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Posted September 26, 2019 6:57pm by

MRO: copy, list

when i call mro in the following code it lists: (<class 'filledlist.FilledList'>, <class 'list'>, <class 'object'>)

why does it not contain copy instead of list? how does it default to list:

thx!

import copy

class FilledList(list):
    def __init__(self, count, value, *args, **kwargs):
        super().__init__()
        for _ in range(count):
            self.append(copy.copy(value))

import filledlist
from filledlist import FilledList
fl = filledlist.FilledList(9,2)
print(len(fl))
print(fl)
print(FilledList.__mro__)

October 9, 2019 3:55pm

ok, Chris so "list" that is being passed to the Class the List object?

October 9, 2019 3:58pm

Yes, in a class definition, all of the parameters must be a class. In this case, “list” refers to the class list

October 10, 2019 5:33pm

ok..I think I figures it out. here is the complete code:

import random


class Die:
    def __init__(self, sides=2, value=0):
        if not sides >= 2:
            raise ValueError("Must have at least 2 sides")
        if not isinstance(sides, int):
            raise ValueError("Sides must be a whole number")
        self.value = value or random.randint(1, sides)

    def __int__(self):
        return self.value

    def __eq__(self, other):
        return int(self) == other

    def __ne__(self, other):
        return int(self) != other

    def __gt__(self, other):
        return int(self) > other

    def __lt__(self, other):
        return int(self) < other

    def __ge__(self, other):
        return int(self) > other or int(self) == other

    def __le__(self, other):
        return int(self) < other or int(self) == other

    def __add__(self, other):
        return int(self) + other

    def __radd__(self, other):
        return int(self) + other




class D6(Die):
    def __init__(self, value=0):
        super().__init__(sides=6, value=value)

Note this code does not include below which is used in previous examples. If you add that line you get values instead of memory.

 def __repr__(self):
        return str(self.value)

Two questions

based on your previous answer, since Hand passes the object list (which has __str__), why do you need the __repr__ in Die(D6 is an extension of Die) ? This is confusing for me
Are __str__ and __repr__ more or less the same?

thx again for your help

October 11, 2019 1:19am

When hand is typed in the REPL, the __repr__ method is called, if referenced by print or the format method then the __str__ method is called. Some times the __repl__ method is defined to call the __str__ method.

>>> class obj:
...    def __str__(self):
...       return 'str string'
...    def __repr__(self):
...       return 'repr string'
... 
>>> o = obj()
>>> o
repr string
>>> print(o)
str string

If no __str__ method is explicitly defined, the default object.__str__ method will call the __repr__ method.

More info in the docs

1 Answer

September 26, 2019 8:36pm

Good question. The use of a built-in or imported function does not make it part of the object.

The methods and attributes of list become part of FilledList along with the methods and attributes of object that are part all objects.

If copy were part of FilledList you would be able to call it using self.copy. Note that there is a list.copy method but this returns a shallow copy of the whole list and does not make a copy of some other argument.

Post back if you need more help. Good luck!!

October 8, 2019 6:20pm

Chris one more question if you dont mind...when I print fl in the following code I actually get the list:

import copy

class FilledList(list):
    def __init__(self, count, value, *args, **kwargs):
        super().__init__()
        for _ in range(count):
            self.append(copy.copy(value))
fl = FilledList(9,2)

print(fl)
[2, 2, 2, 2, 2, 2, 2, 2, 2]

Other times when I print an instance of an object I get a memory address, for ex:

<thieves.Thief object at 0x00D60990>

why do I not get a memory address in the example above

October 8, 2019 6:31pm

The instance fl inherits the __str__ method from list that is used by print to display the items within the filled list. The class Thief does not define or inherit a __str__ method.

Try:

thief_obj.__str__

where thief_obj is the Thief instance, it should return an undefined attribute error.

October 8, 2019 6:34pm

ty!

October 9, 2019 4:44pm

ty again

October 10, 2019 4:02pm

Chris can I bother you with a related question please..

When Kenneth runs the following he gets a list of object memory info

class Hand(list):
    def __init__(self, size=0, die_class=None, *args, **kwargs):
        if not die_class:
            raise ValueError("You must provide a die class")
        super().__init__()

        for _ in range(size):
            self.append(die_class())

import dice, hands
hand=hands.Hand(size=5, die_class=dice.D6)
hand
[<dice.object at .......>, <.........]

When I run it i get values:

import dice, hands
hand=hands.Hand(size=5, die_class=dice.D6)
hand
[1, 1, 3, 5, 6]

Question...based on your previous answer...since list is being passed, and list has a __str__ , why does Kenneth get memory stuff?

October 10, 2019 4:21pm

The list.__str__ method is what gives you the brackets and comma separated items. It would be the lack of a __str__ method in D6 that causes the “dice.object...”.

First to check: does your hand contain a list of D6 instances or a list of int instances. Try also running:

[type(x) for x in hand]

If you get ints then int.__str__ is running.

Otherwise, I’d have to see the D6 code to know more.

October 16, 2019 4:01pm

Chris may I bother you with one more question please. I am trying to understand the difference between appending a value and appending a copy of a value. in the Subclassing built in section Kenneth uses copy.copy because he says if a mutable was passed in and later changed all instances of the mutable would also change because they all reference the same thing. In the code below I am appending multiple versions of the same mutable(I commented out Kenneth's copy.copy), then I try changing the original version as well as one "copy". Neither causes other "copies" to change. Can you explain please? import copy

code:

class FilledList(list):
    def __init__(self, count, value, *args, **kwargs):
        self.value=value
        super().__init__()
        for _ in range(count):
            # self.append(copy.copy(value))
            self.append(value)


fl2 = FilledList(4,[123])
print(len(fl2))
print(fl2)

fl2.value=[567]
print(fl2.value)
print(fl2)

fl2[0] = [345]
print(fl2[0])
print(fl2)

output:

4
[[123], [123], [123], [123]]
[567]
[[123], [123], [123], [123]]
[345]
[[345], [123], [123], [123]]

October 16, 2019 5:26pm

Without using copy every item in the list points to the same object as shown by object id. Replacing the whole item will, of course, not affect the other items. However, changing an item within the mutable object does affect them all. See adjusted code:

import copy

class FilledList(list):
    def __init__(self, count, value, *args, **kwargs):
        self.value=value
        super().__init__()
        for _ in range(count):
            # self.append(copy.copy(value))
            self.append(value)

class FilledListCopy(list):
    def __init__(self, count, value, *args, **kwargs):
        self.value=value
        super().__init__()
        for _ in range(count):
            self.append(copy.copy(value))


fl2 = FilledList(4,[123])
print(len(fl2))
print(fl2)
print([id(x) for x in fl2])

fl2[0][0] = 'a'
print(fl2[0])
print(fl2) 
print([id(x) for x in fl2])


fl2 = FilledListCopy(4,[123])
print(len(fl2))
print(fl2)
print([id(x) for x in fl2])

fl2[0][0] = 'a'
print(fl2[0])
print(fl2) 
print([id(x) for x in fl2])

Output

4
[[123], [123], [123], [123]]
[4722088904, 4722088904, 4722088904, 4722088904]
[‘a’]
[[‘a’], [‘a’], [‘a’], [‘a’]]
[4722088904, 4722088904, 4722088904, 4722088904]
4
[[123], [123], [123], [123]]
[4662704392, 4662704904, 4662705224, 4662714440]
[‘a’]
[[‘a’], [123], [123], [123]]
[4662704392, 4662704904, 4662705224, 4662714440]

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MRO: copy, list

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Chris Freeman

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1 Answer

Chris Freeman

Chris Freeman

daniel steinberg

daniel steinberg

Chris Freeman

Chris Freeman

daniel steinberg

daniel steinberg

daniel steinberg

daniel steinberg

daniel steinberg

daniel steinberg

Chris Freeman

Chris Freeman

daniel steinberg

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Chris Freeman

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