R help - Nov 2005 - repeated %in%

Hi

I have a list of vectors, each one of which should be a subset of the  
previous one.
How do I check that this property actually holds?

Toy problem follows with a list of length 4 but my list can be any  
length


subsets <- list(
                 l1 = 1:10 ,
                 l2 = seq(from=1,to=9,by=2),
                 l3 = c(3,7),
                 l4 = 3
                 )

I need

all(subsets[[4]] %in% subsets[[3]]) &
all(subsets[[3]] %in% subsets[[2]]) &
all(subsets[[2]] %in% subsets[[1]])

I can write a little function to do this:


check.for.inclusion <- function(subsets){
   out <- rep(FALSE,length(subsets)-1)
   for(i in 1:(length(subsets)-1)){
     out[i] <- all(subsets[[i+1]] %in% subsets[[i]])
   }
   return(all(out))
}


how to do it elegantly and/or quickly?




--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
  tel  023-8059-7743

Robin Hankin <r.hankin at noc.soton.ac.uk> writes:
> Hi
> 
> I have a list of vectors, each one of which should be a subset of the  
> previous one.
> How do I check that this property actually holds?
> 
> Toy problem follows with a list of length 4 but my list can be any  
> length
> 
> 
> subsets <- list(
>                  l1 = 1:10 ,
>                  l2 = seq(from=1,to=9,by=2),
>                  l3 = c(3,7),
>                  l4 = 3
>                  )
> 
> I need
> 
> all(subsets[[4]] %in% subsets[[3]]) &
> all(subsets[[3]] %in% subsets[[2]]) &
> all(subsets[[2]] %in% subsets[[1]])
> 
> I can write a little function to do this:
> 
> 
> check.for.inclusion <- function(subsets){
>    out <- rep(FALSE,length(subsets)-1)
>    for(i in 1:(length(subsets)-1)){
>      out[i] <- all(subsets[[i+1]] %in% subsets[[i]])
>    }
>    return(all(out))
> }
> 
> 
> how to do it elegantly and/or quickly?
How about
> !any(sapply(mapply(setdiff,subsets[-1],subsets[-4]),length))
[1] TRUE

-- 
   O__  ---- Peter Dalgaard             ??ster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark          Ph:  (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)                  FAX: (+45) 35327907

Robin Hankin wrote:
> check.for.inclusion <- function(subsets){
>    out <- rep(FALSE,length(subsets)-1)
>    for(i in 1:(length(subsets)-1)){
>      out[i] <- all(subsets[[i+1]] %in% subsets[[i]])
>    }
>    return(all(out))
> }
> 
> 
> how to do it elegantly and/or quickly?
> 
  My first thought was to rewrite that function but drop out if it didnt 
match. But then...

  Assuming the first list element is the longest, then its a one liner:

check2 <- function(l){
   length(unique(unlist(l))) == length(l[[1]])
}

  - basically the set of everything in all the elements is the first 
element, so take unique(everything) and see if its the same size as the 
first element. Unless I've missed something...

  I'd call that elegant, it may also be quicker!

Baz

Whoops

  The code I just posted only tested if all the subsequent elements were 
subsets of the first, it didn't check all the sequential subsets!

  Too cold to think straight here today...

Baz