This is tricky.
From the Dictionary:
If y is an atom, then ,.y is 1 1$y ; otherwise, ,.y is ,"_1 y , the
table formed by ravelling each item of y .
$ ,"_1 i. 0
0 1
So it's right. But why?
In
,"_1 i. 0
the rank of ,"_1 is 0, so the verb is applied to a cell of fills.
That means that
, 0
is executed, and this has shape 1. The shape of the result of
,"_1 i. 0
then becomes this shape from the fill, prefixed by the frame of (i. 0)
with respect to rank 0; since the shape of (i. 0) is 0, this frame is
0
so the result shape is
0 , 1 = 0 1
Henry Rich
Igor Zhuravlov wrote:
> Ravelling items of non-empty list is clear:
> $ ,. 1 2 3
> 3 1
>
> For empty list it's slightly confusing for me:
> $ ,. i.0
> 0 1
> Answer is an empty list of lists of length 1. But shouldn't it be of shape
> (0 0) instead? Empty list has no elements, so, ravelled empty list is empty
> list again:
> $ , i. 0
> 0
> so the answer should be formed as empty list of empty lists.
>
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