similar to: library in R2.0.0 - summary

Hi all I upgraded to 2.0.0 version and did everything as I used to do before. I installed windows binary, copy/paste other than bundled packages. I got e.g. > library(chron) Error in library(chron) : 'chron' is not a valid package -- installed < 2.0.0? so I loaded it from CRAN and everything worked OK except my own personal functions (they are not on CRAN). So I went

Thanks a lot. setHook is Currently not in my knowledge set But it's great to save these Thing so I can look them up When I feel more comfortable. Just to add to that Stata versus R discussion : I believe, anyone who uses any other package than R, is probably missing out in the long run. It's truly unbelievable what has been done here. I feel like I fell asleep for 5 years ( by not using

Hallo all Please help me. I am lost and do not know what is the problem. I have a factor called kvartaly. > attributes(kvartaly) $levels [1] "1Q.04" "2Q.04" "3Q.04" "4Q.04" "1Q.05" "2Q.05" "3Q.05" "4Q.05" $class [1] "factor" > mode(kvartaly) [1] "numeric" > str(kvartaly) Factor w/ 8

Hi scale_colour_gradient(?red?, ?blue?) should do the trick. Actually I found it by Google ggplot colour http://www.cookbook-r.com/Graphs/Colors_(ggplot2)/ http://www.sthda.com/english/wiki/ggplot2-colors-how-to-change-colors-automatically-and-manually#gradient-colors-for-scatter-plots question. So you could find it too and probably far more quickly then myself as I have also other duties. Cheers

>From: =?ISO646-US?Q?Gunther_H=3Fning?= <gunther.hoening at ukmainz.de> >Date: 2006/09/18 Mon AM 06:26:25 CDT >To: 'Petr Pikal' <petr.pikal at precheza.cz> >Cc: r-help at stat.math.ethz.ch >Subject: Re: [R] Question on apply() with more information... I think you want something like below but it probably needs some fixing up because i don't recall the syntax

Hi R users Here is what I got with help from Petr Pikal (Thanks Petr Pikal). I modified Petr Pikal's code to a little to meet my purpose. I created a function to generate a matrix generate.matrix<-function(n.variable) { mat<-matrix(0,n.variable,(n.variable/2)/5+1) #matrix of zeroes dd<-dim(mat) # actual dimensions mat[1:(dd[1]/2),1]<-1 #put 1 in first half of first column

Hi It is preferable to echo your posts to r-help, you usually get more answers and some definitelly superb to mine. It is also better to start a new mail if your question has nothing to do with original subject "Maura E Monville" <maura.monville at gmail.com> napsal dne 01.10.2007 17:44:43: > Unluckily I do not have the privilege of practising with R all day > long. I

Dear all I noticed from NEWS 2.11.0,dev SIGNIFICANT USER-VISIBLE CHANGES o Packages must have been installed under R 2.10.0 or later, as the current help system is the only one now supported. So I tried to follow instructions in manual, Duncan Murdoch presentation and help pages to prepare and accomplish installation of a set of functions I use. However in R 2.11.0dev and too in

------- Forwarded message follows ------- From: Petr Pikal <petr.pikal at precheza.cz> To: Henrik Parn <henrik.parn at bio.ntnu.no> Subject: Re: [R] paired t-test. Need to rearrange data? Date sent: Tue, 08 Aug 2006 16:13:47 +0200 Hi Uff, it takes me a bit headache but this shall do it ?unstack ?table ?t new.list<-unstack(test.data,y~id)

Petr, there was a thinko in your response. tmp <- data.frame(m=factor(letters[1:4]), n=1:4) tmp tmp$m <- factor(tmp$m, levels=c("c","b","a","d")) ## right tmp[order(tmp$m),] tmp <- data.frame(m=factor(letters[1:4]), n=1:4) levels(tmp$m) <- c("c","b","a","d") ## wrong tmp[order(tmp$m),] changing levels

CC to R-help From: PIKAL Petr Sent: Friday, July 28, 2017 2:20 PM To: 'li xiaomei' <lixiaomei0921 at gmail.com> Cc: r-help at r-project.org Subject: RE: [R] Fail to install package in R Hi You should keep conversation within the list, others could bring better answers. I still do not see any error message. I found this answer to seemingly same question.

Hi Petr and Richard; Thanks for your responses and supports. I just faced a different problem. I have the following R codes and work well. p <- ggplot(a, aes(x=Phenotypes, y=Metabolites, size=abs(Beta), colour=factor(sign(Beta)))) + theme(axis.text=element_text(size = 5)) p1<-p+geom_point() p2<-p1+theme(panel.grid.major = element_blank(), panel.grid.minor = element_blank(),

Hi Put fill outside aes p+geom_ribbon(aes(ymin = slope_1*x + intercept_1 - 1/w[2], ymax = slope_1*x + intercept_1 + 1/w[2]), fill = "blue", alpha=0.1) The "hole" is because you have two levels of data (red and blue). To get rid of this you should put new data in ribbon call. Something like newdat <- trainset newdat$z <- factor(0) p+geom_ribbon(data=newdat, aes(ymin =

BTW. I checked help page of contour and maybe it could mention a note about akima package or interp function. Petr ------- Forwarded message follows ------- From: Petr Pikal <petr.pikal at precheza.cz> To: "Abhinav Verma" <abhinav1205 at gmail.com>, r-help at stat.math.ethz.ch Subject: Re: [R] Reading xyz data from a file and plotting a

Hi When you load external file to R, character variables are converted to factors by default and alphabetically sorted. I have limited connection to internet, so I cannot find the answer, you could try it yourself. Maybe you could try not to convert vector with names to factor, which, for plotting issue is not different from factor coding. See ?read.table for details However I am not sure if it

Hi After melt you can change levels of your factor variable. Again with the toy example. > levels(temp$variable) [1] "y1" "y2" "y3" "y4" > levels(temp$variable) <- levels(temp$variable)[c(2,4,1,3)] > levels(temp$variable) [1] "y2" "y4" "y1" "y3" > And you will get graphs with this new levels ordering.

Hemant's problem is that the indicators are not distributed uniformly. With a uniform distribution, categorization gives a reasonably optimal separation of cases. One approach would be to drop categorization and calculate the overall score as the mean of the standardized indicator scores. Whether this is an option I do not know. I did offer an "eyeball" set of breaks in a previous

Hi keep your emails to R help, I do not offer private consultance and others could have different opinion how to solve your problem. Did you even try my suggestion? If not, why not? If yes in what respect it does not comply with your expectations. Cheers Petr From: niharika singhal [mailto:niharikasinghal1990 at gmail.com] Sent: Tuesday, October 10, 2017 2:34 PM To: PIKAL Petr <petr.pikal at

On 15 Jun 2004 at 13:52, Vito Muggeo wrote: > Dear Petr, > Probably I don't understand exactly what you are looking for. > > However your "plot(x,c(y,z))" suggests a broken-line model for the > response "c(y,x)" versus the variables x. Therefore you could estimate > a segmented model to obtain (different) slope (and breakpoint) > estimates. See

Please use a descriptive subject and not tag onto a prior thread for new topics. Assuming the first row is time 1 and the second row is time 2 and so on try: diff(log(P)) On 6/15/06, Wong, Kim <kwong at nymex.com> wrote: > > Hi, thank you all for the help. > > The split function works very well. > > I have an additional question. If I have a matrix of prices (row = 30,