try vs. has_key()

[Jeremy Hylton]

>>>>> "JE" == Jeff Epler <jepler at inetnebr.com> writes:

  >>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)

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

Yeah.  Concatenating two lists results in a new list being created
every time (and you're already creating a new list containing the
value of word).  Two list allocs is definitely more expensive that an
append.

Jeremy