R help - Sep 2009 - Re-order columns

Dear R'sians,
Would really appreciate if you could suggest a more efficient way to order
the columns of a dataset. The column names of the dataset contain indices
separated by a period. Following are examples of my code and the dataset.

oC <- function(tg=x2) {
    lth    <- length(grep("T",names(tg)))
    thix   <- match(paste("T",seq(lth),sep=""),names(tg))
    soi    <-
c(grep("O",names(tg),fixed=T),grep("S",names(tg),fixed=T))
    oi     <-
data.frame(t(data.frame(strsplit(names(tg)[soi],"\\."),stringsAsFactors=F)),stringsAsFactors=F)
    names(oi) <-
c("par","ix1","ix2");row.names(oi) <- NULL
    oi$cnm <- sub("^\\s+((.*\\S)\\s+)?$", "\\2",
paste("
",paste(oi$par,oi$ix1,oi$ix2,"
",sep="."),sep=""))
    oi1    <- oi[order(oi$par,as.integer(oi$ix1),as.integer(oi$ix2)),]
    return(list(oi1=oi1,thix=thix))
 }

x1 <- c("TABLE NO. 1: Gold" ,
" R            T2           T1           O.4.1.       O.3.1.
S.1.1.      " ,
"            0  3.68000E+01  1.41000E+03  4.94000E+03  2.33000E+01
4.90000E+01",
"           43  3.77870E+01  2.46260E+03  4.44227E+03  2.31390E+01
4.84272E+01",
"           -1  3.67870E+01  1.46260E+03  4.44227E+03  2.31390E+01
4.84271E+01")
x2 <- read.table(textConnection(x1),as.is=T,head=T,skip=1)
x3 <- oC(x2)
pcols  <- c(names(x2)[x3$thix],x3$oi1$cnm)
x4 <- x2[,match(c(setdiff(names(x2),pcols),pcols),names(x2))]

Thanks a TON

Santosh

	[[alternative HTML version deleted]]

I'm not sure I understand what you are doing below but to rearrange columns
of a data.frame you can just rearrange the indices.  Is this what you mean?

A couple of examples:

(mydata <- data.frame(aa=as.character(c("a", "b",
"c", "d", "e")), bb = 1:5))

(mydata  <- mydata[,2:1])


df1 <- structure(list(site = c(1, 1, 4, 4, 1, 4), id = structure(c(1L,
2L, 2L, 3L, 1L, 2L), .Label = c("a", "b", "c"),
class = "factor"),
    cata = c(1, 1, 6, 1, 1, NA), catb = c(1, 2, 3, 4, 5, 6),
    doga = c(3, 5, 3, 6, 4, 0), dogb = c(2, 4, 6, 8, 10, 12),
    rata = c(NA, 9, 9, 8, 9, 8), ratb = c(1, 2, 3, 4, 5, 6),
    bata = c(12, 42, NA, 45, 32, 54), batb = c(13, 15, 17, 19,
    21, 23)), .Names = c("site", "id", "cata",
"catb", "doga",
"dogb", "rata", "ratb", "bata",
"batb"), row.names = c("aa",
"bb", "cc", "dd", "ee", "ff"),
class = "data.frame")

df1

(df1 <- df1[,rev(1:10)])


--- On Sat, 9/19/09, Santosh <santosh2005 at gmail.com> wrote:
> From: Santosh <santosh2005 at gmail.com>
> Subject: [R] Re-order columns
> To: r-help at r-project.org
> Received: Saturday, September 19, 2009, 11:25 AM
> Dear R'sians,
> Would really appreciate if you could suggest a more
> efficient way to order
> the columns of a dataset. The column names of the dataset
> contain indices
> separated by a period. Following are examples of my code
> and the dataset.
> 
> oC <- function(tg=x2) {
> ? ? lth? ? <-
> length(grep("T",names(tg)))
> ? ? thix???<-
> match(paste("T",seq(lth),sep=""),names(tg))
> ? ? soi? ? <-
>
c(grep("O",names(tg),fixed=T),grep("S",names(tg),fixed=T))
> ? ? oi? ???<-
>
data.frame(t(data.frame(strsplit(names(tg)[soi],"\\."),stringsAsFactors=F)),stringsAsFactors=F)
> ? ? names(oi) <-
> c("par","ix1","ix2");row.names(oi) <- NULL
> ? ? oi$cnm <- sub("^\\s+((.*\\S)\\s+)?$",
> "\\2", paste("
> ",paste(oi$par,oi$ix1,oi$ix2,"
",sep="."),sep=""))
> ? ? oi1? ? <-
> oi[order(oi$par,as.integer(oi$ix1),as.integer(oi$ix2)),]
> ? ? return(list(oi1=oi1,thix=thix))
>  }
> 
> x1 <- c("TABLE NO. 1: Gold" ,
> " R? ? ? ? ? ? T2?
> ? ? ? ???T1? ?
> ? ? ???O.4.1.? ?
> ???O.3.1.
> S.1.1.? ? ? " ,
> "? ? ? ? ? ? 0?
> 3.68000E+01? 1.41000E+03? 4.94000E+03?
> 2.33000E+01
> 4.90000E+01",
> "? ? ? ? ???43?
> 3.77870E+01? 2.46260E+03? 4.44227E+03?
> 2.31390E+01
> 4.84272E+01",
> "? ? ? ? ???-1?
> 3.67870E+01? 1.46260E+03? 4.44227E+03?
> 2.31390E+01
> 4.84271E+01")
> x2 <-
> read.table(textConnection(x1),as.is=T,head=T,skip=1)
> x3 <- oC(x2)
> pcols? <- c(names(x2)[x3$thix],x3$oi1$cnm)
> x4 <-
> x2[,match(c(setdiff(names(x2),pcols),pcols),names(x2))]
> 
> Thanks a TON
> 
> Santosh
> 
> ??? [[alternative HTML version deleted]]
> 
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