similar to: new function: showcolors {base}

I suggest to add a new function to create a vector of n ``contiguous'' colors with tails in two colors. This function is similar to `cm.colors' but the colors can be choosen by hsv values. This function could be used e.g. as alternative to the default ``col.regions'' in `levelplot'. Perhaps the arguments in the following code could be simplified. Wolfram Fischer #---

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

What is the easiest way to change within vector of strings each letter after a space into a capital letter? E.g.: c( "this is an element of the vector of strings", "second element" ) becomes: c( "This Is An Element Of The Vector Of Strings", "Second Element" ) My reason to try to do this is to get more readable abbreviations. (A suggestion would be to

Hello All I am looking for high precision values for the normal distribution in the tail,(1e-10 and 1 - 1e-10) as the R package that I am using sets any number which is out of this range to these values and then calls the qnorm and qt function. What I have noticed is that the qnorm implementation in R is not symmetric when looking at the tails. This is quite surprising to me, as it is well known

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

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

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 > !

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

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,

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,]

Hi, Problem Summary: I have been trying to add the highlight effect to my table but everytime a new record is added the highlighting gives an rjs error. However when i click the add to cart button on a product that already exists inside the table, the highlight effect works. Is this because the partial must only have only one <tr></tr> in it and the <table> must be on the other

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:

I suggest to add a panel function to levelplot (or perhaps to an other 3d lattice function) which is able to translate the z values into the size of the rectangles. It could be used to display categorical data. I append the proposed code and two examples: - panel.catlevelplot() - example1.catlevelplot.esoph() - example2.catlevelplot.esoph() Wolfram Fischer #------ CODE

Dear list, Here's a function that works fine when I assign one value to i. qx, ax and gr are vectors. ax.to.nax <- function(qx, ax, gr=c(0,1,5,10)){ px <- 1 - qx last.n <- gr[length(gr)] - gr[length(gr) - 1] all.bounds <- c(gr, gr[length(gr)] + last.n) n <- diff(all.bounds) # i <- 1 # this i should loop through the values of # gr: i=0, i=1, i=5 and i=10. # testprod

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

hello, all. i am trying something new and am stuck. i typically go with whatever colors the pie function pops out, but i would like to customize it. please find below a sample dataset. ##TVs per household. TVs <- factor(c(rep("0", 1), rep("1", 16), rep("2", 14), rep("3", 12), rep("4", 3), rep("5", 2), rep("6", 2)))

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 ==

[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 ) )