Hi,
May be this also works:
dist(x)
# 1 2 3
#2 2
#3 6 4
#4 12 10 6
as.matrix(dist(x))
# 1 2 3 4
#1 0 2 6 12
#2 2 0 4 10
#3 6 4 0 6
#4 12 10 6 0
which(dist(x)==min(dist(x)))
#[1] 1
A.K.
----- Original Message -----
From: Ben Bolker <bbol...@gmail.com>
To: r-h...@stat.math.ethz.ch
Cc:
Sent: Tuesday, September 10, 2013 5:39 PM
Subject: Re: [R] Subtracting elements of a vector from each other stepwise
arun <smartpink111 <at> yahoo.com> writes:
>
> Hi,
> Not sure this is what you wanted:
>
> sapply(seq_along(x), function(i) {x1<- x[i]; x2<- x[-i];
x3<-x2[which.min(abs(x1-x2))];c(x1,x3)})
> # [,1] [,2] [,3] [,4]
> #[1,] 17 19 23 29
> #[2,] 19 17 19 23
> A.K.
It's a little inefficient (because it constructs
the distances in both directions), but how about:
x = c(17,19,23,29)
d <- abs(outer(x,x,"-"))
diag(d) <- NA
d[lower.tri(d)] <- NA
which(d==min(d,na.rm=TRUE),arr.ind=TRUE)
?
> ----- Original Message -----
> From: Michael Budnick <mbudnick08 <at> snet.net>
> To: r-help <at> r-project.org
> Cc:
> Sent: Tuesday, September 10, 2013 4:06 PM
> Subject: [R] Subtracting elements of a vector from each other stepwise
>
> I am trying to figure out how to create a loop that will take the
> difference of each member of a vector from each other and also spit out
> which one has the least difference.
>
> I do not want the vector member to subtract from itself or it must be able
> to disregard the 0 obtained from subtracting from itself.
>
> For example:
>
> x = c(17,19,23,29)
[snip]
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