OpenEuphoria

Re: splitting to sublist - HELLPPPP

• Posted by petelomax May 31, 2018
• 1150 views

You were pretty close, I think, just a single > that needed to be >=. Try this:

function restructure(object orgseq)  
  object ors={}, merge={}  
  integer i=1, tstart=1, len=length(orgseq) 
  while i <= len do  
    tstart = i  
    if orgseq[i][1] = 1002 then  
      if i < len and orgseq[i+1][1] = 1002 then  
        while (i <= len) and (orgseq[i][1] = 1002) do  
          i += 1  
        end while       
      end if  
      if tstart <= i then  
        if i > len then  
          ors = append(ors, {1002, tstart-1, len})  
        else  
          ors = append(ors, {1002, tstart-1, i})  
        end if  
      end if  
    elsif i <= len and orgseq[i+1][1] != 1002 then    
      while (i < len) and orgseq[i+1][1] != 1002 do  
        i += 1  
      end while       
      if tstart <= i then  
--      if i > len then  
        if i >= len then  
          ors = append(ors, {1000, tstart, len})  
        else  
          ors = append(ors, {1000, tstart, i-1})  
        end if  
      end if        
    end if  
    i += 1  
  end while  
  return ors  
end function  

I can now imagine why you might be struggling. Here is my attempt to explain the problem, as I understand it:

Given a sequence of 'and'/'or' elements, which must start with 'and', collect any consecutive runs of 'or' along with 
their previous 'and', and any consecutive runs of 'and' excluding those lumped in with an 'or' run, as separate groups. 
eg  { and,and,  and,or,or,  and }   -- (spaces added are mine alone) 
 => {{and,and},{and,or,or},{and}} 

And here is my attempt at that:

function get_next(sequence s, integer done, integer and_or) 
    for i=done+1 to length(s) do 
        if s[i][1]=and_or then return i end if 
    end for 
    return 0 
end function 
 
function restructure(sequence s) 
sequence res = {} 
integer done = 0, next 
    if s[1][1]!=1000 then 
        ?9/0  -- sanity check (must start with 'and') 
    end if 
    while done<length(s) do 
        next = get_next(s,done,1002) 
        if next=0 then exit end if 
        if next>done+2 then 
            -- add a run of {and,and,and} 
            -- (not including the one just before  
            --  the or/1002 that we just found) 
            res = append(res,s[done+1..next-2]) 
            done = next-2 
        end if 
        next = get_next(s,next,1000) 
        if next=0 then exit end if 
        -- add a run of {and,or,or,or} 
        -- (up to but not including the  
        --  and/1000 that we just found) 
        res = append(res,s[done+1..next-1]) 
        done = next-1 
    end while 
    -- add last run 
    res = append(res,s[done+1..len]) 
    return res 
end function 
 

HTH, Pete