Hi Bill and Pascal,
I'd use
(<f;g;h){ i. 5 4 3
56 54 55
59 57 58
50 48 49
53 51 52
20 18 19
23 21 22
14 12 13
17 15 16
44 42 43
47 45 46
38 36 37
41 39 40
32 30 31
35 33 34
26 24 25
29 27 28
8 6 7
11 9 10
2 0 1
5 3 4
1000 timespacex 'f{0 1|: g{ 0 1|: h{ 0 1|:a'
2.141e_6 9600
1000 timespacex '4 1 3 2 0 { 2 3 0 1 {("2) 2 0 1{"1 i.5 4 3'
1.568e_6 7104
1000 timespacex '(<f;g;h){ i. 5 4 3'
1.203e_6 5760
Cheers, bob
> On May 24, 2020, at 08:40, 'Pascal Jasmin' via Programming
> <programm...@jsoftware.com> wrote:
>
> don't know if any faster, but
>
> 4 1 3 2 0 { 2 3 0 1 {("2) 2 0 1{"1 i.5 4 3
>
>
>
>
>
>
> On Sunday, May 24, 2020, 11:35:01 a.m. EDT, bill lam <bbill....@gmail.com>
> wrote:
>
>
>
>
>
> Suppose a I got a cube and selection vectors for each of the 3 axis, The
> following is the way that I do the selection, to transpose the axis to the
> front and then do selection. But I think this method is inefficient because
> it transpose 3 times. Any idea of better ways?
>
> ]a=. i.5 4 3
> 0 1 2
> 3 4 5
> 6 7 8
> 9 10 11
>
> 12 13 14
> 15 16 17
> 18 19 20
> 21 22 23
>
> 24 25 26
> 27 28 29
> 30 31 32
> 33 34 35
>
> 36 37 38
> 39 40 41
> 42 43 44
> 45 46 47
>
> 48 49 50
> 51 52 53
> 54 55 56
> 57 58 59
>
> f=.?.~5 [ g=.?.~4 [ h=. 2 0 1
> f;g;h
> +---------+-------+-----+
> |4 1 3 2 0|2 3 0 1|2 0 1|
> +---------+-------+-----+
> f{0 1|: g{ 0 1|: h{ 0 1|:a
> 56 54 55
> 59 57 58
> 50 48 49
> 53 51 52
>
> 20 18 19
> 23 21 22
> 14 12 13
> 17 15 16
>
> 44 42 43
> 47 45 46
> 38 36 37
> 41 39 40
>
> 32 30 31
> 35 33 34
> 26 24 25
> 29 27 28
>
> 8 6 7
> 11 9 10
> 2 0 1
> 5 3 4
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