Hello Jürgen, Elias
The difference I mention is in
(1/3⊃V)←1
vs
(3⊃V)←1
The second expression is the one discussed in the Mail.
And I agree with your statement Jürgen that, even if not specified, your
solution is correct.
3⊃V is the third elemnt in V and that is being set to 1.
The first expression however is
1/3⊃V which is selecting all elements of the third element of V
then set it's supposed be
1 1 1
1 1 1
1 1 1
as with Dyalog.
Further...
(c is equvalent to 3⊃V)
Expected :
c←3 3⍴0
(1/c)←3
c
3 3 3
3 3 3
3 3 3
Not expected 3⊃V should remain a Matrix
V←0 0 (3 3⍴0)
(1/3⊃V)←1
V
0 0 1
Expected:
c←3 3⍴0
((,3 3⍴1 0 0 0)/, c)←3
c
3 0 0
0 3 0
0 0 3
The "mapping" of 3 3⍴1 0 0 0 gets ignored.
V←0 0 (3 3⍴0)
((,3 3⍴1 0 0 0)/, 3⊃V)←3
V
0 0 3
Expected:
c←3 3⍴0
(↑¨c)←2
c
2 2 2
2 2 2
2 2 2
I would say it's inconsistent.
V←0 0 (3 3⍴0)
(↑¨3⊃V)←1
V
0 0 1 0 0
0 0 0
0 0 0
Expected:
c←3 3⍴0
(↑c)←2
c
2 0 0
0 0 0
0 0 0
differs here too.
V←0 0 (3 3⍴0)
(↑3⊃V)←1
V
0 0 1
Or why go
V←0 0 (3 3⍴0)
c←3⊃V
(1/c)←3
(3⊃V)←c
when
(1/3⊃V)←3
would hold?
Best Regards
Hans-Peter
Am 07.03.23 um 17:39 schrieb Elias Mårtenson:
I think this is because monadic ⊃ is first in Dyalog, as opposed to
disclose in GNU, APL2 (and, incidentally, KAP 😀).
Use monadic ↑ instead to get the effect you want.
Den ons 8 mars 2023 00:34Dr. Jürgen Sauermann
<mail@jürgen-sauermann.de <mailto:mail@j%C3%BCrgen-sauermann.de>> skrev:
Hi Hans-Peter,
I believe that GNU APL, Dyalog, and IBM APL2 only now behave
the same while GNU APL did not before the fix:
T*ryAPL Version 3.6.1 (enter ]State for details)**
**Tue Mar 07 2023 16:59:04**
**Copyright (c) Dyalog Limited 1982-2023**
** Q←0 0 0**
** (3⊃Q)←2 2⍴2**
** Q**
**┌─┬─┬───┐**
**│0│0│2 2│**
**│ │ │2 2│**
**└─┴─┴───┘
** (3⊃Q)←33**
** Q**
**0 0 33*
*GNU APL SVN 1656:*
* Q←0 0 0**
** (3⊃Q)←2 2⍴2**
** Q **
** 0 0 2 2 **
** 2 2
** (3⊃Q)←33**
** Q **
**0 0 33*
The cleanest way to achieve the previous (incompatible)
behaviour in GNU APL is to disclose the left of →
* Q←0 0 0**
** (3⊃Q)←2 2⍴2**
** Q **
** 0 0 2 2 **
** 2 2
** (⊃3⊃Q)←33**
** Q**
** 0 0 33 33 **
** 33 33 *
For dubious reasons this does not work on Dyalog
because they only assign the first item:
* (⊃3⊃Q)←33**
** Q**
**┌─┬─┬────┐**
**│0│0│33 2│**
**│ │ │ 2 2│**
**└─┴─┴────┘**
*
which is not consistent with (Dyalog):
* W←⊂2 2⍴2**
** W**
**┌───┐**
**│2 2│**
**│2 2│**
**└───┘
** (⊃W)←33**
** W**
**33**
*
Although this is a discrepancy between GNU APL and Dyalog,
I believe that Dyalog is wrong here. IBM APL2 says
DOMAIN ERROR in this case.
Best Regards,
Jürgen
On 3/6/23 9:16 PM, Hans-Peter Sorge wrote:
Hello Jürgen,
I agree with your case 1/2 Statement.
The examples I was showing is actually "off by 1".
I was referring to
(1/3⊃V)←1
having
a←1
b←'ABC'
c←3 3⍴⍳9
V←a b c
(3⊃V)
1 2 3
4 5 6
7 8 9
As expected with case 1:
(3⊃V)←1
V
1 ABC 1
V←a b c
Not expected:
(*1/*3⊃V)←1
V
1 ABC 1
Expected:
(1/3⊃V)←1
V
1 ABC 1 1 1
1 1 1
1 1 1
as with
(1/c)←1
c
1 1 1
1 1 1
1 1 1
And that's Dyalog too.
Please restore compatibility:-)
Best Regards
Hans-Peter
Am 06.03.23 um 16:10 schrieb Dr. Jürgen Sauermann:
Gentlemen,
thanks for the discussion, fixed in *SVN 1655*.
Hans-Peter, I am sorry that this change creates an
incompatibility in your code.
My thinking for the old solution was this:
* V←0 0 0** ◊ V←1 ◊ V ∩ case 1.**
**1**
**
** V←0 0 0** ◊ V[]←1 ◊ V ⍝ case 2.**
**1 1 1**
*
This applies to GNU APL, APL2, and Dyalog. The question is then
if (A⊃V) in
(A⊃B)←X should behave like case 1 or like case 2 above. The case
(A⊃B)←X
with nested (A⊃B)is described neither in the "IBM APL2 Language
Reference"
nor in the "ISO 13751" standard, leaving some room for
interpretation.
However, both APL2 and Dyalog agree on case 1 and therefore I
changed
GNU APL to behave the same.
Best Regards,
Jürgen
On 3/4/23 8:25 PM, Hans-Peter Sorge wrote:
Hi,
Works as expected
⊃'Sue' 'Maria' 'Annalisa'
is an array 3 by 8.
⊂⊃'Susan' 'Mary' 'Annalisa'
is an element (⊂) of a 3 by 8 array (⊃'Susan' 'Mary'
'Annalisa' ).
Finally each element in ⊃'Sue' 'Maria' 'Annalisa' gets
assigned an array of ⊃'Susan' 'Mary' 'Annalisa'
Greetings
Hans-Peter
Am 04.03.23 um 16:53 schrieb Mr. Sunday:
Hi,
I have an issue with reassigning an element of a nested
array. Here is an example.
14535:15a:~% apl --version
BUILDTAG:
---------
Project: GNU APL
Version / SVN: 1.8 / SVN: 1651M
Build Date: 2023-03-02 00:25:07 UTC
Build OS: Darwin 21.6.0 x86_64
config.status: default ./configure options
Archive SVN: 1621
var←0 0 0 ⋄ (1⊃var)←5 4 ⋄ (2⊃var)←3 4⍴⍳12 ⋄
(3⊃var)←⊃'Sue' 'Maria' 'Annalisa' ⋄ var ⋄ (3⊃var)←⊂⊃'Susan'
'Mary' 'Annalisa' ⋄ var
┌→────────────────────────────┐
│┌→──┐ ┌→─────────┐ ┌→───────┐│
││5 4│ ↓1 2 3 4│ ↓Sue ││
│└───┘ │5 6 7 8│ │Maria ││
│ │9 10 11 12│ │Annalisa││
│ └──────────┘ └────────┘│
└ϵ────────────────────────────┘
┌→───────────────────────────────────────────────────────────────────────────────────────────────────────────┐
│┌→──┐ ┌→─────────┐
┌→──────────────────────────────────────────────────────────────────────────────────────┐│
││5 4│ ↓1 2 3 4│ ↓┌→───────┐ ┌→───────┐ ┌→───────┐
┌→───────┐ ┌→───────┐ ┌→───────┐ ┌→───────┐ ┌→───────┐││
│└───┘ │5 6 7 8│ │↓Susan │ ↓Susan │ ↓Susan │ ↓Susan
│ ↓Susan │ ↓Susan │ ↓Susan │ ↓Susan │││
│ │9 10 11 12│ ││Mary │ │Mary │ │Mary │ │Mary
│ │Mary │ │Mary │ │Mary │ │Mary │││
│ └──────────┘ ││Annalisa│ │Annalisa│ │Annalisa│
│Annalisa│ │Annalisa│ │Annalisa│ │Annalisa│ │Annalisa│││
│ │└────────┘ └────────┘ └────────┘
└────────┘ └────────┘ └────────┘ └────────┘ └────────┘││
│ │ ││
│ │┌→───────┐ ┌→───────┐ ┌→───────┐
┌→───────┐ ┌→───────┐ ┌→───────┐ ┌→───────┐ ┌→───────┐││
│ │↓Susan │ ↓Susan │ ↓Susan │ ↓Susan
│ ↓Susan │ ↓Susan │ ↓Susan │ ↓Susan │││
│ ││Mary │ │Mary │ │Mary │ │Mary
│ │Mary │ │Mary │ │Mary │ │Mary │││
│ ││Annalisa│ │Annalisa│ │Annalisa│
│Annalisa│ │Annalisa│ │Annalisa│ │Annalisa│ │Annalisa│││
│ │└────────┘ └────────┘ └────────┘
└────────┘ └────────┘ └────────┘ └────────┘ └────────┘││
│ │ ││
│ │┌→───────┐ ┌→───────┐ ┌→───────┐
┌→───────┐ ┌→───────┐ ┌→───────┐ ┌→───────┐ ┌→───────┐││
│ │↓Susan │ ↓Susan │ ↓Susan │ ↓Susan
│ ↓Susan │ ↓Susan │ ↓Susan │ ↓Susan │││
│ ││Mary │ │Mary │ │Mary │ │Mary
│ │Mary │ │Mary │ │Mary │ │Mary │││
│ ││Annalisa│ │Annalisa│ │Annalisa│
│Annalisa│ │Annalisa│ │Annalisa│ │Annalisa│ │Annalisa│││
│ │└────────┘ └────────┘ └────────┘
└────────┘ └────────┘ └────────┘ └────────┘ └────────┘││
│
└ϵ──────────────────────────────────────────────────────────────────────────────────────┘│
└ϵϵ──────────────────────────────────────────────────────────────────────────────────────────────────────────┘
-- Everyday is Sunday.
