Re: APL2 code no longer works

[Dr . Jürgen Sauermann]

    Hi Blake,
      
      thanks, see below,
      
      Best Rgeads,
      Jürgen

On 3/21/22 8:41 PM, Blake McBride wrote:

Greetings,

I wrote the following APL1 code a long time ago.

    ∇
[0]   r←d Parse v
[1]  ⍝ Convert vector v into a matrix breaking at delimiter d
[2]   r←(((0≠⍴v)×⍴r),⌈/r)⍴(,r∘.≥⍳⌈/r←¯1+(r,1+⍴v)-0,r←r/⍳⍴v)\(~r←v∈d)/v←,v
    ∇

It worked in 1981 and it works now.

I believe at some point someone on this list re-wrote the above in GNU APL (APL2) as follows:

    ∇
[0]   r←d Parse2 v
[1]  ⍝ break up vector v according to delimiter d into seperate arrays
[2]   r←1↓1↓¨(1++\r∊d)⊂r←(1↑d),(1↑d),v
    ∇

Interestingly, Parse2 used to work but doesn't anymore.  I suppose the APL2 logic in GNU APL has changed.  Two points:

1.  Can someone help me fix this?  (I don't know too much about APL2.)

    Try ⊃Parse2 instead of Parse2:
    
          8 ⎕CR ',' Parse2
        'One,Two,Three,Four'
      ┌→─────────────────────────┐
      │┌→──┐ ┌→──┐ ┌→────┐ ┌→───┐│
      ││One│ │Two│ │Three│ │Four││
      │└───┘ └───┘ └─────┘ └────┘│
      └ϵ─────────────────────────┘
            
            8 ⎕CR ⊃ ',' Parse2 'One,Two,Three,Four'
      ┌→────┐
      ↓One  │
      │Two  │
      │Three│
      │Four │
      └─────┘
      
      Or change Parse2[2] to:
    
    [2]   r←⊃1↓1↓¨(1++\r∊d)⊂r←(1↑d),(1↑d),v

2.  How good is APL2 if it can morph like this?  (APL1 seems much more straightforward and consistent and less open to interpretation.)

    I am pretty sure that this change was caused by a trouble report
    followed by a bug fix at some point in time.
    The current behaviour of GNU APL looks OK to me (although I am not
    really an APL expert). My humble argument
    would be that in order to get a not-nested result, the ⊂ in line 2
    of Parse2 would require a matching ⊃.
    
    Put differently, the original Parse2 was working around a bug in GNU
    APL which was fixed later on.
    
    BTW, for long v, the following is probably more efficient:
    
    [2] r←⊃1↓1↓¨(1++\r∊d)⊂r←(2/↑d),v

Thanks.

Blake McBride