[Tutor] Algorithm

[Wayne]

On Mon, Aug 24, 2009 at 10:48 AM, kreglet <kreglet at gmail.com> wrote:

>
> Wayne,
>
> The reason I used print sorted is that using just print throws a syntax
> error:
>
> print (lettercount.iteritems(), key=itemgetter(1)) ---> error
> print lettercount.iteritems(), key=itemgetter(1) ---> error
> print sorted(lettercount.iteritems(), key=itemgetter(1)) ---> works


Ah, excuse me - I just glanced at your function and didn't notice you
weren't returning any value. You should be returning lettercount from your
function.


>
>
> I don't know why. Seems to me that any of the above should work.
>
> mainword = countletters('batty')
> cmpword = countletters('bat')
>
> myfunc(cmpword, mainword)
>
> Generates error:
>
> Traceback (most recent call last):
>  File "/home/kreglet/bin/test.py", line 23, in <module>
>    myfunc(cmpword, mainword)
>  File "/home/kreglet/bin/test.py", line 13, in myfunc
>    for letter in cmpword:
> TypeError: 'NoneType' object is not iterable
>
> mainword = countletters('batty')
> print mainword returns None
>
> cmpword = countletters('bat')
> print cmpword returns None
>
> Both mainword and cmpword are passed to the function but since the values
> of
> each are None
> the function doesn't work.
>
> Is this correct?
>

That's correct - your function countletters returns None, so you're
assigning cmpword to None. And since None is not an iterable type, it throws
an error. Try returning a value like I mentioned above and see how that
works.

HTH,
Wayne


>
> thanx,
> kreglet
>
>
> Wayne-68 wrote:
> >
> > On Sun, Aug 23, 2009 at 10:01 PM, kreglet <kreglet at gmail.com> wrote:
> >
> > I would actually not bother sorting your return from countletters - keep
> > it
> > a dictionary.
> >
> > Then you can compare like this:
> >
> > mainword = countletters('batty')
> > cmpword = countletters('bat')
> >
> > def myfunc(cmpword, mainword):
> >   for letter in cmpword:
> >     if mainword.gets(letter):
> >         if cmpword[letter] >mainword[letter]:
> >              return False
> >     else:
> >         return False
> >
> > I think that should work. First you're looping over each letter in
> > cmpword.
> > Because mainword is also a dictionary the order isn't terribly important.
> > Then you check if the letter is in mainword. If it's not, obviously
> > cmpword
> > isn't in mainword, so return False. If the letter is, compare the counts.
> > If
> > cmpword has more letters than mainword, it's not in the word so again
> > return
> > False.
> >
> > HTH,
> > Wayne
> >
> > _______________________________________________
> > Tutor maillist  -  Tutor at python.org
> > http://mail.python.org/mailman/listinfo/tutor
> >
> >
>
> --
> View this message in context:
> http://www.nabble.com/Algorithm-tp25107922p25118434.html
> Sent from the Python - tutor mailing list archive at Nabble.com.
>
> _______________________________________________
> Tutor maillist  -  Tutor at python.org
> http://mail.python.org/mailman/listinfo/tutor
>



-- 
To be considered stupid and to be told so is more painful than being called
gluttonous, mendacious, violent, lascivious, lazy, cowardly: every weakness,
every vice, has found its defenders, its rhetoric, its ennoblement and
exaltation, but stupidity hasn’t. - Primo Levi
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mail.python.org/pipermail/tutor/attachments/20090824/c0e29160/attachment.htm>