Better way to find distances between points in a set?

alldist3 <- as.matrix( dist( cv ) )
which( alldist3 == max( alldist3 ), arr.ind=TRUE )

I was playing around a bit to see how I could find the two points in a set of 
points (or ordered pairs)  furthest from each other.

Here's what I did:
1) created a Nrow by 2col matrix, so each row contains an x,y coordinate 
pair.

2) fed the matrix to a nested mapply (cv is my matrix):

mapply(function(k,l) mapply(function(x,y,a,b)
+ sqrt((x-a)^2+(y-b)^2),cv[,1],cv[,2],k,l),cv[,1],cv[,2])->alldist

Then I just did which.max(alldist) and found the original two points by 
figuring out what row, col the results of which.max referred to.

So, what's a better, or cleaner way to do all this?  That is, is there a 
function in some package that will do anything like my nested mapply thing, 
and is there a better tool than which.max for locating a position in a 
matrix?

thanks
Carl

On Tue, Dec 9, 2008 at 11:52 PM, Charles C. Berry <cberry at tajo.ucsd.edu>wrote:

Charles, I am distressed by the nasty tone of your note.  Though I certainly
agree that poor queries on this list are an annoyance, and that the posting
guide gives useful advice on what to do before posting, I don't think that
poor queries are a reason to presume bad faith for any but the most flagrant
correspondents ("here is my problem set, solve it for me").  And though I
have not been on this list very long, I have been answering questions on
other lists like this for a long time.

What's more, it seems to me that Witthoft has a legitimate problem.  I tried
to solve it myself, and here is what I come up with:

test <- matrix(rnorm(20),ncol=2)

which.max(dist(x),arr.ind=TRUE)
     => Error, though which supports arr.ind, which.max doesn't
        Yes, that is documented, but it violates reasonable expectations.

which(dist(x)==max(dist(x)),arr.ind=TRUE)
     => 21  oops, that is a vector index; what about arr.ind?
        Apparently the dist class (return class of dist fnc.)
        doesn't act as an array here (and which/arr.ind is
        happy to treat a vector as an array without warning).
        Is this a problem in dist, or a problem in which?
        So how do I map back from 21 to 3,7?  Nothing obvious.
        ? `[`, ? array, and ? dist don't seem to help.

So let's convert to a matrix explicitly:

which(as.matrix(dist(x)==max(dist(x))),arr.ind=TRUE)
     =>      row col
         [1,] 21   1
        Hmm, that's not right; apparently the == flattened the
        dist class into its base vector, which which/arr.ind
        helpfully (?) treated as a matrix.

which(as.matrix(dist(x))==max(dist(x)),arr.ind=TRUE)
    =>        row col
          7     7   3
          3     3   7
        Finally getting useful results; but converting to the array
        added in the upper triangular part... oh, well, we need to
        choose *which* maximum anyway.

So, as Witthoft said: what's a better, or cleaner way to do all this?  And
why aren't R's operators more regular in their behavior?  Why doesn't
arr.ind work for which.max?  Why doesn't dist(x)==1 return an object of the
same shape as the original dist (as matrix(...)==1 would have), but instead
flatten it?  Where does one find the function that maps from a list of array
indices to the corresponding vector index and vice versa?

             -s

On Tue, Dec 9, 2008 at 11:52 PM, Charles C. Berry <cberry at tajo.ucsd.edu>wrote:

The 'better way' to do almost anything starts with a reading of the
_posting guide_, which reminds you to
       Do your homework before posting [Reasons whyfor deleted]]...

Oh yes, if you are too lazy to look up the posting guide URL, the function
help.request() will open it for you when you admit that you haven't yet read
it (or lead you thru the further steps to prepare a question to this list if
you say that you have read it).

Charles, I am distressed by the nasty tone of your note.

Though I certainly
agree that poor queries on this list are an annoyance, and that the posting
guide gives useful advice on what to do before posting, I don't think that
poor queries are a reason to presume bad faith for any but the most flagrant
correspondents ("here is my problem set, solve it for me"). And though I
have not been on this list very long, I have been answering questions on
other lists like this for a long time.

What's more, it seems to me that Witthoft has a legitimate problem.

to solve it myself, and here is what I come up with:

test <- matrix(rnorm(20),ncol=2)

which.max(dist(x),arr.ind=TRUE)
    => Error, though which supports arr.ind, which.max doesn't
       Yes, that is documented, but it violates reasonable expectations.

which(dist(x)==max(dist(x)),arr.ind=TRUE)
    => 21  oops, that is a vector index; what about arr.ind?
       Apparently the dist class (return class of dist fnc.)
       doesn't act as an array here (and which/arr.ind is
       happy to treat a vector as an array without warning).
       Is this a problem in dist, or a problem in which?
       So how do I map back from 21 to 3,7?  Nothing obvious.
       ? `[`, ? array, and ? dist don't seem to help.

So let's convert to a matrix explicitly:

which(as.matrix(dist(x)==max(dist(x))),arr.ind=TRUE)
    =>      row col
        [1,] 21   1
       Hmm, that's not right; apparently the == flattened the
       dist class into its base vector, which which/arr.ind
       helpfully (?) treated as a matrix.

which(as.matrix(dist(x))==max(dist(x)),arr.ind=TRUE)
   =>        row col
         7     7   3
         3     3   7
       Finally getting useful results; but converting to the array
       added in the upper triangular part... oh, well, we need to
       choose *which* maximum anyway.

So, as Witthoft said: what's a better, or cleaner way to do all this?

And
why aren't R's operators more regular in their behavior?  Why doesn't
arr.ind work for which.max?  Why doesn't dist(x)==1 return an object of the
same shape as the original dist (as matrix(...)==1 would have), but instead
flatten it?  Where does one find the function that maps from a list of array
indices to the corresponding vector index and vice versa?

            -s