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Of Functions, Objects, and Methods-I NEED HELP PLEASE

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  Of Functions, Objects, and Methods-I NEED HELP PLEASE Cathy James <nambo4jb@gmail.com> - 2011-06-08 15:09 -0500
    Re: Of Functions, Objects, and Methods-I NEED HELP PLEASE John Gordon <gordon@panix.com> - 2011-06-08 20:42 +0000
    Re: Of Functions, Objects, and Methods-I NEED HELP PLEASE Larry Hudson <orgnut@yahoo.com> - 2011-06-08 23:59 -0700
      Re: Of Functions, Objects, and Methods-I NEED HELP PLEASE Ethan Furman <ethan@stoneleaf.us> - 2011-06-09 05:54 -0700
      Re: Of Functions, Objects, and Methods-I NEED HELP PLEASE Ethan Furman <ethan@stoneleaf.us> - 2011-06-09 06:10 -0700
      Re: Of Functions, Objects, and Methods-I NEED HELP PLEASE Dennis Lee Bieber <wlfraed@ix.netcom.com> - 2011-06-09 10:25 -0700
      Re: Of Functions, Objects, and Methods-I NEED HELP PLEASE Larry Hudson <orgnut@yahoo.com> - 2011-06-09 23:07 -0700

#7254 — Of Functions, Objects, and Methods-I NEED HELP PLEASE

I am almost there, but I need a little help:

I would like to

a) print my dogs in the format  index. name: breed as follows:

0. Mimi:Poodle
1.Sunny: Beagle
2. Bunny: German Shepard
I am getting

(0, ('Mimi', 'Poodle')) . Mimi : Poodle instead-what have I done wrong?

b) I would like to append to my list, but my line dogs.dogAppend() is
giving a TypeError:

for i in enumerate (self.dogAppend()):
TypeError: 'list' object is not callable

Any help?

#MY CODE BELOW:

import sys
class Dog():
    def __init__(self, name, breed):
        self.name = name
        self.breed = breed

    def dogAppend(self):
        self.dogAppend = []
        self.dogAppend.append((self.name,self.breed))
        return self.dogAppend


    def display (self):
        for i in enumerate (self.dogAppend()):
            print (i,".",  self.name, ": " + self.breed)

if __name__ == "__main__":
    dogs = Dog(name=input (" Enter Dog Name: "), breed=input ("Enter
Dog Breed: "))
    while not dogs:
        print("Goodbye!!")
        sys.exit()
    else:
        #dogs.dogAppend()
        dogs.display()

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#7259

In <mailman.30.1307563778.11593.python-list@python.org> Cathy James <nambo4jb@gmail.com> writes:

> b) I would like to append to my list, but my line dogs.dogAppend() is
> giving a TypeError:

> for i in enumerate (self.dogAppend()):
> TypeError: 'list' object is not callable

>     def dogAppend(self):
>         self.dogAppend = []

You have a method and a list that are both called dogAppend.  Try naming
one of them something different.

-- 
John Gordon                   A is for Amy, who fell down the stairs
gordon@panix.com              B is for Basil, assaulted by bears
                                -- Edward Gorey, "The Gashlycrumb Tinies"

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#7276

On 06/08/2011 01:09 PM, Cathy James wrote:
> I am almost there, but I need a little help:
>
> I would like to
>
> a) print my dogs in the format  index. name: breed as follows:
>
> 0. Mimi:Poodle
> 1.Sunny: Beagle
> 2. Bunny: German Shepard
> I am getting
>
> (0, ('Mimi', 'Poodle')) . Mimi : Poodle instead-what have I done wrong?
>
> b) I would like to append to my list, but my line dogs.dogAppend() is
> giving a TypeError:
>
> for i in enumerate (self.dogAppend()):
> TypeError: 'list' object is not callable
>
> Any help?
>
> #MY CODE BELOW:
>
> import sys

Not needed in the class definition, move it to the "if __name__..." section.
Actually, it's not needed at all.  That part could be rewritten to avoid the need for sys.exit() 
entirely.

> class Dog():
>      def __init__(self, name, breed):
>          self.name = name
>          self.breed = breed
>
>      def dogAppend(self):
>          self.dogAppend = []
>          self.dogAppend.append((self.name,self.breed))
>          return self.dogAppend

1:  Think about what you're trying to do here...  You seem to want _each_ instance of Dog to 
hold the whole list.  Do you want Dog to be an individual Dog or do you  want it to be a list of 
Dogs?  You can't do both in a single class.
2:  This won't work to append to your list anyway... each time you call dogAppend() it will 
clear out any existing list and result in a list of one element, a tuple of (name, breed).  You 
can do exactly the same thing with a single line:  return [(self.name, self.breed)]

>      def display (self):
>          for i in enumerate (self.dogAppend()):
>              print (i,".",  self.name, ": " + self.breed)
>
Misusing enumerate.  Check the docs.
http://docs.python.org/py3k/library/functions.html?highlight=enumerate#enumerate
Besides that, there are easier ways to print the contents of a list.

> if __name__ == "__main__":
>      dogs = Dog(name=input (" Enter Dog Name: "), breed=input ("Enter
> Dog Breed: "))
>      while not dogs:
>          print("Goodbye!!")
>          sys.exit()
>      else:

else does not belong with while.

>          #dogs.dogAppend()
>          dogs.display()

Here's one possible replacement.  There are many other approaches as well.
(This leaves the individual dogs as a (name, breed) tuple.  It could be modified for other 
definitions of a dog. -- Exercise left for the reader...)    ;-)

class DogKennel:
     def __init__(self):
         self.dogs = []      #  list of dogs

     def addDog(self):
         name = input("Enter dog name:  ")
         if name == "":
             return False
         breed = input("Enter dog breed:  ")
         if breed == "":
             return False
         self.dogs.append((name, breed))   #  Append this dog tuple to the list
         return True

     def display(self):
         i = 1
         for dog in self.dogs:
             print("%d.  %s: %s" % (i, dog[0], dog[1]))
             i += 1;

if __name__ == "__main__":
     dogs = DogKennel()
     while (dogs.addDog()):
         pass
     dogs.display()

      -=- Larry -=-

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#7295

Larry Hudson wrote:
> On 06/08/2011 01:09 PM, Cathy James wrote:
>> Dog Breed: "))
>>      while not dogs:
>>          print("Goodbye!!")
>>          sys.exit()
>>      else:
> 
> else does not belong with while.

else works just fine with while; it is the path taken when the while is 
exhausted, but not broken out of:

--> i = 5
--> while i:
...   print(i)
...   i -= 1
... else:
...   print("blast off!")
...
5
4
3
2
1
blast off!


--> i = 5
--> while i:
...   print(i)
...   i -= 1
...   if i == 3:
...     print('aborting')
...     break
... else:
...   print("blast off!")
...
5
4
aborting

~Ethan~

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#7296

Ethan Furman wrote:
> Larry Hudson wrote:
>> On 06/08/2011 01:09 PM, Cathy James wrote:
>>> Dog Breed: "))
>>>      while not dogs:
>>>          print("Goodbye!!")
>>>          sys.exit()
>>>      else:
>>
>> else does not belong with while.
> 
> else works just fine with while; it is the path taken when the while is 
> exhausted, but not broken out of:

It works with 'for' as well.

~Ethan~

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#7309

On Wed, 08 Jun 2011 23:59:35 -0700, Larry Hudson <orgnut@yahoo.com>
declaimed the following in gmane.comp.python.general:

> > class Dog():
> >      def __init__(self, name, breed):
> >          self.name = name
> >          self.breed = breed
> >
> >      def dogAppend(self):
> >          self.dogAppend = []
> >          self.dogAppend.append((self.name,self.breed))
> >          return self.dogAppend
>
 
> 2:  This won't work to append to your list anyway... each time you call dogAppend() it will 
> clear out any existing list and result in a list of one element, a tuple of (name, breed).  You 
> can do exactly the same thing with a single line:  return [(self.name, self.breed)]
>
	Worse than that... The first time dogAppend is called on an instance
of Dog, it replaces the dogAppend method with a binding to the (single
element) list created by the method.
 
-- 
	Wulfraed                 Dennis Lee Bieber         AF6VN
        wlfraed@ix.netcom.com    HTTP://wlfraed.home.netcom.com/

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#7358

On 06/08/2011 11:59 PM, Larry Hudson wrote:
> On 06/08/2011 01:09 PM, Cathy James wrote:
>> I am almost there, but I need a little help:
>>
>> I would like to
>>
... <deleted text>

> Here's one possible replacement. There are many other approaches as well.
> (This leaves the individual dogs as a (name, breed) tuple. It could be modified for other
> definitions of a dog. -- Exercise left for the reader...) ;-)
>
... <more deleted text>

In thinking about this some more, I thought a dictionary instead of a list would be a better fit 
for this example.  For one thing, it allows accessing the dogs by name instead of an arbitrary 
index number.  But remember that a dictionary is unordered -- it won't be displayed in the same 
order that the entries were made.

Here's this approach, rewritten (and slightly expanded) using a dictionary.

class DogKennel:
     def __init__(self):
         """Dog names/breeds kept in a dictionary"""
         self.dogs = {}

     def addDog(self):
         """Add a single dog to the dictionary"""
         name = input("Enter dog's name:  ")
         if name == "":      #   Abort with empty input
             return False
         breed = input("Enter dog's breed: ")
         if breed == "":     #   Abort here if needed also
             return False
         self.dogs[name] = breed
         return True

     def makeKennel(self):
         """Add multiple dogs (a pack?) to the dictionary"""
         while self.addDog():
             pass

     def getDog(self, name=""):
         """Get the dog's breed by its name"""
         if name in self.dogs:
             return self.dogs[name]
         else:
             return None

     def display(self):
         """Display all the dogs in the kennel (the dictionary)"""
         i = 1
         for dog in self.dogs.keys():
             print("%2d.  %s: %s" % (i, dog, self.dogs[dog]))
             i += 1

#   Note this is a normal function, NOT a member function of DogKennel.
#   It probably should go in the __main__ section to keep it separate from
#   the DogKennel class.  (This would be significant only if you're going
#   to import the DogKennel class into other programs.)
def yesno(prompt = ""):
     """Get a yes or no answer.  Returns True if yes, False if no"""
     if prompt != "":
         prompt = prompt + " (y/n)  "
     while True:
         ans = input(prompt).upper()
         if ans != "":           #   Answer is not empty
             if ans[0] == 'Y':   #   1st char is 'Y'?
                 return True
             if ans[0] == 'N':   #   1st char is 'N'?
                 return False
             #   Otherwise loop back for another go

if __name__ == "__main__":
     dogs = DogKennel()
     dogs.makeKennel()
     dogs.display()
     if yesno("Add more dogs?"):
         dogs.makeKennel()
         print("The kennel now contains")
         dogs.display()
     while True:
         name = input("Which dog do you want?  ")
         if name == "":
             break
         breed = dogs.getDog(name)
         if breed != None:
             print(name, "is a", breed)

#---------------
I hope studying this (and my previous) examples help you understand things better.
Keep at it...  It will sink in with a little effort.     :-)
I also hope my rather verbose answers give you a little insight about the sort of things you 
need to consider when designing your programs.

      -=- Larry -=-

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