Hi,
Not exactly an answer to the question, but rather a cute way to pad an array of
any rank using a cool feature of ;. cut that I ran into today.
From the dictionary:
u;.0 y applies u to y after reversing y along each axis; it is equivalent to (0
_1 */$y) u;.0 y .
Using this:
padop=: &((+$) {. ]) (;.0) (^:2)
2 padop i.2 3
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 1 2 0 0
0 0 3 4 5 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
1 padop i.2 2 2
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 1 0
0 2 3 0
0 0 0 0
0 0 0 0
0 4 5 0
0 6 7 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
Unfortunately padop must be an adverb in order to be tacit (unless there is
maybe some wicked way). The corresponding explicit verb could be used for a
default left argument.
Anyway just an illustration of one of cut’s many powers that I probably often
overlook.
Cheers,
Louis
> On 5 May 2019, at 09:44, 'robert therriault' via Programming
> <[email protected]> wrote:
>
> Hi Linda,
>
> I think that the trick is that the left operand for ^: needs to be a verb.
> this seems to work.
>
> f=:(0 ,.~ 0 ,. 0 ,~ 0 , ])^:[
> 1 f i. 3 4
> 0 0 0 0 0 0
> 0 0 1 2 3 0
> 0 4 5 6 7 0
> 0 8 9 10 11 0
> 0 0 0 0 0 0
> g=: 4 : '(0 ,.~ 0 ,. 0 ,~ 0 , ])^:x y'
> 1 g i. 3 4
> 0 0 0 0 0 0
> 0 0 1 2 3 0
> 0 4 5 6 7 0
> 0 8 9 10 11 0
> 0 0 0 0 0 0
>
> Cheers, bob
>
>> On May 5, 2019, at 12:36 AM, Linda Alvord <[email protected]> wrote:
>>
>> f=:(0 ,.~ 0 ,. 0 ,~ 0 , ])^:[
>
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