[R-sig-eco] altitudinal overlap

[Sargeant, Glen]

Here you go.

> #Data.  Each row gives altitudinal limits for a species
>
> x <- matrix(runif(20),10,2)

> x <- t(apply(x,1,sort))

> dimnames(x) <- list(NULL, c("min","max"))

> species <- paste("species",1:10,sep="")

> x <- data.frame(species,x)

> x
     species        min       max
1   species1 0.20903724 0.8232747
2   species2 0.25137925 0.7794901
3   species3 0.09128804 0.2311824
4   species4 0.07792821 0.8668317
5   species5 0.34627541 0.5385044
6   species6 0.14072631 0.3356345
7   species7 0.52154781 0.9216631
8   species8 0.03126684 0.3382409
9   species9 0.43925482 0.4398011
10 species10 0.41146614 0.4611622

> #Order by min and then by max within min
> x <- x[order(x$min,x$max),]

> x
     species        min       max
8   species8 0.03126684 0.3382409
4   species4 0.07792821 0.8668317
3   species3 0.09128804 0.2311824
6   species6 0.14072631 0.3356345
1   species1 0.20903724 0.8232747
2   species2 0.25137925 0.7794901
5   species5 0.34627541 0.5385044
10 species10 0.41146614 0.4611622
9   species9 0.43925482 0.4398011
7   species7 0.52154781 0.9216631

> #Compare each minimum to preceding maxima.

> #This is where the reordering comes in.
> #The only way two species can overlap is if
> #the minimum for the second one is less than
> #or equal to the maximum for the first one.
> mat <- matrix(NA,nrow(x),nrow(x))

> dimnames(mat) <- list(x$species,x$species)

> for(i in 1:(nrow(x)-1)){
+   for(j in (i+1):nrow(x)){
+     mat[i,j] <- x$min[j] <= x$max[i]
+   }
+ }

> mat
          species8 species4 species3 species6 species1 species2 species5
species10 species9 species7
species8        NA     TRUE     TRUE     TRUE     TRUE     TRUE
FALSE     FALSE    FALSE    FALSE
species4        NA       NA     TRUE     TRUE     TRUE     TRUE
TRUE      TRUE     TRUE     TRUE
species3        NA       NA       NA     TRUE     TRUE    FALSE
FALSE     FALSE    FALSE    FALSE
species6        NA       NA       NA       NA     TRUE     TRUE
FALSE     FALSE    FALSE    FALSE
species1        NA       NA       NA       NA       NA     TRUE
TRUE      TRUE     TRUE     TRUE
species2        NA       NA       NA       NA       NA       NA
TRUE      TRUE     TRUE     TRUE
species5        NA       NA       NA       NA       NA       NA
NA      TRUE     TRUE     TRUE
species10       NA       NA       NA       NA       NA       NA
NA        NA     TRUE    FALSE
species9        NA       NA       NA       NA       NA       NA
NA        NA       NA    FALSE
species7        NA       NA       NA       NA       NA       NA
NA        NA       NA       NA

> apply(mat,1,sum,na.rm=TRUE)
 species8  species4  species3  species6  species1  species2  species5
species10  species9  species7
        5         8         2         2         5         4
3         1         0         0

On Tue, Jul 14, 2015 at 6:37 PM, Karla Shikev <karlashikev at gmail.com> wrote:

> Ok, here's another (a bit more challenging) question.
>
> Imagine that I have a S species and a S x 2 matrix of altitudinal limits
> (min and max elevation). How would you compute the number of co-occurring
> species for each of the S species?
>
> Thanks again for your assistance.
>
> Karla
>
> On Tue, Jul 14, 2015 at 5:14 AM, Stefano Leonardi <
> stefano.leonardi at unipr.it
> > wrote:
>
> >
> > The first think that came to my mind:
> >
> > overlap <- function(v1,v2) {
> > ov <-min(max(v1), max(v2)) - max(min(v1), min(v2))
> > ifelse(ov > 0, ov, 0)
> > }
> >
> > overlap(dat[1,], dat[2,])
> > [1] 0.5
> >
> >
> > Ciao
> > Stefano
> >
> > On 14/07/2015 01:03, Karla Shikev wrote:
> >
> >> Hi there,
> >>
> >> This is a newbie question, and I'm sure there are simple ways to do
> this,
> >> but I've spent my entire afternoon and I couldn't get it to work.
> >>
> >> Imagine that I got the altitudinal range of different species. For
> >> instance:
> >>
> >>  dat<-matrix(c(1,3,2.5,4), ncol=2, byrow=TRUE)
> >>> dat
> >>>
> >>       [,1] [,2]
> >> [1,]  1.0    3
> >> [2,]  2.5    4
> >>
> >>
> >> The first line indicates that this species is found between 1 and 3,
> >> whereas the second species was found between 2.5 and 4.
> >>
> >> I need a simple way to calculate the overlap of their extents (0.5 in
> this
> >> case). This way should provide 0 if there is no overlap, and it should
> >> also
> >> work in the case where one subject is found only within the extent of
> the
> >> second subject.
> >>
> >> Any help will be greatly appreciated.
> >>
> >> Karla--
> >>
> > ======================================================================
> >  Stefano Leonardi
> >  Dipartimento di Bioscienze
> >  Universita` di Parma
> >  Viale Usberti 11a                             Phone : +39-0521-905659
> >  43124 PARMA  (Italy)                          Fax   : +39-0521-905402
> >
> >  Il mio photoblog:                http://stefanoleonardi.wordpress.com
> > ======================================================================
> >
>
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>
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>



-- 
Glen Sargeant, Ph.D.
Research Wildlife Biologist/Statistician
USGS Northern Prairie Wildlife Research Center
E-mail: gsargeant at usgs.gov
Phone: (701) 253-5528

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