No, like h {"1 (<f;g) { y
On 5/24/2020 1:06 PM, Henry Rich wrote:
(<f;g;h) { y is the natural idiom, as Bob says here.
The implementation does the indexing two axes at a time, so that this
is executed just like
h {"2 (<f;g) { y
Henry Rich
On 5/24/2020 12:17 PM, 'robert therriault' via Programming wrote:
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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