try vs. has_key()

[Jeff Epler]

On Wed, 28 Apr 1999 11:13:28 -0400 (EDT), Jeremy Hylton
 <jeremy at cnri.reston.va.us> wrote:
>    d={}
>    for word in words:
>        first_two=word[:2]
>        d[first_two]=d.get(first_two, []).append(word)
>
>This second bit doesn't work because the append method on list objects 
>returns None.  As a result, the first time a new value for first_two
>appears None will be assigned to that key in the dictionary.  The
>second time that value of first_two shows up, None will be returned by 
>d.get.  Then the code will raise an AttributeError, because None
>doesn't have an append method.
>
>The following code would be correct:
>
>    d={}
>    for word in words:
>        first_two=word[:2]
>        d[first_two]= temp = d.get(first_two, [])
>        temp.append(word)

what about
	d[first_two] = d.get(first_two, [])+[word]
?  Or is list construction and addition going to be enough more expensive
than the function call to make this a lose as well?

Jeff