On Sat, Nov 12, 2011 at 10:57 PM, Dave Angel <d...@davea.name> wrote: > On 11/12/2011 09:48 AM, lina wrote: >> >> On Sat, Nov 12, 2011 at 9:22 PM, Dave Angel<d...@davea.name> wrote: >>> >>> On 11/12/2011 03:54 AM, lina wrote: >>>> >>>> <SNIP> >>>> The one I tried : >>>> if longest>= 2: >>>> sublist=L1[x_longest-longest:x_longest] >>>> result=result.append(sublist) >>>> if sublist not in sublists: >>>> sublists.append(sublist) >>>> >>>> the $ python3 CommonSublists.py >>>> atom-pair_1.txt atom-pair_2.txt >>>> Traceback (most recent call last): >>>> File "CommonSublists.py", line 47, in<module> >>>> print(CommonSublist(a,b)) >>>> File "CommonSublists.py", line 24, in CommonSublist >>>> result=result.append(sublist) >>>> AttributeError: 'NoneType' object has no attribute 'append' >>>> >>>> in local domain I set the result=[] >>>> I don't know why it complains its NoneType, since the "result" is >>>> nearly the same as "sublists". >>>> >>> Assuming this snippet is part of a loop, I see the problem: >>> >>> result = result.append(sublist) >>> >>> list.append() returns none. It modifies the list object in place, but it >>> doesn't return anything. So that statement modifies the result object, >>> appending the sublist to it, then it sets it to None. The second time >>> around you see that error. >> >> I am sorry. haha ... still lack of understanding above sentence. >> >>>>> a >> >> ['3', '5', '7', '8', '9'] >>>>> >>>>> d.append(a) >>>>> d >> >> [['3', '5', '7', '8', '9']] >>>>> >>>>> type(a) >> >> <class 'list'> >> >> Sorry and thanks, best regards, >> >> lina >> >>> >>> In general, most methods in the standard library either modify the object >>> they're working on, OR they return something. The append method is in >>> the >>> first category. >>> >>> > > To keep it simple, I'm using three separate variables. d and a are as you > tried to show above. Now what happens when I append? > > Python 2.7.1+ (r271:86832, Apr 11 2011, 18:13:53) > [GCC 4.5.2] on linux2 > Type "help", "copyright", "credits" or "license" for more information. >>>> d = [] >>>> a = [3, 5, 7] >>>> xxx = d.append(a) >>>> print(repr(xxx)) > None >>>> print d > [[3, 5, 7]] > > Notice that d does change as we expected. But xxx, the return value, is > None. The append() method doesn't return any useful value, so don't assign > it to anything.
Thanks, ^_^, now better. I checked, the sublist (list) here can't be as a key of the results (dict). actually I also wish to get the occurence of those sublist in the script, except using external one in command line as uniq -c. ^_^ Have a nice weekend, > > The statement in your code that's wrong is > result = result.append(sublist) > > The final value that goes into result is None, no matter what the earlier > values of result and sublist were. > > -- > > DaveA > _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
