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Merging Lists

SBrickles

I am writing some code which uses lists.
I need a way to be able merge two lists into one such that

list1:

( x "foo"
     (
        ( a "a" "func1()" )
        ( b "b" "func2()" )
        ( c "c" (
                    ( d "d" "func3()" )
                  )
     )
)

and list2: 

( x "foo"
     (
        ( e "e" "func4()" )
        ( c "c" (
                    ( f "f" "func5" )
                  )
     )
)

merge so that you end up with:

( x "foo"
     (
        ( a "a" "func1()" )
        ( b "b" "func2()" )
        ( c "c" (
                    ( d "d" "func3()" )
                    ( f "f" "func5" )
                  )
        ( e "e" "func4()" )
     )
)

The code needs to be able to merge at multiple levels depending on
whether the  'key' element (the car of the list at any level) is the same value or not.

If the item in the second list is of the form (a "a" "func()" ) then it should overwrite the
value when it gets merged with the first list.  So in the example above if the second list
had  ( a "a" "func8()" ) in it then it would have overwritten the ( a "a" "func1()" ) in the first list
- but only if it was at the same level of hierachy  as in the first list.

Any ideas ?

Stephen 

Parents

  • Andrew - Many many thanks for your help and prompt response on this.
    It completely solved my problem !!

    I made the following modification to the SBmergeLists procedure to allow a list
    to be replaced by an item:

    procedure( SBmergeLists(l1 l2)
      let( ( match )
        foreach( l2el l2
          match = assoc( car( l2el ) l1 )
          if( match then
    	if( listp( caddr( match ) ) then
              if( listp( nth( 2 l2el ) ) then
      	    SBmergeLists( caddr( match ) caddr( l2el ) )
              else
                rplaca( member( match l1 ) l2el )
              )
            else
              rplaca( cddr( match ) caddr( l2el ) )
            )
          else
            rplacd(last( l1 ) list( l2el ) )
          )
        )
        l1
      )
    )

    This allows the case below to work as well:

    list1=
    '( x "foo"
         (
            ( a "a" "func1()" )
            ( b "b" "func2()" )
            ( c "c" (
                        ( d "d" "func3()" )
                    )
            )
         )
    )

    list2=
    '( x "foo"
         (
            ( c "c" "func4()" )
         )
    ) 

    Resulting in:

    pp( list1 )
    ( x "foo"
         (
            ( a "a" "func1()" )
            ( b "b" "func2()" )
            ( c "c" "func4()" )
         )
    )

    The opposite case of turning an item into a list works fine as well.

    Again, many thanks,

    Stephen.

     
Reply

  • Andrew - Many many thanks for your help and prompt response on this.
    It completely solved my problem !!

    I made the following modification to the SBmergeLists procedure to allow a list
    to be replaced by an item:

    procedure( SBmergeLists(l1 l2)
      let( ( match )
        foreach( l2el l2
          match = assoc( car( l2el ) l1 )
          if( match then
    	if( listp( caddr( match ) ) then
              if( listp( nth( 2 l2el ) ) then
      	    SBmergeLists( caddr( match ) caddr( l2el ) )
              else
                rplaca( member( match l1 ) l2el )
              )
            else
              rplaca( cddr( match ) caddr( l2el ) )
            )
          else
            rplacd(last( l1 ) list( l2el ) )
          )
        )
        l1
      )
    )

    This allows the case below to work as well:

    list1=
    '( x "foo"
         (
            ( a "a" "func1()" )
            ( b "b" "func2()" )
            ( c "c" (
                        ( d "d" "func3()" )
                    )
            )
         )
    )

    list2=
    '( x "foo"
         (
            ( c "c" "func4()" )
         )
    ) 

    Resulting in:

    pp( list1 )
    ( x "foo"
         (
            ( a "a" "func1()" )
            ( b "b" "func2()" )
            ( c "c" "func4()" )
         )
    )

    The opposite case of turning an item into a list works fine as well.

    Again, many thanks,

    Stephen.

     
Children
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