Fast Efficient way to transfer an object to another list

[Tim Chase]

On 04/30/2010 10:35 PM, Steven D'Aprano wrote:
> If you know there is one, and only one, item with that stock code:
>
> def transfer_stock(stock_code, old_list, new_list):
>      """ Transfer a stock from one list to another """
>      i = old_list.index(stock_code)  # search
>      new_list.append(old_list[i])  # copy
>      del old_list[i]  # delete
>      return new_list

This could be written as

   def move(code, source, dest):
     dest.append(source.pop(source.index(code)))
     return dest

depending on how one thinks.  I tend to prefer

   lst.pop(idx)

over

   tmp = lst[idx]
   del lst[idx]

only using the latter if "idx" is a range/slice.

Though since the function mutates the arguments, I'd be tempted 
to return None like list.sort() does for the same rationale.

If more than one item is in the source, it will move/remove the 
first leaving the remainder; if no matching item is in the source 
it will appropriately raise a ValueError.

-tkc