similar to: accolade

mtex[3]<-"I need\'s a \"double quote\" with no backslash" mtex[3] [1] "I need's a \"double quote\" with no backslash" so how is it done? Thanks in advance, graham lawrence _________________________________________________________________ -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list --

Dear list, # I have a DF like this: sleep$b <- c(rep(8,10), rep(9,10)) sleep$me <- with(sleep, ave(extra, group, FUN = mean)) sleep # I would like to create a new variable # holding the b-th value of group 1 and 2. # This is not what I want, it takes always the '8' from group '1' # and not the '9' sleep$gr <- with(sleep, ave(extra, group, FUN = function(x)

> > In addition, if you go to 365, you are NOT BUYING the software, you're > renting the service. You will be paying every year, and a service contract > will cost, and, presumably, cost more every year. > Still a lot better than trying to run your own hodge-podge of nightmares on Linux.

> > > Still a lot better than trying to run your own hodge-podge of nightmares > > on Linux. > > Beg pardon? Did I make a mistake on the email address? I thought this went > to the CentOS general discussion list. > I specifically meant setting up and running email services on linux is not for the feint of heart and delivers little real value considering the plethora of

Dear list, this works fine: x <- split(iris, iris$Species) x1 <- lapply(x, function(L) transform(L, g = L[,1:4] * 3)) but I would like to multiply each Species with another factor: setosa by 2, versicolor by 3 and virginica by 4. I've tried mapply but without success. Any thoughts? Thanks for any idea! Patrick

Sorry, but I'm just an ignorant Linux user! I have a silly question. Situation: Now that R 2.1.0 is out (Thank you, to all concerned!) I'm about to go off and download it and all the packages. What this means in practice is trotting off to my "local" (and brand new) Internet Cafe with some CDs, where I can get the lot over a fast link and burn it onto a CD or two. However,

I have this old memory that the kernels used are a hodge podge of backports etc. So for a kernel-smp-2.6.9-55.0.9.EL should I use 0.4.10? Quoted from http://ivtvdriver.org/index.php/Download: The latest stable releases can be found here. Currently this is version 0.4.10 for kernels <= 2.6.15, version 0.6.7 for kernel 2.6.16, version 0.7.4 for kernel 2.6.17, version 0.10.6 for kernels >=

Dear R-users, again two question... # Question 1 Adding two lines of text to a plot, I am using: # ------------------------------- plot(k[,1], k[,2], pch=16, ylim=range((min(k[,2])-0.2):(max(k[,2])+1))) a <- paste("Cor.:" ,cor(k[,2],k[,1])) b <- paste(nrow(k), "Countries") text(90, max((k[,2])-0.51), a) text(90, max((k[,2])-0.83), b) #

Hi there, I have a data.frame (pwt6) which I would like to transform: country year gdp MEX 1950 2 MEX 1951 5 BOL 1950 4 BOL 1951 12 ITA 1950 45 ITA 1951 2 This should be the result: year MEX.gdp BOL.gdp ITA.gdp 1950 2 4 45 1951 5 12 2 Right now I have this code (better - no code): country.label<-names(table(pwt6$country)) result<-data.frame(year=NULL) for(i in country.label) ?

Hello, I am having a problem with the get.hist.quote command (tseries library) in the Windows version. This problem is not happening is the Linux version (Mandrake 8.2). Attached is the error message, for an example included in the help file. Also the R.Version() details is attached. Please, do you know if there is a workaround ? Thanks, Carlos. ++++++++++++++++++++++++ ERROR MESSAGE

Hello, I have a csv-file which looks like: #### pwt6_r.csv #### code;year;rgdpch AGO;1998;1234 ALB;1998;3576 ARG;1998;#NA SVN;1996;13439 SWE;1996;21492 AGO;1960;#NA ALB;1960;2345 ARG;1960;4634 #### pwt6_r.csv #### To import this file i call: pwt6<-read.csv("d:/pwt6_r.csv",header=T,na.strings="#NA",sep=";") Now I want to generate a new data.frame which include

Hi, I want to read some Times Series of GDP from OECD-Countrys. First I call: > oecd96<-ts(read.csv("oecd96.csv",header=T,sep=";"),start=1950,freq=1) > summary(oecd96) gdpcausb gdpcautb gdpcbelb gdpccanb Min. : 8567 Min. : 4533 Min. : 6616 Min. : 8966 1st Qu.:10771 1st Qu.: 8717 1st Qu.: 9440 1st Qu.:11694 Median

Dear Team, Don't understand output below. Please advise. Windows98 and rw1041(patched) from 1/25/02. Graham Lawrence C:\>PATH=C:\bin;C:\mingw32;C:\Perl;C:\HtmlWork;C:\R\rw1041\bin;C:\Tcl;C:\Tcl\bin C:\>Rcmd INSTALL C:\R\lattice_0.4-0.tar.gz Bad command or file name C:\>Rcmd C:\>Rcmd Install Bad command or file name C:\>dir C:\R\rw1041\bin Volume in drive C has no

On Fri, January 9, 2015 17:36, John R Pierce wrote: > On 1/9/2015 2:32 PM, Always Learning wrote: >> Enterprise, in the RHEL context, suggests stability or have I >> misunderstood the USA definition of "Enterprise" ? > > > Enterprise to me implies large business Enterprise literally means 'undertaking'. It has been used euphemistically since the later

Dear r-help, I'm having this DF df <- data.frame(id = 1:6, xout = c(1234, 2134, 234, 456, 324, 345), xin= c(NA, 34,67,87,34, NA)) and would like to calculate the fraction (xin_t / xout_t-1) The result should be: # NA, 2.76, 3.14, 37.18, 7.46, NA I am sure there is a solution using zoo... but I don't know how... Thanks for any help! Patrick

Dear list, I am trying to split a string using regexp: x <- "2 Value 34 a-c 45 t" strsplit(x, "[0-9]") [[1]] [1] "" " Value " "" " a-c " "" " t" But I don't want to lose the digits (pattern), the result should be: [[1]] [1] "2" " Value " "34" " a-c "

Dear list, I'm trying to replace NA-values with the preceding values in that column. This code works, but I am sure there is a more elegant way... df <- data.frame(id = c("A1", NA, NA, NA, "B1", NA, NA, "C1", NA, NA, NA, NA), value = c(1:12)) rn <- c(rownames(df[!is.na(df$id),]), nrow(df)+1) rn <-

Dear list, I want to apply the "table" function to every pair of variables in df and the return should be a list. setwd(123) asd <- data.frame(a1=sample(1:4, 20, replace=TRUE), a2=sample(1:4, 20, replace=TRUE), a3=sample(1:4, 20, replace=TRUE), a4=sample(1:4, 20, replace=TRUE)) with(asd, table(a1, a2)) with(asd, table(a1,

Hello, I am trying to get a new vector 'x1' based on the not NA-values in column 'a' and 'b'. I found a way but I am sure this is not the best solution. So any ideas on how to "optimize" this would be great! m <- factor(c("a1", "a1", "a2", "b1", "b2", "b3", "d1", "d1"), ordered

Hello all, I have a df like this: w <- c(1.20, 1.34, 2.34, 3.12, 2.89, 4.67, 2.43, 2.89, 1.99, 3.45, 2.01, 2.23, 1.45, 1.59) g <- rep(c("a", "b"), each=7) df <- data.frame(g, w) df # 1. Mean for each group tapply(df$w, df$g, function(x) mean(x)) # 2. Range for each group - fix value 0.15 tapply(df$w, df$g, function(x) x[(x > mean(x) -