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On 01/06/2011 11:57 PM, (Ted Harding) wrote:
> On 06-Jan-11 22:16:38, array chip wrote:
>> Hi, I am seeking help on designing an algorithm to identify the
>> locations of stretches of 1s in a vector of 0s and 1s. Below is
>> an simple example:
>>
>>> dat<-as.data.frame(cbind(a=c(F,F,T,T,T,T,F,F,T,T,F,T,T,T,T,F,F,F,F,T)
>> ,b=c(4,12,13,16,18,20,28,30,34,46,47,49,61,73,77,84,87,90,95,97)))
>>
>>> dat
>> a b
>> 1 0 4
>> 2 0 12
>> 3 1 13
>> 4 1 16
>> 5 1 18
>> 6 1 20
>> 7 0 28
>> 8 0 30
>> 9 1 34
>> 10 1 46
>> 11 0 47
>> 12 1 49
>> 13 1 61
>> 14 1 73
>> 15 1 77
>> 16 0 84
>> 17 0 87
>> 18 0 90
>> 19 0 95
>> 20 1 97
>>
>> In this dataset, "b" is sorted and denotes the location for each
>> number in "a".
>> So I would like to find the starting & ending locations for each
>> stretch of 1s within "a", also counting the number of 1s in each
>> stretch as well.
>> Hope the results from the algorithm would be:
>>
>> stretch start end No.of.1s
>> 1 13 20 4
>> 2 34 46 2
>> 3 49 77 4
>> 4 97 97 1
>>
>> I can imagine using for loops can do the job, but I feel it's not a
>> clever way to do this. Is there an efficient algorithm that can do
>> this fast?
>>
>> Thanks for any suggestions.
>> John
>
> The basic information you need can be got using rle() ("run length
> encoding"). See '?rle'. In your example:
>
> rle(dat$a)
> # Run Length Encoding
> # lengths: int [1:8] 2 4 2 2 1 4 4 1
> # values : num [1:8] 0 1 0 1 0 1 0 1
> ## Note: F -> 0, T -> 1
>
> The following has a somewhat twisted logic at the end, and may
> be flawed, but you can probably adapt it!
>
> L <- rle(dat$a)$lengths
> V <- rle(dat$a)$values
> pos <- c(1,cumsum(L))
> V1 <- c(-1,V)
> 1+pos[V1==0]
> # [1] 3 9 12 20
> ## Positions in the series dat$a where each run of "T" (i.e. 1)
> ## starts
A different approach would be to use the diff() function:
Where
> diff(dat$a)
[1] 0 1 0 0 0 -1 0 1 0 -1 1 0 0 0 -1 0 0 0 1
is not equal 0, the value is changing from 0 to 1 or one to 0.
The indices of the first new value can be found by:
> which(diff(dat$a)!=0) + 1
[1] 3 7 9 11 12 16 20
where it is changing from 0 to 1 is at
> which(diff(dat$a)==1) + 1
[1] 3 9 12 20
where it is changing from 1 to 0 is at
> which(diff(dat$a)==-1) + 1
[1] 7 11 16
By taking into consideration if the first value and the last values are
0 or 1, you can calculate the length.
Cheers,
Rainer
>
> Hoping this helps,
> Ted.
>
> --------------------------------------------------------------------
> E-Mail: (Ted Harding) <ted.hard...@wlandres.net>
> Fax-to-email: +44 (0)870 094 0861
> Date: 06-Jan-11 Time: 22:57:44
> ------------------------------ XFMail ------------------------------
>
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- --
Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation
Biology, UCT), Dipl. Phys. (Germany)
Centre of Excellence for Invasion Biology
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