One more thing, is it cheap to hash immutable maps vs mutable ones?
On Saturday, June 27, 2015 at 10:44:44 AM UTC-4, Jacob Goodson wrote:
>
> 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]>
>> 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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>>
>>
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