Re: [R] sscanf equivalent
On Fri, 7 Oct 2005, Prof Brian Ripley wrote: On Fri, 7 Oct 2005, Paul Roebuck wrote: I have a data file from which I need to read portions of data but data location/quantity can change from file to file. I wrote some code and have a working solution but it seems wasteful to have to do it this way. Here's the contrived incomplete code. datalines - readLines(datafile.pathname) # marker will appear on line preceding and following # actual data offset.data - grep(marker, datalines) datalines - NULL # grab first column of each assoc dataline data - scan(datafile.pathname, what = numeric(0), skip = offset.data[1], nlines = offset.data[2]-offset.data[1]-1, flush = TRUE, multi.line = FALSE, quiet = TRUE) # output is vector of values Originally wrote code to parse data from 'datalines' using sub and strsplit methods but it was woefully slower and more complex than using scan method. What is desired is a means of invoking method like scan but with existing data instead of filename. Why not use a text connection? I tried that but result was far slower than the method above. R file.info(datafile.pathname)$size [1] 944850 R system.time(datalines-readLines(datafile.pathname), TRUE)[3] [1] 0.59 R length(datalines) [1] 67931 R system.time(tconn-textConnection(datalines), TRUE)[3] [1] 52.97 Once a textConnection object was created, the scan method invocation using it took less than half the time of the corresponding filename-based invocation. Problem is that this was only taking a second to perform the scan using the filename-based invocation. And since grep method doesn't accept textConnection as argument, I still require the otherwise unused 'datalines' variable and its associated memory. Even if grep supported such, the timing increased even more not having the variable. R system.time(tconn-textConnection(readLines(datafile.pathname)), TRUE)[3] [1] 66.61 Any other thoughts? # R version 2.1.1, 2005-06-20, powerpc-apple-darwin7.9.0 -- SIGSIG -- signature too long (core dumped) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to get the remaining vector after sampling a subset?
Hi , I have a vector,for example, x=rnorm(100) Then i rendom choose 20 of them. chosen=sample(x,20). And i want to get the remain values in x. Is there a quick way to go? Thanks in advance. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to get the remaining vector after sampling a subset?
Xiao Shi wrote: Hi , I have a vector,for example, x=rnorm(100) Then i rendom choose 20 of them. chosen=sample(x,20). And i want to get the remain values in x. Is there a quick way to go? Thanks in advance. [[alternative HTML version deleted]] How about: x - rnorm(100) y - sample(x, 20) z - x[!x %in% y] But probably a safer way is to sample the indicies: x - rnorm(100) w - sample(length(x), 20) y - x[w] z - x[-w] HTH, --sundar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to get the remaining vector after sampling a subset?
one way is to use: x[!x %in% chosen] I hope it helps. Best, Dimitris - Original Message - From: Xiao Shi [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Sunday, October 09, 2005 11:26 AM Subject: [R] How to get the remaining vector after sampling a subset? Hi , I have a vector,for example, x=rnorm(100) Then i rendom choose 20 of them. chosen=sample(x,20). And i want to get the remain values in x. Is there a quick way to go? Thanks in advance. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] lm.ridge
On Sat, 8 Oct 2005, Erin Hodgess wrote: Dear R People: I have a question about the lm.ridge function, please. In the example, there is one set of output values in the select function but another in the comment section. Am I missing something please? The values in the examples were computed in S-PLUS. Apparently the dataset in R is not the same as that in S-PLUS. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Rmetrics fMultivar how to?
Hi Everybody, I am a total beginner at this so please bear with me. I downloaded by hand the file WIG20.txt (Warsaw Stock Exchange Index of 20 most important stocks). The format is this: Name,Date,Open,High,Low,Close,Volume WIG20,19940414,1000.00,1000.00,1000.00,1000.00,71600.000 WIG20,19940418,1050.50,1050.50,1050.50,1050.50,99950.000 WIG20,19940419,1124.90,1124.90,1124.90,1124.90,138059.000 WIG20,19940421,1304.80,1304.80,1304.80,1304.80,154151.000 WIG20,19940425,1350.10,1350.10,1350.10,1350.10,228438.000 WIG20,19940426,1216.20,1216.20,1216.20,1216.20,16618.000 WIG20,19940428,1096.70,1096.70,1096.70,1096.70,32685.000 WIG20,19940505,1138.10,1138.10,1138.10,1138.10,113777.000 WIG20,19940506,1077.60,1077.60,1077.60,1077.60,137910.000 WIG20,19940509,1035.60,1035.60,1035.60,1035.60,97091.000 I read the data in with this command: tabelka = read.table(/home/kb2qzv/WIG20.txt,sep=,,header=TRUE) And I can see that 'tabelka' has 2840 rows in it which is correct (this many sessions since the Exchange started operating). Now I would like to utilize some of the functions available from RMetrics and see how useful they might be. I can image that one way to know this would be to plot a candlestick chart of the wig20 index and below it a TA indicator, like one of these: macdTA MACD Indicator cdsTA MACD Signal Line cdoTA MACD Oscillator vohlTA High/Low Volatility Trouble is there is no easy way (for me as a beginner) to start with. Questions: A) how to plot a candlestick chart with my data? there seems to be no 'type=candlestick' parameter in plot() function. B) how to create a second pane where an indicator can be drawn. thanks for any answers or tips. -- Benedict Cyfrowy klucz publiczny / Digital public key http://agrypa1.atspace.com/klucze/kb2qzv_wp.pl-public.asc __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] background color of xyplot
Dear All, I am wondering if there is a way to change the color of the panels of the xyplot (lattice package) from gray to white .. Because the printing of the xyplot's graph is not visible with the gray color ... I've seen the xyplot help but without any success Thanks lot in advance, Bernard, - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] background color of xyplot
see ?trellis.device trellis.device(color=F) Depth - equal.count(quakes$depth, number=8, overlap=.1) xyplot(lat ~ long | Depth, data = quakes) will get what you want. === 2005-10-09 22:32:10 您在来信中写道：=== Dear All, I am wondering if there is a way to change the color of the panels of the xyplot (lattice package) from gray to white .. Because the printing of the xyplot's graph is not visible with the gray color ... I've seen the xyplot help but without any success Thanks lot in advance, Bernard, - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html = = = = = = = = = = = = = = = = = = = = 2005-10-09 -- Deparment of Sociology Fudan University __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] color for points
Le 8 Octobre 2005 05:35, [EMAIL PROTECTED] a écrit :
Sam R. Smith a écrit :
Hi,
I have the following code to randomly generate the points:
csr -function(n=60){
x=runif(n)
y=runif(n)
f=cbind(x,y)
}
plot(csr())
I wonder how to code to make the first twenty points to be BLUE; second
twenty points to be RED; the last twenty points to be GREEN?
mynewfct = function(n=60)
{
x=runif(n)
y=runif(n)
f=cbind(x,y)
plot(f[1:20] , col='blue');
par(new=T);
plot(f[21:40] , col='red');
par(new=T);
plot(f[41:60] , col='green');
}
hih
This function will plot 60 points, no matter the value of argument 'n'.
Building also on Jim Holtman's comment, may I rather suggest
csr - function(n=60)
{
x - cbind(runif(n), runif(n))
plot(x, col=rep(c(blue, red, green), each=20, length.out=n))
x # why not return the values...
}
HTH
--
Vincent Goulet, Professeur agrégé
École d'actuariat
Université Laval, Québec
[EMAIL PROTECTED] http://vgoulet.act.ulaval.ca
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Re: [R] Glmm for multiple outcomes
Does the following help: n.subjects - 3 J - 4 K - 5 n.ijk - rep(2, each=n.subjects*J*K) x - rep(1:K, n.subjects, each=J) subj - factor(rep(1:n.subjects, each=K*J)) sa.subject - 1 sb.subject - 1 set.seed(2) a.subj - rep(sa.subject*rnorm(n.subjects), each=K*J) b.subj - rep(sb.subject*rnorm(n.subjects), each=K*J) Z - a.subj+b.subj*x library(boot) Y - (rbinom(n.subjects*K*J, n.ijk, inv.logit(Z)) /n.ijk) Dat - data.frame(subj=subj, x=x, y=Y) library(lme4) fit - lmer(y~x+(x|subj), Dat) Linear mixed-effects model fit by REML Formula: y ~ x + (x | subj) Data: Dat AIC BIClogLik MLdeviance REMLdeviance 51.63172 64.19779 -19.8158633.1066 39.63172 Random effects: Groups NameVariance Std.Dev. Corr subj (Intercept) 0.0446346 0.211269 x 0.0032613 0.057108 1.000 Residual 0.0879438 0.296553 # of obs: 60, groups: subj, 3 Fixed effects: Estimate Std. Error DF t value Pr(|t|) (Intercept) 0.35 0.151459 58 2.3109 0.02442 * x0.03 0.042661 58 0.7814 0.43777 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 spencer graves Abderrahim Oulhaj wrote: Dear All, I wonder if there is an efficient way to fit the generalized linear mixed model for multivariate outcomes. More specifically, Suppose that for a given subject i and at a given time j we observe a multivariate outcome Yij = (Y_ij1, Y_ij2, ..., Y_ijK). where Y_ijk is a binomial(n_ijk, p_ijk). One way to jointly model the data is to use the following specification: g(p_ijk) = beta_0k + b_0ik + (beta_1k + b_1ik)*x_ijk with k = 1,2 , K , g is a specified link function and (b_0ik,b_1ik) k=1,...K are random effects ... I my case, the glmmPQL converges only and give good results when k is less than 3 (i.e. for a small number of random effects). I also used the gee (generalized estimating equations) to estimate the fixed effects and the same probleme ariseed with k. Is there any help? Thank you in advance, Abderrahim Oulhaj [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Insert value from same column of another row (lag across observations)
I know I've done this before, but it's been a while and I can't find quite what I need in the help files or archives. I have a text field in a very large data frame. I'd like to add a column that represents the value from an existing field, from the next record (the data are sorted). I'm trying to represent what happens tomorrow, so the today row would have an NA value (unknown). So if x is: x [,1] [1,] today [2,] yesterday [3,] the day before [4,] two days before any y is cbind(x,c(1,2,3,4)) [,1] [,2] [1,] today 1 [2,] yesterday 2 [3,] the day before 3 [4,] two days before 4 z should become [,1] [,2] [,3] [1,] today 1 NA [2,] yesterday 2 1 [3,] the day before 3 2 [4,] two days before 4 3 I've tried z-cbind(y,c(NA,y[-1,2])) but I get [1,] today 1 NA [2,] yesterday 2 2 [3,] the day before 3 3 [4,] two days before 4 4 So the lagging doesn't work. Any ideas? -- --- David L. Van Brunt, Ph.D. mailto:[EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] enter a survey design in survey2.9
Hi dears, I expect that Mr Thomas Lumley will read this message. I have data from a complexe stratified survey. The population is divide in 12 regions and a region consist to and urban area and rural one. there to region just with urbain area. stratification variable is a combinaison of region and area type (urban/rural) In rural area, subdivision are sample with probabilties proporionnal to size in population then enuration area are sample in selected division and finally households are selected in those EA. In urban area, EA are directly selected and finally household are selected. to schematise we have: (12 regions) each region is divised in two regions / Urbain and rural. this are strata in Rural : PSU are subdivision , SSU are EA and TSU are households in Urban : PSU are EA , SSU are households. I use svydesign function as follow : esi-svydesign(id=~subdiv+EA+HHID,strata=~REGION+AREATYP,fpc=~FPC1+FPC2FPC3,weig=~pw,nest=T) FPC1: number of subdivision in each strata FPC2: number of EA in each subdivision FPC3: number of HH in each EA. pw : sampling weights but I have this error message : erron in data.frame(strata, 1:i,...) I dont understand why ! Can someone help me ? Sincerly. - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error using a data frame as the start parameter in mle()
I will attempt a brief comment on this post, as I've seen no replies. It's difficult to comment on this, as you do not really provide enough information to permit someone like me to understand your problem. Only one portion of this problem is the fact that I can't find your function mle in one of the standard libraries, and I don't know if it is part of a contributed library or something you wrote. However, I've encountered problems roughly like what you describe, and I've solved them in three different ways: (a) Making a local copy inside the wrapper function and then pass to the standard function like mle only arguments locally defined. (b) Using assign with (I think) something like pos=0. I don't remember exactly for that, and the help file for assign and several searches with RSiteSearch didn't help me. If you want to try this, I suggest you look for information on lexical scoping with assign, etc. (c) Passing additional references via ... if mle has such in its function definition. If you still would like more help from this group, PLEASE do read the posting guide (www.R-project.org/posting-guide.html). I believe people who follow that guide on average get more useful answers quicker. Viel Glück! spencer graves Coryn Bailer-Jones wrote: Dear R-Users, I am trying to use mle() to optimize two (or more) parameters, but I want to specify those parmeters in a data frame rather than having to spell them out separately in the start variable of mle(). My call is mle(negll, start=list(aps=init), fixed=list(measphot=newphot, formod=formod, Nbands=Nbands), method=BFGS) where negll is a function I have written which uses the function predict.loess(). negll works fine when called directly. The parameter I am trying to optimize, aps, is a data frame containing two parameters, e.g. init teff logg 1 8000 4.5 When I run mle I get the following error message Error in predict.loess(formod[[band]], aps) : Argument aps is missing, with no default As negll does work fine, I presume I am incorrectly passing aps into mle(). Note that mle() works fine if I rewrite negll to work on a scalar aps and then I use start=list(aps=500), for example. Can anyone help me with this? Incidentally, I am only using a data frame for aps because I am using loess(), and this seems to require a formula with named variables in a data frame (here logg and teff). I can't get it work with arrays: temp - loess(formula = photd[, band] ~ gridaps[, 1] * gridaps[, 2]) predict(temp, c(4,8000)) Error in predict.loess(temp, c(4, 8000)) : newdata does not contain the variables needed Thanks in advance for any clues. Coryn. --- Coryn Bailer-Jones[EMAIL PROTECTED] Max-Planck-Institut fuer Astronomie http://www.mpia-hd.mpg.de/homes/calj/ Koenigstuhl 17tel: +49 6221 528-224(direct) D-69117 Heidelberg +49 6221 528-0 (reception) Germany fax: +49 6221 528-246 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Matrix calculations in R--erroneous?
On Fri, 7 Oct 2005, Peter Muhlberger wrote: The max function won't do the trick because I need the entire matrix. I could do one cell at a time, but this is part of a ML routine that needs to be evaluated hundreds of thousands of times, so I can't afford to slow it down that much. pmax, then? -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] acf.plot() question
When I run the acf() function using the acf(ts.union(mdeaths,
fdeaths)) example, the acf() function calls the acf.plot()
function to generate this plot...
http://members.cox.net/ddebarr/images/acf_example.png
The plot in the lower right-hand corner is labeled fdeaths mdeaths,
but the negative lags appear to belong to mdeaths fdeaths [which
correspond to the positive lags of fdeaths mdeaths].
Am I missing something, or should the plot in the lower right-hand
corner be labeled mdeaths fdeaths (instead of fdeaths mdeaths)?
Note: The unit of measure for the lags is years.
autocorrelation - function(x, y, lags) {
+ n - length(x)
+ x.bar - mean(x)
+ y.bar - mean(y)
+ c - array(0, length(lags))
+ i - 1
+ for (t in lags) {
+ s - seq(max(1, 1 - t), min(n - t, n))
+ c[i] - sum((x[s + t] - x.bar) * (y[s] - y.bar)) / n
+ i - i + 1
+ }
+ x.sd - sqrt(sum((x - x.bar) ^ 2) / n)
+ y.sd - sqrt(sum((y - y.bar) ^ 2) / n)
+ return(c / (x.sd * y.sd))
+ }
autocorrelation(mdeaths, fdeaths, -15:15)
[1] 0.015054983 0.365626026 0.615427121 0.708206289 0.621895801
[6] 0.340005447 -0.024534195 -0.381671430 -0.611793479 -0.677803477
[11] -0.604031174 -0.349468396 0.019759425 0.405200639 0.744309322
[16] 0.976241251 0.735668532 0.364241839 -0.010675725 -0.382920620
[21] -0.622386979 -0.688538519 -0.610583980 -0.383338305 -0.018112073
[26] 0.391983088 0.656592111 0.721397236 0.639104375 0.361352626
[31] -0.003385423
autocorrelation(fdeaths, mdeaths, -15:15)
[1] -0.003385423 0.361352626 0.639104375 0.721397236 0.656592111
[6] 0.391983088 -0.018112073 -0.383338305 -0.610583980 -0.688538519
[11] -0.622386979 -0.382920620 -0.010675725 0.364241839 0.735668532
[16] 0.976241251 0.744309322 0.405200639 0.019759425 -0.349468396
[21] -0.604031174 -0.677803477 -0.611793479 -0.381671430 -0.024534195
[26] 0.340005447 0.621895801 0.708206289 0.615427121 0.365626026
[31] 0.015054983
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[R] greek symbols using pch
Hi R-users, In a plot, can I specify pch to be a greek symbol? (I looked at show.pch() in the Hmisc package but couldn't see the right symbols in there). If not, I guess I can get around this using text(x,y,expression()). cheers!, Matt. Dr Matt Fischer Postdoctoral Fellow - IPILPS ANSTO - Institute for Nuclear Geophysiology Building 21A PMB 1 Menai NSW 2234 Ph: +61 2 9717 9686 Fax: +61 2 9717 3599 Mobile: 0428 363 146 http://www.ansto.gov.au/nugeo/ http://ipilps.ansto.gov.au/ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] possible bug in image() ??
Hi everyone. The function image() seems not to be correctly plotting some matrices that I give it. (I’m using R 2.1.1) The following code creates a matrix (denoted x) with 1500 rows and 3 columns, with all entries 0 or 1, and then plots the image of this matrix. cc=runif(n=1500,min=0.1,max=1.2) ccc=ceiling(cc-1) dd=runif(n=1500,min=0.1,max=1.2) ddd=ceiling(dd-1) ee=runif(n=1500,min=0.1,max=1.2) eee=ceiling(ee-1) x=matrix(data=c(ccc,ddd,eee),nrow=1500) image(x) ..where the first column in x (vector ccc) is depicted horizontally along the bottom of the image. However, when I overplot the non-zero elements of the vectors ccc, ddd and eee onto their respective horizontal positions on the image,.. points(seq(0,1,1/(1500-1)),(ccc)^-1*0) points(seq(0,1,1/(1500-1)),(ddd)^-1*0.5) points(seq(0,1,1/(1500-1)),(eee)^-1*1) …the locations of the 1’s in image do not always match the locations of the 1’s in ccc,ddd, and eee (although they are mostly correct). Do other people find this problem?? I've tried with other matrices, and the results only seem in error when the matrix is large, say with more than 1000 rows. Cheers, Gareth Davies __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] greek symbols using pch
On Mon, 2005-10-10 at 12:02 +1000, FISCHER, Matthew wrote: Hi R-users, In a plot, can I specify pch to be a greek symbol? (I looked at show.pch() in the Hmisc package but couldn't see the right symbols in there). If not, I guess I can get around this using text(x,y,expression()). cheers!, Matt. You will need to use text() in order to plot greek characters as plotting symbols. The 'pch' argument (or par(pch)) must be a single character or an integer specifying one of the symbols as seen in ?points or as you noted in Frank's show.pch(). See ?plotmath for more information on plotting mathematical expressions. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] plot
After I made a new plot, the old plot can not be found. How can I check all the plots I have made? - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] plot
Select History Recording. You can see previous plots using 'page up'. Murray -Original Message- From: Sam R. Smith [mailto:[EMAIL PROTECTED] Sent: Monday, 10 October 2005 1:27 PM To: r-help@stat.math.ethz.ch Subject: [R] plot After I made a new plot, the old plot can not be found. How can I check all the plots I have made? - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] sqlFetch on MySQL-DB
Dear all, I successfully set up a local MySQL-database. Connecting via RODBC is not problem, the same in fetching 3 of 4 tables. But trying to connect to table 4 fails. author-sqlFetch(test,author) Error in fromchar(unclass(x)) : character string is not in a standard unambiguous format In principle I understand that error message, but I don't know any solution. Thanks for any help, Bernd version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status Under development (unstable) major2 minor3.0 year 2005 month10 day 06 svn rev 35759 language R __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] line
There are four points with coordinates: 2,3;4,9;1,6;3,10. How to use Python to draw one perpendicular bisector between (2,3) and (4,9); the other perpendicular bisector between (1,6)ºÍ(3,10); then, makes the output like: l1 a b c l2 a b c (Note: l indicates the perpendicular bisector with equation ax + by = c.) Plus the intersection coordinates of the two perpendicular bisectors: x,y - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] line
Sorry for the typo, ask for R code. Sam R. Smith [EMAIL PROTECTED] wrote:There are four points with coordinates: 2,3;4,9;1,6;3,10. How to use Python to draw one perpendicular bisector between (2,3) and (4,9); the other perpendicular bisector between (1,6)ºÍ(3,10); then, makes the output like: l1 a b c l2 a b c (Note: l indicates the perpendicular bisector with equation ax + by = c.) Plus the intersection coordinates of the two perpendicular bisectors: x,y - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
