Re: Reassignment of a nested array element

[Dr . Jürgen Sauermann]

    Hi,
      
      I fixed a discrepancy between (3⊃V)← and  (1/3⊃V)←. SVN 1657.
        We know have:
        
              V←1 'bc' (3 3⍴⍳9) ◊ 8 ⎕CR V
          ┌→─────────────┐
          │1 ┌→─┐ ┌→────┐│
          │  │bc│ ↓1 2 3││
          │  └──┘ │4 5 6││
          │       │7 8 9││
          │       └─────┘│
          └ϵ─────────────┘
                (3⊃V)←1           ◊ 8 ⎕CR V
          ┌→───────┐
          │1 ┌→─┐ 1│
          │  │bc│  │
          │  └──┘  │
          └ϵ───────┘
                (1/3⊃V)←1         ◊ 8 ⎕CR V
          ┌→───────┐
          │1 ┌→─┐ 1│
          │  │bc│  │
          │  └──┘  │
          └ϵ───────┘
        
        However, I don't quite get why, as suggested below, the results
        of
     (3⊃V)← and of (1/3⊃V)←
            should differ?
            
            For me 1/X is the same as X (except for a
            length 1 axis added by 1/ when X is
            scalar) and therefore  (3⊃V)←1 and
      (1/3⊃V)←1 should
        IMHO yield the same.
        
        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 SVN 1657
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.