.309414 2.352778 2.228675 2.055096 2.030887 2.069624 1.988072 2.055172 1.857718 1.991638 1.128701 0 0.91546 2.413222 2.456586 2.332483 2.158904 2.134695 2.173432 2.09188 2. 15898 1.961526 2.095446 1.232509 0.91546 0 1 -0.046373 -0.033044 0.003902 0.466827 0.389327 0.131989 -0. 063346 -0.118093 0.016386 -0.140215 0.34511 0.233074 -0.046373 1 0.000717 0.832227 0.077053 -0.106815 -0.050838 0. 936431 0.297387 0.031358 -0.127265 0.31245 -0.053169 -0.033044 0.000717 1 0.011265 -0.018602 -0.023669 -0.028744 -0. 007588 0.148827 -0.007067 0.005362 -0.029172 -0.017367 0.003902 0.832227 0.011265 1 0.071247 -0.101225 0.097117 0. 829167 0.294396 0.017809 -0.129199 0.278831 -0.106427 0.466827 0.077053 -0.018602 0.071247 1 0.078241 0.140679 0. 070069 0.055744 0.013194 -0.196092 -0.027692 0.028857 0.389327 -0.106815 -0.023669 -0.101225 0.078241 1 0.140146 -0. 086806 0.079 0.194904 0.068378 0.213288 0.194006 0.131989 -0.050838 -0.028744 0.097117 0.140679 0.140146 1 -0. 075079 0.137652 0.065127 -0.072768 -0.120751 -0.101777 -0.063346 0.936431 -0.007588 0.829167 0.070069 -0.086806 -0.075079 1 0.312154 0.023108 -0.122502
[R] simulation-binomial
hello, i want to do a binomial simulation, by taking 200 var. from one group (x) and 300 from another (y). the prob. for disease=.6 in both groups. x - rbinom(200, 1, .6) y - rbinom(300, 1, .6) if the person is from group x - the probability to find the disease, assuming the person is sick, is .95, if he is from group Y its .80. i want to know the joint probability: p(the person has the disease and tested sick)=P(D+,T+). my problem is how to write the conditional prob. Thanks for your help, also reference on this subject (binomial simulation) would be great. Sigalit. [[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] test for contingency table when there are many zeros
Here is my table tt A B 1 297 398 2 470 376 3 30 23 4 3 3 5 0 0 b/c two cells are zero, I can't use chisq.test() in R which gives the following output; chisq.test(tt) Pearson's Chi-squared test data: tt X-squared = NaN, df = 4, p-value = NA Warning message: Chi-squared approximation may be incorrect in: chisq.test(tt) What function should I use then? Any suggestion? [[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] Find out the workspace name
?getwd() ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be Do not put your faith in what statistics say until you have carefully considered what they do not say. ~William W. Watt A statistical analysis, properly conducted, is a delicate dissection of uncertainties, a surgery of suppositions. ~M.J.Moroney -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens Luis Ridao Cruz Verzonden: woensdag 8 augustus 2007 11:39 Aan: r-help@stat.math.ethz.ch Onderwerp: [R] Find out the workspace name R-help, Sometimes there might be several R sessions open at the same time. In Windows no name appears in the R main bar (just R Console) Is it possible to know the name of the workspace. I ussually write it on the script I am working on but I wish to know without having to search in a text file. Thanks in advance __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Find out the workspace name
On Wed, 8 Aug 2007, ONKELINX, Thierry wrote: ?getwd() and ?setWindowTitle, which even has this as the first example. help.search(window title) gets you there. [mailto:[EMAIL PROTECTED] Namens Luis Ridao Cruz Sometimes there might be several R sessions open at the same time. In Windows no name appears in the R main bar (just R Console) Is it possible to know the name of the workspace. I ussually write it on the script I am working on but I wish to know without having to search in a text file. -- 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.
[R] Namespace problem
Hi All, I have some questions about making a R-package. I would like to use a namespace. The package contains analysis functions and also a graphical user interface. I would like to allow user to use only the analysis function and not the GUI functions which are mainly bindings to graphical elements. I have exported analysis functions using export directives in a namespace file but when I want to use the GUI, it seems that all the GUI functions are unknown. Is there a way to solve this problem ? thx a lot __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] saving output
Lynn Disney wrote:
I have a question about how to save the output of a logistic regression
model in some format (delimited, preferably) so that I can manipulate it
to make nice tables in excel. I have tried print, save, write, none seem
to do what I want them to. Does anyone have any ideas?
Hi Lynn,
This is an interesting idea that might be useful in the prettyR package.
You have just volunteered to be a test pilot.
delim.table-function(x,con=,delim=\t) {
if(nchar(con)) {
con-file(con,w)
open.con-TRUE
}
else open.con-FALSE
column.names-names(x)
if(is.null(column.names)) column.names-colnames(x)
have.col.names-!is.null(column.names)
row.names-rownames(x)
have.row.names-!is.null(row.names)
xdim-dim(x)
if(have.col.names) {
cat(delim,file=con)
for(col in 1:xdim[2])
cat(column.names[col],delim,sep=,file=con)
cat(\n,file=con)
}
for(row in 1:xdim[1]) {
if(have.row.names) cat(row.names[row],file=con)
cat(delim,file=con)
for(col in 1:xdim[2]) cat(x[row,col],delim,sep=,file=con)
cat(\n,file=con)
}
if(open.con) close(con)
}
test.df-data.frame(a=sample(0:1,100,TRUE),b=rnorm(100),c=rnorm(100))
delim.table(
summary(glm(a~b*c,test.df,family=binomial))$coefficients,
con=test.csv)
This output should import into Excel with no trouble (it does in
OpenOffice Calc). Either use sink() or specify a filename as in the example.
Jim
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Re: [R] Interaction factor and numeric variable versus separate
Thanks to all for the very helpful replies the reference to a chapter
in MASS!
Sven
On Tue, 2007-08-07 at 12:07 -0400, Gabor Grothendieck wrote:
Also check this post
https://stat.ethz.ch/pipermail/r-help/2007-May/132866.html
for a number of formulations.
On 8/7/07, Ted Harding [EMAIL PROTECTED] wrote:
On 07-Aug-07 15:34:13, Gabor Grothendieck wrote:
In the single model all three levels share the same intercept which
means that the slope must change to accomodate it
whereas in the three separate models they each have their own
intercept.
I think this arose because of the formulation of the model with
interaction as:
summary(lm(y~x:f, data=d))
If it has been formulated as
summary(lm(y~x*f, data=d))
there would be three separate intercepts, and three different slopes
(and the differences would be the same as the differences for the
separate models).
Ted.
Try looking at it graphically and note how the black dotted lines
are all forced to go through the same intercept, i.e. the same point
on the y axis, whereas the red dashed lines are each able to
fit their portion of the data using both the intercept and the slope.
y.lm - lm(y~x:f, data=d)
plot(y ~ x, d, col = as.numeric(d$f), xlim = c(-5, 20))
for(i in 1:3) {
abline(a = coef(y.lm)[1], b = coef(y.lm)[1+i], lty = dotted)
abline(lm(y ~ x, d[as.numeric(d$f) == i,]), col = red, lty =
dashed)
}
grid()
On 8/7/07, Sven Garbade [EMAIL PROTECTED] wrote:
Dear list members,
I have problems to interpret the coefficients from a lm model
involving
the interaction of a numeric and factor variable compared to separate
lm
models for each level of the factor variable.
## data:
y1 - rnorm(20) + 6.8
y2 - rnorm(20) + (1:20*1.7 + 1)
y3 - rnorm(20) + (1:20*6.7 + 3.7)
y - c(y1,y2,y3)
x - rep(1:20,3)
f - gl(3,20, labels=paste(lev, 1:3, sep=))
d - data.frame(x=x,y=y, f=f)
## plot
# xyplot(y~x|f)
## lm model with interaction
summary(lm(y~x:f, data=d))
Call:
lm(formula = y ~ x:f, data = d)
Residuals:
Min 1Q Median 3Q Max
-2.8109 -0.8302 0.2542 0.6737 3.5383
Coefficients:
Estimate Std. Error t value Pr(|t|)
(Intercept) 3.687990.41045 8.985 1.91e-12 ***
x:flev1 0.208850.04145 5.039 5.21e-06 ***
x:flev2 1.496700.04145 36.109 2e-16 ***
x:flev3 6.708150.04145 161.838 2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.53 on 56 degrees of freedom
Multiple R-Squared: 0.9984, Adjusted R-squared: 0.9984
F-statistic: 1.191e+04 on 3 and 56 DF, p-value: 2.2e-16
## separate lm fits
lapply(by(d, d$f, function(x) lm(y ~ x, data=x)), coef)
$lev1
(Intercept) x
6.77022860 -0.01667528
$lev2
(Intercept) x
1.0190781.691982
$lev3
(Intercept) x
3.2746566.738396
Can anybody give me a hint why the coefficients for the slopes
(especially for lev1) are so different and how the coefficients from
the
lm model with interaction are related to the separate fits?
Thanks, Sven
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[R] tapply grand mean
Hi R-users, I have a data.frame like this (modificated from https://stat.ethz.ch/pipermail/r-help/2007-August/138124.html). y1 - rnorm(20) + 6.8 y2 - rnorm(20) + (1:20*1.7 + 1) y3 - rnorm(20) + (1:20*6.7 + 3.7) y - c(y1,y2,y3) x - rep(1:5,12) f - gl(3,20, labels=paste(lev, 1:3, sep=)) d - data.frame(x=x,y=y, f=f) and this is how I can calculate mean of these levels. tapply(d$y, list(d$x, d$f), mean) But how can I calculate the mean of d$x 1 and 2 and the grand mean of d$x 1, 2, 3, 4, 5 (within d$f) into a table? Regards, Lauri [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] tapply grand mean
Lauri Nikkinen wrote: Hi R-users, I have a data.frame like this (modificated from https://stat.ethz.ch/pipermail/r-help/2007-August/138124.html). y1 - rnorm(20) + 6.8 y2 - rnorm(20) + (1:20*1.7 + 1) y3 - rnorm(20) + (1:20*6.7 + 3.7) y - c(y1,y2,y3) x - rep(1:5,12) f - gl(3,20, labels=paste(lev, 1:3, sep=)) d - data.frame(x=x,y=y, f=f) and this is how I can calculate mean of these levels. tapply(d$y, list(d$x, d$f), mean) But how can I calculate the mean of d$x 1 and 2 and the grand mean of d$x 1, 2, 3, 4, 5 (within d$f) into a table? You might like the tables produced by summary.formula() in the Hmisc package: library(Hmisc) summary(y ~ x + f, data = d, fun=mean, method=cross, overall=TRUE) UseMethod by x, f +-+ |N| |y| +-+ +---+-+-+-+-+ | x | lev1 | lev2 | lev3 | ALL | +---+-+-+-+-+ |1 | 4 | 4 | 4 |12 | | | 6.452326|15.861256|61.393455|27.902346| +---+-+-+-+-+ |2 | 4 | 4 | 4 |12 | | | 7.403041|17.296270|68.208299|30.969203| +---+-+-+-+-+ |3 | 4 | 4 | 4 |12 | | | 6.117648|17.976864|73.479837|32.524783| +---+-+-+-+-+ |4 | 4 | 4 | 4 |12 | | | 7.831390|19.696998|80.323382|35.950590| +---+-+-+-+-+ |5 | 4 | 4 | 4 |12 | | | 6.746213|21.101952|87.430087|38.426084| +---+-+-+-+-+ |ALL|20 |20 |20 |60 | | | 6.910124|18.386668|74.167012|33.154601| +---+-+-+-+-+ summary(y ~ I(x %in% c(1,2)) + f, data = d, fun=mean, method=cross, overall=TRUE) UseMethod by I(x %in% c(1, 2)), f +-+ |N| |y| +-+ +-+-+-+-+-+ |I(x %in% c(1, 2))| lev1 | lev2 | lev3 | ALL | +-+-+-+-+-+ | FALSE |12 |12 |12 |36 | | | 6.898417|19.591938|80.411102|35.633819| +-+-+-+-+-+ | TRUE | 8 | 8 | 8 |24 | | | 6.927684|16.578763|64.800877|29.435774| +-+-+-+-+-+ | ALL|20 |20 |20 |60 | | | 6.910124|18.386668|74.167012|33.154601| +-+-+-+-+-+ Regards, Lauri [[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. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] tapply grand mean
Thanks Chuck but I would fancy the output made by tapply because the idea is to make a barplot based on those values. -Lauri 2007/8/8, Chuck Cleland [EMAIL PROTECTED]: Lauri Nikkinen wrote: Hi R-users, I have a data.frame like this (modificated from https://stat.ethz.ch/pipermail/r-help/2007-August/138124.html). y1 - rnorm(20) + 6.8 y2 - rnorm(20) + (1:20*1.7 + 1) y3 - rnorm(20) + (1:20*6.7 + 3.7) y - c(y1,y2,y3) x - rep(1:5,12) f - gl(3,20, labels=paste(lev, 1:3, sep=)) d - data.frame(x=x,y=y, f=f) and this is how I can calculate mean of these levels. tapply(d$y, list(d$x, d$f), mean) But how can I calculate the mean of d$x 1 and 2 and the grand mean of d$x 1, 2, 3, 4, 5 (within d$f) into a table? You might like the tables produced by summary.formula() in the Hmisc package: library(Hmisc) summary(y ~ x + f, data = d, fun=mean, method=cross, overall=TRUE) UseMethod by x, f +-+ |N| |y| +-+ +---+-+-+-+-+ | x | lev1 | lev2 | lev3 | ALL | +---+-+-+-+-+ |1 | 4 | 4 | 4 |12 | | | 6.452326|15.861256|61.393455|27.902346| +---+-+-+-+-+ |2 | 4 | 4 | 4 |12 | | | 7.403041|17.296270|68.208299|30.969203| +---+-+-+-+-+ |3 | 4 | 4 | 4 |12 | | | 6.117648|17.976864|73.479837|32.524783| +---+-+-+-+-+ |4 | 4 | 4 | 4 |12 | | | 7.831390|19.696998|80.323382|35.950590| +---+-+-+-+-+ |5 | 4 | 4 | 4 |12 | | | 6.746213|21.101952|87.430087|38.426084| +---+-+-+-+-+ |ALL|20 |20 |20 |60 | | | 6.910124|18.386668|74.167012|33.154601| +---+-+-+-+-+ summary(y ~ I(x %in% c(1,2)) + f, data = d, fun=mean, method=cross, overall=TRUE) UseMethod by I(x %in% c(1, 2)), f +-+ |N| |y| +-+ +-+-+-+-+-+ |I(x %in% c(1, 2))| lev1 | lev2 | lev3 | ALL | +-+-+-+-+-+ | FALSE |12 |12 |12 |36 | | | 6.898417|19.591938|80.411102|35.633819| +-+-+-+-+-+ | TRUE | 8 | 8 | 8 |24 | | | 6.927684|16.578763|64.800877|29.435774| +-+-+-+-+-+ | ALL|20 |20 |20 |60 | | | 6.910124|18.386668|74.167012|33.154601| +-+-+-+-+-+ Regards, Lauri [[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. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 [[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] tapply grand mean
Lauri Nikkinen wrote: Thanks Chuck but I would fancy the output made by tapply because the idea is to make a barplot based on those values. -Lauri sum1 - summary(y ~ x + f, data = d, fun=mean, method=cross, overall=TRUE) df - data.frame(x = sum1$x, f = sum1$f, y = sum1$S) df xf y 11 lev1 6.452326 22 lev1 7.403041 33 lev1 6.117648 44 lev1 7.831390 55 lev1 6.746213 6 ALL lev1 6.910124 71 lev2 15.861256 82 lev2 17.296270 93 lev2 17.976864 10 4 lev2 19.696998 11 5 lev2 21.101952 12 ALL lev2 18.386668 13 1 lev3 61.393455 14 2 lev3 68.208299 15 3 lev3 73.479837 16 4 lev3 80.323382 17 5 lev3 87.430087 18 ALL lev3 74.167012 19 1 ALL 27.902346 20 2 ALL 30.969203 21 3 ALL 32.524783 22 4 ALL 35.950590 23 5 ALL 38.426084 24 ALL ALL 33.154601 library(lattice) barchart(y ~ x | f, data = df, layout=c(4,1,1)) OR barchart(S ~ x | f, data = sum1, layout=c(4,1,1)) 2007/8/8, Chuck Cleland [EMAIL PROTECTED] mailto:[EMAIL PROTECTED]: Lauri Nikkinen wrote: Hi R-users, I have a data.frame like this (modificated from https://stat.ethz.ch/pipermail/r-help/2007-August/138124.html). y1 - rnorm(20) + 6.8 y2 - rnorm(20) + (1:20* 1.7 + 1) y3 - rnorm(20) + (1:20*6.7 + 3.7) y - c(y1,y2,y3) x - rep(1:5,12) f - gl(3,20, labels=paste(lev, 1:3, sep=)) d - data.frame(x=x,y=y, f=f) and this is how I can calculate mean of these levels. tapply(d$y, list(d$x, d$f), mean) But how can I calculate the mean of d$x 1 and 2 and the grand mean of d$x 1, 2, 3, 4, 5 (within d$f) into a table? You might like the tables produced by summary.formula() in the Hmisc package: library(Hmisc) summary(y ~ x + f, data = d, fun=mean, method=cross, overall=TRUE) UseMethod by x, f +-+ |N| |y| +-+ +---+-+-+-+-+ | x | lev1 | lev2 | lev3 | ALL | +---+-+-+-+-+ |1 | 4 | 4 | 4 |12 | | | 6.452326|15.861256|61.393455|27.902346| +---+-+-+-+-+ |2 | 4 | 4 | 4 |12 | | | 7.403041|17.296270|68.208299|30.969203| +---+-+-+-+-+ |3 | 4 | 4 | 4 |12 | | | 6.117648|17.976864|73.479837|32.524783| +---+-+-+-+-+ |4 | 4 | 4 | 4 |12 | | | 7.831390|19.696998|80.323382|35.950590| +---+-+-+-+-+ |5 | 4 | 4 | 4 |12 | | | 6.746213|21.101952|87.430087|38.426084| +---+-+-+-+-+ |ALL|20 |20 |20 |60 | | | 6.910124|18.386668|74.167012|33.154601| +---+-+-+-+-+ summary(y ~ I(x %in% c(1,2)) + f, data = d, fun=mean, method=cross, overall=TRUE) UseMethod by I(x %in% c(1, 2)), f +-+ |N| |y| +-+ +-+-+-+-+-+ |I(x %in% c(1, 2))| lev1 | lev2 | lev3 | ALL | +-+-+-+-+-+ | FALSE |12 |12 |12 |36 | | | 6.898417|19.591938|80.411102|35.633819| +-+-+-+-+-+ | TRUE | 8 | 8 | 8 |24 | | | 6.927684|16.578763|64.800877|29.435774| +-+-+-+-+-+ | ALL|20 |20 |20 |60 | | | 6.910124|18.386668|74.167012|33.154601| +-+-+-+-+-+ Regards, Lauri [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailto:R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
Re: [R] Changing font in boxplots
I don't know if boxplot will accept a font argument.m From ?boxplot it is not clear. You may need to set the par() command before the boxplot Example: par(font.lab=4) boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family,las=1, cex.axis=1) --- G Iossa, School Biological Sciences [EMAIL PROTECTED] wrote: Hi all, I am very new to R and this might be a simple question but I have looked everywhere you suggest before writing to you. I am trying to change font type from san-serif to a serif (Times New Romans) on all labels and axis of my boxplot. I have used this function in other plots before, e.g.: plot(residuals~lnlifespan, data=mydata, pch=psymb, font=6, xlab=ln reproductive lifespan, ylab=residuals ln mass, font.lab=6, cex=1.5, cex.axis=1.5, cex.lab=1.5) and found that font.lab or font.axis=6 gives Times font. However, when I try for boxplot: boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family, font.axis=6, font=6, par(las=1), cex.axis=1) it does not work (R does not give any warning messages). I have also tried family=Times but without success. Any idea of why is not doing it and what I can do to get Times font on my boxplot? I run R on Windows. Thanks a lot, Graziella * Dr. Graziella Iossa Mammal Research Unit School Biological Sciences University of Bristol Woodland Road Bristol BS8 1UG, UK E-mail: [EMAIL PROTECTED] Tel 0044 (0)117 9288918 Fax 0044 (0)117 3317985 http://www.bio.bris.ac.uk/research/mammal/index.html http://www.bio.bris.ac.uk/people/Iossa.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Mixture of Normals with Large Data
BertG == Bert Gunter [EMAIL PROTECTED] on Tue, 7 Aug 2007 16:18:18 -0700 writes: TV Have you considered the situation of wanting to TV characterize probability densities of prevalence TV estimates based on a complex random sample of some TV large population. BertG No -- and I stand by my statement. The empirical BertG distribution of the data themselves are the best BertG characterization of the density. You and others are BertG free to disagree. I do agree with you Bert. From a practical point of view however, you'd still want to use an approximation to the data ECDF, since the full ecdf is just too large an object to handle conveniently. One simple quite small and probably sufficient such approximation maybe using the equivalent of quantile(x, probs = (0:1000)/1000) which is pretty related to just working with a binned version of the original data; something others have proposed as well. Martin BertG On 8/7/07, Bert Gunter [EMAIL PROTECTED] BertG wrote: Why would anyone want to fit a mixture of normals with 110 million observations?? Any questions about the distribution that you would care to ask can be answered directly from the data. Of course, any test of BertG normality (or anything else) would be rejected. More to the point, the data are certainly not a random sample of anything. There will be all kinds of systematic nonrandom structure in them. This is clearly a situation where the researcher needs to think more carefully BertG about the substantive questions of interest and how the data may shed light on them, instead of arbitrarily and perhaps reflexively throwing some silly statistical methodology at them. Bert Gunter Genentech Nonclinical Statistics -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Tim Victor Sent: Tuesday, August 07, 2007 3:02 PM To: r-help@stat.math.ethz.ch Subject: Re: [R] Mixture of Normals with Large Data I wasn't aware of this literature, thanks for the references. On 8/5/07, RAVI VARADHAN [EMAIL PROTECTED] wrote: Another possibility is to use data squashing methods. Relevant papers are: (1) DuMouchel et al. (1999), (2) Madigan et al. (2002), and (3) Owen (1999). Ravi. Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: [EMAIL PROTECTED] - Original Message - From: Charles C. Berry [EMAIL PROTECTED] Date: Saturday, August 4, 2007 8:01 pm Subject: Re: [R] Mixture of Normals with Large Data To: [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch On Sat, 4 Aug 2007, Tim Victor wrote: All: I am trying to fit a mixture of 2 normals with 110 million observations. Iam running R 2.5.1 on a box with 1gb RAM running 32-bit windows and I continue to run out of memory. Does anyone have any suggestions. If the first few million observations can be regarded as a SRS of the rest, then just use them. Or read in blocks of a convenient size and sample some observations from each block. You can repeat this process a few times to see if the results are sufficiently accurate. Otherwise, read in blocks of a convenient size (perhaps 1 million observations at a time), quantize the data to a manageable number of intervals - maybe a few thousand - and tabulate it. Add the counts over all the blocks. Then use mle() to fit a multinomial likelihood whose probabilities are the masses associated with each bin under a mixture of normals law. Chuck Thanks so much, Tim [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list PLEASE do read the posting guideand provide commented, minimal, self-contained, reproducible code. Charles C. Berry (858) 534-2098 Dept of Family/Preventive Medicine E UC San Diego La Jolla, San Diego 92093-0901 __ R-help@stat.math.ethz.ch mailing list PLEASE do read the posting guide and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list
[R] Changing font in boxplots
Hi all, I am very new to R and this might be a simple question but I have looked everywhere you suggest before writing to you. I am trying to change font type from san-serif to a serif (Times New Romans) on all labels and axis of my boxplot. I have used this function in other plots before, e.g.: plot(residuals~lnlifespan, data=mydata, pch=psymb, font=6, xlab=ln reproductive lifespan, ylab=residuals ln mass, font.lab=6, cex=1.5, cex.axis=1.5, cex.lab=1.5) and found that font.lab or font.axis=6 gives Times font. However, when I try for boxplot: boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family, font.axis=6, font=6, par(las=1), cex.axis=1) it does not work (R does not give any warning messages). I have also tried family=Times but without success. Any idea of why is not doing it and what I can do to get Times font on my boxplot? I run R on Windows. Thanks a lot, Graziella * Dr. Graziella Iossa Mammal Research Unit School Biological Sciences University of Bristol Woodland Road Bristol BS8 1UG, UK E-mail: [EMAIL PROTECTED] Tel 0044 (0)117 9288918 Fax 0044 (0)117 3317985 http://www.bio.bris.ac.uk/research/mammal/index.html http://www.bio.bris.ac.uk/people/Iossa.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Changing font in boxplots
Hi John, Thanks so much for such a quick reply. I have tried to set all to Times font running par(font.lab=6) (not 4, maybe this is a local setting on my machine?) but now the boxplot shown has the x and y labels in Times New Roman and the x and y axis still in Arial. Any idea why R is not setting those in Times? Thanks a lot for your advice, Graziella --On 08 August 2007 09:16 -0400 John Kane [EMAIL PROTECTED] wrote: I don't know if boxplot will accept a font argument.m From ?boxplot it is not clear. You may need to set the par() command before the boxplot Example: par(font.lab=4) boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family,las=1, cex.axis=1) --- G Iossa, School Biological Sciences [EMAIL PROTECTED] wrote: Hi all, I am very new to R and this might be a simple question but I have looked everywhere you suggest before writing to you. I am trying to change font type from san-serif to a serif (Times New Romans) on all labels and axis of my boxplot. I have used this function in other plots before, e.g.: plot(residuals~lnlifespan, data=mydata, pch=psymb, font=6, xlab=ln reproductive lifespan, ylab=residuals ln mass, font.lab=6, cex=1.5, cex.axis=1.5, cex.lab=1.5) and found that font.lab or font.axis=6 gives Times font. However, when I try for boxplot: boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family, font.axis=6, font=6, par(las=1), cex.axis=1) it does not work (R does not give any warning messages). I have also tried family=Times but without success. Any idea of why is not doing it and what I can do to get Times font on my boxplot? I run R on Windows. Thanks a lot, Graziella * Dr. Graziella Iossa Mammal Research Unit School Biological Sciences University of Bristol Woodland Road Bristol BS8 1UG, UK E-mail: [EMAIL PROTECTED] Tel 0044 (0)117 9288918 Fax 0044 (0)117 3317985 http://www.bio.bris.ac.uk/research/mammal/index.html http://www.bio.bris.ac.uk/people/Iossa.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Ask a question on any topic and get answers from real people. Go to Yahoo! Answers and share what you know at http://ca.answers.yahoo.com * Dr. Graziella Iossa Mammal Research Unit School Biological Sciences University of Bristol Woodland Road Bristol BS8 1UG, UK E-mail: [EMAIL PROTECTED] Tel 0044 (0)117 9288918 Fax 0044 (0)117 3317985 http://www.bio.bris.ac.uk/research/mammal/index.html http://www.bio.bris.ac.uk/people/Iossa.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Successively eliminating most frequent elemets
Dear experts,
I have a 10x2 matrix T containing random integers. I would like to delete
pairs (rows) iteratively, which contain the most frequent element either in the
first or second column:
T - matrix(trunc(runif(20)*10), nrow=10, ncol=2)
G - matrix(0, nrow=6, ncol=2)
for (i in (1:6)){
print(** Start iteration ~i~ ***)
print(Current matrix:)
print(T)
m - append(T[,1], T[,2])
print(Concatenated columns:)
print(m)
# build frequency table
F - data.matrix(as.data.frame(table(m)))
dimnames(F)-NULL
# pick up the most frequent element: sort decreasing and take is from the top
F - F[order(F[,2], decreasing=TRUE),]
print(Freq. table:)
print(F[1:5,])
todel - F[1,1] #rows containing the most frequent element will be deleted
G[i,1] - todel
G[i,2] - F[1,2]
print(todel=~todel)
# eliminate rows containing the most frequent element
# either the first or the second column contains this element
id - which(T[,1]==todel)
print(Indexes of rows to be deleted:)
print(id)
if (length(id)0){
T - T[-1*id, ]
}
id - which(T[,2]==todel)
print(Indexes of rows to be deleted:)
print(id)
if (length(id)0){
T - T[-1*id, ]
}
print(nrow(T)=~nrow(T))
}
print(Result matrix:)
print(G)
The output of the first two iterations looks like as follows. As one can see,
the frequency table in the second iteration still contains the element deleted
in the first iteration! Is this a bug or what am I doing here wrong?
Any help greatly appreciated!
[1] ** Start iteration 1 ***
[1] Current matrix:
[,1] [,2]
[1,]22
[2,]67
[3,]99
[4,]35
[5,]40
[6,]79
[7,]57
[8,]17
[9,]96
[10,]33
[1] Concatenated columns:
[1] 2 6 9 3 4 7 5 1 9 3 2 7 9 5 0 9 7 7 6 3
[1] Freq. table:
[,1] [,2]
[1,]84
[2,]94
[3,]43
[4,]32
[5,]62
[1] todel=8
[1] Indexes of rows to be deleted:
integer(0)
[1] Indexes of rows to be deleted:
integer(0)
[1] nrow(T)=10
[1] ** Start iteration 2 ***
[1] Current matrix:
[,1] [,2]
[1,]22
[2,]67
[3,]99
[4,]35
[5,]40
[6,]79
[7,]57
[8,]17
[9,]96
[10,]33
[1] Concatenated columns:
[1] 2 6 9 3 4 7 5 1 9 3 2 7 9 5 0 9 7 7 6 3
[1] Freq. table:
[,1] [,2]
[1,]84
[2,]94
[3,]43
[4,]32
[5,]62
[1] todel=8
[1] Indexes of rows to be deleted:
integer(0)
[1] Indexes of rows to be deleted:
integer(0)
[1] nrow(T)=10
[1] ** Start iteration 3 ***
[1] Current matrix:
...
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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] Changing font in boxplots
On Wed, 8 Aug 2007, G Iossa, School Biological Sciences wrote: Hi John, Thanks so much for such a quick reply. I have tried to set all to Times font running par(font.lab=6) (not 4, maybe this is a local setting on my machine?) '6' is a setting specific to certain devices on Windows. You should really be using font families (which are quite new and so not used in many of the introductions). par(family=serif) will change the default for all the text on subsequent plots to be in a serif font, which on the windows() device is (by default) Times. The R posting guide does ask you to tell us your OS, so that points like this do not have to be guessed at. but now the boxplot shown has the x and y labels in Times New Roman and the x and y axis still in Arial. Any idea why R is not setting those in Times? Because you did not ask it to. The font of axis annotation is set by font.axis, not font.lab (which is controls title()'s xlab and ylab and nothing in axis()). See ?axis and ?par, both of which make this clear. John Kane has claimed that what inline pars are used by boxplot() is 'not clear from ?boxplot', but the lack of clarity is his, not in the documentation. ?boxplot refers you to ?bxp, and that spells out exactly which inline pars are used. Thanks a lot for your advice, Graziella --On 08 August 2007 09:16 -0400 John Kane [EMAIL PROTECTED] wrote: I don't know if boxplot will accept a font argument.m From ?boxplot it is not clear. You may need to set the par() command before the boxplot Example: par(font.lab=4) boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family,las=1, cex.axis=1) --- G Iossa, School Biological Sciences [EMAIL PROTECTED] wrote: Hi all, I am very new to R and this might be a simple question but I have looked everywhere you suggest before writing to you. I am trying to change font type from san-serif to a serif (Times New Romans) on all labels and axis of my boxplot. I have used this function in other plots before, e.g.: plot(residuals~lnlifespan, data=mydata, pch=psymb, font=6, xlab=ln reproductive lifespan, ylab=residuals ln mass, font.lab=6, cex=1.5, cex.axis=1.5, cex.lab=1.5) and found that font.lab or font.axis=6 gives Times font. However, when I try for boxplot: boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family, font.axis=6, font=6, par(las=1), cex.axis=1) it does not work (R does not give any warning messages). I have also tried family=Times but without success. Any idea of why is not doing it and what I can do to get Times font on my boxplot? I run R on Windows. Thanks a lot, Graziella * Dr. Graziella Iossa Mammal Research Unit School Biological Sciences University of Bristol Woodland Road Bristol BS8 1UG, UK E-mail: [EMAIL PROTECTED] Tel 0044 (0)117 9288918 Fax 0044 (0)117 3317985 http://www.bio.bris.ac.uk/research/mammal/index.html http://www.bio.bris.ac.uk/people/Iossa.htm -- 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] Changing font in boxplots
Hi John, Thanks so much for such a quick reply. I have tried to set all to Times font running par(font.lab=6) (not 4, maybe this is a local setting on my machine?) but now the boxplot shown has the x and y labels in Times New Roman and the x and y axis still in Arial. Any idea why R is not setting those in Times? Try par(font.axis=6) Thanks a lot for your advice, Graziella --On 08 August 2007 09:16 -0400 John Kane [EMAIL PROTECTED] wrote: I don't know if boxplot will accept a font argument.m From ?boxplot it is not clear. You may need to set the par() command before the boxplot Example: par(font.lab=4) boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family,las=1, cex.axis=1) --- G Iossa, School Biological Sciences [EMAIL PROTECTED] wrote: Hi all, I am very new to R and this might be a simple question but I have looked everywhere you suggest before writing to you. I am trying to change font type from san-serif to a serif (Times New Romans) on all labels and axis of my boxplot. I have used this function in other plots before, e.g.: plot(residuals~lnlifespan, data=mydata, pch=psymb, font=6, xlab=ln reproductive lifespan, ylab=residuals ln mass, font.lab=6, cex=1.5, cex.axis=1.5, cex.lab=1.5) and found that font.lab or font.axis=6 gives Times font. However, when I try for boxplot: boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family, font.axis=6, font=6, par(las=1), cex.axis=1) it does not work (R does not give any warning messages). I have also tried family=Times but without success. Any idea of why is not doing it and what I can do to get Times font on my boxplot? I run R on Windows. Thanks a lot, Graziella * Dr. Graziella Iossa Mammal Research Unit School Biological Sciences University of Bristol Woodland Road Bristol BS8 1UG, UK E-mail: [EMAIL PROTECTED] Tel 0044 (0)117 9288918 Fax 0044 (0)117 3317985 http://www.bio.bris.ac.uk/research/mammal/index.html http://www.bio.bris.ac.uk/people/Iossa.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Ask a question on any topic and get answers from real people. Go to Yahoo! Answers and share what you know at http://ca.answers.yahoo.com * Dr. Graziella Iossa Mammal Research Unit School Biological Sciences University of Bristol Woodland Road Bristol BS8 1UG, UK E-mail: [EMAIL PROTECTED] Tel 0044 (0)117 9288918 Fax 0044 (0)117 3317985 http://www.bio.bris.ac.uk/research/mammal/index.html http://www.bio.bris.ac.uk/people/Iossa.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. --- Scanned by M+ Guardian Messaging Firewall --- HTH Rob __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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: Cannot Coerce POSIXt to POSIXct when building package
A newbie here - please forgive me if this is a basic question. We have an in house package built in R 2.2.1 (yes we're a little behind the times at our firm)and would like to rebuild it using R 2.5.1. However, when I try and build the package from source, I keep getting this error: Error in as(slotVal, slotClass, strict = FALSE) : no method or default for coercing POSIXt to POSIXct Error : unable to load R code in package 'Mango' Error: package/namespace load failed for 'Mango' I tried defining a new method as.POSIXct in the package to coerce POSIXt to POSIXct and then added the as.POSIXct method to the NAMSPACE file. The build still doesn't work (I get the same error message). Any idea what I am doing wrong? The coercion statement looks like this and works in R GUI: #from is a vector of dates in the format %d-%b-%Y) from - as.POSIXct(strptime(from, format = %d%b%Y), tz = GMT) Here is my environment info: R version 2.5.1 (2007-06-27) i386-pc-mingw32 locale: LC_COLLATE=English_United Kingdom.1252;LC_CTYPE=English_United Kingdom.1252;LC_MONETARY=English_United Kingdom.1252;LC_NUMERIC=C;LC_TIME=English_United Kingdom.1252 attached base packages: [1] tcltk stats graphics grDevices utils datasets methods base other attached packages: fSeries nnet mgcv fBasics fCalendar fEcofin spatial MASS 251.70 7.2-34 1.3-25 251.70 251.70 251.70 7.2-34 7.2-34 I would sincerely appreciate any help. [[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] Changing font in boxplots
Thanks everyone for your replies. And sorry if I have not been clear enough in my query (although I did say that I run R on Windows). Typing either par(family=serif) or par(font.axis=6) solved the problem. Much obliged, Graziella --On 08 August 2007 14:58 +0100 Prof Brian Ripley [EMAIL PROTECTED] wrote: On Wed, 8 Aug 2007, G Iossa, School Biological Sciences wrote: Hi John, Thanks so much for such a quick reply. I have tried to set all to Times font running par(font.lab=6) (not 4, maybe this is a local setting on my machine?) '6' is a setting specific to certain devices on Windows. You should really be using font families (which are quite new and so not used in many of the introductions). par(family=serif) will change the default for all the text on subsequent plots to be in a serif font, which on the windows() device is (by default) Times. The R posting guide does ask you to tell us your OS, so that points like this do not have to be guessed at. but now the boxplot shown has the x and y labels in Times New Roman and the x and y axis still in Arial. Any idea why R is not setting those in Times? Because you did not ask it to. The font of axis annotation is set by font.axis, not font.lab (which is controls title()'s xlab and ylab and nothing in axis()). See ?axis and ?par, both of which make this clear. John Kane has claimed that what inline pars are used by boxplot() is 'not clear from ?boxplot', but the lack of clarity is his, not in the documentation. ?boxplot refers you to ?bxp, and that spells out exactly which inline pars are used. Thanks a lot for your advice, Graziella --On 08 August 2007 09:16 -0400 John Kane [EMAIL PROTECTED] wrote: I don't know if boxplot will accept a font argument.m From ?boxplot it is not clear. You may need to set the par() command before the boxplot Example: par(font.lab=4) boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family,las=1, cex.axis=1) --- G Iossa, School Biological Sciences [EMAIL PROTECTED] wrote: Hi all, I am very new to R and this might be a simple question but I have looked everywhere you suggest before writing to you. I am trying to change font type from san-serif to a serif (Times New Romans) on all labels and axis of my boxplot. I have used this function in other plots before, e.g.: plot(residuals~lnlifespan, data=mydata, pch=psymb, font=6, xlab=ln reproductive lifespan, ylab=residuals ln mass, font.lab=6, cex=1.5, cex.axis=1.5, cex.lab=1.5) and found that font.lab or font.axis=6 gives Times font. However, when I try for boxplot: boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family, font.axis=6, font=6, par(las=1), cex.axis=1) it does not work (R does not give any warning messages). I have also tried family=Times but without success. Any idea of why is not doing it and what I can do to get Times font on my boxplot? I run R on Windows. Thanks a lot, Graziella * Dr. Graziella Iossa Mammal Research Unit School Biological Sciences University of Bristol Woodland Road Bristol BS8 1UG, UK E-mail: [EMAIL PROTECTED] Tel 0044 (0)117 9288918 Fax 0044 (0)117 3317985 http://www.bio.bris.ac.uk/research/mammal/index.html http://www.bio.bris.ac.uk/people/Iossa.htm -- 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 * Dr. Graziella Iossa Mammal Research Unit School Biological Sciences University of Bristol Woodland Road Bristol BS8 1UG, UK E-mail: [EMAIL PROTECTED] Tel 0044 (0)117 9288918 Fax 0044 (0)117 3317985 http://www.bio.bris.ac.uk/research/mammal/index.html http://www.bio.bris.ac.uk/people/Iossa.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error: Cannot Coerce POSIXt to POSIXct when building package
A newbie here - please forgive me if this is a basic question. We have an in house package built in R 2.2.1 (yes we're a little behind the times at our firm)and would like to rebuild it using R 2.5.1. However, when I try and build the package from source, I keep getting this error: Error in as(slotVal, slotClass, strict = FALSE) : no method or default for coercing POSIXt to POSIXct Error : unable to load R code in package 'Mango' Error: package/namespace load failed for 'Mango' I tried defining a new method as.POSIXct in the package to coerce POSIXt to POSIXct and then added the as.POSIXct method to the NAMSPACE file. The build still doesn't work (I get the same error message). Any idea what I am doing wrong? The coercion statement looks like this and works in R GUI: #from is a vector of dates in the format %d-%b-%Y) from - as.POSIXct(strptime(from, format = %d%b%Y), tz = GMT) Here is my environment info: R version 2.5.1 (2007-06-27) i386-pc-mingw32 locale: LC_COLLATE=English_United Kingdom.1252;LC_CTYPE=English_United Kingdom.1252;LC_MONETARY=English_United Kingdom.1252;LC_NUMERIC=C;LC_TIME=English_United Kingdom.1252 attached base packages: [1] tcltk stats graphics grDevices utils datasets methods base other attached packages: fSeries nnet mgcv fBasics fCalendar fEcofin spatial MASS 251.70 7.2-34 1.3-25 251.70 251.70 251.70 7.2-34 7.2-34 I would sincerely appreciate any help. [[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] Changing font in boxplots
--- Prof Brian Ripley [EMAIL PROTECTED] wrote: On Wed, 8 Aug 2007, G Iossa, School Biological Sciences wrote: Hi John, Thanks so much for such a quick reply. I have tried to set all to Times font running par(font.lab=6) (not 4, maybe this is a local setting on my machine?) '6' is a setting specific to certain devices on Windows. You should really be using font families (which are quite new and so not used in many of the introductions). par(family=serif) will change the default for all the text on subsequent plots to be in a serif font, which on the windows() device is (by default) Times. Thanks, that works beautifully. I had seen the 'family= once but never needed it and forgot about it. The R posting guide does ask you to tell us your OS, so that points like this do not have to be guessed at. but now the boxplot shown has the x and y labels in Times New Roman and the x and y axis still in Arial. Any idea why R is not setting those in Times? Because you did not ask it to. The font of axis annotation is set by font.axis, not font.lab (which is controls title()'s xlab and ylab and nothing in axis()). See ?axis and ?par, both of which make this clear. John Kane has claimed that what inline pars are used by boxplot() is 'not clear from ?boxplot', but the lack of clarity is his, not in the documentation. ?boxplot refers you to ?bxp, and that spells out exactly which inline pars are used. Thanks a lot for your advice, Graziella --On 08 August 2007 09:16 -0400 John Kane [EMAIL PROTECTED] wrote: I don't know if boxplot will accept a font argument.m From ?boxplot it is not clear. You may need to set the par() command before the boxplot Example: par(font.lab=4) boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family,las=1, cex.axis=1) --- G Iossa, School Biological Sciences [EMAIL PROTECTED] wrote: Hi all, I am very new to R and this might be a simple question but I have looked everywhere you suggest before writing to you. I am trying to change font type from san-serif to a serif (Times New Romans) on all labels and axis of my boxplot. I have used this function in other plots before, e.g.: plot(residuals~lnlifespan, data=mydata, pch=psymb, font=6, xlab=ln reproductive lifespan, ylab=residuals ln mass, font.lab=6, cex=1.5, cex.axis=1.5, cex.lab=1.5) and found that font.lab or font.axis=6 gives Times font. However, when I try for boxplot: boxplot(mass ~ family, data=mydata, ylab=mass %, xlab=family, font.axis=6, font=6, par(las=1), cex.axis=1) it does not work (R does not give any warning messages). I have also tried family=Times but without success. Any idea of why is not doing it and what I can do to get Times font on my boxplot? I run R on Windows. Thanks a lot, Graziella * Dr. Graziella Iossa Mammal Research Unit School Biological Sciences University of Bristol Woodland Road Bristol BS8 1UG, UK E-mail: [EMAIL PROTECTED] Tel 0044 (0)117 9288918 Fax 0044 (0)117 3317985 http://www.bio.bris.ac.uk/research/mammal/index.html http://www.bio.bris.ac.uk/people/Iossa.htm -- 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] Error: Cannot Coerce POSIXt to POSIXct when building package
On Wed, 8 Aug 2007, Praveen Kanakamedala wrote: A newbie here - please forgive me if this is a basic question. We have an in house package built in R 2.2.1 (yes we're a little behind the times at our firm)and would like to rebuild it using R 2.5.1. However, when I try and build the package from source, I keep getting this error: Error in as(slotVal, slotClass, strict = FALSE) : no method or default for coercing POSIXt to POSIXct Error : unable to load R code in package 'Mango' Error: package/namespace load failed for 'Mango' I tried defining a new method as.POSIXct in the package to coerce POSIXt to POSIXct and then added the as.POSIXct method to the NAMSPACE file. The build still doesn't work (I get the same error message). Any idea what I am doing wrong? The coercion statement looks like this and works in R GUI: How did you get this? There should be no objects of class 'POSIXt' alone, and I get e.g. now - Sys.time() as(now, POSIXct) Error in asMethod(object) : explicit coercion of old-style class (POSIXt, POSIXct) is not defined That can be fixed (see ?as), but you seem to have a malformed object in one of your slots. As often applies, PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. #from is a vector of dates in the format %d-%b-%Y) from - as.POSIXct(strptime(from, format = %d%b%Y), tz = GMT) Here is my environment info: R version 2.5.1 (2007-06-27) i386-pc-mingw32 locale: LC_COLLATE=English_United Kingdom.1252;LC_CTYPE=English_United Kingdom.1252;LC_MONETARY=English_United Kingdom.1252;LC_NUMERIC=C;LC_TIME=English_United Kingdom.1252 attached base packages: [1] tcltk stats graphics grDevices utils datasets methods base other attached packages: fSeries nnet mgcv fBasics fCalendar fEcofin spatial MASS 251.70 7.2-34 1.3-25 251.70 251.70 251.70 7.2-34 7.2-34 I would sincerely appreciate any help. [[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. -- 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] small sample techniques
Indeed, I understand what you say. The df of freedom for the dummy example is n1+n2-2 = 8. But when I run the t.test I get it as 5.08, am I missing something? -Original Message- From: Moshe Olshansky [mailto:[EMAIL PROTECTED] Sent: Tuesday, August 07, 2007 9:05 PM To: Nair, Murlidharan T; r-help@stat.math.ethz.ch Subject: Re: [R] small sample techniques Hi Nair, If the two populations are normal the t-test gives you the exact result for whatever the sample size is (the sample size will affect the number of degrees of freedom). When the populations are not normal and the sample size is large it is still OK to use t-test (because of the Central Limit Theorem) but this is not necessarily true for the small sample size. You could use simulation to find the relevant probabilities. --- Nair, Murlidharan T [EMAIL PROTECTED] wrote: If my sample size is small is there a particular switch option that I need to use with t.test so that it calculates the t ratio correctly? Here is a dummy example? á =0.05 Mean pain reduction for A =27; B =31 and SD are SDA=9 SDB=12 drgA.p-rnorm(5,27,9); drgB.p-rnorm(5,31,12) t.test(drgA.p,drgB.p) # what do I need to give as additional parameter here? I can do it manually but was looking for a switch option that I need to specify for t.test. Thanks ../Murli [[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-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Problems with nls-function
Dear all I have got some problems with a least-squares regression using the function nls. I want to estimate h, k and X of the following formula by using nls : exp(2*200*(q^2-4*h/k-0.25+(2/k-0.5+4*h^2/k^2)*log(abs((k*q^2+2*h*q-1)/(0.25*k-h-1)/((-k*q^2-2*h*q+1)*X) y as defined by c(0.009747 0.001949 0.00 0.003899 0.00 0.00 0.005848 0.001949) q as defined by c(-0.7500 -0.6875 -0.5625 -0.4875 -0.4625 -0.4375 -0.4125 -0.3875) (length of the real q and y is 46; too long to post them here) i tought the correct using of nls would be: Mic-nls(y~function, start = list(k=1.0,h=0.1,X=exp(10)) But it doesn`t work. i tryed an easier formula like : Mic-nls(y~h*exp(2*k*200*(q^2)), start=list(h=0.1,k=1,X=10)) The result was the same. Isn`t nls the function i should use to solve this regression problem? Which things did i make wrong? Thank you very much in advance Michael _ In 5 Schritten zur eigenen Homepage. Jetzt Domain sichern und gestalten! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error: Cannot Coerce POSIXt to POSIXct when building package
Thank you very much for your response. I think I found the offending code. It's somewhere in here. setClass(timeSeries, representation( Data = matrix, positions = character, format = character, FinCenter = character, units = character, recordIDs = data.frame, title = character, documentation = character), prototype = list( Data = matrix(NA, dimnames = list(31-Dec-2006, timeSeries)), positions = 31-Dec-2006, format = %d-%b-%Y, FinCenter = myFinCenter(), units = Series, recordIDs = data.frame(), title = Time Series Object, documentation = paste(Created at, myFinCenter(), now()@Data) ) ) myFinCenter - function() return(London) Funny enough...this worked in R-2.2.1 but doesn't work in R2.2.1 On 8/8/07, Prof Brian Ripley [EMAIL PROTECTED] wrote: On Wed, 8 Aug 2007, Praveen Kanakamedala wrote: A newbie here - please forgive me if this is a basic question. We have an in house package built in R 2.2.1 (yes we're a little behind the times at our firm)and would like to rebuild it using R 2.5.1. However, when I try and build the package from source, I keep getting this error: Error in as(slotVal, slotClass, strict = FALSE) : no method or default for coercing POSIXt to POSIXct Error : unable to load R code in package 'Mango' Error: package/namespace load failed for 'Mango' I tried defining a new method as.POSIXct in the package to coerce POSIXt to POSIXct and then added the as.POSIXct method to the NAMSPACE file. The build still doesn't work (I get the same error message). Any idea what I am doing wrong? The coercion statement looks like this and works in R GUI: How did you get this? There should be no objects of class 'POSIXt' alone, and I get e.g. now - Sys.time() as(now, POSIXct) Error in asMethod(object) : explicit coercion of old-style class (POSIXt, POSIXct) is not defined That can be fixed (see ?as), but you seem to have a malformed object in one of your slots. As often applies, PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. #from is a vector of dates in the format %d-%b-%Y) from - as.POSIXct(strptime(from, format = %d%b%Y), tz = GMT) Here is my environment info: R version 2.5.1 (2007-06-27) i386-pc-mingw32 locale: LC_COLLATE=English_United Kingdom.1252;LC_CTYPE=English_United Kingdom.1252;LC_MONETARY=English_United Kingdom.1252;LC_NUMERIC=C;LC_TIME=English_United Kingdom.1252 attached base packages: [1] tcltk stats graphics grDevices utils datasets methods base other attached packages: fSeries nnet mgcv fBasics fCalendar fEcofin spatial MASS 251.70 7.2-34 1.3-25 251.70 251.70 251.70 7.2-34 7.2-34 I would sincerely appreciate any help. [[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. -- 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 [[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] fit.contrast - contrast testing with lme
I need some help on contrast testing with lme. It seems fit.contrast in 'gmodels' package works fine for simple contrasting among levels of a factor such as fit.contrast(fit.lme, Trust, c(1,-1)) Estimate Std. Error t-value Pr(|t|) Trust c=( 1 -1 ) -0.001442638 0.0005836959 -2.471557 0.01566063 However if I want to compare some interaction terms, it does not work as I wanted: fit.contrast(fit.lme, Trust:Sex, c(1,-1,0,0)) Error in `contrasts-`(`*tmp*`, value = c(0.5, -0.5, 0, 0, -0.353553390593274, : contrasts apply only to factors or, fit.contrast(fit.lme, TrustU:Sex, c(1,-1)) Error in `contrasts-`(`*tmp*`, value = c(0.5, -0.5)) : contrasts apply only to factors Is there a way I can run such tests? Also how can run an F test for a hypothesis like the following with either fit.contrast or lme? H0: FreqHi - FreqLo = 0, and FreqNo - FreqLo = 0 I also tried to run contrasting directly in lme, but could not get it work: anova(fit.lme, L=c(TrustT:Sex:Freq=1, TrustU:Sex:Freq=-1)) Error in anova.lme(fit.lme, L = c(TrustT:Sex:Freq = 1, TrustU:Sex:Freq = -1)) : Effects TrustT:Sex:Freq, TrustU:Sex:Freq not matched I'm confused with the error message, and could not figure out what it means. Here is more information: fit.lme - lme(Beta ~ Trust*Sex*Freq, random = ~1|Subj, Model) summary(fit.lme) Linear mixed-effects model fit by REML Data: Model AIC BIC logLik -825.4663 -791.4348 426.7331 Random effects: Formula: ~1 | Subj (Intercept)Residual StdDev: 0.001144573 0.001167392 Fixed effects: Beta ~ Trust * Sex * Freq ValueStd.Error DFt-value p-value (Intercept) 0.0001090007 0.0005780194 77 0.1885762 0.8509 TrustU 0.0014426378 0.0005836959 77 2.4715572 0.0157 SexM0.0008230359 0.0005836959 77 1.4100423 0.1626 FreqLo 0.0001998191 0.0005836959 77 0.3423343 0.7330 FreqNo 0.0004900107 0.0005836959 77 0.8394965 0.4038 TrustU:SexM-0.0012598266 0.0008254707 77 -1.5261918 0.1311 TrustU:FreqLo -0.0012383346 0.0008254707 77 -1.5001558 0.1377 TrustU:FreqNo -0.0009141543 0.0008254707 77 -1.1074341 0.2716 SexM:FreqLo-0.0008469211 0.0008254707 77 -1.0259857 0.3081 SexM:FreqNo 0.0006361012 0.0008254707 77 0.7705922 0.4433 TrustU:SexM:FreqLo 0.0013272173 0.0011673918 77 1.1369082 0.2591 TrustU:SexM:FreqNo 0.0006241524 0.0011673918 77 0.5346555 0.5944 ... anova(lme(Beta ~ Trust*Sex*Freq, random = ~1|Subj, Model)) numDF denDF F-value p-value (Intercept)177 4.813938 0.0312 Trust 177 3.113293 0.0816 Sex177 3.535774 0.0638 Freq 277 6.083832 0.0035 Trust:Sex 177 1.634858 0.2049 Trust:Freq 277 0.678558 0.5104 Sex:Freq 277 2.165059 0.1217 Trust:Sex:Freq 277 0.647042 0.5264 Your help is highly appreciated. Thanks, Gang __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Mixture of Normals with Large Data
On Wed, 8 Aug 2007, Martin Maechler wrote: BertG == Bert Gunter [EMAIL PROTECTED] on Tue, 7 Aug 2007 16:18:18 -0700 writes: TV Have you considered the situation of wanting to TV characterize probability densities of prevalence TV estimates based on a complex random sample of some TV large population. BertG No -- and I stand by my statement. The empirical BertG distribution of the data themselves are the best BertG characterization of the density. You and others are BertG free to disagree. I do agree with you Bert. From a practical point of view however, you'd still want to use an approximation to the data ECDF, since the full ecdf is just too large an object to handle conveniently. One simple quite small and probably sufficient such approximation maybe using the equivalent of quantile(x, probs = (0:1000)/1000) which is pretty related to just working with a binned version of the original data; something others have proposed as well. I have done Normal (actually logNormal) mixture fitting to pretty large data (particle counts by size) for summary purposes. In that case it would not have done just as well to use quantiles as I had many sets of data (every three hours for several months) and the locations of the mixture components drift around over time. The location, scale, and mass of the four mixture components really were the best summaries. This was the application that constrOptim() was written for. -thomas Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] small sample techniques
On Wed, 8 Aug 2007, Nair, Murlidharan T wrote: Indeed, I understand what you say. The df of freedom for the dummy example is n1+n2-2 = 8. But when I run the t.test I get it as 5.08, am I missing something? Yes. You are probably looking for the version of the t.test that assumes equal variances (the original one), so you need var.equal=TRUE. -thomas -Original Message- From: Moshe Olshansky [mailto:[EMAIL PROTECTED] Sent: Tuesday, August 07, 2007 9:05 PM To: Nair, Murlidharan T; r-help@stat.math.ethz.ch Subject: Re: [R] small sample techniques Hi Nair, If the two populations are normal the t-test gives you the exact result for whatever the sample size is (the sample size will affect the number of degrees of freedom). When the populations are not normal and the sample size is large it is still OK to use t-test (because of the Central Limit Theorem) but this is not necessarily true for the small sample size. You could use simulation to find the relevant probabilities. --- Nair, Murlidharan T [EMAIL PROTECTED] wrote: If my sample size is small is there a particular switch option that I need to use with t.test so that it calculates the t ratio correctly? Here is a dummy example? á =0.05 Mean pain reduction for A =27; B =31 and SD are SDA=9 SDB=12 drgA.p-rnorm(5,27,9); drgB.p-rnorm(5,31,12) t.test(drgA.p,drgB.p) # what do I need to give as additional parameter here? I can do it manually but was looking for a switch option that I need to specify for t.test. Thanks ../Murli [[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-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] simulation-binomial
That depends on what you meant by writing the conditional probability. Bayes rule says that the probability of testing positive when one has the disease is calculated as follows: Pr(T+ | D+)=(Pr(D+ | T+)*Pr(T+))/Pr(D+) is that what you mean? Kyle H. Ambert Department of Behavioral Neuroscience Oregon Health Science University On 8/8/07, sigalit mangut-leiba [EMAIL PROTECTED] wrote: hello, i want to do a binomial simulation, by taking 200 var. from one group (x) and 300 from another (y). the prob. for disease=.6 in both groups. x - rbinom(200, 1, .6) y - rbinom(300, 1, .6) if the person is from group x - the probability to find the disease, assuming the person is sick, is .95, if he is from group Y its .80. i want to know the joint probability: p(the person has the disease and tested sick)=P(D+,T+). my problem is how to write the conditional prob. Thanks for your help, also reference on this subject (binomial simulation) would be great. Sigalit. [[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. [[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] prediction using gam
I am fitting a two dimensional smoother in gam, say junk = gam(y~s(x1,x2)), to a response variable y that is always positive and pretty well behaved, both x1 and x2 are contained within [0,1]. I then create a new dataset for prediction with values of (x1,x2) within the range of the original data. predict(junk,newdata,type=response) My predicted values are a bit strange (some negative and some large positive values). When I plot the predicted surface, it looks well behaved with no strange dips/etc. Could it be a problem with the predict command? Is there a safe version of predict for higher-dimensional smoothers in gam? Elizabeth Johnson Research Associate Johns Hopkins Unviersity Department of Biostatistics [[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] Problems with nls-function
On 8/8/07, Michael Petram [EMAIL PROTECTED] wrote: Dear all I have got some problems with a least-squares regression using the function nls. I want to estimate h, k and X of the following formula by using nls : exp(2*200*(q^2-4*h/k-0.25+(2/k-0.5+4*h^2/k^2)*log(abs((k*q^2+2*h*q-1)/(0.25*k-h-1)/((-k*q^2-2*h*q+1)*X) y as defined by c(0.009747 0.001949 0.00 0.003899 0.00 0.00 0.005848 0.001949) q as defined by c(-0.7500 -0.6875 -0.5625 -0.4875 -0.4625 -0.4375 -0.4125 -0.3875) (length of the real q and y is 46; too long to post them here) i tought the correct using of nls would be: Mic-nls(y~function, start = list(k=1.0,h=0.1,X=exp(10)) But it doesn`t work. i tryed an easier formula like : Mic-nls(y~h*exp(2*k*200*(q^2)), start=list(h=0.1,k=1,X=10)) The result was the same. Isn`t nls the function i should use to solve this regression problem? Which things did i make wrong? X is not in the model so by including it in your starting values you make it non-idenfiable. Get rid of it. Also use better starting values. Here we use grid search to get them: g - expand.grid(h = 1:100/100, k = 1:100/100) st - g[which.min(apply(g, 1, function(x) x[1] * exp(2*x[2]*200*q^2))),] nls(y~h*exp(2*k*200*(q^2)), start = st) Nonlinear regression model model: y ~ h * exp(2 * k * 200 * (q^2)) data: parent.frame() h k 0.0003533 0.0139700 residual sum-of-squares: 5.039e-05 Number of iterations to convergence: 16 Achieved convergence tolerance: 8.883e-06 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] R2WinBUGS results not different with different runs
Gregor Gorjanc wrote: toby909 at gmail.com writes: Is this a specialty with R2WinBUGS? Does it have something to do with the seed value? Isnt the seed value reset everytime I restart winbugs? I always have the same seed if I start WinBUGS multiple times. So you get exactly the same chain, numerically, when rerunning the same model, with the same number of iterations, everything the same.? Wouldnt that be problematic if every researcher in the world who uses winbugs uses the same sequence of random numbers? R's random numbers are different each time, because the seed is linked to the clock in your PC. T Gregor __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help using gPath
Hi everyone,I'm trying to figure out how to use gPath and the documentation is not very helpful :( I have the following plot object: plot-surrounds:: background plot.gTree.378:: background guide.gTree.355:: (background.rect.345, minor-horizontal.segments.347, minor-vertical.segments.349, major-horizontal.segments.351, major-vertical.segments.353) guide.gTree.356:: (background.rect.345, minor-horizontal.segments.347, minor-vertical.segments.349, major-horizontal.segments.351, major-vertical.segments.353) yaxis.gTree.338:: ticks.segments.321 labels.gTree.335:: (label.text.324, label.text.326, label.text.328, label.text.330, label.text.332, label.text.334) xaxis.gTree.339:: ticks.segments.309 labels.gTree.315:: (label.text.312, label.text.314) xaxis.gTree.340:: ticks.segments.309 labels.gTree.315:: (label.text.312, label.text.314) strip.gTree.364:: (background.rect.361, label.text.363) strip.gTree.370:: (background.rect.367, label.text.369) guide.rect.357 guide.rect.358 boxplots.gTree.283:: geom_boxplot.gTree.273:: (GRID.segments.267, GRID.segments.268, geom_bar.rect.270, geom_bar.rect.272) geom_boxplot.gTree.281:: (GRID.segments.275, GRID.segments.276, geom_bar.rect.278, geom_bar.rect.280) boxplots.gTree.301:: geom_boxplot.gTree.291:: (GRID.segments.285, GRID.segments.286, geom_bar.rect.288, geom_bar.rect.290) geom_boxplot.gTree.299:: (GRID.segments.293, GRID.segments.294, geom_bar.rect.296, geom_bar.rect.298) geom_jitter.points.303 geom_jitter.points.305 guide.rect.357 guide.rect.358 ylabel.text.382 xlabel.text.380 title Could someone be so kind and create the proper call to grid.gedit() to access a couple of different aspects of this graph? I tried: grid.gedit(gPath(ylabel.text.382,labels), gp=gpar(fontsize=16)) # error I'd like to change the margins on the label for the yaxis (not the tick marks) to put more space between the label and the tick marks. I'd also like to remove the left border on the first panel. I'd like to adjust the size of the font for the axis labels independently of the tick marks. I'd like to change the color of the lines that make up the boxplots. Plus, I'd like to change the margins of the strip labels. If you could show me a couple of examples I'm sure I cold get the rest working. Thanks so much, emilio [[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] Logistic Models, Cross Validation, error tables , Monte Carlo Simulation?
Thanks Mr. Ellison, Your remark helped solve my error table problem. However, I found a new one. Now that I have my error tables, i realised that it is no good statistical practise to validate a model, based one one error table. So i should use a tool like K-fold CV. ex: binary_model - glm (y_binary~ x_value, family = binomial,data=dataset) cv.binary(binary_model,rand=NULL, nfolds=1000, print.details=TRUE) This is no problem for the binary model, for the odds model this is not the case. Do you know a tool that can do this, or perhapes a way to implement it in a monte carlo simulation? (i added the way i solved the error table problem below) Kind regards, Tom. ERROR TABLE DILEMMA For a binary model there is no problem (here y has an outcome of 0 or 1) ex:pred_binary_model=(expit(predict(binary_model,tsample))P) table_binary_model=table(pred_binary_model,tsample[,2]) TER_binary_model=sum(diag(table_binary_model[,]))/sum(table_binary_model) (error table1) pred_binary_model 0 1 FALSE 28 95 TRUE4 114 [1] 0.5892116 -- of correct classified cases Here there are 28 + 114 correctly predicted test cases, this results in 58.9% correct classified cases. A few more misclassified cases does not result in big departures from this 58.9%. When i preform this on categorical data, i have to use frequency tables. This predicts the number of successes and the number of failures, per interval.(odds per interval) So the error table does contain an outcome of odds for every given interval. ex: (error table2) oddsPt pred_percent_model 0.00 0.16 0.37 0.84 0.051 0 0 0 0.160 1 0 0 0.340 0 1 0 0.780 0 0 1 [1] 1 -- of correct classified cases As you can see, one misclassification will take disastrous proportions. (~25% difference) The output of error table2 is interpretable, but it is not ideal, and oversensitive to misclassification. I was able to solve this later problem by extracting the model coefficients, and then using them in a function. Based on this function, i was able to write an error table equal to table 1. Disclaimer: click here [[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] rational or float?
Hi All, I am using the function fractions() and cognates from MASS. I would like to be able to tell if some calculations I am doing on some rationals are transformed into floats and then retransformed into rationals. For instance I suspect that as.fractions(1/8) + as.fractions(1/4) does transform into floats and back, while I know 1/as.fractions(8) + 1/as.fractions(4) does not. Since I am using sum() and doing a number of multiplications I would like to know so I can intervene. For instance, what's happening here? pun [,1] [,2] [1,]11 [2,]44 library(MASS) pun[1,]/as.fractions(pun[2,]) [1] 1/4 1/4 sum(pun[1,]/as.fractions(pun[2,])) [1] 1/2 Best, Federico -- Federico C. F. Calboli Department of Epidemiology and Public Health Imperial College, St. Mary's Campus Norfolk Place, London W2 1PG Tel +44 (0)20 75941602 Fax +44 (0)20 75943193 f.calboli [.a.t] imperial.ac.uk f.calboli [.a.t] gmail.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] Change in R**2 for block entry regression
Hi all, I'm demonstrating a block entry regression using R for my regression class. For each block, I get the R**2 and the associated F. I do this with separate regressions adding the next block in and then get the results by writing separate summary() statements for each regression. Is there a more convenient way to do this and also to get the change in R**2 and associated F test for the change? Thanks in advance. 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] homepage: http://www.education.wisc.edu/edpsych/facstaff/kaplan/kaplan.htm Phone: 608-262-0836 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Rcmdr window border lost
Dear all, I have recently lost my Rcmdr window borders (all my other programs have borders)! I am unsure of what I have done, although I have recently update.packages() in R... How can I reclaim them? I am using: Ubuntu Linux (Feisty) R version 2.5.1 R Commander Version 1.3-5 I have deleted the folder: /usr/local/lib/R/site-library/Rcmdr and reinstalled Rcmdr with: install.packages(Rcmdr, dep=TRUE) This has not solved my problem though. Maybe I need to reinstall something else as well? Thanks in advance, Andy __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] test for contingency table when there are many zeros
fisher.test(tt) Francisco gallon li wrote: Here is my table tt A B 1 297 398 2 470 376 3 30 23 4 3 3 5 0 0 b/c two cells are zero, I can't use chisq.test() in R which gives the following output; chisq.test(tt) Pearson's Chi-squared test data: tt X-squared = NaN, df = 4, p-value = NA Warning message: Chi-squared approximation may be incorrect in: chisq.test(tt) What function should I use then? Any suggestion? [[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-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] prediction using gam
Without seeing the data and results it's hard to say. mgcv::predict.gam is already `safe' so that's not the issue. It's also pretty heavily tested, so a problem with that function wouldn't be the first place I'd look. How `large positive' are the predictions relative to the observed response? How well do the x1,x2 cover the unit square? Smoothers often do not extrapolate well even over quite modest distances... Slight negative predictions are not really surprising, given that the model you have fitted allows -ve fitted values. You could fix this by using a log link (with gaussian or Gamma family). If you think the results are not right, I can take a look at what's happening if you send me the data, R and mgcv version numbers and exact commands generating the problem (`off line'). I would not use the data for any other purpose of course. It may take a little while to get to however, as I've one or two local difficulties here at the moment. best, Simon On Wednesday 08 August 2007 16:58, Johnson, Elizabeth wrote: I am fitting a two dimensional smoother in gam, say junk = gam(y~s(x1,x2)), to a response variable y that is always positive and pretty well behaved, both x1 and x2 are contained within [0,1]. I then create a new dataset for prediction with values of (x1,x2) within the range of the original data. predict(junk,newdata,type=response) My predicted values are a bit strange (some negative and some large positive values). When I plot the predicted surface, it looks well behaved with no strange dips/etc. Could it be a problem with the predict command? Is there a safe version of predict for higher-dimensional smoothers in gam? Elizabeth Johnson Research Associate Johns Hopkins Unviersity Department of Biostatistics [[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. -- Simon Wood, Mathematical Sciences, University of Bath, Bath, BA2 7AY UK +44 1225 386603 www.maths.bath.ac.uk/~sw283 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] cointegration analysis
Hello, I tried to use urca package (R) for cointegration analysis. The data matrix to be investigated for cointegration contains 8 columns (variables). Both procedures, Phillips Ouliaris test and Johansen's procedures give errors (error in evaluating the argument 'object' in selecting a method for function 'summary' respectiv too many variables, critical values cannot be computed). What can I do? With regards, Dorina LAZAR Department of Statistics, Forecasting, Mathematics Babes Bolyai University, Faculty of Economic Science Teodor Mihali 58-60, 400591 Cluj-Napoca, Romania __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Relocating Axis Label/Title --2
Apologies for the previous mail (I sent it off too early by mistake). This is the correct example: rm(list=ls()) D_mean-seq(-5,5,length=100) y-exp(-D_mean^2/5) pdf(my.pdf) plot(D_mean,y,type=l,yaxt=n,lty=2,lwd=2,col=black, ylab = list(expression(paste(dN/dlogD[agg], [*cm^-3*]))), xlab = expression(paste(D[agg], [nm])), cex.lab=1.2 ) axis(2, mgp=c(0, 0.2, -2)) dev.off() With mgp() I can tune the distance between the ticks and the tick labels, but how can I move the axis label? I would like to move the one along y to visualize correctly the exponent 3. Kind Regards Lorenzo On 08/08/07, Lorenzo Isella [EMAIL PROTECTED] wrote: Dear All, I am experiencing some problems with relocating an axis title. I visited the following link before posting: http://tolstoy.newcastle.edu.au/R/help/05/05/5283.html But this is not entirely what I would like to do Consider the example below: rm(list=ls()) D_mean-seq(-5,5,length=100) y-exp(-D_mean^2/5) pdf(my.pdf) plot(D_mean,y,type=l,yaxt=n,lty=2,lwd=2,col=black, ylab = list(expression(paste(dN/dlogD[agg], [*cm^-3*]))), xlab = expression(paste(D[agg], [nm])), cex.lab=1.2 ) title(2, mgp=c(0, .3, 0)) dev.off() I have the problem that the 3 in cubic centimeters (on the y axis) is somehow cut in the pdf file I generate. Everything would be fine if I could shift a bit the title of the y axis. It must be trivial, but so far I have not managed to do it. Any suggestions? Many thanks Lorenzo I tried playing with the mgp parameter, but I managed to move the __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] simulation-binomial
I have the probability: P(T+ / D+) i want to find P(T+,D+) which is: P(T+ / D+)*P(D+) and i have those probabilities. i dont know how to write this in R. something like this: (say p2 is the conditional prob. and p1 is the joint prob.) p2 - p1/.6 x - rbinom(200, 1, .6) y - rbinom(300, 1, .6) if (x) p2==0.95 if (y) p2==0.80 i don't know how to write the if condition. thank you for your reply, sigalit. On 8/8/07, Kyle Henderson [EMAIL PROTECTED] wrote: That depends on what you meant by writing the conditional probability. Bayes rule says that the probability of testing positive when one has the disease is calculated as follows: Pr(T+ | D+)=(Pr(D+ | T+)*Pr(T+))/Pr(D+) is that what you mean? Kyle H. Ambert Department of Behavioral Neuroscience Oregon Health Science University On 8/8/07, sigalit mangut-leiba [EMAIL PROTECTED] wrote: hello, i want to do a binomial simulation, by taking 200 var. from one group (x) and 300 from another (y). the prob. for disease=.6 in both groups. x - rbinom(200, 1, .6) y - rbinom(300, 1, .6) if the person is from group x - the probability to find the disease, assuming the person is sick, is .95, if he is from group Y its .80. i want to know the joint probability: p(the person has the disease and tested sick)=P(D+,T+). my problem is how to write the conditional prob. Thanks for your help, also reference on this subject (binomial simulation) would be great. Sigalit. [[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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Relocating Axis Label/Title
Have a look at mar in ?par. You might want to try something like mar=c(5, 5, 4, 2) + 0.1 rather than the default of c(5, 4, 4, 2) + 0.1 . --- Lorenzo Isella [EMAIL PROTECTED] wrote: Dear All, I am experiencing some problems with relocating an axis title. I visited the following link before posting: http://tolstoy.newcastle.edu.au/R/help/05/05/5283.html But this is not entirely what I would like to do Consider the example below: rm(list=ls()) D_mean-seq(-5,5,length=100) y-exp(-D_mean^2/5) pdf(my.pdf) plot(D_mean,y,type=l,yaxt=n,lty=2,lwd=2,col=black, ylab = list(expression(paste(dN/dlogD[agg], [*cm^-3*]))), xlab = expression(paste(D[agg], [nm])), cex.lab=1.2 ) title(2, mgp=c(0, .3, 0)) dev.off() I have the problem that the 3 in cubic centimeters (on the y axis) is somehow cut in the pdf file I generate. Everything would be fine if I could shift a bit the title of the y axis. It must be trivial, but so far I have not managed to do it. Any suggestions? Many thanks Lorenzo I tried playing with the mgp parameter, but I managed to move the __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] how to include bar values in a barplot?
Do you mean like this? my.values=10:15 x - barplot(my.values, ylim=c(0,11)) text(x, my.values, labels=my.values, pos=3) It is very bad practice and OOo should have its fingers slapped for perpetuating such a form. --- Donatas G. [EMAIL PROTECTED] wrote: How do I include bar values in a barplot (or other R graphics, where this could be applicable)? To make sure I am clear I am attaching a barplot created with OpenOffice.org which has barplot values written on top of each barplot. -- Donatas Glodenis http://dg.lapas.info __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Regsubsets statistics
Dear R-help, I have used the regsubsets function from the leaps package to do subset selection of a logistic regression model with 6 independent variables and all possible ^2 interactions. As I want to get information about the statistics behind the selection output, I´ve intensively searched the mailing list to find answers to following questions: 1. What should I do to get the statistics behind the selection (e.g. BIC)? summary.regsubsets(object) just returns * meaning in or meaning out. For the plot function generates BICs, it is obviously that these values must be computed and available somewhere, but where? Is it possible to directly get AIC values instead of BIC? 2. As to the plot function, I´ve encountered a problem with setting the ylim argument. I fear that this (nice!) particular plot function ignores many of these additional arguments. How can I nevertheless change this setting? 3. For it is not explicitly mentioned in the manual, can I really use regsubsets for logistic regression? Thanks a lot for your help, I hope these questions aren´t too general. Best wishes Markus Brugger _ Sie suchen E-Mails, Dokumente oder Fotos? Die neue MSN Suche Toolbar mit [[trailing spam removed]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] small sample techniques
About using t tests and confidence intervals for large samples - large may need to be very large. The old pre-computer-age rule of n = 30 is inadequate. For example, for an exponential distribution, the actual size of a nominal 2.5% one-sided t-test is not accurate to within 10% (i.e. between 2.25% 2.75%) until n is around 5000. The error (actual - nominal size) decreases very slowly, at the rate 1/sqrt(n). In practice, real distributions may be even more skewed than the exponential distribution, even though they appear less skewed, if they have long tails. In this case the sample size would need to be even larger for t procedures to be reasonably accurate. An alternative is to use bootstrapping. Bootstrap procedures that decrease at the rate 1/n include bootstrap t, BCa, and bootstrap tilting. Moshe Olshansky [EMAIL PROTECTED] wrote: If the two populations are normal the t-test gives you the exact result for whatever the sample size is (the sample size will affect the number of degrees of freedom). When the populations are not normal and the sample size is large it is still OK to use t-test (because of the Central Limit Theorem) but this is not necessarily true for the small sample size. You could use simulation to find the relevant probabilities. ... | Tim Hesterberg Senior Research Scientist | | [EMAIL PROTECTED] Insightful Corp.| | (206)802-23191700 Westlake Ave. N, Suite 500 | | (206)283-8691 (fax) Seattle, WA 98109-3044, U.S.A. | | www.insightful.com/Hesterberg | Short course - Bootstrap Methods and Permutation Tests Oct 10-11 San Francisco, 3-4 Oct UK. http://www.insightful.com/services/training.asp __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] converting character string to an expression
Hi Everyone, I would simply like to coerce a character string into an expression: something like: as.expression(paste(letters[1:3], collapse=+)) but I can't seem to get rid of the quotes. The only way I can get it to work is using as.formula: as.expression(as.formula(paste(~, paste(letters[1:3], collapse=+ but this requires the expression to have a tilde, which it will not always have. Thanks, Jarrod __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] mice package
Hi all, I am trying to run the mice package (for multiple imputation) on a data frame that is 5174 x 100. When I run mice(frame), I get the following response: Error in solve.default(t(xobs) %*% xobs) : Lapack routine dgesv: system is exactly singular and execution stops. I'm no expert at matrix algebra, so if someone could explain this to me and what I can do to get around it, I'd be very appreciative. Thanks. Best, -N. -- Nathan A. Paxton Ph.D. Candidate Dept. of Government, Harvard University Resident Tutor John Winthrop House, Harvard University napaxton AT fas DOT harvard DOT edu http://www.fas.harvard.edu/~napaxton === When you have to stay eight years away from California, you live in a perpetual state of homesickness. - Ronald Reagan The most courageous act is still to think for yourself. Aloud. -Coco Chanel __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] simulation-binomial
Does this do what you want? x2 - rbinom( 200, 1, ifelse(x, .95, p1/.6) ) y2 - rbinom( 300, 1, ifelse(y, .8, p1/.6) ) 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 sigalit mangut-leiba Sent: Wednesday, August 08, 2007 11:18 AM To: r-help Subject: Re: [R] simulation-binomial I have the probability: P(T+ / D+) i want to find P(T+,D+) which is: P(T+ / D+)*P(D+) and i have those probabilities. i dont know how to write this in R. something like this: (say p2 is the conditional prob. and p1 is the joint prob.) p2 - p1/.6 x - rbinom(200, 1, .6) y - rbinom(300, 1, .6) if (x) p2==0.95 if (y) p2==0.80 i don't know how to write the if condition. thank you for your reply, sigalit. On 8/8/07, Kyle Henderson [EMAIL PROTECTED] wrote: That depends on what you meant by writing the conditional probability. Bayes rule says that the probability of testing positive when one has the disease is calculated as follows: Pr(T+ | D+)=(Pr(D+ | T+)*Pr(T+))/Pr(D+) is that what you mean? Kyle H. Ambert Department of Behavioral Neuroscience Oregon Health Science University On 8/8/07, sigalit mangut-leiba [EMAIL PROTECTED] wrote: hello, i want to do a binomial simulation, by taking 200 var. from one group (x) and 300 from another (y). the prob. for disease=.6 in both groups. x - rbinom(200, 1, .6) y - rbinom(300, 1, .6) if the person is from group x - the probability to find the disease, assuming the person is sick, is .95, if he is from group Y its .80. i want to know the joint probability: p(the person has the disease and tested sick)=P(D+,T+). my problem is how to write the conditional prob. Thanks for your help, also reference on this subject (binomial simulation) would be great. Sigalit. [[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.htmlhttp://www.r-project.org /posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] S4 methods: unable to find an inherited method
Hi Paul,
H. Paul Benton [EMAIL PROTECTED] writes:
I consider myself pretty new to the whole OO based programming so
I'm sorry if I'm doing something stupid.
These sorts of questions might be better sent to the R-devel list...
xml-read.metlin(url)
Error in function (classes, fdef, mtable) :
unable to find an inherited method for function read.metlin,
for signature url
So the error message is telling you that it can't find a method for
the read.metlin generic that matches the class of the xml argument you
passed in.
You defined:
setMethod(read.metlin, xcmsRaw, function(xml) {
#Parsing the METLIN XML File
reading-readLines(xml)
#do rest of script
})
So there is a method for read.metlin when the xml argument is an
xcmsRaw object. As you show, you passed in an object with class
url.
url
description
http://metlin.scripps.edu/download/MSMS_test.XML;
class
url
Any help as to why I'm getting the inherited method error would be
great.
You either need to add a method specialized on the url class (warning,
url is not an S4 class, there will be tricks required). Or you need
to pass in an xcmsRaw object.
Cheers,
+ seth
--
Seth Falcon | Computational Biology | Fred Hutchinson Cancer Research Center
BioC: http://bioconductor.org/
Blog: http://userprimary.net/user/
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Re: [R] simulation-binomial
i will use that, thank you. sigalit. On 8/8/07, Greg Snow [EMAIL PROTECTED] wrote: Does this do what you want? x2 - rbinom( 200, 1, ifelse(x, .95, p1/.6) ) y2 - rbinom( 300, 1, ifelse(y, .8, p1/.6) ) 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 sigalit mangut-leiba Sent: Wednesday, August 08, 2007 11:18 AM To: r-help Subject: Re: [R] simulation-binomial I have the probability: P(T+ / D+) i want to find P(T+,D+) which is: P(T+ / D+)*P(D+) and i have those probabilities. i dont know how to write this in R. something like this: (say p2 is the conditional prob. and p1 is the joint prob.) p2 - p1/.6 x - rbinom(200, 1, .6) y - rbinom(300, 1, .6) if (x) p2==0.95 if (y) p2==0.80 i don't know how to write the if condition. thank you for your reply, sigalit. On 8/8/07, Kyle Henderson [EMAIL PROTECTED] wrote: That depends on what you meant by writing the conditional probability. Bayes rule says that the probability of testing positive when one has the disease is calculated as follows: Pr(T+ | D+)=(Pr(D+ | T+)*Pr(T+))/Pr(D+) is that what you mean? Kyle H. Ambert Department of Behavioral Neuroscience Oregon Health Science University On 8/8/07, sigalit mangut-leiba [EMAIL PROTECTED] wrote: hello, i want to do a binomial simulation, by taking 200 var. from one group (x) and 300 from another (y). the prob. for disease=.6 in both groups. x - rbinom(200, 1, .6) y - rbinom(300, 1, .6) if the person is from group x - the probability to find the disease, assuming the person is sick, is .95, if he is from group Y its .80. i want to know the joint probability: p(the person has the disease and tested sick)=P(D+,T+). my problem is how to write the conditional prob. Thanks for your help, also reference on this subject (binomial simulation) would be great. Sigalit. [[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.htmlhttp://www.r-project.org /posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[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] Namespace problem
Dimitri Mahieux wrote: Hi All, I have some questions about making a R-package. I would like to use a namespace. The package contains analysis functions and also a graphical user interface. I would like to allow user to use only the analysis function and not the GUI functions which are mainly bindings to graphical elements. I have exported analysis functions using export directives in a namespace file but when I want to use the GUI, it seems that all the GUI functions are unknown. Is there a way to solve this problem ? If you don't export the functions, they won't normally be visible outside your package, but you can still use the ::: notation to get them, e.g. mypackage:::myfunction gets myfunction even if it was not exported. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Test (please ignore)
Please excuse this -- I need to test whether I can get through to R-help! (Have failed repeatedly today). Ted. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] converting character string to an expression
I think you're looking for parse(text=paste(letters[1:3], collapse=+)) --- Jarrod Hadfield [EMAIL PROTECTED] wrote: Hi Everyone, I would simply like to coerce a character string into an expression: something like: as.expression(paste(letters[1:3], collapse=+)) but I can't seem to get rid of the quotes. The only way I can get it to work is using as.formula: as.expression(as.formula(paste(~, paste(letters[1:3], collapse=+ but this requires the expression to have a tilde, which it will not always have. Thanks, Jarrod __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] ksvm-kernel
HI
I am new to R.
I have one problem in the predict function of the kernlab.
I want to use ksvm and predict with kernelmatrix (S4 method for signature
'kernelMatrix')
#executing the following sentences
library(kernlab)
# identity kernel
k - function(x,y) {
n-length(x)
cont-0
for(i in 1:n){
if(x[i]==y[i]){
cont-cont+1
}
}
cont
}
class(k) - kernel
data(promotergene)
ind - sample(1:dim(promotergene)[1],20)
genetrain - promotergene[-ind, -1]
genetest - promotergene[ind,-1 ]
kx - kernelMatrix(k, as.matrix(genetrain))
y-as.vector(promotergene[-ind,1 ])
#y-as.factor(promotergene[-ind,1 ])
y
gene1 - ksvm(kx, y, type=C-svc)
gene1
genetype - predict(gene1,genetest)
Error en as.matrix(Z) : objeto Z no encontrado
#genetest1-as.matrix(genetest)
#genetype - predict(gene1,genetest1)
genetype
thank you,
nelsonhernandez
[EMAIL PROTECTED]
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and provide commented, minimal, self-contained, reproducible code.
[R] Systematically biased count data regression model
Dear all, I am attempting to explain patterns of arthropod family richness (count data) using a regression model. It seems to be able to do a pretty good job as an explanatory model (i.e. demonstrating relationships between dependent and independent variables), but it has systematic problems as a predictive model: It is biased high at low observed values of family richness and biased low at high observed values of family richness (see attached pdf). I have tried diverse kinds of reasonable regression models mostly as in Zeileis, et al. (2007), as well as transforming my variables, both with only small improvements. Do you have suggestions for making a model that would perform better as a predictive model? Thank you for your time. Sincerely, Matthew Bowser STEP student USFWS Kenai National Wildlife Refuge Soldotna, Alaska, USA M.Sc. student University of Alaska Fairbanks Fairbankse, Alaska, USA Reference Zeileis, A., C. Kleiber, and S. Jackman, 2007. Regression models for count data in R. Technical Report 53, Department of Statistics and Mathematics, Wirtschaftsuniversität Wien, Wien, Austria. URL http://cran.r-project.org/doc/vignettes/pscl/countreg.pdf. Code `data` - structure(list(D = c(4, 5, 12, 4, 9, 15, 4, 8, 3, 9, 6, 17, 4, 9, 6, 9, 3, 9, 7, 11, 17, 3, 10, 8, 9, 6, 7, 9, 7, 5, 15, 15, 12, 9, 10, 4, 4, 15, 7, 7, 12, 7, 12, 7, 7, 7, 5, 14, 7, 13, 1, 9, 2, 13, 6, 8, 2, 10, 5, 14, 4, 13, 5, 17, 12, 13, 7, 12, 5, 6, 10, 6, 6, 10, 4, 4, 12, 10, 3, 4, 4, 6, 7, 15, 1, 8, 8, 5, 12, 0, 5, 7, 4, 9, 6, 10, 5, 7, 7, 14, 3, 8, 15, 14, 7, 8, 7, 8, 8, 10, 9, 2, 7, 8, 2, 6, 7, 9, 3, 20, 10, 10, 4, 2, 8, 10, 10, 8, 8, 12, 8, 6, 16, 10, 5, 1, 1, 5, 3, 11, 4, 9, 16, 3, 1, 6, 5, 5, 7, 11, 11, 5, 7, 5, 3, 2, 3, 0, 3, 0, 4, 1, 12, 16, 9, 0, 7, 0, 11, 7, 9, 4, 16, 9, 10, 0, 1, 9, 15, 6, 8, 6, 4, 6, 7, 5, 7, 14, 16, 5, 8, 1, 8, 2, 10, 9, 6, 11, 3, 16, 3, 6, 8, 12, 5, 1, 1, 3, 3, 1, 5, 15, 4, 2, 2, 6, 5, 0, 0, 0, 3, 0, 16, 0, 9, 0, 0, 8, 1, 2, 2, 3, 4, 17, 4, 1, 4, 6, 4, 3, 15, 2, 2, 13, 1, 9, 7, 7, 13, 10, 11, 2, 15, 7), Day = c(159, 159, 159, 159, 166, 175, 161, 168, 161, 166, 161, 166, 161, 161, 161, 175, 161, 175, 161, 165, 176, 161, 163, 161, 168, 161, 161, 161, 161, 161, 165, 176, 175, 176, 163, 175, 163, 168, 163, 176, 176, 165, 176, 175, 161, 163, 163, 168, 163, 175, 167, 176, 167, 165, 165, 169, 165, 169, 165, 161, 165, 175, 165, 176, 175, 167, 167, 175, 167, 164, 167, 164, 181, 164, 167, 164, 176, 164, 167, 164, 167, 164, 167, 175, 167, 173, 176, 173, 178, 167, 173, 172, 173, 178, 178, 172, 181, 182, 173, 162, 162, 173, 178, 173, 172, 162, 173, 162, 173, 162, 173, 170, 178, 166, 166, 162, 166, 177, 166, 170, 166, 172, 172, 166, 172, 166, 174, 162, 164, 162, 170, 164, 170, 164, 170, 164, 177, 164, 164, 174, 174, 162, 170, 162, 172, 162, 165, 162, 165, 177, 172, 162, 170, 162, 170, 174, 165, 174, 166, 172, 174, 172, 174, 170, 170, 165, 170, 174, 174, 172, 174, 172, 174, 165, 170, 165, 170, 174, 172, 174, 172, 175, 175, 170, 171, 174, 174, 174, 172, 175, 171, 175, 174, 174, 174, 175, 172, 171, 171, 174, 160, 175, 160, 171, 170, 175, 170, 170, 160, 160, 160, 171, 171, 171, 171, 160, 160, 160, 171, 171, 176, 171, 176, 176, 171, 176, 171, 176, 176, 176, 176, 159, 166, 159, 159, 166, 168, 169, 159, 168, 169, 166, 163, 180, 163, 165, 164, 180, 166, 166, 164, 164, 177, 166), NDVI = c(0.187, 0.2, 0.379, 0.253, 0.356, 0.341, 0.268, 0.431, 0.282, 0.181, 0.243, 0.327, 0.26, 0.232, 0.438, 0.275, 0.169, 0.288, 0.138, 0.404, 0.386, 0.194, 0.266, 0.23, 0.333, 0.234, 0.258, 0.333, 0.234, 0.096, 0.354, 0.394, 0.304, 0.162, 0.565, 0.348, 0.345, 0.226, 0.316, 0.312, 0.333, 0.28, 0.325, 0.243, 0.194, 0.29, 0.221, 0.217, 0.122, 0.289, 0.475, 0.048, 0.416, 0.481, 0.159, 0.238, 0.183, 0.28, 0.32, 0.288, 0.24, 0.287, 0.363, 0.367, 0.24, 0.55, 0.441, 0.34, 0.295, 0.23, 0.32, 0.184, 0.306, 0.232, 0.289, 0.341, 0.221, 0.333, 0.17, 0.139, 0.2, 0.204, 0.301, 0.253, -0.08, 0.309, 0.232, 0.23, 0.239, -0.12, 0.26, 0.285, 0.45, 0.348, 0.396, 0.311, 0.318, 0.31, 0.261, 0.441, 0.147, 0.283, 0.339, 0.224, 0.5, 0.265, 0.2, 0.287, 0.398, 0.116, 0.292, 0.045, 0.137, 0.542, 0.171, 0.38, 0.469, 0.325, 0.139, 0.166, 0.247, 0.253, 0.466, 0.26, 0.288, 0.34, 0.288, 0.26, 0.178, 0.274, 0.358, 0.285, 0.225, 0.162, 0.223, 0.301, -0.398, -0.2, 0.239, 0.228, 0.255, 0.166, 0.306, 0.28, 0.279, 0.208, 0.377, 0.413, 0.489, 0.417, 0.333, 0.208, 0.232, 0.431, 0.283, 0.241, 0.105, 0.18, -0.172, -0.374, 0.25, 0.043, 0.215, 0.204, 0.19, 0.177, -0.106, -0.143, 0.062, 0.462, 0.256, 0.229, 0.314, 0.415, 0.307, 0.238, -0.35, 0.34, 0.275, 0.097, 0.353, 0.214, 0.435, 0.055, -0.289, 0.239, 0.186, 0.135, 0.259, 0.268, 0.258, 0.032, 0.489, 0.389, 0.298, 0.164, 0.325, 0.254, -0.059, 0.524, 0.539, 0.25, 0.175, 0.326, 0.302, -0.047, -0.301, -0.149, 0.358, 0.495, 0.311, 0.235, 0.558, -0.156, 0, 0.146, 0.329, -0.069, -0.352, -0.356, -0.206, -0.179, 0.467, -0.325, 0.39, -0.399, -0.165, 0.267, -0.334, -0.17, 0.58, 0.228, 0.234, 0.351, 0.3, -0.018, 0.125, 0.176, 0.322, 0.246,
[R] R memory usage
Hi All, I have two questions in terms of the memory usage in R (sorry if the questions are naive, I am not familiar with this at all). 1) I am running R in a linux cluster. By reading the R helps, it seems there are no default upper limits for vsize or nsize. Is this right? Is there an upper limit for whole memory usage? How can I know the default in my specific linux environment? And can I increase the default? 2) I use R to read in several big files (~200Mb each), and then I run: gc() I get: used (Mb) gc trigger (Mb) max used Ncells 23083130 616.4 51411332 1372.9 51411332 Vcells 106644603 813.7 240815267 1837.3 227550003 (Mb) 1372.9 1736.1 What do columns of used, gc trigger and max used mean? It seems to me I have used 616Mb of Ncells and 813.7Mb of Vcells. Comparing with the numbers of max used, I still should have enough memory. But when I try object.size(area.results) ## area.results is a big data.frame I get an error message: Error: cannot allocate vector of size 32768 Kb Why is that? Looks like I am running out of memory. Is there a way to solve this problem? Thank you very much! Jun __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Relocating Axis Label/Title --2
You don't need to move anything. Just allocate more room for what you have already. Try this: rm(list=ls()) D_mean-seq(-5,5,length=100) y-exp(-D_mean^2/5) pdf(my.pdf) ### par(mar = c(4.5, 4.5, 1, 1) + 0.1)# ### plot(D_mean,y,type=l,yaxt=n,lty=2,lwd=2,col=black, ylab = list(expression(paste(dN/dlogD[agg], [*cm^-3*]))), xlab = expression(paste(D[agg], [nm])), cex.lab=1.2 ) axis(2, mgp=c(0, 0.2, -2)) dev.off() Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Lorenzo Isella Sent: Wednesday, August 08, 2007 12:53 PM To: r-help@stat.math.ethz.ch Subject: [R] Relocating Axis Label/Title --2 Apologies for the previous mail (I sent it off too early by mistake). This is the correct example: rm(list=ls()) D_mean-seq(-5,5,length=100) y-exp(-D_mean^2/5) pdf(my.pdf) plot(D_mean,y,type=l,yaxt=n,lty=2,lwd=2,col=black, ylab = list(expression(paste(dN/dlogD[agg], [*cm^-3*]))), xlab = expression(paste(D[agg], [nm])), cex.lab=1.2 ) axis(2, mgp=c(0, 0.2, -2)) dev.off() With mgp() I can tune the distance between the ticks and the tick labels, but how can I move the axis label? I would like to move the one along y to visualize correctly the exponent 3. Kind Regards Lorenzo On 08/08/07, Lorenzo Isella [EMAIL PROTECTED] wrote: Dear All, I am experiencing some problems with relocating an axis title. I visited the following link before posting: http://tolstoy.newcastle.edu.au/R/help/05/05/5283.html But this is not entirely what I would like to do Consider the example below: rm(list=ls()) D_mean-seq(-5,5,length=100) y-exp(-D_mean^2/5) pdf(my.pdf) plot(D_mean,y,type=l,yaxt=n,lty=2,lwd=2,col=black, ylab = list(expression(paste(dN/dlogD[agg], [*cm^-3*]))), xlab = expression(paste(D[agg], [nm])), cex.lab=1.2 ) title(2, mgp=c(0, .3, 0)) dev.off() I have the problem that the 3 in cubic centimeters (on the y axis) is somehow cut in the pdf file I generate. Everything would be fine if I could shift a bit the title of the y axis. It must be trivial, but so far I have not managed to do it. Any suggestions? Many thanks Lorenzo I tried playing with the mgp parameter, but I managed to move the __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Relocating Axis Label/Title --2 (Second try.)
I hate it when the line feeds get lost and the message becomes unintelligible. I'm sorry. You don't need to move anything. Just allocate more room for what you have already. Try this: rm(list=ls()) D_mean-seq(-5,5,length=100) y-exp(-D_mean^2/5) pdf(my.pdf) ### par(mar = c(4.5, 4.5, 1, 1) + 0.1) ### plot(D_mean,y,type=l,yaxt=n,lty=2,lwd=2,col=black, ylab = list(expression(paste(dN/dlogD[agg], [*cm^-3*]))), xlab = expression(paste(D[agg], [nm])), cex.lab=1.2 ) axis(2, mgp=c(0, 0.2, -2)) dev.off() Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Lorenzo Isella Sent: Wednesday, August 08, 2007 12:53 PM To: r-help@stat.math.ethz.ch Subject: [R] Relocating Axis Label/Title --2 Apologies for the previous mail (I sent it off too early by mistake). This is the correct example: rm(list=ls()) D_mean-seq(-5,5,length=100) y-exp(-D_mean^2/5) pdf(my.pdf) plot(D_mean,y,type=l,yaxt=n,lty=2,lwd=2,col=black, ylab = list(expression(paste(dN/dlogD[agg], [*cm^-3*]))), xlab = expression(paste(D[agg], [nm])), cex.lab=1.2 ) axis(2, mgp=c(0, 0.2, -2)) dev.off() With mgp() I can tune the distance between the ticks and the tick labels, but how can I move the axis label? I would like to move the one along y to visualize correctly the exponent 3. Kind Regards Lorenzo On 08/08/07, Lorenzo Isella [EMAIL PROTECTED] wrote: Dear All, I am experiencing some problems with relocating an axis title. I visited the following link before posting: http://tolstoy.newcastle.edu.au/R/help/05/05/5283.html But this is not entirely what I would like to do Consider the example below: rm(list=ls()) D_mean-seq(-5,5,length=100) y-exp(-D_mean^2/5) pdf(my.pdf) plot(D_mean,y,type=l,yaxt=n,lty=2,lwd=2,col=black, ylab = list(expression(paste(dN/dlogD[agg], [*cm^-3*]))), xlab = expression(paste(D[agg], [nm])), cex.lab=1.2 ) title(2, mgp=c(0, .3, 0)) dev.off() I have the problem that the 3 in cubic centimeters (on the y axis) is somehow cut in the pdf file I generate. Everything would be fine if I could shift a bit the title of the y axis. It must be trivial, but so far I have not managed to do it. Any suggestions? Many thanks Lorenzo I tried playing with the mgp parameter, but I managed to move the __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] small sample techniques
As Thomas Lumley noted, there exist several versions of t-test. If you use t1 - t.test(x,y) then no assumption is made of x and y having equal variance and of the two sample sizes being equal and then an approximate t-test is used with an approximate number of degrees of freedom (and this is what you got). If you use t2 - t.test(x,y,var.equal=TRUE) then equal variance is assumed and you get 8 degrees of freedom. If you use t3 - t.test(x,y,paired=TRUE) then equal sample sizes are assumed and the number of degrees of freedom is 4 (5-1). --- Nair, Murlidharan T [EMAIL PROTECTED] wrote: Indeed, I understand what you say. The df of freedom for the dummy example is n1+n2-2 = 8. But when I run the t.test I get it as 5.08, am I missing something? -Original Message- From: Moshe Olshansky [mailto:[EMAIL PROTECTED] Sent: Tuesday, August 07, 2007 9:05 PM To: Nair, Murlidharan T; r-help@stat.math.ethz.ch Subject: Re: [R] small sample techniques Hi Nair, If the two populations are normal the t-test gives you the exact result for whatever the sample size is (the sample size will affect the number of degrees of freedom). When the populations are not normal and the sample size is large it is still OK to use t-test (because of the Central Limit Theorem) but this is not necessarily true for the small sample size. You could use simulation to find the relevant probabilities. --- Nair, Murlidharan T [EMAIL PROTECTED] wrote: If my sample size is small is there a particular switch option that I need to use with t.test so that it calculates the t ratio correctly? Here is a dummy example? á =0.05 Mean pain reduction for A =27; B =31 and SD are SDA=9 SDB=12 drgA.p-rnorm(5,27,9); drgB.p-rnorm(5,31,12) t.test(drgA.p,drgB.p) # what do I need to give as additional parameter here? I can do it manually but was looking for a switch option that I need to specify for t.test. Thanks ../Murli [[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-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] small sample techniques
On 9/08/2007, at 2:57 PM, Moshe Olshansky wrote:
As Thomas Lumley noted, there exist several versions
of t-test.
snip
If you use t3 - t.test(x,y,paired=TRUE) then equal
sample sizes are assumed and the number of degrees of
freedom is 4 (5-1).
This is seriously misleading. The assumption is not that the sample
sizes
are equal, but rather that there is ***just one sample***, namely
the sample of differences.
More explicitly the assumptions are that
x_i - y_i
are i.i.d. Gaussian with mean mu and variance sigma^2.
One is trying to conduct inference about mu, of course.
It should also be noted that it is a crucial assumption for the
``non-paired''
t-test that the two samples be ***independent*** of each other, as
well as
being Gaussian.
None of this is however germane to Nair's original question; it is
clear
that he is interested in a two-independent-sample t-test.
cheers,
Rolf Turner
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] small sample techniques
Well, this an explanation of what is done in the paired t-test (and why the number of df is as it is). I was too lazy to write all this. It is nice that some list members are less lazy! --- Rolf Turner [EMAIL PROTECTED] wrote: On 9/08/2007, at 2:57 PM, Moshe Olshansky wrote: As Thomas Lumley noted, there exist several versions of t-test. snip If you use t3 - t.test(x,y,paired=TRUE) then equal sample sizes are assumed and the number of degrees of freedom is 4 (5-1). This is seriously misleading. The assumption is not that the sample sizes are equal, but rather that there is ***just one sample***, namely the sample of differences. More explicitly the assumptions are that x_i - y_i are i.i.d. Gaussian with mean mu and variance sigma^2. One is trying to conduct inference about mu, of course. It should also be noted that it is a crucial assumption for the ``non-paired'' t-test that the two samples be ***independent*** of each other, as well as being Gaussian. None of this is however germane to Nair's original question; it is clear that he is interested in a two-independent-sample t-test. cheers, Rolf Turner ## Attention: This e-mail message is privileged and confidential. If you are not the intended recipient please delete the message and notify the sender. Any views or opinions presented are solely those of the author. This e-mail has been scanned and cleared by MailMarshal www.marshalsoftware.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] passing variables
Hi am new to jmeter... hear am explaining what my problem is? i have recorded the fallowing operations in my application by using proxy server...i login to the application by using username and password...after getting the home page i have created a new brand by giving brandname and description for that brand...after recording these operations what i want to do is iwant to create 5 user and all those 5 users need to login to the application and create new brands for each user... for this what i need to do...how to send variablesplease tell me how to get this -- This message was sent on behalf of [EMAIL PROTECTED] at openSubscriber.com http://www.opensubscriber.com/messages/r-help@stat.math.ethz.ch/topic.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Reading time/date string
Hello everyone, Can anyone tell me what function I should use to read time/date strings and turn them into a form such that I can easily calculate the difference of two? The strings I've got look like 10:17:07 02 Aug 2007. If I could calculate the number of seconds between them I'd be very happy! Cheers, Matthew __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] tcltk error on Linux
I am having trouble getting tcltk package to load on openSuse 10.2 running R-devel. I have specifically put my /usr/share/tcl directory in my PATH, but R doesn't seem to see it. I also have installed tk on my system. Any ideas on what the problem is? Also, note that I have some warning messages on starting up R, not sure what they mean or if they are pertinent. Thanks, Mark Warning messages: 1: In .updateMethodsInTable(fdef, where, attach) : Couldn't find methods table for conditional, package Category may be out of date 2: In .updateMethodsInTable(fdef, where, attach) : Methods list for generic conditional not found require(tcltk) Loading required package: tcltk Error in firstlib(which.lib.loc, package) : Tcl/Tk support is not available on this system sessionInfo() R version 2.6.0 Under development (unstable) (2007-08-01 r42387) i686-pc-linux-gnu locale: LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=en_US.UTF-8;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C attached base packages: [1] splines tools stats graphics grDevices utils datasets [8] methods base other attached packages: [1] affycoretools_1.9.3annaffy_1.9.1 xtable_1.5-0 [4] gcrma_2.9.1matchprobes_1.9.10 biomaRt_1.11.4 [7] RCurl_0.8-1XML_1.9-0 GOstats_2.3.8 [10] Category_2.3.19genefilter_1.15.9 survival_2.32 [13] KEGG_1.17.0RBGL_1.13.3annotate_1.15.3 [16] AnnotationDbi_0.0.88 RSQLite_0.6-0 DBI_0.2-3 [19] GO_1.17.0 limma_2.11.9 affy_1.15.7 [22] preprocessCore_0.99.12 affyio_1.5.6 Biobase_1.15.23 [25] graph_1.15.10 loaded via a namespace (and not attached): [1] cluster_1.11.7 rcompgen_0.1-15 -- --- Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry Indiana University School of Medicine 15032 Hunter Court, Westfield, IN 46074 (317) 490-5129 Work, Mobile VoiceMail (317) 663-0513 Home (no voice mail please) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] plot table with sapply - labeling problems
Hi List, I am trying to label a barplot group with variable names when using sapply unsucessfully. I can't seem to extract the names for the indiviual plots: test-as.data.frame(cbind(round(runif(50,0,5)),round(runif(50,0,3)),roun d(runif(50,0,4 sapply(test, table)-vardist sapply(test, function(x) round(table(x)/sum(table(x))*100,1) )-vardist1 par(mfrow=c(1,3)) sapply(vardist1, function(x) barplot(x, ylim=c(0,100),main=Varset1,xlab=names(x))) par(mfrow=c(1,1)) Names don't show up although names(vardist) works. Also I would like to put a single Title on this plot instead of repeating Varset three times. Any hints appreciated. Thanx Herry __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Reading time/date string
Look at some of these functions... DateTimeClasses(base) Date-Time Classes as.POSIXct(base)Date-time Conversion Functions cut.POSIXt(base)Convert a Date or Date-Time Object to a Factor format.Date(base) Date Conversion Functions to and from Character Mark --- Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry Indiana University School of Medicine 15032 Hunter Court, Westfield, IN 46074 (317) 490-5129 Work, Mobile VoiceMail (317) 663-0513 Home (no voice mail please) ** Matthew Walker wrote: Hello everyone, Can anyone tell me what function I should use to read time/date strings and turn them into a form such that I can easily calculate the difference of two? The strings I've got look like 10:17:07 02 Aug 2007. If I could calculate the number of seconds between them I'd be very happy! Cheers, Matthew __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] R2WinBUGS results not different with different runs
toby909 at gmail.com writes: Gregor Gorjanc wrote: toby909 at gmail.com writes: Is this a specialty with R2WinBUGS? Does it have something to do with the seed value? Isnt the seed value reset everytime I restart winbugs? I always have the same seed if I start WinBUGS multiple times. So you get exactly the same chain, numerically, when rerunning the same model, with the same number of iterations, everything the same.? I just tried now with Gelman's schools example from bugs() help page. I runned the same job twice, with exactly the same initial values. Notes that WinBUGS uses always the same starting seed. I got this schools.sim$last.values [[1]] [[1]]$theta [1] 23.700 10.060 12.760 13.090 1.693 14.390 7.599 3.961 [[1]]$mu.theta [1] 25.03 [[1]]$sigma.theta [1] 20.08 schools.sim2$last.values [[1]] [[1]]$theta [1] 23.700 10.060 12.760 13.090 1.693 14.390 7.599 3.961 [[1]]$mu.theta [1] 25.03 [[1]]$sigma.theta [1] 20.08 Wouldnt that be problematic if every researcher in the world who uses winbugs uses the same sequence of random numbers? R's random numbers are different each time, because the seed is linked to the clock in your PC. You can reset the seed if you want. Having the same seed is also nice for repeatability. Gregor __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
