similar to: How to change letters after space into capital letters

I have: d <- sample(10:100, 9) o <- order(d) r <- d[o] How I can get d (in the original order), knowing only r and o? Thanks - Wolfram

Is there a simple possibility to become directly a matrix from a call of sub() on a matrix? --------- START OF LOGFILE ---------------- # R 1.4.1 > a <- matrix( letters[1:6], 2, 3 ) # a is a matrix > print( a ) [,1] [,2] [,3] [1,] "a" "c" "e" [2,] "b" "d" "f" > b <- sub( '(.)', '-\\1-', a )

What is the reason, that the levels of the factor returned by cut() are not marked as ordered levels? > is.ordered( cut( breaks=3, sample(10 ) ) ) FALSE > help(factor) ... If 'ordered' is 'TRUE', the factor levels are assumed to be ordered. ... Wolfram

I propose to add a function that allows to display colors selected by a text pattern or by color vectors in a plot. Wolfram Fischer #--- showcolors.R showcolors <- function( col = "red" , index = NULL , pie = TRUE , lwd = 6 , cex = 1.0 , main = NULL , sub = NULL , ... ){ n.colors <- length( col ) if( n.colors > 1 ){ main <- deparse( substitute( col ) )

I would propose to add " See also: `nchar' for counting the number of character in character vectors. " to the helpfile of length(), because it is rather difficult to find nchar() if one has only search terms as "length", "len", "strlen" in mind. Sincerly Wolfram Fischer

I wanted to test if there exists already a name (which is incidentally a substring of another name) in a dataframe. I did e.g.: > data(swiss) > names(swiss) [1] "Fertility" "Agriculture" "Examination" "Education" [5] "Catholic" "Infant.Mortality" > ! is.null(swiss$EduX) [1] FALSE > !

Is there a function in R that corresponds to the function ``map'' of perl? It could be called like: vector.a <- map( vector.b, FUN, args.for.FUN ) It should execute for each element ele.b of vector.b: FUN( vector.b, args.for.FUN) It should return a vector (or data.frame) of the results of the calls of FUN. It nearly works using: apply( data.frame( vector.b ), 1, FUN,

I downloaded: http://www.bioconductor.org/data/metaData/hgu95av2_1.7.0.tar.gz described as: Package: hgu95av2 Title: A data package containing annotation data for hgu95av2 Version: 1.7.0 Created: Wed Jan 12 16:57:23 2005 Author: Lin,Chenwei Description: Annotation data file for hgu95av2 assembled using data from public data repositories Maintainer:

x.fun <- function( formula, data ) dotplot( formula, data ) x.grp <- function( formula, groups, data ) dotplot( formula, groups, data ) data( barley ) > x.fun( variety ~ yield | site, data=barley ) # no problem > dotplot( variety ~ yield | site, groups=year, data=barley ) # no problem > x.grp( variety ~ yield | site, groups=year, data=barley ) object "year" not found

If I want to plot different widths of line segments or arrows I have to program loops to plot each line apartly. Is that right? n <- 6 x <- 1:n y <- rnorm(1:n) q <- ( x %% 3 + 1 ) * 2 plot( x, y, cex=q ) for( i in 1:(n-1) ) lines( x[i:(i+1)], y[i:(i+1)], lwd=q[i], col=q[i] ) Would it not be possible to make plotting functions accept vectors of line widths (as they

[R 2.0.0 on Linux] I tried: > df <- data.frame( grp1=factor( c('A' ,'A' ,'A' ,'D', 'D' ) ) , grp2=factor( c('a1','a2','a2','d1','d1') ) ) > df grp1 grp2 val 1 A a1 1 2 A a2 2 3 A a2 4 4 D d1 8 5 D d1 16 I got: > with( df, ave( val, grp1, grp2, FUN=sum ) )

Question: How can a repeat a dataframe n times in the direction of the rows? (Or: How I can rbind a dataframe n times ?) Example Data: x <- data.frame( alpha=letters[1:3], num=1:3 ) n <- 4 Complicated solution: To rbind a dataframe n times I can program a loop: xr <- x for( i in 2:n ) xr <- rbind( xr, n ) Observation: To cbind a dataframe n times I

On some errors during the plot of a lattice/grid graphics, there is written a message like "Error using packet 1: missing value where TRUE/FALSE needed" into the concering panel and the next panel is plotted. Which option I could use to stop the execution to have a look at the error by a traceback? Regards - Wolfram

R 1.8.1: ___COMMAND____________________________________________ levelplot( yield ~ year * variety | site, barley ) ___ERROR_MESSAGE______________________________________ Error in Summary.factor(..., na.rm = na.rm) : "min" not meaningful for factors ___COMMENT____________________________________________ levelplot( yield ~ as.numeric(year) * as.numeric(variety) | site,

Problem: - Shell does not echo anymore after using CTRL-C (to interrupt calculations) once during a R session. Procedure: - [start session] R - [input] CTRL-C - [end session] q() Some details: - `stty` before: speed 38400 baud; line = 0; lnext = <undef>; - `stty` after: speed 38400 baud; line = 0; lnext = <undef>; min = 1; time = 0; -icrnl -icanon -echo I

[R 1.6.1] PROBLEM If text() is called with an argument of type POSIXlt, I get the following error message: "Error in as.double.default(x) : (list) object cannot be coerced to vector type 14" REMARK plot() accepts this data. EXAMPLE data( Lifeboats, package='vcd' ) attach(Lifeboats) plot( launch, total ) text( launch, total,

PROBLEM ``./configure'' creates empty Makefiles and the following error messages if there is a ``,'' in the name of the working directory. REMARK As I have seen, ``,'' is used as separator for the sed statements in config.status. Would it be possible to use instead a character which cannot be used in a pathname, e.g. ``;''? ERROR MESSAGES ... checking for lpr...

I have two lists: xa <- list( X=c(1,2,3), Y=c(4,5,6), Z=c(7,8,9) ) xb <- with( barley, tapply( X=seq(1:nrow(barley)), INDEX=site , FUN=function(z)yield[z])) I can convert xa to a dataframe easily with: as.data.frame(xa) But if i try the same with xb I get: as.data.frame(xb) Error in as.data.frame.default(xb) : can't coerce array into a data.frame What

See if the following helps: > m <- outer(letters[1:5], 1:4, paste, sep="") > m [,1] [,2] [,3] [,4] [1,] "a1" "a2" "a3" "a4" [2,] "b1" "b2" "b3" "b4" [3,] "c1" "c2" "c3" "c4" [4,] "d1" "d2" "d3" "d4" [5,]

How to show panels for factor levels of conditioning variables which do have no values? E.g. there are panels for "Grand Rapids" when they have values: data( barley ) with( barley, dotplot(variety ~ yield | year * site, layout=c(6,2) ) ) There are no panels for "Grand Rapids" when there are no values for "Grand Rapids": my.barley <- subset( barley, ! ( site ==