Martin Maechler wrote:
>>>>>> "MS" == Marc Schwartz <[EMAIL PROTECTED]>
>>>>>> on Thu, 20 Dec 2007 16:33:54 -0600 writes:
>
> MS> On Thu, 2007-12-20 at 22:43 +0100, Johannes Graumann wrote:
> >> Hi all,
> >>
> >> Does anybody have a magic trick handy to isolate directly consecutive
> >> integers from something like this:
> >> c(1,2,3,4,7,8,9,10,12,13)
> >>
> >> The result should be, that groups 1-4, 7-10 and 12-13 are consecutive
> >> integers ...
> >>
> >> Thanks for any hints, Joh
>
> MS> Not fully tested, but here is one possible approach:
>
> >> Vec
> MS> [1] 1 2 3 4 7 8 9 10 12 13
>
> MS> Breaks <- c(0, which(diff(Vec) != 1), length(Vec))
>
> >> Breaks
> MS> [1] 0 4 8 10
>
> >> sapply(seq(length(Breaks) - 1),
> MS> function(i) Vec[(Breaks[i] + 1):Breaks[i+1]])
> MS> [[1]]
> MS> [1] 1 2 3 4
>
> MS> [[2]]
> MS> [1] 7 8 9 10
>
> MS> [[3]]
> MS> [1] 12 13
>
>
>
> MS> For a quick test, I tried it on another vector:
>
>
> MS> set.seed(1)
> MS> Vec <- sort(sample(20, 15))
>
> >> Vec
> MS> [1] 1 2 3 4 5 6 8 9 10 11 14 15 16 19 20
>
> MS> Breaks <- c(0, which(diff(Vec) != 1), length(Vec))
>
> >> Breaks
> MS> [1] 0 6 10 13 15
>
> >> sapply(seq(length(Breaks) - 1),
> MS> function(i) Vec[(Breaks[i] + 1):Breaks[i+1]])
> MS> [[1]]
> MS> [1] 1 2 3 4 5 6
>
> MS> [[2]]
> MS> [1] 8 9 10 11
>
> MS> [[3]]
> MS> [1] 14 15 16
>
> MS> [[4]]
> MS> [1] 19 20
>
> Seems ok, but ``only works for increasing sequences''.
> More than 12 years ago, I had encountered the same problem and
> solved it like this:
>
> In package 'sfsmisc', there has been the function inv.seq(),
> named for "inversion of seq()",
> which does this too, currently returning an expression,
> but returning a call in the development version of sfsmisc:
>
> Its definition is currently
>
> inv.seq <- function(i) {
> ## Purpose: 'Inverse seq': Return a short expression for the 'index' `i'
> ## --------------------------------------------------------------------
> ## Arguments: i: vector of (usually increasing) integers.
> ## --------------------------------------------------------------------
> ## Author: Martin Maechler, Date: 3 Oct 95, 18:08
> ## --------------------------------------------------------------------
> ## EXAMPLES: cat(rr <- inv.seq(c(3:12, 20:24, 27, 30:33)),"\n"); eval(rr)
> ## r2 <- inv.seq(c(20:13, 3:12, -1:-4, 27, 30:31)); eval(r2); r2
> li <- length(i <- as.integer(i))
> if(li == 0) return(expression(NULL))
> else if(li == 1) return(as.expression(i))
> ##-- now have: length(i) >= 2
> di1 <- abs(diff(i)) == 1 #-- those are just simple sequences n1:n2 !
> s1 <- i[!c(FALSE,di1)] # beginnings
> s2 <- i[!c(di1,FALSE)] # endings
>
> ## using text & parse {cheap and dirty} :
> mkseq <- function(i,j) if(i == j) i else paste(i,":",j, sep="")
> parse(text =
> paste("c(", paste(mapply(mkseq, s1,s2), collapse = ","), ")", sep =
> ""),
> srcfile = NULL)[[1]]
> }
>
> with example code
>
> > v <- c(1:10,11,6,5,4,0,1)
> > (iv <- inv.seq(v))
> c(1:11, 6:4, 0:1)
> > stopifnot(identical(eval(iv), as.integer(v)))
> > iv[[2]]
> 1:11
> > str(iv)
> language c(1:11, 6:4, 0:1)
> > str(iv[[2]])
> language 1:11
> >
>
>
> Now, given that this stems from 1995, I should be excused for
> using parse(text = *) [see fortune(106) if you don't understand].
>
> However, doing this differently by constructing the resulting
> language object directly {using substitute(), as.symbol(),
> as.expression() ... etc}
> seems not quite trivial.
>
> So here's the Friday afternoon / Christmas break quizz:
>
> What's the most elegant way
> to replace the last statements in inv.seq()
> ------------------------------------------------------------------------
> ## using text & parse {cheap and dirty} :
> mkseq <- function(i,j) if(i == j) i else paste(i,":",j, sep="")
> parse(text =
> paste("c(", paste(mapply(mkseq, s1,s2), collapse = ","), ")", sep =
> ""),
> srcfile = NULL)[[1]]
> ------------------------------------------------------------------------
>
> by code that does not use parse (or source() or similar) ???
>
> I don't have an answer yet, at least not at all an elegant one.
> And maybe, the solution to the quiz is that there is no elegant
> solution.
How about this ? :
> i <- c(1, 10, 12)
> j <- c(5, 10, 14)
> mkseq <- function(i, j) if (i==j) i else call(':', i, j)
> as.call(c(list(as.name('c')), mapply(i, j, FUN=mkseq)))
c(1:5, 10, 12:14)
> eval(.Last.value)
[1] 1 2 3 4 5 10 12 13 14
>
-- Tony Plate
>
> Martin
>
>
> MS> HTH,
>
> MS> Marc Schwartz
>
> ______________________________________________
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>
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