similar to: subsets in data frame

Hi, Laura: How about the following: > sapply(x1.df, class) d x y "factor" "numeric" "numeric" > (byLH <- by(x1.df[-1], x1.df$d, var)) x1.df$d: 1/1/2005 x y x 1.272688 -0.783719 y -0.783719 0.884866 ------------------------------------------------------------- x1.df$d: 1/15/2005 x y x

Hi R People: Is there a way to put an abline line for its objects on a plot, please? I have an its object, ibm2, which runs from the January 2 through May 28. >ibm2 ibm 2004-01-02 91.55 2004-01-05 93.05 2004-01-06 93.06 2004-01-07 92.78 2004-01-08 93.04 2004-01-09 91.21 2004-01-12 91.55 2004-01-13 89.70 2004-01-14 90.31 2004-01-15 94.02 . . . I plot the data. No

Hi R People: Here is a simple set of commands: >x1 <- 1:10 >x2 <- sqrt(x1) >plot(x1) >points(x2,col="red",pch=3) >legend(2,8,legend=c("First","Second"),col=c("black","red"),pch=c(1,3)) > Fine. Now, I would like to put the legend box on the outside of the plot itself, perhaps in the lower left hand corner. I've been

Dear R People: I used the "Surv" function to produce the following object: >a <- Surv(1:4,2:5,c(0,1,1,0)) a [1] (1,2+] (2,3 ] (3,4 ] (4,5+] >str(a) Error in "[.Surv"(object, 1:ile) : subscript out of bounds > Why does str(a) give an error, please? Or did I do something wrong? Thanks in advance. R Version 1.9.1 Windows Sincerely, Laura Holt mailto: lauraholt_983

Hi there! Is ROracle available for Windows, please? I found a download site, but it's really for UNIX/Linux. Here is a "thought question", please: Why do the vector elements start at location 1 rather than zero, as C does? Thanks in advance! R Version 1.9.1 Windows Sincerely, Laura Holt mailto: lauraholt_983 at hotmail.com

Dear R People: I have a data frame with some columns that are numeric and some which are factors. There are zeros in the numeric columns and I would like to change them to NAs. However, there are zeros in some of the factor columns, and I would like them to be left alone. Is there a "global" way to do this, please? I was thinking about use the results from "str" but am

Hi again R People: I have the following time series: >x.ts Jan Feb Mar Apr May Jun Jul Aug Sep 2000 0.95 0.66 0.83 0.66 -1.45 -1.25 0.33 1.03 -0.48 2001 2.55 1.21 -1.10 -0.63 0.01 -2.20 -0.51 1.12 1.11 2002 -1.37 0.55 -0.63 -0.56 0.92 -1.73 0.59 0.77 0.30 2003 0.55 -0.01 -0.54 2.27 -1.29 -0.23 0.09 -0.50 -0.61 Oct Nov Dec 2000 0.84 -1.35

Dear R : Sorry for the off topic question, but does anyone know the reference for the -2 Ln Lambda following a Chi Square distribution, please? Possibly one of Bartlett's? Thanks in advance! Sincerely, Laura Holt mailto: lauraholt_983 at hotmail.com

Hi R people: I would like to learn about classes, methods, S3 and S4. Which book would be the most helpful for this info, please: the green one or the white(and blue) one? Or is there something that would be even better, please? Thanks in advance. Sincerely, Laura Holt mailto: lauraholt_983 at hotmail.com R 2.1.0 Windows. trying to learn

Hi R People: Several of you pointed out that using "tapply" on a data frame will work on the iris data frame. I'm still having a problem. The iris data frame has 150 rows, 5 variables. The first 4 are numeric, while the last is a factor, which has the Species names. I can use tapply for 1 variable at a time: >tapply(iris[,1],iris[,5],mean) setosa versicolor virginica

Dear R People: I have finally seen the error of my ways and have decided to use ESS for R and S + stuff. However, I have a question right from the beginning. I'm somewhat confused by the installation instructions. Do I install XEMACS or ESS first, please? Windows XP R Version 2.0.1 (S + 6.2) Thanks so much! Sincerely, Laura Holt mailto: lauraholt_983 at hotmail.com

Dear R People: I would like to generate a vector/variable name from within a loop to be passed to a table function. This is what I have so far: >assign("p1",paste("raw3.df$",rw2$V1[3],sep="")) >p1 [1] "raw3.df$CITIZEN" > Essentially, I want to use the raw3.df$CITIZEN along with another value to generate a table. However, I'm stuck here. I

Dear R People: I have used the command round(x,3) to produce values with 3 places to the right of the decimal. Is there any command to remove the leading zero before the decimal point, please: that is, if I have 0.375, how do I produce just .375, please? Thanks in advance R 2.0.1 for Windows Sincerely, Laura Holt mailto: lauraholt_983 at hotmail.com

Dear R People: Here is an interesting question: >library(chron) >xt <- seq.dates(from="01/01/2004",by="days",length=5) >xt [1] 01/01/04 01/02/04 01/03/04 01/04/04 01/05/04 > #Fine so far >as.POSIXct(xt) [1] "2003-12-31 18:00:00 Central Standard Time" [2] "2004-01-01 18:00:00 Central Standard Time" [3] "2004-01-02 18:00:00 Central

Dear R People: Here is a function to minimized: >mfun1 function(x,a) { x[1] <- a[1]*x[2] + a[3] - a[2]*(a[1]-a[2])*a[3] x[2] <- a[1]*x[1] - a[2]*a[3] return(x) } Here is my first try: >nlm(mfun1,c(1,1)) Error in f(x, ...) : Argument "a" is missing, with no default > >nlm(mfun1,c(1,1),a=c(0.8,0.5,1)) Error in nlm(mfun1, c(1, 1), a = c(0.8, 0.5, 1)) :

Dear R: First of all, nothing is wrong! I just had a question about the following" "Natural language support but running in an English locale" What does than mean, please? Thanks in Advance R 2.1.0 Windows Laura Holt mailto: lauraholt_983 at hotmail.com

Hi R People: I got the following error from using the "latex" command from Hmisc; >latex(bbm) 'latex' is not recognized as an internal or external command, operable program or batch file. Warning message: cmd execution failed with error code 1 in: shell(cmd, wait = TRUE, intern = output) > I was printing a matrix to a Latex File. This worked fine in Version 1.8. This

Dear R People: I'm trying to use the Rcmd SHLIB to produce a dll. Rcmd SHLIB -o test2.dll test2.f make[1]: *** [libR.a] Error 255 make: *** [libR] Error 2 Where do I go to find out about the "make" errors, please? I suspect that I might be missing something. I have the tools for creating new packages, but maybe I left out something. This is for R version 1.9.0 on Windows.

Hello R People: When using the arima function with the seasonal option, are the seasonal options only good for monthly and quarterly data, please? Also, I believe that weekly and daily data are not appropriate for seasonal parm estimation via arima. Is that correct, please? Thanks, Sincerely, Laura Holt mailto: lauraholt_983 at hotmail.com download!

Dear R People: I have an its object. The original its is the Dow Jones Industrial average from March 23, 1990 to March 23, 2000. I did the following; dj1 = original series >dj2 <- log(dj1) >ddj2 <- diff(dj2) > Ok so far. Now I would like to plot this series with a certain y limit: >min(ddj2) [1] -0.07454905 >max(ddj2) [1] 0.04860535 >plot(ddj2,ylim=c(-0.1,0.1)) Error