Thanks.
What about the perf of something like this...
({{:x 1 :y 2} :point} {:x 1 :y 2})
vs
({1 2} 1)
and what if the map was mutable as a key?
On Saturday, June 27, 2015 at 10:08:19 AM UTC-4, Andy Fingerhut wrote:
>
> (let [foo {:x 1}]
> (= foo foo))
>
> is fast, because they are identical, and = is fast for things that are
> identical.
>
> In general, two maps that are = are often not identical.
>
> If two maps have different numbers of elements, = should quickly return
> false because of the way equiv() is implemented for maps, here:
> https://github.com/clojure/clojure/blob/master/src/jvm/clojure/lang/APersistentMap.java#L73-L94
>
> There is a check whether the maps are the same size, and if not, return
> false without checking all key/value pairs.
>
> If they have equal number of key/value pairs, then it will iterate through
> all key/value pairs of one, checking whether the key of one is present in
> the other, and if so, whether their associated values are equiv() or not.
>
> Those checks can be fast if the keys are quickly comparable, and the
> values are quickly comparable.
>
> If you take one persistent map m1 and remove a few keys and/or add a few
> keys to produce m2, then all of the unchanged keys and values they have in
> common will be identical, and very quickly comparable. The slowest part of
> comparing them would be comparing their non-identical parts.
>
> Andy
>
> On Sat, Jun 27, 2015 at 5:01 AM, Jacob Goodson <[email protected]
> <javascript:>> wrote:
>
>> I was wondering something...
>>
>> Would (= {:x 1} {:x 1}) be just as fast as (= 1 1) since maps are
>> immutable? The idea is that since clojure data structures are just values
>> can they be compared much faster than there mutable counter parts? Thanks.
>>
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