Re: [R] Histogram ploting
Mateusz == Mateusz £oskot [EMAIL PROTECTED]
on Sun, 18 Apr 2004 17:13:34 +0200 writes:
Mateusz Hi Christophe, On 4/18/2004 3:17 PM, Christophe
Mateusz Pallier wrote:
The 'hist' function works on the raw data. In your data
set example, you have already computed the number of data
points in each bin.
Mateusz Yes, you are right. I evidently misunderstood the
Mateusz hist function usage described in manuals.
What you really want is probably a barplot of N You could
display your data:
plot(Class,N,'h')
Mateusz Yes, that's right. Thank you very much.
well, I think you did have real histogram data,
and in teaching about graphics I do emphasize the difference
between a barplot {in R: plot of table(); space between bars}
and a histogram {continuous x; no space between bars}.
In this case, I'd rather construct an object of class
'histogram' and plot() it, i.e., call the plot.histogram method:
(mids - seq(12.5, 47.5, by = 5))
N - c(3,10, 12,8, 7,3, 4,2)
## Construct breaks from mids in general
## (here, simply br - seq(10,50,by=5) is easier)
dx - mean(diff(mids))
br - (mids[-1] + mids[-length(mids)])/2
(br - c(br[1] - dx, br, br[length(br)] + dx))
his - list(breaks=br, counts=N, mids = mids)
class(his) - histogram
plot(his, main = Histogram of my stuff)
Regards,
Martin
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RE: [R] outliers using Random Forest
Dear Andy, Thanks for your quick answer. I increased the number of trees and the outlyingness measure got more stable. But still I do not know if I am working with the raw measure or with the normalized measure mentioned in the Breiman's Wald lecture. The normalized measure nout is nout=(nout-med)/mean(abs(nout-med)) where med is the median of the class containing the case correponding to nout. Best regards Edgar Acuna On Sun, 18 Apr 2004, Liaw, Andy wrote: The thing to do is probably: 1. Use fairly large number of trees (e.g., 1000). 2. Run a few times and average the results. The reason for the instability is sort of two fold: 1. The random forest algorithm itself is based on randomization. That's why it's probably a good idea to have 500-1000 trees to get more stable proximity measures (of which the outlying measures are based on). 2. If you are running randomForest in unsupervised mode (i.e., not giving it the class labels), then the program treats the data as class 1, creates a synthetic class 2, and run the classification algorithm to get the proximity measures. You probably need to run the algorithm a few times so that the result will be based on several simulated data, instead of just one. HTH, Andy From: Edgar Acuna Hello, Does anybody know if the outscale option of randomForest yields the standarized version of the outlier measure for each case? or the results are only the raw values. Also I have notice that this measure presents very high variability. I mean if I repeat the experiment I am getting very different values for this measure and it is hard to flag the outliers. This does not happen with two other criteria than I am using: LOF and Bay's Orca. I am getting several cases that can be considered as outliers with both approaches. I run my experiments using Bupa and Diabetes available at UCI Machine database repository. Thanks in advance for any response. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Notice: This e-mail message, together with any attachments, contains information of Merck Co., Inc. (One Merck Drive, Whitehouse Station, New Jersey, USA 08889), and/or its affiliates (which may be known outside the United States as Merck Frosst, Merck Sharp Dohme or MSD and in Japan as Banyu) that may be confidential, proprietary copyrighted and/or legally privileged. It is intended solely for the use of the individual or entity named on this message. If you are not the intended recipient, and have received this message in error, please notify us immediately by reply e-mail and then delete it from your system. -- __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Turning windows screen buffering on and off
The buffered output is a nice thing, and if the user want to use it, then fine! However, it should be nice to know if it is set ON or OFF, and to temporarily change it for some outputs in R scripts. I think this is the primary request. Then, your request appears to me as a secondary one: to set buffered output ON or OFF at statup of Rgui (by defining it in the preference panel). Personnally, I would really like to have both. Best, Philippe Grosjean ...°})) ) ) ) ) ) ( ( ( ( ( Prof. Philippe Grosjean \ ___ ) \/ECO\ ( Numerical Ecology of Aquatic Systems /\___/ ) Mons-Hainaut University, Pentagone / ___ /( 8, Av. du Champ de Mars, 7000 Mons, Belgium /NUM\/ ) \___/\ ( phone: + 32.65.37.34.97, fax: + 32.65.37.33.12 \ ) email: [EMAIL PROTECTED] ) ) ) ) ) SciViews project coordinator (http://www.sciviews.org) ( ( ( ( ( ... -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED] Sent: Friday, 16 April, 2004 18:09 To: [EMAIL PROTECTED] Subject: RE: [R] Turning windows screen buffering on and off Well, I too would like to be able to set buffered output to false for Rgui Windows without user intervention. Maybe it could be set via the Edit - Gui Preferences so that it can be saved and set at startup. The GUI Preferences are part of the interface and not the standard language definition right?...so that seems to be a good spot to set something like that. Benjamin Stabler Transportation Planning Analysis Unit Oregon Department of Transportation 555 13th Street NE, Suite 2 Salem, OR 97301 Ph: 503-986-4104 - Message: 75 Date: Fri, 16 Apr 2004 08:59:43 +0100 (BST) From: Prof Brian Ripley [EMAIL PROTECTED] Subject: RE: [R] Turning windows screen buffering on and off To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Message-ID: [EMAIL PROTECTED] Content-Type: TEXT/PLAIN; charset=US-ASCII No, and options() really is part of the R/S language not the interface. See the rw-FAQ Q6.3 for how to manage the buffering more effectively. (Hint: you need to put calls to flush.console() in your code.) On Fri, 16 Apr 2004 [EMAIL PROTECTED] wrote: I meant via a function or something like: options( buffered.output = FALSE) Sorry, I should have made that clearer. Cheers Toby -Original Message- From: Roger D. Peng [mailto:[EMAIL PROTECTED] Sent: Friday, April 16, 2004 1:11 PM To: Patterson, Toby (Marine, Hobart) Cc: [EMAIL PROTECTED] Subject: Re: [R] Turning windows screen buffering on and off Ctrl-W. -roger [EMAIL PROTECTED] wrote: All, Does anyone know if there is an option I can set to turn screen-buffered output on and off with the win32 rgui? (Apart from the point and click method). I am running some simulations where it is useful to watch output but it gets mildly tiresome having to manually switch things on and off via the gui. Thanks Toby. version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major1 minor8.1 year 2003 month11 day 21 language R __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- 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 272860 (secr) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] SE for combined data
Dear all I have just had the question from a colleague. I know that it is not directly related to R (I will probably use R to do the analysis), but I hope someone can give us some insight: Thanks, AJ Smit I sampled populations of a seaweed in the intertidal in order to estimate the standing biomass of that seaweed at that site. Due to clumped distribution patterns, I chose a stratified sampling system, as follows. In each of three subjectively defined biomass classes (low, medium and high biomass density), four quadrats (usually) were haphazardly placed, and the biomass in those quadrats harvested. This provided an estimate of the biomass density present in that biomass density class. The area of ground covered by that biomass density class was also estimated, and, by combining the estimated biomass density and the area covered by that biomass density class, the total biomass in that biomass density class was estimated. When the estimated biomass in the three biomass density classes was combined, I had a figure for the standing biomass for that site.. So, for each biomass density class, I have a number of biomass density estimates (usually, but not always, four), and an estimate of the area covered by that class. I repeated this at a number of sites. Biomass density classes were not necessarily the same between sites. Given that I can calculate measures of variation for each biomass density class, is there a way to combine these data, presumably weighted by the area covered by each biomass density class, and calculate the standard error for the final biomass estimate at each site? Thanks Neil -- ~~~ Dr Albertus J. Smit Department of Botany University of Cape Town Private Bag Rondebosch 7700 Cape Town SOUTH AFRICA Tel. +27 21 689 3032 Fax +27 21 650 4041 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Box-Ljung p-value - Test for Independence
1) A small p-value is evidence that there is dependence.
So you want to see large p-values. But a large p-value
is not really evidence of independence -- merely a lack
of evidence of dependence.
Hello Aroon,
additionally to the points voiced by Patrick, only if your series at hand is
normally distributed you can infer from uncorrelatedness (Ljung-Box test) to
independence. These two are equivalent in the case of normally distributed
random variables. Uncorrelated rv are not necessarily independent, whereas
the opposite is true. Hence, you want to utilise Jarque-Bera-test, or some
test for normality, too.
HTH,
Bernhard
You might be able to get a hint of the power of your test
(which is what you really care about) from the working
paper about Ljung-Box on the Burns Statistics website.
2) The statistic is really an average of the lags up to the
stated lag. So if the dependence is all at lag 5, tests with
lags below 5 have no power, the lag 5 test has maximum
power, and the power decreases as the lag of the test
increases above 5.
Patrick Burns
Burns Statistics
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and A Guide for the Unwilling S User)
Aroon Nataraj wrote:
Hi all
I'm using the Box-Ljung test (from within R) to test if a
time-series
in independently distributed.
2 questions:
1) p-value returned by Box-Ljung:
IF I want to test if the time-series is independant at say 0.05
sig-level (it means that prob of erroneously accepting that the
time-series is independent is 0.05 right?)
-- then do I consider time-series as independant when
-- p-value (from Box-Ljung) 0.05
OR
-- p-value 0.05
Or should i be using (0.95) instead of (0.05) for this case. I'm
confused about this (this is a goodness-of-fit test?).
2) Box-Ljung takes a lag argument, say lag=k.
Does it check all lags upto k
OR
Does it check only AT k (i.e acf val is small at only k?)
thank you in advance. I apologize if the questions are very basic.
Aroon
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[R] solve 2 var problem
Hi I'm getting started with R and i have difficulties finding how to solve this problem in R : Find x,y satisfying Trace(K) C, K positive where ( K=x.A+y.B), [A,B,K square Matrix in R(n x n), x,y in R] Thanks in advance, Ramzi [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] trend turning points
does anybody know of a nice test to detect trend turning points in time series? Possibly with reference? You can look at the function breakpoints() in the package strucchange and the function segmented() in the package segmented which do segmentation of (generalized) linear regression models. The former tries to fit fully segmented regression models, the latter broken line trends. References are given on the respective help pages. A suitable test for a change in trend in linear regression models is the OLS-based CUSUM test with a Cramer-von Mises functional of Kraemer Ploberger (1996, JoE) which is available via efp() in strucchange and associated methods. I have seen strucchange with great interest but somehow the combination of the R and the statistics knowledge that is required to access it feels like a bottleneck :-) Could someone perhaps write a simple R program which illustrates the facilities and usage? -- Ajay Shah Consultant [EMAIL PROTECTED] Department of Economic Affairs http://www.mayin.org/ajayshah Ministry of Finance, New Delhi __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Non-Linear Regression (Cobb-Douglas and C.E.S)
Dear all,
For estimating Cobb-Douglad production Function [ Y = ALPHA *
(L^(BETA1)) *
(K^(BETA2)) ], i want to use nls function (without
linearizing it). But
how can i get initial values?
options(prompt= R )
R Y - c(59.6, 63.9, 73.5, 75.6, 77.3, 82.8, 83.6, 84.9,
90.3, 80.5,
73.5, 60.3, 58.2, 64.4, 75.4, 85, 92.7, 85.4, 92.3, 101.2,
113.3, 107.8,
105.2, 107.1, 108.8, 131.4, 130.9, 134.7, 129.1, 147.8, 152.1, 154.3,
159.9) # production
R L - c(39.4, 41.4, 43.9, 43.3, 44.5, 45.8, 45.9, 46.4,
47.6, 45.5,
42.6, 39.3, 39.6, 42.7, 44.2, 47.1, 48.2, 46.4, 47.8, 49.6,
54.1, 59.1,
64.9, 66, 64.4, 58.9, 59.3, 60.2, 58.7, 60, 63.8, 64.9, 66) #
employment
R K - c(236.2, 240.2, 248.9, 254.5, 264.1, 273.9, 282.6,
290.2, 299.4,
303.3, 303.4, 297.1, 290.1, 285.4, 287.8, 292.1, 300.3, 301.4, 305.6,
313.3, 327.4, 339, 347.1, 353.5, 354.1, 359.4, 359.3, 365.2,
363.2, 373.7,
386, 396.5, 408) # capital
R klein - cbind(Y,L,K)
R klein.data-data.frame(klein)
R coef(lm(log(Y)~log(L)+log(K)))
# i used these linearized model's estimated parameters as
initial values
(Intercept) log(L) log(K)
-3.6529493 1.0376775 0.7187662
R nls(Y~ALPHA * (L^(BETA1)) * (K^(BETA2)),
data=klein.data, start =
c(ALPHA=-3.6529493,BETA1=1.0376775,BETA2=0.7187662), trace = T)
6852786785 : -3.6529493 1.0376775 0.7187662
1515217 : -0.02903916 1.04258165 0.71279051
467521.8 : -0.02987718 1.67381193 -0.05609925
346945.7 : -0.5570735 10.2050667 -10.2087997
Error in numericDeriv(form[[3]], names(ind), env) :
Missing value or an Infinity produced when
evaluating the model
1. What went wrong? I think the initial values are not good
enough: How can
i make a grid search?
2. How can i estimate C.E.S Production Function [ Y = GAMA *
((DELTA*K^(-BETA)) + ((1-DELTA)*L^(-BETA)))^(-PHI/BETA) ]
using the same
data? How to get the initial value?
Dear James, Wettenhall,
as far as the CES production function is concerned, you might want to
utilise the Kmenta approximation. The following link elucidates this
approach and other feasible estimation techniques.
http://www.cu.lu/crea/projets/mod-L/prod.pdf
HTH,
Bernhard
N.B.: The data file is available at
http://www.angelfire.com/ab5/get5/klein.txt
Any response / help / comment / suggestion / idea / web-link / replies will
be greatly appreciated.
Thanks in advance for your time.
___
Mohammad Ehsanul Karim [EMAIL PROTECTED]
Institute of Statistical Research and Training
University of Dhaka, Dhaka- 1000, Bangladesh
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Re: [R] trend turning points
On Mon, 19 Apr 2004 14:27:00 +0530 Ajay Shah wrote: does anybody know of a nice test to detect trend turning points in time series? Possibly with reference? You can look at the function breakpoints() in the package strucchange and the function segmented() in the package segmented which do segmentation of (generalized) linear regression models. The former tries to fit fully segmented regression models, the latter broken line trends. References are given on the respective help pages. A suitable test for a change in trend in linear regression models is the OLS-based CUSUM test with a Cramer-von Mises functional of Kraemer Ploberger (1996, JoE) which is available via efp() in strucchange and associated methods. I have seen strucchange with great interest but somehow the combination of the R and the statistics knowledge that is required to access it feels like a bottleneck :-) Could someone perhaps write a simple R program which illustrates the facilities and usage? The package has an accompanying vignette which explains the ideas behind it and how to use its inference functions. The breakpoints() facilities are also documented in a paper by Zeileis, Kleiber, Kraemer, Hornik (2003, CSDA). A mini-example would be: ## load package and data R data(Nile) R library(strucchange) ## fit, visualize and test OLS-based CUSUM test ## with a Cramer-von Mises functional R ocus - efp(Nile ~ 1, type = OLS-CUSUM) R plot(ocus, functional = meanL2) R sctest(ocus, functional = meanL2) ## estimate breakpoints R bp - breakpoints(Nile ~ 1) R plot(bp) R summary(bp) hth, Z -- Ajay Shah Consultant [EMAIL PROTECTED] Department of Economic Affairs http://www.mayin.org/ajayshah Ministry of Finance, New Delhi __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How can I get the length of returning array in c++ calling R code?
Hello, Now I would like to call R code from c++ and R code return a dynamic length of integer array. I used length(eval(VECTOR_ELT(expression,1),R_GlobalEnv)) to return the length of result array in c++, but it did not work. What can I do to get the length? Thank you very much! Baiyi Song CEI Dortmund Univercity __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] looking up value from a pair of vectors
Hello R community. Can anyone inform me how to solve this short problem? I have a dataset that I want to recode according to a pair of short numerical vectors. Here is a short example: # Start R code map1 - matrix(ncol=3, byrow=T,c( 197796,label0,1, 197797,label1,2, 197798,label2,3, 197799,label3,4, 197800,label4,5)) xx - as.integer(map1[,1]) lab - map1[,2] yy - as.integer(map1[,3] data1a - c(197797, 197798, 197799, 197800, 197797, 197797, 197797, 197797) data1b - some.function.or.combinaion.of.few(data1a,xx,yy) # desired result for data1b data1b 2 3 4 5 2 2 2 2 # End R code data1a is always longer than xx. all numerical entries in data1a are members of xx. (in the more general problem, the the number 197796 may or may not be included in data1a) value in data1a that are not members of xx would produce a Na Thanks in advance for all replies, Stefan Jonsson __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] looking up value from a pair of vectors
match(data1a, xx) [1] 2 3 4 5 2 2 2 2 On Mon, 19 Apr 2004, Stefann Jonsso wrote: Hello R community. Can anyone inform me how to solve this short problem? I have a dataset that I want to recode according to a pair of short numerical vectors. Here is a short example: # Start R code map1 - matrix(ncol=3, byrow=T,c( 197796,label0,1, 197797,label1,2, 197798,label2,3, 197799,label3,4, 197800,label4,5)) xx - as.integer(map1[,1]) lab - map1[,2] yy - as.integer(map1[,3] data1a - c(197797, 197798, 197799, 197800, 197797, 197797, 197797, 197797) data1b - some.function.or.combinaion.of.few(data1a,xx,yy) # desired result for data1b data1b 2 3 4 5 2 2 2 2 # End R code data1a is always longer than xx. all numerical entries in data1a are members of xx. (in the more general problem, the the number 197796 may or may not be included in data1a) value in data1a that are not members of xx would produce a Na -- 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 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] looking up value from a pair of vectors
(m - match(data1a, map1[ ,1])) [1] 2 3 4 5 2 2 2 2 map1[m ,2] # to return labels [1] label1 label2 label3 label4 label1 label1 label1 label1 On Mon, 2004-04-19 at 11:34, Stefann Jonsso wrote: Hello R community. Can anyone inform me how to solve this short problem? I have a dataset that I want to recode according to a pair of short numerical vectors. Here is a short example: # Start R code map1 - matrix(ncol=3, byrow=T,c( 197796,label0,1, 197797,label1,2, 197798,label2,3, 197799,label3,4, 197800,label4,5)) xx - as.integer(map1[,1]) lab - map1[,2] yy - as.integer(map1[,3] data1a - c(197797, 197798, 197799, 197800, 197797, 197797, 197797, 197797) data1b - some.function.or.combinaion.of.few(data1a,xx,yy) # desired result for data1b data1b 2 3 4 5 2 2 2 2 # End R code data1a is always longer than xx. all numerical entries in data1a are members of xx. (in the more general problem, the the number 197796 may or may not be included in data1a) value in data1a that are not members of xx would produce a Na Thanks in advance for all replies, Stefan Jonsson __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] New unique name
On 19-Apr-04 Erich Neuwirth wrote: In some languages there is a function gensym() which returns a new unique name (in the current environment). This is quite helpful when one has to do temporary assignments. I could not find such a function in R. Is there one? If you are running R on a Unix-like system (such as Linux) then you could use the mktemp function (see 'man mktemp'). In R, system(mktemp -u tmpXX) will return a unique identifier (and will not, because of the -u flag, create the file). The identifier will be of the form (e.g.) tmpdPT6Nw, i.e. with the 6 X's replaced by random characters. A bit sledgehammer for nut, but as least it meets your needs! (Strictly speaking, uniqueness in repeated use is not absolutely guaranteed with the -u option, since strict uniqueness depends on 'mktemp' checking for an existing file with the same name. However, since there are some (26+26+10)^6 10^10 combinations to randomly choose from, you should be pretty safe.) Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 167 1972 Date: 19-Apr-04 Time: 12:00:35 -- XFMail -- __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R analog of Matlab eigs function
Hi, I was wondering if anyone knew of an implementation of a function similar to eigs in Matlab (full description here: http://www.mathworks.com/access/helpdesk/help/techdoc/ref/eigs.html). This function differs from the standard eigen in that it computes a *few* eigenvectors for cases in which your matrix is very large and/or you don't need all the eigenvectors. (It uses the Arnoldi-Lanczos iterative method, as implemented in C in ARPACK). For example, this is the case for classical multidimensional scaling when you only need the first 2 eigenvectors. I feel almost certain that something like this is probably somewhere in R, possibly even as a (hidden) subroutine within a function like cmdscale. However, I didn't turn anything up after quite a bit of googling and help.search(). Would appreciate any help with this. Thanks in advance, Balaji Srinivasan __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] One inflated Poisson or Negative Binomal regression
Dr. Flom, I was searching the web for any examples of one-inflated negative binomial regression, and ran across your post. Fittingly, I am working on the analysis of data from the NIDA Cooperative Agreement where I had the pleasure of working with Sherry Deren and other folks at NDRI. NBR does a poor job of modeling number of sex partners. (I am using Stata.) Did you have any luck modeling a one-inflated negative binomial? Brian Weir Program Design Evaluation Services Multnomah County Health Department [EMAIL PROTECTED] [EMAIL PROTECTED] [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] New unique name
(Ted Harding) wrote:
A bit sledgehammer for nut, but as least it meets your needs!
You could even use uuidgen to create a universally unique ID, and then
use make.names to R-ify it:
make.names(system(uuidgen,intern=T))
[1] aa33d26c.88a5.4eab.94ba.5073c4718ffe
make.names(system(uuidgen,intern=T))
[1] d5aea7a0.81e9.4690.8d48.d74fd2b50a83
make.names(system(uuidgen,intern=T))
[1] X2570d3e0.6b07.42be.a9c6.3701ac82b4f0
Of course the requirement was to make an object with a name that
didn't already exist, and there's no guarantee that you haven't already
created an object called X2570d3e0.6b07.42be.a9c6.3701ac82b4f0 for
some reason.
So that's no good.
How about getting an alphabetically-sorted list of all the current
objects and then using the last one augmented with something:
makeUnique - function(){
allObjects=sort(ls(pos=seq(1:length(search()
paste(allObjects[length(allObjects)],-temp,sep='')
}
I'm sure someone can come up with a bigger sledgehammer, or a reason
why this wont always be unique. What's the longest name of an object in
R? That'll break it... Pah.
Baz
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Re: [R] New unique name
On Mon, 19 Apr 2004, Barry Rowlingson wrote:
(Ted Harding) wrote:
A bit sledgehammer for nut, but as least it meets your needs!
You could even use uuidgen to create a universally unique ID, and then
use make.names to R-ify it:
make.names(system(uuidgen,intern=T))
[1] aa33d26c.88a5.4eab.94ba.5073c4718ffe
make.names(system(uuidgen,intern=T))
[1] d5aea7a0.81e9.4690.8d48.d74fd2b50a83
make.names(system(uuidgen,intern=T))
[1] X2570d3e0.6b07.42be.a9c6.3701ac82b4f0
Of course the requirement was to make an object with a name that
didn't already exist, and there's no guarantee that you haven't already
created an object called X2570d3e0.6b07.42be.a9c6.3701ac82b4f0 for
some reason.
So that's no good.
How about getting an alphabetically-sorted list of all the current
objects and then using the last one augmented with something:
makeUnique - function(){
allObjects=sort(ls(pos=seq(1:length(search()
paste(allObjects[length(allObjects)],-temp,sep='')
}
I'm sure someone can come up with a bigger sledgehammer, or a reason
why this wont always be unique. What's the longest name of an object in
R? That'll break it... Pah.
There is more to the environment than the search path, which is what makes
this tricky. I suspect Erich really only wanted to avoid objects in the
current frame (which is not on the search path except for interactive use
of R), but he didn't tell us too precisely.
--
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 272860 (secr)
Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] geoR - help for bayesian modelling
Hi, I am trying to do a bayesian prediction for soil pollution data above a certain threshold, using geoR. Everything is working fine until i am doing the krig.bayes ( I am using as a guide the geoR tutorial from the web page http://www.est.ufpr.br/geoR/geoRdoc/geoRintro.html#starting). I tried to do the prediction on a grid 67 by 113 cells and my computer is freezing to death. At larger numbers of cells it tells me after a while that it reaches the max. memory of 511 Mb. My computer has only 512 Mb of RAM. What RAM capacity should i look for to do a 150 x 250 cell grid??? (I tried the modelling on a 1 Gb RAM computer and it didn't work either). I am interested to do a modelling where my resolution is 5 m x 5 m (150 x 250 grid cell). If i want to do the prediction on my initial data locations (well, actually the prediction points are shifted 1 m in X and respectively Y direction, so the raw data coordinates don't coincide with the prediction coordinates) i am getting the following error using the command: zn.bayes - krige.bayes(zn.gdata, loc = xy, model = model.control(cov.model = exponential, lambda = 0), prior = prior.control(phi.prior =exponential, phi = 89.1894), output=output.control(n.predictive=2, mean.var = TRUE, quantile = c(0.025,0.25, 0.5, 0.75, 0.975), threshold = c(300))) Error in cond.sim(env.loc = base.env, env.iter = iter.env, loc.coincide = get(loc.coincide, : chol: matrix not pos def, diag[13]= -1.279220e-018 I will really appreciate any suggestion you may have. Thank you so much, Monica __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to write an S4 method for sum or a Summary generic
If I have a class Foo, then i can write an S3 method for sum for it:
setClass(Foo,representation(a=integer));aFoo=new(Foo,a=c(1:3,NA))
sum.Foo - function(x,na.rm){print(x);print(na.rm);sum([EMAIL PROTECTED],na.rm=na.rm)}
sum(aFoo)
But how do I write an S4 method for this? All my attempts to do so have
foundered. For example
setMethod(sum,signature(Foo,logical),
function(x,na.rm){print(x);print(na.rm);sum([EMAIL PROTECTED],na.rm=na.rm)}
creates a method which seems to despatch on na.rm=Foo:
getMethods(sum)
na.rm = ANY:
function (..., na.rm = FALSE)
.Internal(sum(..., na.rm = na.rm))
na.rm = Foo:
function (..., na.rm = FALSE)
{
.local - function (x, na.rm)
{
print(x)
print(na.rm)
sum([EMAIL PROTECTED], na.rm = na.rm)
}
.local(..., na.rm = na.rm)
}
na.rm = missing:
function (..., na.rm = FALSE)
.Internal(sum(..., na.rm = na.rm))
##:(inherited from na.rm = ANY)
Pages 350-352 of the Green book discuss at some length how to write a
generic function for Summary group generics which uses tail recursion to
allow the correct method to be called on each member of a ... argument list.
But it gives no examples of what individual method functions need to look
like. Any ideas or a place to look for working code?
Jonathan Swinton, Statistical Scientist, Computational Biology, Pathway
Analysis, Global Sciences and Information, AstraZeneca.
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Re: [R] New unique name and fixing getAnywhere()
# what about
gensym - function(root=GeneratedSymbolname, pool=c(letters, LETTERS,
0:1), n=16, sep=_)
{
todo - TRUE
while (todo){
symbolname - paste(root, paste(sample(pool, n, TRUE), collapse=),
sep=sep)
todo - length(getAnywhere(symbolname)$objs)
}
symbolname
}
# but this requires a slightly changed version of getAnywhere()
# which currently finds: getAnywhere(find)
# but does not find: symbolname - find; getAnywhere(symbolname)
# (BTW current getAnywhere() has returnvalue$objs whereas the documentation
says returnvalue$funs)
# the following patch avoids this problem and is more aligned with get()
getAnywhere - function(x)
{
stopifnot(is.character(x))
objs - list()
where - character(0)
visible - logical(0)
if (length(pos - find(x, numeric = TRUE))) {
objs - lapply(pos, function(pos, x) get(x, pos = pos),
x = x)
where - names(pos)
visible - rep.int(TRUE, length(pos))
}
if (length(grep(., x, fixed = TRUE))) {
np - length(parts - strsplit(x, ., fixed = TRUE)[[1]])
for (i in 2:np) {
gen - paste(parts[1:(i - 1)], collapse = .)
cl - paste(parts[i:np], collapse = .)
if (!is.null(f - getS3method(gen, cl, TRUE))) {
ev - topenv(environment(f), NULL)
nmev - if (isNamespace(ev))
getNamespaceName(ev)
else NULL
objs - c(objs, f)
msg - paste(registered S3 method for, gen)
if (!is.null(nmev))
msg - paste(msg, from namespace, nmev)
where - c(where, msg)
visible - c(visible, FALSE)
}
}
}
for (i in loadedNamespaces()) {
ns - asNamespace(i)
if (exists(x, envir = ns, inherits = FALSE)) {
f - get(x, envir = ns, inherits = FALSE)
objs - c(objs, f)
where - c(where, paste(namespace, i, sep = :))
visible - c(visible, FALSE)
}
}
ln - length(objs)
dups - rep.int(FALSE, ln)
objs2 - lapply(objs, function(x) {
if (is.function(x))
environment(x) - NULL
x
})
if (ln 1)
for (i in 2:ln) for (j in 1:(i - 1)) if (identical(objs2[[i]],
objs2[[j]])) {
dups[i] - TRUE
break
}
res - list(name = x, objs = objs, where = where, visible = visible,
dups = dups)
class(res) - getAnywhere
res
}
--
Ab sofort DSL-Tarif ohne Grundgebühr: http://www.gmx.net/info
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RE: [R] outliers using Random Forest
From: Edgar Acuna [mailto:[EMAIL PROTECTED]
Dear Andy,
Thanks for your quick answer. I increased the number of trees and the
outlyingness measure got more stable. But still I do not know if I am
working with the raw measure or with the normalized measure mentioned
in the Breiman's Wald lecture. The normalized measure nout is
nout=(nout-med)/mean(abs(nout-med))
where med is the median of the class containing the case correponding
to nout.
Looking at the Fortran subroutine `locateout' in rfsub.f, yes, they are
normalized. (That part of the code is not changed from Breiman Cutler's
original.)
Andy
Best regards
Edgar Acuna
On Sun, 18 Apr 2004, Liaw, Andy wrote:
The thing to do is probably:
1. Use fairly large number of trees (e.g., 1000).
2. Run a few times and average the results.
The reason for the instability is sort of two fold:
1. The random forest algorithm itself is based on
randomization. That's why
it's probably a good idea to have 500-1000 trees to get more stable
proximity measures (of which the outlying measures are based on).
2. If you are running randomForest in unsupervised mode
(i.e., not giving it
the class labels), then the program treats the data as
class 1, creates a
synthetic class 2, and run the classification algorithm to get the
proximity measures. You probably need to run the algorithm
a few times so
that the result will be based on several simulated data,
instead of just
one.
HTH,
Andy
From: Edgar Acuna
Hello,
Does anybody know if the outscale option of randomForest
yields the
standarized version of the outlier measure for each case? or
the results
are only the raw values. Also I have notice that this
measure presents
very high variability. I mean if I repeat the experiment I am
getting very
different values for this measure and it is hard to flag
the outliers.
This does not happen with two other criteria than I am
using: LOF and
Bay's Orca. I am getting several cases that can be considered
as outliers
with both approaches.
I run my experiments using Bupa and Diabetes available at
UCI Machine database repository.
Thanks in advance for any response.
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[R] looking for something like ave I can pass non numeric to
Hi, I have been trying to calculate summary error and coding statistics on a by subject basis and seem to be writing a lot of code to do a simple thing. I won't go into my messy version. What I am asking is if anyone knows of a single command that could take the following data and get error rates over a given vector. Given data from ss with,,, subja rt code 1 1 200 good 1 1 321 good 1 2 457 good 1 2 384 bad 2 1 228 good 2 1 343 bad I would like to do a calculation something like ave returns. An artificial example that does not work where I have passed the code field to ave is below. ss$FAcount- ave(ss$code, ss$subj, function(x) length(x[x==good])/length(x)) __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R-1.9.0: make error on slackware-current!
I have been able to reproduce this. It appears that _some_ XFree 4.4.0 Linux installations need #include stdio.h so please add that before the X11/X.h call in src/modules/X11/dataentry.c It seems nothing to do with _Xconst. On Sun, 18 Apr 2004, Prof Brian Ripley wrote: On Sun, 18 Apr 2004, Timothy Tatar wrote: It appears that the #If NeedFunctionPrototypes compiler directive has been removed from Xlib.h and Xutil.h in Xfree86 4.4. All the prototypes containing the offending _Xconst are now being processed. R 1.8.1, which built successfully under XFree86 4.3, fails under XFree86 4.4 with the same error messages. Yes, we do know. However, 1.9.0 _has_ been built against XFree 4.4, and also XFree 8.3 with NeedFunctionPrototypes defined to be 1. _Xconst should be defined, so the problem seems to be with `slackware-current' (whatever that is) rather than XFree 4.4. Does adding #ifndef _Xconst #define _Xconst const #endif /* _Xconst */ help? (That's inside #if NeedFunctionPrototypes in the headers I have.) -- 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 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R-1.9.0: make error on slackware-current!
Sorry, a time skew caused a problem here: removing the line #define NeedFunctionPrototypes 0 is also needed. On Mon, 19 Apr 2004, Prof Brian Ripley wrote: I have been able to reproduce this. It appears that _some_ XFree 4.4.0 Linux installations need #include stdio.h so please add that before the X11/X.h call in src/modules/X11/dataentry.c It seems nothing to do with _Xconst. On Sun, 18 Apr 2004, Prof Brian Ripley wrote: On Sun, 18 Apr 2004, Timothy Tatar wrote: It appears that the #If NeedFunctionPrototypes compiler directive has been removed from Xlib.h and Xutil.h in Xfree86 4.4. All the prototypes containing the offending _Xconst are now being processed. R 1.8.1, which built successfully under XFree86 4.3, fails under XFree86 4.4 with the same error messages. Yes, we do know. However, 1.9.0 _has_ been built against XFree 4.4, and also XFree 8.3 with NeedFunctionPrototypes defined to be 1. _Xconst should be defined, so the problem seems to be with `slackware-current' (whatever that is) rather than XFree 4.4. Does adding #ifndef _Xconst #define _Xconst const #endif /* _Xconst */ help? (That's inside #if NeedFunctionPrototypes in the headers I have.) -- 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 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] solve 2 var problem
trace(K) is just sum(diag(K)). Assuming by `.' you mean multiplication (I'll use `*' below), trace(x*A + y*B) = x*trace(A) + y*trace(B) Let a = trace(A) = sum(diag(A)) and b = trace(B) = sum(diag(B)). You are then looking for (x, y) that satisfy a*x + b*y C which ought to be trivial. I don't think you need R for that. Andy From: Ramzi TEMANNI Hi I'm getting started with R and i have difficulties finding how to solve this problem in R : Find x,y satisfying Trace(K) C, K positive where ( K=x.A+y.B), [A,B,K square Matrix in R(n x n), x,y in R] Thanks in advance, Ramzi [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] histogram y-scaling
Hello all, Relative to WinXP R1.8 I have two histograms to plot, and for comparison purposes I want them to have the same Y-scaling. I tried to find the size of the bin with the maximum count before generating the histogram, but this did not work (see below). What is a better way? par(mfrow=c(2,1)) # set up for plotting in 2 rows and 1 column x1-seq(-0.5,58.5,1) # make a range of x values for histogram I thought the following lines would allow me to capture the results of the hist function and determine the max bin count for scaling *before* making the plot, but R cleverly saw around my method and plots it anyway. With this code I get two plots. q=hist(mt1,x1) # stick results in a variable... alas also plots cts=q$counts # get the bin counts mct1=max(cts)# how many values in the bin with the most values hist(mt1,x1) # generate histogram plot # go on with histogram #2... Thanks, =Randy= R. Zelick email: [EMAIL PROTECTED] Department of Biology voice: 503-725-3086 Portland State University fax: 503-725-3888 mailing: P.O. Box 751 Portland, OR 97207 shipping: 1719 SW 10th Ave, Room 246 Portland, OR 97201 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] histogram y-scaling
?hist reveals argument plot=TRUE, so try plot=FALSE. On Mon, 19 Apr 2004, Randy Zelick wrote: Hello all, Relative to WinXP R1.8 No such thing. There is R 1.8.0 and R 1.8.1 but not R 1.8. I have two histograms to plot, and for comparison purposes I want them to have the same Y-scaling. I tried to find the size of the bin with the maximum count before generating the histogram, but this did not work (see below). What is a better way? par(mfrow=c(2,1)) # set up for plotting in 2 rows and 1 column x1-seq(-0.5,58.5,1) # make a range of x values for histogram I thought the following lines would allow me to capture the results of the hist function and determine the max bin count for scaling *before* making the plot, but R cleverly saw around my method and plots it anyway. With this code I get two plots. q=hist(mt1,x1) # stick results in a variable... alas also plots cts=q$counts # get the bin counts mct1=max(cts)# how many values in the bin with the most values hist(mt1,x1) # generate histogram plot # go on with histogram #2... Something like q1 - hist(mt1, x1, plot = FALSE) q2 - hist(mt2, x2, plot = FALSE) mctl - max(q1$counts, q2$counts) plot(q1, ylim=c(0, mctl)) plot(q2, ylim=c(0, mctl)) -- 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 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] New unique name
On Mon, 19 Apr 2004, Erich Neuwirth wrote:
Since I have not been precise enough, let me explain in more detail.
I am executing some R code in the COM server.
For making function calls with parameters existing as ranges in Excel,
I need to assign values for the arguments of the function call to
temporary R variables.
This is the only way of transferring large matrices quickly from Excel
to R.
Then, I construct a string which is the function call to be executed.
The string contains the names of the newly created R variables
for the function arguments.
This string then is sent to R and executed as a command.
I used the following for local variables in a macro
gensym-function(base=.v.,envir=parent.frame()){
repeat{
nm-paste(base,paste(sample(letters,7,replace=TRUE),
collapse=),sep=.)
if (!exists(nm,envir=envir))
break
}
as.name(nm)
}
You wouldn't need the as.name() conversion, since you want a string.
A sufficiently clever person could come up with code that this could
conflict with (eg, if your code relied on the non-existence of
.v..agdefge), but in normal circumstances it should be fine.
-thomas
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Re: [R] Turning windows screen buffering on and off
On Mon, 19 Apr 2004 10:05:48 +0200, Philippe Grosjean [EMAIL PROTECTED] wrote : The buffered output is a nice thing, and if the user want to use it, then fine! However, it should be nice to know if it is set ON or OFF, and to temporarily change it for some outputs in R scripts. I think this is the primary request. Then, your request appears to me as a secondary one: to set buffered output ON or OFF at statup of Rgui (by defining it in the preference panel). Personnally, I would really like to have both. In the long term, I'd like it if all of Rgui could be moved to a package. From most users' points of view things wouldn't change much, but it would mean it was more practical for someone like you to write your own front-end, either by modifying the standard one or by starting from scratch. Duncan Murdoch __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R-estimators
Dear all, could you please suggest me the package I have to install in order to use ranks statistics in regression models (R-Estimators, wilcoxon scores ...) Thanks in advance Paolo Radaelli [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] New unique name
I think the function below will tell you all the names that are in
scope at the point where it is called. You can then make up something
that's not in the list.
Duncan Murdoch
all.names - function() {
env - new.env(parent=parent.frame())
result - ls(env=env)
while (!is.null(env)) {
env - parent.env(env)
result - c(result, ls(env=env))
}
sort(unique(result))
}
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[R] Menu addditions to Rcmdr v0.9-6
1) In general, I would appreciate help in adding functions to the Rcmdr
menu system. I've been able to modify the menus themselves and source test
code but I can't get R functions to execute from the menu. My latest proof
of concept code follows:
Three lines added to Rcmdr-menus.txt:
menujunkMenutopMenu
itemjunkMenucommand Print stuff... function()
stuff ()
itemtopMenu cascade JunkjunkMenu
stuff.R:
stuff -function(){
c(This is stuff to print...)
}
2) A file called compareModels.demo is mentioned in the documentation
but I haven't been able to find it in the Windows library downloaded from
CRAN. Where should I look?
Thanks!
Chuck
Charles E. White, Senior Biostatistician, MS
Walter Reed Army Institute of Research
503 Robert Grant Ave., Room 1w102
Silver Spring, MD 20910-1557
301 319-9781
Personal/Professional Site:
http://users.starpower.net/cwhite571/professional/
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Re: [R] New unique name
One possibility would be the Ruuid library in BioConductor. I'm not
sure if the uuid library is available (or compilable) under MS
Windows, though. I'll be checking in a bit (weeks) if no one else
does.
Erich Neuwirth [EMAIL PROTECTED] writes:
In some languages there is a function
gensym()
which returns a new unique name (in the current environment).
This is quite helpful when one has to do temporary assignments.
I could not find such a function in R.
Is there one?
--
Erich Neuwirth, Computer Supported Didactics Working Group
Visit our SunSITE at http://sunsite.univie.ac.at
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[EMAIL PROTECTED]http://www.analytics.washington.edu/
Biomedical and Health Informatics University of Washington
Biostatistics, SCHARP/HVTN Fred Hutchinson Cancer Research Center
UW (Tu/Th/F): 206-616-7630 FAX=206-543-3461 | Voicemail is unreliable
FHCRC (M/W): 206-667-7025 FAX=206-667-4812 | use Email
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[R] classification and association rules in R
hello, I am looking for a classification or/and association rules in R. However, after searching in CRAN, nothing found. Is anyone know if they are available in R? By the way, I heard that there are some people developing a better search interface for R (or CRAN?). Where are the related information I can get? Thanks. Regards, Rong-En Fan __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Turning windows screen buffering on and off
Duncan Murdoch wrote: In the long term, I'd like it if all of Rgui could be moved to a package. From most users' points of view things wouldn't change much, but it would mean it was more practical for someone like you to write your own front-end, either by modifying the standard one or by starting from scratch. I totally agree. Recent developments in SciViews (which will be publicly released in May) focus on GUI extentions *for* Rgui, not *in replacement* of it as it was the case for SciViews Insider. I have difficulties with specific aspects like buffered output, whose state is hard to know from R code... This would be a lot more easier if all Rgui extensions could be accessed through R code. A dedicated package is an excellent idea. OK, that said, I go back to my work, because I still have a lot to finish for UseR! conference... Best regards, Philippe Grosjean ...?})) ) ) ) ) ) ( ( ( ( ( Prof. Philippe Grosjean \ ___ ) \/ECO\ ( Numerical Ecology of Aquatic Systems /\___/ ) Mons-Hainaut University, Pentagone / ___ /( 8, Av. du Champ de Mars, 7000 Mons, Belgium /NUM\/ ) \___/\ ( phone: + 32.65.37.34.97, fax: + 32.65.37.33.12 \ ) email: [EMAIL PROTECTED] ) ) ) ) ) SciViews project coordinator (http://www.sciviews.org) ( ( ( ( ( ... __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] using subscripts in a plot title with 2 lines
Hi,
I'm making a plot in which the title takes up two lines. The title
contains a subscript but when I look at the plot it does'nt seem tocome
out properly. The code I'm using is:
xtxt = expression(paste('Observed -log( ', IC[50], ' )'))
ytxt = expression(paste('Predicted -log( ', IC[50], ' )'))
mtxt = expression(paste('Plot of Observed vs. Predicted -log(',
IC[50], ') Values for the PDGFR\nDataset'))
plot(tset$V2, tset$V3, xlim = c(-2,2), ylim = c(-2,2),
pch = 19, col = blue,
main=mtxt,
xlab = xtxt,
ylab = ytxt)
Removing the IC[50] term makes it come out OK.
Is there any way to get around this?
Thanks,
---
Rajarshi Guha [EMAIL PROTECTED] http://jijo.cjb.net
GPG Fingerprint: 0CCA 8EE2 2EEB 25E2 AB04 06F7 1BB9 E634 9B87 56EE
---
Does Ramanujan know Polish?
-- E.B. Ross
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Re: [R] How to write an S4 method for sum or a Summary generic
Swinton, Jonathan [EMAIL PROTECTED] writes:
If I have a class Foo, then i can write an S3 method for sum for it:
setClass(Foo,representation(a=integer));aFoo=new(Foo,a=c(1:3,NA))
sum.Foo - function(x,na.rm){print(x);print(na.rm);sum([EMAIL
PROTECTED],na.rm=na.rm)}
sum(aFoo)
But how do I write an S4 method for this? All my attempts to do so have
foundered. For example
setMethod(sum,signature(Foo,logical),
function(x,na.rm){print(x);print(na.rm);sum([EMAIL PROTECTED],na.rm=na.rm)}
creates a method which seems to despatch on na.rm=Foo:
There is no x argument in the generic so you can't dispatch on it.
sum
function (..., na.rm = FALSE)
.Internal(sum(..., na.rm = na.rm))
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[R] survival analysis question
Dear R users, I have survival estimates for 2 groups according to a factorial covariate (covar=1 or 2) and I d like to only plot the difference in term of survival between these 2 groups over time with 95%CI. Could someone send me a short script to do that since I m not an old R programmer ? thanks Philippe [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] One inflated Poisson or Negative Binomal regression
Dear Peter:
I notice there is a R code for a Zero-inflated Poisson/NB process on the
Stanford Political Science Computational Lab (Prof. Simon Jackman) web
page. If I were wanting to do a one-inflated model, I would start with
that because, at least to my eye, it is very easy to follow. Mind you,
I did not try this myself, but I bet you could make it go. In the file
zeroinfl.r, look at the function:
zeroinflNegBin - function(parms){
it is pretty clear you'd have to supply a probability model for the
outcomes valued 1 and then fit them into the overall likelihood.
pj
http://pscl.stanford.edu/content.html
Peter Flom wrote:
Hello
I am interested in Poisson or (ideally) Negative Binomial regression
with an inflated number of 1 responses
I have seen JK Lindsey's fmr function in the gnlm library, which fits
zero inflated Poisson (ZIP) or zero inflated negative binomial
regression, but the help file states that for ' Poisson or related
distributions the mixture involves the zero category'.
I had thought of perhaps subtracting 1 from all the counts and then
fitting the ZIP or ZINB models, and then adding 1, but am not sure if
this is legitimate, or if there is some better method.
Contextual details:
The dependent variable is number of primary sexual partners in the last
year. The independent variables include a) Being married or in a
committed relationship b) using hard drugs c) sex d) age
N is c. 500
Not surprisingly, there are a large number of 1 responses, especially
for those who are married or in a relationship. More surprisingly, the
mean number of partners is the same (1.05 vs. 1.02) for people in and
not in relationships, but the variances are very different, mostly
because those in a relationhsip are much more likely to say exactly 1.
Thanks in advance
Peter
Peter L. Flom, PhD
Assistant Director, Statistics and Data Analysis Core
Center for Drug Use and HIV Research
--
Paul E. Johnson email: [EMAIL PROTECTED]
Dept. of Political Sciencehttp://lark.cc.ku.edu/~pauljohn
1541 Lilac Lane, Rm 504
University of Kansas Office: (785) 864-9086
Lawrence, Kansas 66044-3177 FAX: (785) 864-5700
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Re: [R] Unexpected behaviour of identical
Swinton, Jonathan [EMAIL PROTECTED] writes:
# works as expected
ac - c('A','B');
identical(ac,ac[1:2])
[1] TRUE
#but
af - factor(ac)
identical(af,af[1:2])
[1] FALSE
Any opinions?
Hmm, surprising indeed. The proximate cause would seem to be that
names(attributes(af))
[1] levels class
names(attributes(af[1:2]))
[1] class levels
--
O__ Peter Dalgaard Blegdamsvej 3
c/ /'_ --- Dept. of Biostatistics 2200 Cph. N
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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[R] question about linear models.
hi there, i have the following table with two factors A, B each respectively with 3 and 4 levels (unbalanced design) S1 samples A B 1 1.3398553 0 0 2 0.8455924 0 0 3 1.0290893 0 0 4 1.2720512 0 0 5 1.2071754 0 0 6 1.1859539 0 0 7 2.7399659 2 3 8 1.2476911 2 3 9 2.6389479 2 2 10 1.6914068 1 2 11 2.2260561 2 1 12 1.2955187 1 1 13 1.6526140 1 3 14 2.3159151 2 3 15 2.3905009 1 2 16 2.9520105 2 2 17 1.9478868 1 1 18 1.9936118 1 1 19 1.3775338 1 3 20 1.9638190 2 2 21 1.4697860 1 2 22 2.2028858 2 3 23 2.4024771 2 1 24 1.9935864 1 1 i fit two different models fit1-aov(samples~A + B,data=S1,contrasts = list(A = contr.treatment, B = contr.treatment)) fit2-aov(samples~A,data=S1,contrasts = list(A = contr.treatment)) fit3-aov(samples~B,data=S1,contrasts = list(B = contr.treatment)) and using anova(fit1,fit2) Analysis of Variance Table Model 1: samples ~ A + B Model 2: samples ~ A Res.Df RSS Df Sum of Sq F Pr(F) 1 19 2.74820 2 21 3.14667 -2 -0.39847 1.3774 0.2763 i get B as not significant and anova(fit1,fit3) Analysis of Variance Table Model 1: samples ~ A + B Model 2: samples ~ B Res.Df RSS Df Sum of Sq F Pr(F) 1 19 2.7482 2 20 4.2391 -1 -1.4909 10.308 0.004604 ** A as significant. however if i do anova(fit3) Analysis of Variance Table Response: samples Df Sum Sq Mean Sq F value Pr(F) B 3 3.7241 1.2414 5.8567 0.004854 ** Residuals 20 4.2391 0.2120 i get B as significant and anova(fit2) Analysis of Variance Table Response: samples Df Sum Sq Mean Sq F valuePr(F) A 2 4.8165 2.4083 16.072 5.835e-05 *** Residuals 21 3.1467 0.1498 A as significant. Should i conclude that A is significant and B is not or rather that both factors are significant ? all the best __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] using subscripts in a plot title with 2 lines
If I understood your problem correctly, one ugly way of getting around this
is to use call mtext() twice to put the two lines in the title area, using
the line argument to mtext to control the location of the lines. E.g.,
something like:
plot(1, main=)
mtext(expression(IC[50]), line=3, side=3, cex=1.2)
mtext(of My Data, line=1.5, side=3, cex=1.2)
HTH,
Andy
From: Rajarshi Guha
Hi,
I'm making a plot in which the title takes up two lines. The title
contains a subscript but when I look at the plot it does'nt
seem tocome
out properly. The code I'm using is:
xtxt = expression(paste('Observed -log( ', IC[50], ' )'))
ytxt = expression(paste('Predicted -log( ', IC[50], ' )'))
mtxt = expression(paste('Plot of Observed vs. Predicted -log(',
IC[50], ') Values for the PDGFR\nDataset'))
plot(tset$V2, tset$V3, xlim = c(-2,2), ylim = c(-2,2),
pch = 19, col = blue,
main=mtxt,
xlab = xtxt,
ylab = ytxt)
Removing the IC[50] term makes it come out OK.
Is there any way to get around this?
Thanks,
---
Rajarshi Guha [EMAIL PROTECTED] http://jijo.cjb.net
GPG Fingerprint: 0CCA 8EE2 2EEB 25E2 AB04 06F7 1BB9 E634 9B87 56EE
---
Does Ramanujan know Polish?
-- E.B. Ross
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[R] [R-pkgs] New package: mcgibbsit, an MCMC run length diagnostic
Package: mcgibbsit
Title: Warnes and Raftery's MCGibbsit MCMC diagnostic
Version: 1.0
Author: Gregory R. Warnes [EMAIL PROTECTED]
Description:
mcgibbsit provides an implementation of Warnes Raftery's MCGibbsit
run-length diagnostic for a set of (not-necessarily independent) MCMC
sampers. It combines the estimate error-bounding approach of Raftery
and Lewis with evaulate between verses within chain approach
of Gelman and Rubin.
Maintainer: Gregory R. Warnes [EMAIL PROTECTED]
License: GPL
Depends: coda
References:
Warnes GR. The Normal Kernel Coupler: An adaptive Markov Chain Monte Carlo
method for efficiently sampling from multi-modal distributions
http://www.analytics.washington.edu/statcomp/projects/mcmc/nkc/, Ph.D.
thesis, Department of Biostatistics, University of Washington,
http://www.biostat.washington.edu/ October 2000. (See Chapter 3, Using
the Normal Kernel Coupler)
Gregory R. Warnes
Manager, Non-Clinical Statistics
Pfizer Global Research and Development
Tel: 860-715-3536
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[R] axis label character bolder
Hi, how can I make the character label of the axes of a plot darker (bolder), but not in a larger size? __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] axis label character bolder
If you simply want a bold font, have you tried setting font.axis and/or font.lab in par()? For example, par(font.axis=2, font.lab=2) plot(rnorm(10)) That gives me bold fonts for both components. version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status Patched major1 minor8.1 year 2003 month12 day 04 language R array chip wrote: Hi, how can I make the character label of the axes of a plot darker (bolder), but not in a larger size? -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 452-1424 (M, W, F) fax: (917) 438-0894 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Size of R user base
I have been trying to determine the size of the R user base, and was asked to share my findings with this mailing list. Although I still don't have any definite estimate of this number, I do have some interesting and indicative information: 1. It appears that there are about 100,000 S-PLUS users. Rationale: According to Insightful's 2002 Annual Report, over 100,000 people use Insightful software; since license revenues from S-PLUS and add-on modules accounted for nearly all of their license revenues in 2002, and their other products are much more costly than S-PLUS, it seems that the great majority of users of Insightful software are S-PLUS users. Conclusion: S-PLUS costs $3500 (Windows) or $4500 (Linux/Unix) for an individual copy; R is free. This suggests that there may be more R users than S-PLUS users, which suggests 100,000 R users. Does anyone has any other information that would give some notion as to the RELATIVE numbers of R and S-PLUS users? 2. At least one R book has achieved sales of just over 5,000 copies. (I could not find sales figures for other R books, as it appears that publishers are closed-mouthed about such figures. And no, I can't reveal which particular book this was, so don't ask.) Conclusion: Very few books sell to more than 12% of the population of potential buyers, and most books have a far lower penetration -- 1% or less is not uncommon. A 12% penetration for the book in question implies 42,000 R users; a more reasonable 5% penetration implies 100,000 users. A low 1% penetration implies 500,000 users. 3. There are a total of 3225 unique subscribers to the three R mailing lists. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] fill between lines
hi: is it possible to color areas between two functions? for example, x- 1:100; plot(x, x^2, type=l); lines(x,0.5*x^2, type=l); # better plotwithfill(x, x^2, 0.5*x^2, color=c(yellow, red); where the first color is used if f(x)=x^2 g(x)=0.5*x^2, and the second for the reverse. Help appreciated. Regards, / ivo __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] fill between lines
Nothing easily that I can think of. You need to plot each `piece' separately using polygon(). I.e., 1. Find the `pieces'. 2. Determine the color the piece should be in. 3. Draw the piece with polygon(). Step 1 above is probably the trickiest part, but quite doable, I think. Best, Andy From: ivo welch hi: is it possible to color areas between two functions? for example, x- 1:100; plot(x, x^2, type=l); lines(x,0.5*x^2, type=l); # better plotwithfill(x, x^2, 0.5*x^2, color=c(yellow, red); where the first color is used if f(x)=x^2 g(x)=0.5*x^2, and the second for the reverse. Help appreciated. Regards, / ivo __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] fill between lines
If one function dominates the other, you can use polygon(), x - 1:100 p - rbind(cbind(x, 0.5*x^2), cbind(rev(x), rev(x^2))) matplot(x, cbind(x^2, 0.5*x^2), type = l, col = 1, lty = 1) polygon(p, col = 2) or some variant of that. -roger ivo welch wrote: hi: is it possible to color areas between two functions? for example, x- 1:100; plot(x, x^2, type=l); lines(x,0.5*x^2, type=l); # better plotwithfill(x, x^2, 0.5*x^2, color=c(yellow, red); where the first color is used if f(x)=x^2 g(x)=0.5*x^2, and the second for the reverse. Help appreciated. Regards, / ivo __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] thanks for fill
Hi: thanks for the polygon recommendations. these work like a charm for me. minor documentation bug: ?plot.default states lwd is not yet supported for postscript, but it does seem to work. regards, /iaw __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] error message in mle function
I am getting an error message concerning the estimation of confidence intervals when fitting a mixed model and don't know what the problem is, or its solution. Just to provide context: the model is describing the effects of age, exp(age), harvest age, and climate variables on bighorn horn annular length. The data structure is repeated measures (between individuals, within individuals over time). Id is a random effect (there are between 3-11 horn measurements per ram, one horn measurement per age, over the 25 year period in the dataset). The mixed effect results is unable to provide confidence intervals for the fixed and random effects because: of an error in the variance-covariance structure. The error says that the intervals are non-positive definitive. Bill Shipley [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Informations about merge function
Hello, I want to have some informations about the merge function: indeed, I want te merge many data frames but I don't want to have missing value NA for the rows but the number 0 for the rows who're not in all tables! I didn't see any options to do that!! and when I use function like na.omit and na.action to deal missing value (after merging all tables) and changing them in 0, after three hours I have no longer the results because I've got a data frame with 31 colonnes and more than 2 rows. Do you have a solution for me, Thank you for answering, Nicolas BOUGET. Accédez au courrier électronique de La Poste : www.laposte.net ; 3615 LAPOSTENET (0,34/mn) ; tél : 08 92 68 13 50 (0,34/mn) [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] specifying as.svrepdesign with odd number PSUs
Is there a way to create a BRR svrepdesign from a survey design when the number of PSUs is odd in one or more stratum? Creating a JKn svrepdesign with that condition works okay, but when I tried to create a svrepdesign with type=BRR I get an error and this message: Can't split with odd numbers of PSUs in a stratum I get that message when I tell it to merge the PSUs. Maybe I'm just not doing it correctly. I tried svb - as.svrepdesign(sd, type=BRR) svb - as.svrepdesign(sd, type=BRR, brrweights (large=merge)) svb - as.svrepdesign(sd, type=BRR, large=merge)) where sd is the survey design object. None of these attempts made it past the error message. Does BRR require the number of PSUs be even? I was hoping it would randomly assign the last PSU (if the number is odd) to one or the other halves. Maybe that's naive. The JKn method didn't seem to mind the odd-number of PSUs. This worked okay. svr - as.svrepdesign(sd, type=JKn). Thanks. Fred - [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] specifying as.svrepdesign with odd number PSUs
Is there a way to create a BRR svrepdesign from a survey design when the number of PSUs is odd in one or more stratum? Creating a JKn svrepdesign with that condition works okay, but when I tried to create a svrepdesign with type=BRR I get an error and this message: Can't split with odd numbers of PSUs in a stratum I get that message when I tell it to merge the PSUs. Maybe I'm just not doing it correctly. I tried svb - as.svrepdesign(sd, type=BRR) svb - as.svrepdesign(sd, type=BRR, brrweights (large=merge)) svb - as.svrepdesign(sd, type=BRR, large=merge)) where sd is the survey design object. None of these attempts made it past the error message. Does BRR require the number of PSUs be even? I was hoping it would randomly assign the last PSU (if the number is odd) to one or the other halves. Maybe that's naive. The JKn method didn't seem to mind the odd-number of PSUs. This worked okay. svr - as.svrepdesign(sd, type=JKn). Thanks. Fred - [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Size of R user base
A very intriguing commentary! Some comments to modulate these estimates. On 19-Apr-04 Kevin S. Van Horn wrote: 1. It appears that there are about 100,000 S-PLUS users. Rationale: According to Insightful's 2002 Annual Report, over 100,000 people use Insightful software; since license revenues from S-PLUS and add-on modules accounted for nearly all of their license revenues in 2002, and their other products are much more costly than S-PLUS, it seems that the great majority of users of Insightful software are S-PLUS users. Conclusion: S-PLUS costs $3500 (Windows) or $4500 (Linux/Unix) for an individual copy; R is free. This suggests that there may be more R users than S-PLUS users, which suggests 100,000 R users. Does anyone has any other information that would give some notion as to the RELATIVE numbers of R and S-PLUS users? There is one major factor in here. The number of Windows users in the world is much higher than the number of Unix/Linux users, especially in the corporate sector. Organisations whose work needs R/S-PLUS and whose IT is Windows based will (I believe) mostly go for S-PLUS (I could expand in my reasons for believing this). Therefore I suspect that in the 2-way table Windows Unix/Linux S-PLUS N11 N12 R N21 N22 you are likely to find that N11/N21 N12/N22. Certainly N11+N21 N12+N22. This tends to imply N11+N12 N12+N22. The relative cost of S-PLUS vs R is not likely to be a factor in the choice, for most corporate users. Therefore I would lower your estimate, here, of R usage quite a bit (though I can't guess by how much). 2. At least one R book has achieved sales of just over 5,000 copies. (I could not find sales figures for other R books, as it appears that publishers are closed-mouthed about such figures. And no, I can't reveal which particular book this was, so don't ask.) Conclusion: Very few books sell to more than 12% of the population of potential buyers, and most books have a far lower penetration -- 1% or less is not uncommon. A 12% penetration for the book in question implies 42,000 R users; a more reasonable 5% penetration implies 100,000 users. A low 1% penetration implies 500,000 users. Comment: More R users are likely to buy a book on R than S-PLUS users are likely to buy a book on S-PLUS. S-PLUS users who do buy a book may in fact buy a book on R rather than S-PLUS, if that book is well known to be good. (I'm assuming that the R book you refer to is R-specific rather than written for both R and S-PLUS or for S-PLUS with R variations; otherwise you have to take off the S-PLUS-only purchasers) 3. There are a total of 3225 unique subscribers to the three R mailing lists. I think this may be the most directly informative piece of data (though still on the soft side). People who use R are likely to become aware of the mailing lists, and to subscribe. So I suspect that this number exceeds say 20-40% of R users (you can't be precise with this sort of intuitive guess). This would suggest 7000-16000 R users. You might perhaps double or triple this to allow for groups where one member of the group subscribes as the spokesman for the rest. Maybe also inflate a bit to allow for R users who don't think they need to consult mailing lists (who are they??). Hmmm! Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 167 1972 Date: 19-Apr-04 Time: 21:22:51 -- XFMail -- __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] question about linear models.
This would make a good exam question! First, look at the distribution of levels: B=0 B=1 B=2 B=3 A=06------ A=1 -- 4 3 2 A=2 -- 2 3 4 And then look at the mean values within combinations of levels: B=0 B=1 B=2 B=3 A=0 1.15 ------ | 1.15 A=1 -- 1.81 1.85 1.52 | 1.76 A=2 -- 2.31 2.52 2.13 | 2.30 +-- 1.15 1.98 2.18 1.93 | 1.81 (Residual SE after fitting A+B = 0.38) First, it is clear that (A=0) vs (A0) is exactly associated with (B=0) vs (B0). Therefore any difference between means for (A=0) vs (A0) is fully confounded with (B=0) vs (B0). Clearly (from table of means) there *is* a difference here (significant as it turns out), so fitting A alone will give a significant result as will fitting B alone. Further (table of means) the response increases almost linearly with A (about 0.6/level), while it does not change much for (B=1/2/3). So almost all if the variation with respect to B is accounted for by the difference between (B=0) and (B0) which is totally confounded with A. Therefore, once you have fitted A, fitting B as an additional variate will not change the fit significantly. However, if you fit B first followed by adding A, you first (B fit) take out the difference between (B=0) vs (B0), equivalent to (A=0) vs (A0). However, from inspection of table of means, while there is little differfence between (B=1)/(B=2)/(B=3) nevertheless there is a systematic difference at each level of B between (A=1) and (A=2) -- 0.5, 0.67 and 0.61 respectively. This shows up as an effect of A after fitting B. So, in summary, there is a significant effect of A alone (due to the constant increase per increment in level); there is a significant effect of B alone (due to the contrast between (B=0) and (B0) equivalent to the contrast between (A=0) and (A0)); however, once the effect of A has been allowed for you only have the contrast between levels (B=1)/(B=2)/(B=3) of B which do not differ enough to be significant. On the other hand, fitting B first still leaves a constant effect of A at each of the levels of B which shows up as significant for A after fitting B. You do not have enough data to detect as significant the sort of differences between levels of B=1/2/3. Best wishes, Ted. == On 19-Apr-04 [EMAIL PROTECTED] wrote: i have the following table with two factors A, B each respectively with 3 and 4 levels (unbalanced design) S1 samples A B 1 1.3398553 0 0 2 0.8455924 0 0 3 1.0290893 0 0 4 1.2720512 0 0 5 1.2071754 0 0 6 1.1859539 0 0 7 2.7399659 2 3 8 1.2476911 2 3 9 2.6389479 2 2 10 1.6914068 1 2 11 2.2260561 2 1 12 1.2955187 1 1 13 1.6526140 1 3 14 2.3159151 2 3 15 2.3905009 1 2 16 2.9520105 2 2 17 1.9478868 1 1 18 1.9936118 1 1 19 1.3775338 1 3 20 1.9638190 2 2 21 1.4697860 1 2 22 2.2028858 2 3 23 2.4024771 2 1 24 1.9935864 1 1 i fit two different models fit1-aov(samples~A + B,data=S1,contrasts = list(A = contr.treatment, B = contr.treatment)) fit2-aov(samples~A,data=S1,contrasts = list(A = contr.treatment)) fit3-aov(samples~B,data=S1,contrasts = list(B = contr.treatment)) and using anova(fit1,fit2) Analysis of Variance Table Model 1: samples ~ A + B Model 2: samples ~ A Res.Df RSS Df Sum of Sq F Pr(F) 1 19 2.74820 2 21 3.14667 -2 -0.39847 1.3774 0.2763 i get B as not significant and anova(fit1,fit3) Analysis of Variance Table Model 1: samples ~ A + B Model 2: samples ~ B Res.Df RSS Df Sum of Sq F Pr(F) 1 19 2.7482 2 20 4.2391 -1 -1.4909 10.308 0.004604 ** A as significant. however if i do anova(fit3) Analysis of Variance Table Response: samples Df Sum Sq Mean Sq F value Pr(F) B 3 3.7241 1.2414 5.8567 0.004854 ** Residuals 20 4.2391 0.2120 i get B as significant and anova(fit2) Analysis of Variance Table Response: samples Df Sum Sq Mean Sq F valuePr(F) A 2 4.8165 2.4083 16.072 5.835e-05 *** Residuals 21 3.1467 0.1498 A as significant. Should i conclude that A is significant and B is not or rather that both factors are significant ? all the best __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 167 1972 Date: 19-Apr-04 Time: 20:28:31 -- XFMail -- __ [EMAIL
Re: [R] Size of R user base
(Ted Harding) wrote: 1. It appears that there are about 100,000 S-PLUS users. [...] Does anyone has any other information that would give some notion as to the RELATIVE numbers of R and S-PLUS users? There is one major factor in here. The number of Windows users in the world is much higher than the number of Unix/Linux users, especially in the corporate sector. Organisations whose work needs R/S-PLUS and whose IT is Windows based will (I believe) mostly go for S-PLUS (I could expand in my reasons for believing this). But R is available for Windows, too. I've downloaded and installed both the Linux and Windows versions; neither task was difficult, and the Windows version had a rather nicer interface. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] fill between lines
It could be approximated with line segments, using the segments function. Something like two calls similar to segments(x,x^2, x, 0.5*x^2) one call where f g, the other call f g, using a different value for the 'col' argument Make x very dense, say seq(1,100,len=1000) instead of 1:100, and it will look like it's filled. -Don At 2:16 PM -0400 4/19/04, ivo welch wrote: hi: is it possible to color areas between two functions? for example, x- 1:100; plot(x, x^2, type=l); lines(x,0.5*x^2, type=l); # better plotwithfill(x, x^2, 0.5*x^2, color=c(yellow, red); where the first color is used if f(x)=x^2 g(x)=0.5*x^2, and the second for the reverse. Help appreciated. Regards, / ivo __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- -- Don MacQueen Environmental Protection Department Lawrence Livermore National Laboratory Livermore, CA, USA __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Size of R user base
On Mon, 19 Apr 2004, Kevin S. Van Horn wrote: 2. At least one R book has achieved sales of just over 5,000 copies. (I could not find sales figures for other R books, as it appears that publishers are closed-mouthed about such figures. And no, I can't reveal which particular book this was, so don't ask.) Some of us know quite accurately, though. Conclusion: Very few books sell to more than 12% of the population of potential buyers, and most books have a far lower penetration -- 1% or Where did you get that 12% from? less is not uncommon. A 12% penetration for the book in question implies 42,000 R users; a more reasonable 5% penetration implies 100,000 users. A low 1% penetration implies 500,000 users. One S book has sold half your number of S-PLUS users, although some sales are known to be to R users. I have big problems with the definition. What is an `R user'? Someone who has ever used R, even for a one-hour practical class? Someone who has used R in the last 3 months? Even given a definition, I would not be able to give you an accurate answer for our site, for either S-PLUS or R. (There are machines with each installed that I strongly suspect are unused.) -- 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 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Size of R user base
On Mon, Apr 19, 2004 at 10:27:25PM +0100, Prof Brian Ripley wrote: less is not uncommon. A 12% penetration for the book in question implies 42,000 R users; a more reasonable 5% penetration implies 100,000 users. A low 1% penetration implies 500,000 users. One S book has sold half your number of S-PLUS users, although some sales are known to be to R users. But then you also need to control for different editions and serial buyers. I happen to have purchased three different editions of a certain S-Plus / R book now in its 4th edition. My preference goes with the numbering scheme attributed to a tribe on some island in the Pacific which consists of a 'factor' with four levels: 'one', 'two', 'three', and 'lots'. Hence, I'd go with 'lots of R users'. Dirk -- The relationship between the computed price and reality is as yet unknown. -- From the pac(8) manual page __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Size of R user base
On Mon, 19-Apr-2004 at 04:47PM -0500, Dirk Eddelbuettel wrote: | On Mon, Apr 19, 2004 at 10:27:25PM +0100, Prof Brian Ripley wrote: | less is not uncommon. A 12% penetration for the book in question | implies 42,000 R users; a more reasonable 5% penetration implies 100,000 | users. A low 1% penetration implies 500,000 users. | | One S book has sold half your number of S-PLUS users, although some sales | are known to be to R users. | | But then you also need to control for different editions and serial buyers. | I happen to have purchased three different editions of a certain S-Plus / R | book now in its 4th edition. | | My preference goes with the numbering scheme attributed to a tribe on some | island in the Pacific which consists of a 'factor' with four levels: 'one', | 'two', 'three', and 'lots'. Hence, I'd go with 'lots of R users'. So how will you distinguish your 4th edition of a certain S-Plus / R book from what comes next? -- Patrick Connolly HortResearch Mt Albert Auckland New Zealand Ph: +64-9 815 4200 x 7188 ~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~ I have the world`s largest collection of seashells. I keep it on all the beaches of the world ... Perhaps you`ve seen it. ---Steven Wright ~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Size of R user base
Patrick Connolly [EMAIL PROTECTED] writes:
On Mon, 19-Apr-2004 at 04:47PM -0500, Dirk Eddelbuettel wrote:
| On Mon, Apr 19, 2004 at 10:27:25PM +0100, Prof Brian Ripley wrote:
| less is not uncommon. A 12% penetration for the book in question
| implies 42,000 R users; a more reasonable 5% penetration implies 100,000
| users. A low 1% penetration implies 500,000 users.
|
| One S book has sold half your number of S-PLUS users, although some sales
| are known to be to R users.
|
| But then you also need to control for different editions and serial buyers.
| I happen to have purchased three different editions of a certain S-Plus / R
| book now in its 4th edition.
|
| My preference goes with the numbering scheme attributed to a tribe on some
| island in the Pacific which consists of a 'factor' with four levels: 'one',
| 'two', 'three', and 'lots'. Hence, I'd go with 'lots of R users'.
So how will you distinguish your 4th edition of a certain S-Plus / R
book from what comes next?
It's all lots, and I hope it stays (i.e. continues) that way.
best,
-tony
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Re: [R] Size of R user base
On 19-Apr-04 Kevin S. Van Horn wrote: (Ted Harding) wrote: There is one major factor in here. The number of Windows users in the world is much higher than the number of Unix/Linux users, especially in the corporate sector. Organisations whose work needs R/S-PLUS and whose IT is Windows based will (I believe) mostly go for S-PLUS (I could expand in my reasons for believing this). But R is available for Windows, too. I've downloaded and installed both the Linux and Windows versions; neither task was difficult, and the Windows version had a rather nicer interface. Thanks. I know that there is R for Windows, and I'm not disputing your comparison between R for Linux and R for Windows, but my argument was directed at the choices people would make between S-PLUS for Windows and R for Windows, coupled with the fact that Windows is much more prevalent than Unix/Linux. Though they have the choice, I argue that many (especially corporate but by no means only these) Windows-based organisations would go for S-PLUS rather than R (for all sorts of reasons, ranging from To install S-PLUS just plug in the CD and click to the manuals for S-PLUS and add-ons which get the user, albeit potentially brainlessly, from installation to data-analysis much more readily than the R documentation which does demand considerable study, thought, and development of understanding). Not to mention that (in theory) S-PLUS as a commercial product could be presumed to come with guarantees and support and, quite possibly erroneously, expected to be a better-tested, more reliable, quality product ... it's what they pay those $$000s for, isn't it? So the gross preponderance of Windows, and the motives of many Windows-based organisations, will (I argue) lead to a preponderance of S-PLUS over R. This preponderance would, I suspect, be less marked (could just possibly swing the other way) for Unix users. But the net effect overall would, I believe, be that S-PLUS would outnumber R. Cheers, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 167 1972 Date: 19-Apr-04 Time: 22:47:59 -- XFMail -- __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] classification and association rules in R
Rong-En Fan wrote: By the way, I heard that there are some people developing a better search interface for R (or CRAN?). Where are the related information I can get? Strangely enough, by following the Search link on CRAN. Jason __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Size of R user base
Prof Brian Ripley wrote: Conclusion: Very few books sell to more than 12% of the population of potential buyers, and most books have a far lower penetration -- 1% or Where did you get that 12% from? A booklet on assessing financial feasibility in nonfiction book publishing. That's a general figure, so perhaps it doesn't apply if the book in question is a must-have, definitive reference for the group in question... like the book you mention (if it's the one I think it is). I have big problems with the definition. What is an `R user'? Someone who has ever used R, even for a one-hour practical class? Someone who has used R in the last 3 months? Good question. I guess I'd lean more towards your second definition, with the added caveat of and expects to use it again in the next 3 months. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Size of R user base
Dirk Eddelbuettel wrote: My preference goes with the numbering scheme attributed to a tribe on some island in the Pacific which consists of a 'factor' with four levels: 'one', 'two', 'three', and 'lots'. Australia. I've been there. Nice place. ;) Jason (who is an Australian) __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] error message in mle function
Is one of your variance components essentially estimating zero? If you check the numbers, delete the smallest one, and then do anova comparing the two fits, you might find that the one you deleted was not statistically significant. If I'm not mistaken, lme estimates the logarithms of the variance components. When one of them is zero, the logarithm wants to go to (-Inf). In that case, lme will still return an answer. However, intervals complains, because the observed information matrix for the parameter estimates is singular in the direction of log(variance component) that wants to go to (-Inf). Doug Bates, the developer of lme has taught many people about how to fix that problem using profile likelihood. I have not seen lme4 yet, so the problem may already have been fixed. If any if this is inaccurate, I hope Doug will correct me. hope this helps. Spencer Graves Bill Shipley wrote: I am getting an error message concerning the estimation of confidence intervals when fitting a mixed model and don't know what the problem is, or its solution. Just to provide context: the model is describing the effects of age, exp(age), harvest age, and climate variables on bighorn horn annular length. The data structure is repeated measures (between individuals, within individuals over time). Id is a random effect (there are between 3-11 horn measurements per ram, one horn measurement per age, over the 25 year period in the dataset). The mixed effect results is unable to provide confidence intervals for the fixed and random effects because: of an error in the variance-covariance structure. The error says that the intervals are non-positive definitive. Bill Shipley [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] specifying as.svrepdesign with odd number PSUs
On Mon, 19 Apr 2004, Fred Rohde wrote: Is there a way to create a BRR svrepdesign from a survey design when the number of PSUs is odd in one or more stratum? Creating a JKn svrepdesign with that condition works okay, but when I tried to create a svrepdesign with type=BRR I get an error and this message: Can't split with odd numbers of PSUs in a stratum Oh dear, how very frustrating, (or words to that effect). The parameters aren't being passed on to brrweights(). You aren't supposed to be able to split a stratum with an odd number of clusters, because the result won't be balanced.You are supposed to be able to merge two strata with an odd number of clusters, but it isn't working. Note that it should only work when the number of odd strata is even. BRR is, fundamentally, intended for strata of size 2 and I think there's something slightly dodgy about attempts to get around this. -thomas __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] using subscripts in a plot title with 2 lines
Hi
Liaw, Andy wrote:
If I understood your problem correctly, one ugly way of getting around this
is to use call mtext() twice to put the two lines in the title area, using
the line argument to mtext to control the location of the lines. E.g.,
something like:
plot(1, main=)
mtext(expression(IC[50]), line=3, side=3, cex=1.2)
mtext(of My Data, line=1.5, side=3, cex=1.2)
An alternative that keeps everything as one expression is to use atop()
within the expression rather than the '\n'. For example:
plot(1, main=)
mtext(expression(paste(atop(paste('Plot of Observed vs. Predicted
-log(', IC[50], ') Values for the PDGFR'),'Dataset'))), side=3, line=1.5)
But this may still look too ugly :)
In general, '\n' will not be handled very well within mathematical
expressions.
Paul
From: Rajarshi Guha
Hi,
I'm making a plot in which the title takes up two lines. The title
contains a subscript but when I look at the plot it does'nt
seem tocome
out properly. The code I'm using is:
xtxt = expression(paste('Observed -log( ', IC[50], ' )'))
ytxt = expression(paste('Predicted -log( ', IC[50], ' )'))
mtxt = expression(paste('Plot of Observed vs. Predicted -log(',
IC[50], ') Values for the PDGFR\nDataset'))
plot(tset$V2, tset$V3, xlim = c(-2,2), ylim = c(-2,2),
pch = 19, col = blue,
main=mtxt,
xlab = xtxt,
ylab = ytxt)
Removing the IC[50] term makes it come out OK.
Is there any way to get around this?
Thanks,
---
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RE: [R] Can't find memory.size()
From: Shin, Daehyok
I try memory.size function to find out available memory size,
but surprisingly R complains it can't find the function.
?memory.size also failed.
Is it not in the base library?
Not really. help.search(memory.size) in R-1.9.0 says:
memory.size(utils) Report on Memory Allocation
so it's in `utils', which I believe is loaded by default.
If so, why can't my R find it?
I am using the binary 1.9.0 version for Mandrake 9.1.
Because Linux is not Windows...
Andy
Thanks in advance.
Daehyok Shin
Terrestrial Hydrological Ecosystem Modellers
Geography Department
University of North Carolina-Chapel Hill
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only small things with great love.
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[R] Error while loading R2HTML in Rprofile
Hi List, When loading library(R2HTML) in Rprofile I get the following error: Error in firstlib(which.lib.loc, package) : couldn't find function ps.options Error in library(R2HTML) : .First.lib failed [Previously saved workspace restored] % ps.options is there % Loading the library from the command prompt works fine, other libraries load fine through Rprofile. % This only happens since R1.9.0. Any ideas? Thanks Herry System: Windows XP on dual processor dell, 1gig ram. Alexander Herr - Herry Northern Futures Davies Laboratory, CSIRO PMB, Aitkenvale, QLD 4814 Phone (07) 4753 8510 Fax (07) 4753 8650 Home: http://herry.ausbats.org.au Webadmin ABS: http://ausbats.org.au Sustainable Ecosystems: http://www.cse.csiro.au/ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Rank - Descending order
Dear All, Is there any simple way to way to produce rank, for a given list, but in a descending order? E.G: x = list(a=c(1,5,2,4)); rank(x$a); produces 1,4,2,3 However I am looking for a way to generate (4,1,3,2). It would be particularly nice if the proposed solution has all the niceties of rank function (like NA handling and ties.method functionality) TIA Manoj __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Rank - Descending order
Here are a couple to try: z - c(1,5,2,4) rank(-rank(z)) # If z is numeric this can be simplified to: rank(-z) I haven't checked what happens to NAs and ties.method. Manoj - Hachibushu Capital Wanzare at HCJP.com writes: : : Dear All, : Is there any simple way to way to produce rank, for a given : list, but in a descending order? : : E.G: : x = list(a=c(1,5,2,4)); : rank(x$a); produces 1,4,2,3 : : However I am looking for a way to generate (4,1,3,2). : : It would be particularly nice if the proposed solution has all : the niceties of rank function (like NA handling and ties.method : functionality) : : TIA : : Manoj __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [Way OT] [R] Size of R user base
Jason Turner [EMAIL PROTECTED] writes: And the Huli of Papua New Guinea use '15' to mean a very large number and '15 times 15 samting (something)' to mean something close to infinity. Dirk Eddelbuettel wrote: My preference goes with the numbering scheme attributed to a tribe on some island in the Pacific which consists of a 'factor' with four levels: 'one', 'two', 'three', and 'lots'. Australia. I've been there. Nice place. ;) Jason (who is an Australian) -- David Whiting Dar es Salaam, Tanzania __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
