Dyadic gerund amend looks like this

   x u0`u1`u2} y

and works like this:
   (x u0 y)  (x u1 y)}  x u2 y

(Monadic gerund amend is very different.)

Does that help?

Thanks,

-- 
Raul


On Wed, Jan 28, 2015 at 12:59 PM, Joe Bogner <[email protected]> wrote:
> On Tue, Jan 27, 2015 at 11:45 PM, Raul Miller <[email protected]> wrote:
>> I guess this is how I'd write that:
>>
>>    f=: (3##\@])`(,@])`[} 0 1 2 +/~I.
>>
>> Is that easy enough to read, or should I spell out how it works?
>>
>
> This was enjoyable to decode. It looks simple but there are a few
> things that I haven't used extensively.
>
> I still don't understand the gerund  item amend.
>
> Here is my interpretation for others who may be interested
>
> * 1. First I realized it's a hook
>
> I tried using trace and dissect and couldn't figure it out from those
> outputs, so then I tried
>
> (arr f arr) -: (f arr)
> 1
>
> http://www.jsoftware.com/jwiki/Vocabulary/hook
>
> * 2. Knowing it was a hook I started to break it down
>
> The spacing threw me off a bit initially
>
> NB. gerund amend http://www.jsoftware.com/jwiki/Vocabulary/curlyrt#dyadic
> u=: (3##\@])`(,@])`[ }
> v=: 0 1 2 +/~I.
>
>    ((u v) arr) -: (f arr)
> 1
>
>    (arr g (v arr)) -: (f arr)
> 1
>
> * 3. Item Amend
>
> I haven't yet figured out how this part works
>
> Here's a simpler version to look at
>
>    (12#0) ((3 # #\@])`(,@])`[}) ((1,2,3),(5,6,7),:(9,10,11))
> 0 1 1 1 0 2 2 2 0 3 3 3
>
>
> The first gerund replicates 3 times the number which corresponds to #
> of rows in the prefix
>
>    #\ ((1,2,3),(5,6,7),:(9,10,11))
> 1 2 3
>
> The second gerund appears to yield the concatenation
>
> The third gerund yields the left
>
> I don't understand the order these gerunds get applied
>
> The dictionary say: "If m is a gerund, one of its elements determines
> the index argument to the adverb } , and the others modify the
> arguments x and y :"[1]
>
> The phrase that's confusing me is "one of its elements" -- which one?
>
> [1] http://www.jsoftware.com/help/dictionary/d530n.htm
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