Well, it can't be the best, because it has a bug. Calling it with "a"
instead of "aa" gives me a NPE:
x=> (subseq (sorted-map-by string-like-coll-comparator (vec "ab") 1
(vec "n") 2) > (vec "a"))
NullPointerException clojure.lang.AFunction.compare (AFunction.java:
59)
The following comparator should do the trick:
(defn string-like-coll-comparator
[coll1 coll2]
(or (first (drop-while zero? (map compare coll1 coll2)))
(- (count coll1) (count coll2))))
x=> (subseq (sorted-map-by string-like-coll-comparator (vec "ab") 1
(vec "n") 2) > (vec "a"))
([[\a \b] 1] [[\n] 2])
So is this then the best way to get the sorting behavior I need?
On Jul 10, 6:09 pm, Eugen Dück <eu...@dueck.org> wrote:
> Say I have this sorted map, using strings as keys:
>
> x=> (sorted-map "ab" 1 "n" 2)
> {"ab" 1, "n" 2}
>
> When I do a subseq
>
> x=> (subseq (sorted-map "ab" 1 "n" 2) > "aa")
> (["ab" 1] ["n" 2])
>
> I get back both entries.
>
> Now if I do the same subseq on the same map, except that I turn all
> strings into character collections, I only get the "ab" entry:
>
> x=> (subseq (sorted-map (vec "ab") 1 (vec "n") 2) > (vec "aa"))
> ([[\a \b] 1])
>
> I guess that's fine, the length of collections is probably used first
> and only if it differs would the actual entries be compared.
>
> Using my own comparator to simulate the string comparison I can get
> the string-like behavior:
>
> (defn string-like-coll-comparator
> [coll1 coll2]
> (first (drop-while zero? (map compare coll1 coll2))))
>
> x=> (subseq (sorted-map-by string-like-coll-comparator (vec "ab") 1
> (vec "n") 2) > (vec "aa"))
> ([[\a \b] 1] [[\n] 2])
>
> Here's my question - is this the best way to get what I want?
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