[R] Efficient way to find consecutive integers in vector?
Martin Maechler
maechler at stat.math.ethz.ch
Fri Dec 21 15:54:22 CET 2007
>>>>> "MS" == Marc Schwartz <marc_schwartz at comcast.net>
>>>>> 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.
Martin
MS> HTH,
MS> Marc Schwartz
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