Yes - the point is that
_5 {. vector NB. whether sparse or not
returns a NEW vector whose indices range from 0 to 4, whether sparse or not
Mike
On 20/12/2017 23:31, robert therriault wrote:
Henry,
With further thought, I suppose that what is happening is that the array
remains a sparse type and that the indices now reflect the new shape after the
selection. Which does make sense when I think about it, but I guess I had it in
my mind that the indices would remain attached to the values and not be
recalculated. At least this is the way that I view this at the moment and look
forward to being corrected if I am mistaken.
Cheers, bob
On Dec 20, 2017, at 3:22 PM, robert therriault <[email protected]> wrote:
Henry,
I think Brian may be referring to the indices shown in the sparse
representation.
t=: $. 0 0 1 0 2.5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 999 _ 0 0 0
t NB. Index values as expected
2 │ 1
4 │ 2.5
19 │ 2
21 │ 999
22 │ _
_5{. t NB. I would have expected 21 and 22 to match the values in the
original
0 │ 999
1 │ _
5 {. t
2 │ 1
4 │ 2.5
19 { t
2
5 {. 20}.t NB. It looks like it is reindexing after dropping the first 20
values
1 │ 999
2 │ _
Cheers, bob
On Dec 20, 2017, at 3:06 PM, Henry Rich <[email protected]> wrote:
No, _5 {. z has shape 5.
Henry Rich
On 12/20/2017 6:04 PM, Brian Schott wrote:
Negative take seems screwy.
$ y=: 0 0 1 0 2.5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 999 _ 0 0 0
26
_5{. z NB. Should the 0 and 1 be large numbers?
0 │ 999
1 │ _
5{. z
2 │ 1
4 │ 2.5
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