[Tutor] joining selected items in list

[Steven D'Aprano]

David wrote:
> Dear Tutors,
> 
> I have a list that I wish to reorganise into fewer list items.
> What happens is that some of the items belong together:
> not ['keine', 'Antwort'] but ['Keine Antwort'].
> 
> I am not aware of any list methods that can help me here, and would thus
> be grateful for a hint or two.

This solves like a good case for one of the oldest, most general techniques in 
use: an accumulator.

You construct a new list, from the old list, but using a temporary accumulator 
first. The basic idea is:

for each item in old list:
     if data in the accumulator is complete:
         move the completed accumulator data to the new list
         clear the accumulator so it is ready to start collecting fresh data
     otherwise:
         move item from old list to the accumulator


There's a million variations on this, depending on whether you test the 
accumulator data at the start of the loop or the end of the loop, how and 
where you process items, etc.

In your case, it sounds like it will be even easier, since you need collect 
*at most* two items:

set a temporary variable to empty  # the accumulator
for each item in old list:
     if the temporary variable is not empty:
         combine the temp variable and the current item
         move them to the new list
     otherwise:
         if item is complete on its own:
             move it straight into the new list
         otherwise:
             move it to a temporary variable



Converting this to Python is trivial:


tmp = ''
for item in old_list:
     if tmp:
         new_list.append(tmp + item)
     else:
         if stands_alone(item):  # You must write this function.
             new_list.append(item)
         else:
             tmp = item


Define this function first:

def stands_alone(item):
     decide whether item stands alone, and if so, return True
     otherwise return False


That decision function is the hard part. Is there some rule that tells you 
whether or not a word in the list belongs with the next word?




-- 
Steven