From: jepler at inetnebr.com (Jeff Epler)
Date: Fri, 30 Apr 1999 01:13:10 GMT
Subject: try vs. has_key()
References: <aahzFAM4oJ.M7M@netcom.com> <yWOT2.6007$8m5.9320@newsr1.twcny.rr.com> <Pine.SUN.3.95-heb-2.07.990423140345.21577A-100000@sunset.ma.huji.ac.il> <7g51q6$1pt$1@vvs.superst.iae.nl> <aahzFAvFCz.3Lr@netcom.com> <14119.8145.293039.667256@bitdiddle.cnri.reston.va.us>
Message-ID: <slrn7ii0vb.tdc.jepler@craie.inetnebr.com>
X-UID: 120                                                  

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