this works:

(apply pmap + (partition-all 5 (range 1 20)))

And then reducing the overall result:

(reduce + (apply pmap + (partition-all 5 (range 1 20))))


Leonardo Borges
www.leonardoborges.com


On Sun, Feb 3, 2013 at 12:35 AM, Leandro Moreira
<leandro.rhc...@gmail.com> wrote:
> Hi Jim,
>
> Thanks for your help & time, how do I apply instead of reduce?
>
> I was looking the documentation of pmap and it says
>
> (pmap fn coll & colls)
>
> So I think we can use pmap this way:
>
> (pmap + '(1 2 3) '(4 5 6) '(7 8 9) )
>
> Which gives us the result: (12 15 18)
>
> My intention it's only to use range and partition to substitute that, but I
> think partition-all doesn't work like separated argument as list, instead.
> When we type:
>
> (partition-all 5 (range 1 20))
> Results in this:
> ((1 2 3 4 5) (6 7 8 9 10) ...
>
> There is any way of transforming (maybe a macro or destruct way) this "list"
> in several colls argument as expected by pmap signature (fn coll & colls)?
>
>
> On Saturday, February 2, 2013 11:10:35 AM UTC-2, Jim foo.bar wrote:
>>
>> Use this:
>>
>> (pmap #(reduce + %) (partition-all 5 (range 1 20)))
>>
>> OR as you yourself said you can use 'apply' instead of reduce
>>
>> Jim
>>
>>
>>
>> On 02/02/13 12:31, Leandro Moreira wrote:
>>
>> Hi there,
>>
>> I have this:
>>
>> user=> (partition-all 5 (range 1 20))
>> ((1 2 3 4 5) (6 7 8 9 10) (11 12 13 14 15) (16 17 18 19))
>>
>> And I would like to apply the pmap over the partitions, something like:
>> (the line bellow doesn't work)
>> user=> (pmap + (partition-all 5 (range 1 20)))
>>
>> I would like to have this (pmap + (1 2 3 4 5) (6 7 8 9 10) (11 12 13 14
>> 15) (16 17 18 19)) instead of this (pmap + ((1 2 3 4 5) (6 7 8 9 10) (11 12
>> 13 14 15) (16 17 18 19)))
>>
>> How can I apply pmap over a partition-all result?
>> If it's possible, by doing that am I distributing the load better than if
>> I used the pmap with the whole list?
>>
>> Thanks
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