#(1 1 1 2 2 2 3 3 3 4 4 4) . 3 -> #(#(1 1 1) (2 2 2) (3 3 3) #(4 4 4))

Hi guys 

Do they group together because they have the same value or do they just get grouped into subgroups of size three or do they get grouped as runs?

Hi Stef,
 
from the (incomplete) sample it is unclear what is really needed:
 
 1. Both return the same result: #(#(1 1 1) (2 2 2) (3 3 3) #(4 4 4))  which is syntactically equal to #(#(1 1 1) #(2 2 2) #(3 3 3) #(4 4 4))

    So why did you leave out the # for the contained sub-array with 2's and 3's?

 2. When the result is equal in both examples why give a different argument like 3 and 4? 

 3. What about edge case inputs like:   #(1 1 1 2 2 2 4 4 4)          not having sucessors/predecessor order
                                        #(2 2 2 1 1 1 3 3 3 4 4 4)    with different order (not ascending)
                                        #(1 1 1 1 2 2 2 3 3 3 4 4 4)  where 1 is included more often than the other   

 4. ...


Nonetheless: the closest that might possibly be of use is #groupByRuns: message


    #(1 1 1 2 2 2 3 3 3 4 4 4) groupByRuns: [:e | e ]    ->  #(#(1 1 1) #(2 2 2) #(3 3 3) #(4 4 4))

Have fun
T.
  

Gesendet: Mittwoch, 03. Juni 2020 um 22:53 Uhr
Von: "Stéphane Ducasse" <[hidden email]>
An: "Pharo Development List" <[hidden email]>
Betreff: [Pharo-dev] #(1 1 1 2 2 2 3 3 3 4 4 4) . 3 -> #(#(1 1 1) (2 2 2) (3 3 3) #(4 4 4))

Hi guys 
 
do you know if we have around a method doing the following?
 
#(1 1 1 2 2 2 3 3 3 4 4 4) . 3 -> #(#(1 1 1) (2 2 2) (3 3 3) #(4 4 4)) 
 
it could also be another one
 
#(1 1 1 2 2 2 3 3 3 4 4 4) . 4 -> #(#(1 1 1) (2 2 2) (3 3 3) #(4 4 4)) 
 
S
 

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