Fast Efficient way to transfer an object to another list

[MRAB]

Tim Chase wrote:
> 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.
> 
It would be more efficient if instead of deleting or popping the item
you moved the last one into its place:

     if idx == len(source) - 1:
         item = source.pop()
     else:
         item = source[idx]
         source[idx] = source.pop()

assuming that the order of the items in the list doesn't matter.