The use of "partial" is unnecessary because "apply" takes any number
of arguments and expands its last argument.

(apply map vector [[1 2 3] [4 5 6] [7 8 9]])

is equivalent to

(map vector [1 2 3] [4 5 6] [7 8 9])

and results in

([1 4 7] [2 5 8] [3 6 9])

On Sun, Nov 14, 2010 at 8:30 AM, Moritz Ulrich
<ulrich.mor...@googlemail.com> wrote:
>> (apply (partial map vector) [[1 2 3] [4 5 6] [7 8 9]])
> ([1 4 7] [2 5 8] [3 6 9])
> On Sun, Nov 14, 2010 at 2:16 PM, garf <gary.overg...@gmail.com> wrote:
>>
>> (map vector [1 2 3] [4 5 6] [7 8 9])   gives back ([1 4 7] [2 5 8] [3
>> 6 9])
>> but if I have a function that returns the list:
>>   '([1 2 3] [4 5 6] [7 8 9])
>> and call map vector with the list, then I no longer get  ([1 4 7] [2 5
>> 8] [3 6 9])
>>
>> i.e.
>> (def x '([1 2 3] [4 5 6] [7 8 9]))
>>  (map vector x)
>>   ==>([[1 2 3]] [[4 5 6]] [[7 8 9]])
>>
>> so is there a way to effectively expand the argument x (map vector x)
>> to be
>>  (map vector [1 2 3] [4 5 6] [7 8 9])
>>
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>
> --
> Moritz Ulrich
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>
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