similar to: How to Un-group a grouped data set?

Hi: Would it be fair to say (or completely out to lunch) that a local variable is local to a method and an instance variable is local to an object. Thus an instance variable is a sort of limited global variable or super-powered local variable. Its scope is larger than a local but smaller than a global. Or that it is a global within a limited domain (the object). or is it a whole lot

Hi everybody, Hope your are not all on holyday because I've got a problem that is going to drive me crazy... I would like to remove some rows from a dataframe. The rows correspond to some specific indexes which I can get by looking at the name in the first column of my dataset. But I manage to get only the opposite of what I really want (function #1) #Function#1:

Hi, I use Sweave extensively in my consulting work. When submitting reports to the scientists I work with I like to use the citation function to reference any packages I use, to give proper acknowledgement. I noted in the documentation that a citation inherits from bibentry, and indeed, > citr<- citation() > class(citr) [1] "citation" "bibentry" However, following

Hi All, I am currently preparing some form of slideshow introducing R and its capabilities for some colleagues. The thing will be about 30 mins, and I'd like to have some "pretty pictures" and some "amazing facts" (I'm trying to sell, obviously :)). Can I ask if it's possible to easily retrieve a gross figure of the number of functions in R considering the

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Estimada comunidad, tengo que hacer un grafico con 4 paneles (4 graficos pegados juntos) y sobre este tengo que dibujar 4 rectas ajustadas a los datos de cada panel. Intente hacerlo con lattice usando el siguiente codigo : library(lattice) library(latticeExtra) xYplot(Cbind(mf[,3],mf[,5],mf[,6])~dia|factor(sol),data=mf,xlab="Time

I downloaded the latest version of R (1.5.0) and issued a read.table() command that used to work with (1.3.1). Unfortunately I erased 1.3.1 from my hard drive after I downloaded R 1.5.0. Here's the command and the error message I get under 1.5.0: > golub1<-read.table("C:/Microarrays/code/data_set_ALL_AML_train.txt",sep="\t",q

Hi the following example I think demonstrates the inconsistency > f<-function(x) x > length(f) [1] 1 > is.na(f) [1] FALSE Warning message: In is.na(f) : is.na() applied to non-(list or vector) of type 'closure' The documentation states: Arguments x an R object to be tested. value a suitable index vector for use with x. And nowhere in the details was it implied (to me

Hi, I would like to create data.frame and increment their name to finally merge them. Does anybody has a solution ? Something like that ? n<-6 for (i in 1:n) { m[i]<-data.frame(name,value) m<-merge(m[i]:m[n]) } Nolwenn ******************************************** Nolwenn Le Meur INSERM U533 Facult? de m?decine 1, rue Gaston Veil 44035 Nantes Cedex 1 France Tel:

Hi I've records like this df= x panel 4 1 93 2 21 3 83 4 75 1 87 2 87 3 78 4 50 1 76 2 86 3 65 4 84 1 40 2 39 3 26 4 i want to create histogram out of it . i want all the mid and count values for panel wise my code is histoutput = ddply(df,.(df[2]),hist) i'm not able to get the required result. please help me using for loop takes a lot of time if there are more records ----- Thanks

Hi, Is there a command that will give the total size of an R object in bytes? thanks Nicholas -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at

Hi, I am having trouble with the survival library, particualrily the coxph function. the following works coxph(jtree9$cph.call,z,rep(1,dim(z)[1])) Call: coxph(formula = jtree9$cph.call, data = z, weights = rep(1, dim(z)[1])) coef exp(coef) se(coef) z p SM 0.2574 1.294 0.0786 3.274 1.1e-03 Sex -0.1283 0.880 0.1809 -0.709

Hi, Am I hitting some limit in match? Consider the following example: > tst<-seq(100,125,by=.2)%in%seq(0,800,by=.1) > sum(tst) [1] 76 > seq(100,125,by=.2) [1] 100.0 100.2 100.4 100.6 100.8 101.0 101.2 101.4 101.6 101.8 102.0 102.2 [13] 102.4 102.6 102.8 103.0 103.2 103.4 103.6 103.8 104.0 104.2 104.4 104.6 [25] 104.8 105.0 105.2 105.4 105.6 105.8 106.0 106.2 106.4 106.6 106.8

Hi, Am I hitting some limit in match? Consider the following example: > tst<-seq(100,125,by=.2)%in%seq(0,800,by=.1) > sum(tst) [1] 76 > seq(100,125,by=.2) [1] 100.0 100.2 100.4 100.6 100.8 101.0 101.2 101.4 101.6 101.8 102.0 102.2 [13] 102.4 102.6 102.8 103.0 103.2 103.4 103.6 103.8 104.0 104.2 104.4 104.6 [25] 104.8 105.0 105.2 105.4 105.6 105.8 106.0 106.2 106.4 106.6 106.8

HI, I would like to know if it's possible to create a postscript file with multiple graphs . I'm creating some graphs by the means of a loop and I want to save each graph in the same file (splitting making the device by the number of graphs). Do I have to use the par(matrix()) option, the layout() function or the split.screen() one ? Thanks Nolwenn

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Hello, I am trying to read a database exported from SAS. It is form of csv, and the date format reads '02-Feb-99'. I used following code to convert character to date format, db$dob<-as.Date(db$dob, format="%d-%b-%y"). but it doesn't work, only seems NA. What's wrong with this code? Thank you in advance. Best, Dong-Hee Koh [[alternative HTML version deleted]]

Hello, Using base R only, the following seems to do what you want. with(mydf, ave(speed, date_time, type, FUN = weighted.mean, w = n_vehicles)) Hope this helps, Rui Barradas Em 09-11-2017 13:16, Massimo Bressan escreveu: > Hello > > an update about my question: I worked out the following solution (with the package "dplyr") > > library(dplyr) > > mydf%>% >

Hi, I am writing some basic smoothers in R for cleaning some spectral data. I wanted to see if I could get close to matlab for speed, so I was trying to compare SparseM with Matrix to see which could do the choleski decomposition the fastest. Here is the function using SparseM difsm <- function(y, lambda, d){ # Smoothing with a finite difference penalty # y: signal to be smoothed #

Given the data structure below, how can I create a bar plot for the values of disc for each area grouped by year? bar <-structure(list(year = c(2003, 2003, 2003, 2003, 2003, 2003, 2003, 2007, 2007, 2007, 2007, 2007, 2007, 2007), area = structure(c(6L, 4L, 1L, 2L, 3L, 5L, 7L, 6L, 4L, 1L, 2L, 3L, 5L, 7L), .Label = c("AK", "ID", "MT", "NW",

Hi, consider the following > a<-gl(3,3,9) > a [1] 1 1 1 2 2 2 3 3 3 Levels: 1 2 3 > levels(a)<-3:1 > a [1] 3 3 3 2 2 2 1 1 1 Levels: 3 2 1 > a<-gl(3,3,9) > factor(a,levels=3:1) [1] 1 1 1 2 2 2 3 3 3 Levels: 3 2 1 It is probably something obvious I missed, but reading the documentation of factor, and levels I would have thought that both should produce the same output as