Because
(partial map first) return function
=> (-> [[1 2] [3 4] [5]] (partial map first))
#
you should call function to get result
=> (-> [[1 2] [3 4] [5]] ((partial map first)) flatten)
(1 3 5)
2013/6/8 Matt Smith
> Newbie question here. This code:
>
> (println (flatten(map first '([1 2] [3
Got it. Thanks!
On Friday, June 7, 2013 3:29:55 PM UTC-6, Jonathan Fischer Friberg wrote:
>
> On Fri, Jun 7, 2013 at 11:13 PM, Matt Smith
> > wrote:
>
>> (-> '([1 2] [3 4] [5])
>> (partial map first)
>> flatten
>> )
>
>
> Because this becomes
>
> (flatten (partial '([1 2] [3 4] [5]
On Fri, Jun 7, 2013 at 10:13 PM, Matt Smith wrote:
> (println
> (-> '([1 2] [3 4] [5])
> (partial map first)
> flatten
> ))
>
> when expanded becomes this:
(println
(flatten (partial '([1 2] [3 4] [5]) map first)))
It looks like what you really want is to drop the partial a
On Fri, Jun 7, 2013 at 11:13 PM, Matt Smith wrote:
> (-> '([1 2] [3 4] [5])
> (partial map first)
> flatten
> )
Because this becomes
(flatten (partial '([1 2] [3 4] [5]) map first))
I think I understand how you thought; "(partial map first) becomes a
function, then I call this
Newbie question here. This code:
(println (flatten(map first '([1 2] [3 4] [5]
(def mapfirst (partial map first))
(println
(-> '([1 2] [3 4] [5])
mapfirst
flatten
))
(println
(-> '([1 2] [3 4] [5])
(partial map first)
flatten
))
prints out:
> (1 3 5
