[R] Comparing entire row sets at once efficiently

[Gabor Grothendieck]

If Comparison and Candidates each have no duplicated rows (which
is the situation in the example) then try this:

tail(!duplicated(rbind(Comparison, Candidates)), nrow(Candidates))


On 9/28/06, Dirk Eddelbuettel <edd at debian.org> wrote:
>
> Dear useRs,
>
> I am having a hard time coming up with a nice and efficient solution to
> a problem on entires matrices or data.frames. In spirit, this is similar to
> what setdiff() and setequal() do, but I need it in more dimensions.
>
> Here's a brief description.
>
>  * given a set of factors or sequences, expand.grid() gives me the set
>    of permutations in a data.frame;
>
>    in my case all arguments are numeric so I could convert the data frame to
>    a matrix
>
>    let's call this one Candidates
>
>  * I have a second matrix (or data frame) to compare to; this second
>    set may be a subset of the first, or a superset but it guaranted to
>    contain the same columns
>
>    let's call this one Comparison
>
>  * I want know which rows in Candidates are not yet in Comparison.
>
> A toy example:
>
> > Comparison <- matrix(1:30, ncol=5)
> > Candidates <- Comparison[c(2,4), ]
> > checkRow <- function(r, M) { any( (r[1] == M[,1]) & (r[2] == M[,2]) & (r[3] == M[,3]) & (r[4] == M[,4]) ) }
> > checkRow( Candidates[1,], Comparison)
> [1] TRUE
> > falseRow <- Candidates[1,]
> > falseRow[2] <- 42
> > checkRow( falseRow, Comparison)
> [1] FALSE
> >
>
> The checkRow function works but is a) klunky, b) hardcodes the dimension and
> c) works only on one row at a time.
>
> There must be better ways, at least for a) and b).  What am I missing?
>
> Feel free to reply off-list and I'd gladly summarize back to the list. If you
> don't want your reply (or email) summarized back, please indicate.
>
> Thanks, Dirk
>
>
>
> --
> Hell, there are no rules here - we're trying to accomplish something.
>                                                  -- Thomas A. Edison
>
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