Raul -
So back to square One, totally different approach. (Most of this
again for my reference.)
I reviewed (x #: y) by the 'time' example
hms=. 24 60 60 & #:
hms 3600 1825 1201 930
1 0 0
0 30 25
0 20 1
0 15 30
Each atom on the right produces a list (3 items each in this case,
according to the given dividers 60 60 24).
This lead to understanding this structure
2 3 #: i.7
0 0
0 1
0 2
1 0
1 1
1 2
0 0
Replacing the right argument list by a table
i. 2 3
0 1 2
3 4 5
2 3 #: i. 2 3
0 0
0 1
0 2
1 0
1 1
1 2
each table element produced a 2-item-list (2 rows, 2 result blocks).
Given this table
i. 3 3
0 1 2
3 4 5
6 7 8
one can retrieve the 2nd row or 1st column respectively depending on
rank used on 'From' (x { y):
1 {"2 i. 3 3 NB. rank 2 gets the row
3 4 5
0 {"1 i. 3 3 NB. rank 1 gets the column
0 3 6
If there are several (layout-identical) matrices, it will pick from
the same place in each matrix
0 {"1 (2 3 #: i. 2 3) NB. 1st column from each
0 0 0
1 1 1
In the original exercise (Bo's question) we would be after the 2nd column:
1 {"1 (3 5 5 #: i. 3 5 5) NB. 2nd column from each
0 0 0 0 0
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
0 0 0 0 0
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
0 0 0 0 0
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
Going tacid (step by step) I arrived at these lines:
1 {"1 (#: i.) 3 5 5
(1 & {"1 @ #: i.) 3 5 5
which is very near to your solutiion
(1 & { @ #: i.) 3 5 5
(I had a look at (#:"#:) and (#: b.0) but can't figure out the relevance.)
Q: Why is it that one may skip the rank notation at this time
which had seemed so crucial in the beginning..?
Thanks
-M
At 2016-06-10 10:51, you wrote:
Try this: 3 5 5 #: i. 3 5 5 Then try this: 1 {"1 (3 5 5 #: i.
3 5 5) Then try this: #:"#: I hope that helps, Thanks, -- Raul On
Fri, Jun 10, 2016 at 6:38 AM, Martin Kreuzer <[email protected]>
wrote: > Raul - > > Following this thread, I managed to grasp
(reproduce) the expression (0 2 1 > |: 3 5 5 $ i. 5) which Bo found
satisfactory. > > Challenged by your remark "But probably no easier
to read." I have tried to > sort of reconstruct your approach: > >
First I read up on dyadic Antibase (x #: y) and found the remainder
example; > thus > > > ,(i. 10) ;(5 #: i. 10) > 0 1 2 3 4 5 6 7 8
9 > 0 1 2 3 4 0 1 2 3 4 > > Producing this square matrix > > i. 5
5 > 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 > > and applying the above I got > > 5 #: i.
5 5 > 0 1 2 3 4 > 0 1 2 3 4 > 0 1 2 3 4 > 0 1 2 3 4 > 0 1 2 3 4 > >
and (5 #: i. 3 5 5) got me three blocks of those. > > Changing the
axis preference (?) switched rows and columns > > 1 0 |: 5 #: i.
5 5 > 0 0 0 0 0 > 1 1 1 1 1 > 2 2 2 2 2 > 3 3 3 3 3 > 4 4 4 4 4 > >
and in the case of the three blocks this would be written as (0 2 1
|: 5 #: > i. 3 5 5) as the number of blocks remains untouched. > >
What seemed to me a shortcut for the special case of a
three-dimensinal > arrangement of square matrices this gives the
same result: > > 1 |: 5 #: i. 3 5 5 > 0 0 0 0 0 > 1 1 1 1 1 > 2 2
2 2 2 > 3 3 3 3 3 > 4 4 4 4 4 > > 0 0 0 0 0 > 1 1 1 1 1 > 2 2 2 2
2 > 3 3 3 3 3 > 4 4 4 4 4 > > 0 0 0 0 0 > 1 1 1 1 1 > 2 2 2 2 2 > 3
3 3 3 3 > 4 4 4 4 4 > > Using brackets I could write that as > (1
|: 5 #: i.) 3 5 5 > replacing the (5) by grabbing it from the list
like (1 { 3 5 5) > I continued to > (1 |: 1 & { #: i.) 3 5 5 >
which looked promising. > > It looked to me as you were taking
advantage of the (1) being mentioned > there twice and therefore
combining, but ... > Q: Could you enlighten me on this final
step..? > > Thanks > -M > > > > > At 2016-06-10 06:51, you
wrote: >> >> I just stumbled across this. It occurs to me
that (1&{@#:i.)3 5 5 would >> be one character shorter. But
probably no easier to read. Thanks, -- Raul On >> Wed, Jun 8, 2016
at 5:31 PM, 'Bo Jacoby' via Programming >>
<[email protected]> wrote: > Thanks everyone! > This: > 0
2 1|:3 >> 5 5$i.5 > > produces what I wanted. > The result is
however destroyed in the >> process of emailing it
to [email protected] . Line feeds are >> deleted. I
don't know why. > Problem is solved. Thanks again. > Bo. > > >> Den
21:10 onsdag den 8. juni 2016 skrev Cliff Reiter >>
<[email protected]>: > > > > Or > ,./": 3 5$"1 0 i.5 > > 0 0 0
0 01 1 1 >> 1 12 2 2 2 23 3 3 3 34 4 4 4 4 > > 0 0 0 0 01 1 1 1 12 2
2 2 23 3 3 3 34 4 4 >> 4 4 > > 0 0 0 0 01 1 1 1 12 2 2 2 23 3 3 3 34
4 4 4 4 > > But a strange >> thing to want to build. > > On 6/8/2016
1:37 PM, robert therriault wrote: >> >> Maybe this? >> >> 5( 3
# ,:@,@":@:(#/"0)) i. 5 >> 0 0 0 0 01 1 1 1 12 >> 2 2 2 23 3 3 3 34
4 4 4 4 >> 0 0 0 0 01 1 1 1 12 2 2 2 23 3 3 3 34 4 4 4 4 >> >> 0 0 0
0 01 1 1 1 12 2 2 2 23 3 3 3 34 4 4 4 4 >> >> But I am confused
as >> well about the request for a shape 3 5 5 of what appears as a
shape 3 45 >> literal matrix. I do have '01' in mine though.
:-) >> >> Cheers, bob >> >>> >> On Jun 8, 2016, at 10:29 AM, Raul
Miller <[email protected]> wrote: >>> >> >>> Did you mean
something like this? >>> >>> (<.0.8*1+i.25) 10&#./."1] >> 3#,:5#
i.5 >>> 0 0 0 0 1 1 1 1 12 2 2 2 23 3 3 3 34 4 4 4 4 >>> 0 0 0 0 1
1 >> 1 1 12 2 2 2 23 3 3 3 34 4 4 4 4 >>> 0 0 0 0 1 1 1 1 12 2 2 2
23 3 3 3 34 4 >> 4 4 4 >>> >>> Except, that doesn't get you those
leading zeros for the '01' >> column, >>> so maybe instead it needs
to be character? >>> >>> But a >> character array would not have
anything to do with that 3 5 5 >>> shape you >> suggested, so for
that, and guessing what you want, maybe it >>> should be >>
something like this? >>> >>> <.25%~i.3 5 5 >>> 0 0 0 0 0 >>> 0 0
0 0 0 >> >>> 0 0 0 0 0 >>> 0 0 0 0 0 >>> 0 0 0 0 0 >>> >>> 1 1 1 1
1 >>> 1 1 1 1 1 >> >>> 1 1 1 1 1 >>> 1 1 1 1 1 >>> 1 1 1 1 1 >>> >>>
2 2 2 2 2 >>> 2 2 2 2 2 >> >>> 2 2 2 2 2 >>> 2 2 2 2 2 >>> 2 2 2 2
2 >>> >>> Except that that doesn't >> look at all like what you
asked for. A 5 3 4 >>> shape gets a little
closer: >> >>> >>> <.12%~i.5 3 4 >>> 0 0 0 0 >>> 0 0 0 0 >>> 0 0
0 0 >>> >>> 1 1 1 1 >> >>> 1 1 1 1 >>> 1 1 1 1 >>> >>> 2 2 2 2 >>> 2
2 2 2 >>> 2 2 2 2 >>> >>> 3 3 >> 3 3 >>> 3 3 3 3 >>> 3 3 3 3 >>> >>>
4 4 4 4 >>> 4 4 4 4 >>> 4 4 4 4 >>> >>> >> But all of these have
conflicts with some aspect of your original >>> >> request, and I
can't figure out what it is that you really wanted. >>> >>> I >>
hope this helps? >>> >>> Thanks, >>> >>> -- >>> Raul >>> >>> >>> On
Wed, Jun >> 8, 2016 at 5:22 AM, 'Bo Jacoby' via Programming >>> >>
<[email protected]> wrote: >>>> Dear J'ers. >>>> Please tell
me how >> to program the 3 5 5 array below. >>>> I am experimenting
rather than >> understanding. I expect the answer to be quite
elementary. >>>> Thanks! Bo. >> >>>> >>>> 0 0 0 0 01 1 1 1 12 2 2 2
23 3 3 3 34 4 4 4 4 >>>> 0 0 0 0 01 1 1 >> 1 12 2 2 2 23 3 3 3 34 4
4 4 4 >>>> 0 0 0 0 01 1 1 1 12 2 2 2 23 3 3 3 34 4 >> 4 4
4 >>>> >>>> >>
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