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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