> Here's a definition for at which works exactly like @
>
> at=: 2 :'([: u v)"v
Rather works almost exactly?
('*'"_) @ ((+: @ *:) (d.1)) (0 1 2)
*
('*'"_) @ ((+: at *:) (d.1)) (0 1 2)
***
((+: @ *:) (d.1)) b.0
_ _ _
((+: at *:) (d.1)) b.0
0 0 0
On Wed, Jan 2, 2013 at 10:04 AM, Raul Miller <[email protected]> wrote:
> Here's a definition for at which works exactly like @
>
> at=: 2 :'([: u v)"v
>
> For example:'
> <at i. at > 2 3 4
>
> Now, looking at that, you may think that this means that [: is somehow
> superior to @ but note that [: is not necessary
>
> at=: 2 :'u@:v"v'
>
> Note that in J, @: and @ (without the colon) are different words.
>
> It's also possible to replace @: with an explicit definition, but
> that's probably best left for another time.
>
> FYI,
>
> --
> Raul
>
> On Wed, Jan 2, 2013 at 3:02 AM, Linda Alvord <[email protected]> wrote:
>> Here’s another, hopefully simpler, example of my problem with @
>>
>> ]a=:?>:i.6
>>
>> 0 1 1 3 1 5
>>
>> ]a=:=a
>>
>> 1 0 0 0 0 0
>>
>> 0 1 1 0 1 0
>>
>> 0 0 0 1 0 0
>>
>> 0 0 0 0 0 1
>>
>> ]b=:?>:i.6
>>
>> 0 1 0 3 1 3
>>
>> a #@# b
>>
>> 1 3 1 1
>>
>>
>>
>> There must be some way to get this result without @
>>
>>
>>
>> Here are some of the things I have tried:
>>
>>
>>
>> NB. x u@v y ↔ u x v y (vocabulary definition of @
>>
>> #a#b
>>
>> 4
>>
>>
>>
>> NB. ([: f g)"({. g b. 0) <=> f@g (someone on the forum provided this in
>> an earlier struggle I had with @ )
>>
>> a(([:##)"({.# b. 0))b
>>
>> 4
>>
>>
>>
>> Both of these definitions appear to describe @:
>>
>> a #@:# b
>>
>> 4
>>
>>
>>
>> Are changes ever made to the definitions in the vocabulary when they are
>> unclear or incorrect?
>>
>>
>>
>> Linda
>>
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