There is clj-diff for levenshtein and standard diff:
https://github.com/brentonashworth/clj-diff

Jonathan

On Mon, May 30, 2011 at 10:19 PM, joshua-choi <rbysam...@gmail.com> wrote:

> Thanks for the reply. Levenshtein distance would be especially useful
> for when I need to compare general strings, with a general amount of
> edits between them.
>
> Unfortunately, the problem isn't so much determining whether any two
> strings are one edit apart: that can be done just by checking if the
> first two, first and third, or last two characters of one string
> equals those of the other.
>
> The problem is which strings you compare with to determine which ones
> are one edit apart. Right now, I have to compare every string with
> every other string, but it seems really wasteful, seemingly being in
> O(n^2) time. I just can't figure out how to make it more efficient,
> though.
>
> I'm hoping that this a common problem with a common solution, like
> using some data structure I haven't thought of.
>
> On May 30, 1:01 pm, Laurent PETIT <laurent.pe...@gmail.com> wrote:
> > Hi,
> >
> > there's the levenshtein distance algorithm which will help you
> > determine which string is one "edit" close to another (since all your
> > strings are of length 3, then the distance will inevitably be a single
> > replacement if of size one).
> >
> > Don't know if that helps, anyway here's a compact functional
> > implementation:https://gist.github.com/828413
> >
> > 2011/5/30 joshua-choi <rbysam...@gmail.com>:
> >
> >
> >
> >
> >
> >
> >
> > > Let's say that I have a set of strings, each three English letters
> > > long.
> >
> > > How can I determine which strings differ only at one location (e.g.
> > > "xxe" and "xbe")?
> >
> > > Right now, I'm writing a loop that sequentially compares every string
> > > to every other string. I think that there's a better way, but I don't
> > > know where to start.
> >
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