In class object, how to auto update attributes?

Question

I have a class which has multiple attributes that are related, for example:

class SomeClass：
    def __init__(self, n=0):
        self.list = range(n)
        self.listsquare = [ x**2 for x in self.list ]

If I make an object normally that would no problem, with

a = SomeClass(10)

I will get 2 lists, a.list and a.listsquare.

Now if I want to make a empty object first, and assign one attribute to it, I want the other attributes to be automatically updated, for example if I do

b = SomeClass()
b.list = range(5,10)

I want b.listsquare to be automatically updated, and also the other way around (assign b.listsquare and auto update b.list). Is this possible? Is Class the right choice for this?

---

Thanks to you all, but I'm completely overwhelmed by all the different answers. Can anyone give a complete solution so I can learn write my own?

I would like to achieve a class Foo with 3 attributes length, list and listsquare such that:

1. If I do a = Foo(3), I get a.length = 3, a.list = [0, 1, 2], a.listsquare = [0, 1, 4].
2. If I do b = Foo().list = [5, 6], I get b.length = 2, b.listsquare = [25, 36].
3. If I do c = Foo().listsquare = [4, 9], I get c.length = 2, c.list = [2, 3].

Solution 1

if updating one property due to an update on another property is what you're looking for (instead of recomputing the value of the downstream property on access) use property setters:

class SomeClass(object):
    def __init__(self, n):
        self.list = range(0, n)

    @property
    def list(self):
        return self._list
    @list.setter
    def list(self, val):
        self._list = val
        self._listsquare = [x**2 for x in self._list ]

    @property
    def listsquare(self):
        return self._listsquare
    @listsquare.setter
    def listsquare(self, val):
        self.list = [int(pow(x, 0.5)) for x in val]

>>> c = SomeClass(5)
>>> c.listsquare
[0, 1, 4, 9, 16]
>>> c.list
[0, 1, 2, 3, 4]
>>> c.list = range(0,6)
>>> c.list
[0, 1, 2, 3, 4, 5]
>>> c.listsquare
[0, 1, 4, 9, 16, 25]
>>> c.listsquare = [x**2 for x in range(0,10)]
>>> c.list
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Solution 2

Absolutely. But use a property instead.

class SomeClass(object):
  def __init__(self, n=5):
    self.mylist = range(n)

  @property
  def listsquare(self):
    return [ x**2 for x in self.mylist ]

a = SomeClass()
a.mylist = [4, 5, 8]
print a.listsquare

Caching of the property value is left as an exercise for the reader.

Solution 3

Ignacio's @property solution is great but it recalculates the list every time you reference listsquare - that could get expensive. Mathew's solution is great, but now you have function calls. You can combine these with the 'property' function. Here I define a getter and a setter for my_list (I just couldn't call it 'list'!) that generates listsquare:

class SomeClass(object):

    def __init__(self, n=5):
        self.my_list = range(n)

    def get_my_list(self):
        return self._my_list

    def set_my_list(self, val):
        self._my_list = val
        # generate listsquare when my_list is updated
        self.my_listsquare = [x**2 for x in self._my_list]

    # now my_list can be used as a variable
    my_list = property(get_my_list, set_my_list, None, 'this list is squared')

x = SomeClass(3)
print x.my_list, x.my_listsquare
x.my_list = range(10)
print x.my_list, x.my_listsquare

This outputs:

[0, 1, 2] [0, 1, 4]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9] [0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

Solution 4

You can also just use setter methods, like this:

class SomeClass:
    def __init__(self, n=5):
        self.set_list(range(n))

    def set_list(self, n):
        self.list = n
        self.listsquare = [ x**2 for x in self.list ]

b = SomeClass()
b.set_list(range(5,10))

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Comments

[LWZ]

Great! What about the other way around? Can I define a.listsquare and get a.mylist?

[Ignacio Vazquez-Abrams]

Yes. Just add a setter to your property.

[LWZ]

Does this mean I need to define 3 properties (size, list, listsquare) so when I update one of them the other 2 will update?

[Ali-Akber Saifee]

that is correct. i've updated the above answer to work the way you wanted - the size attribute feel necessary anymore.

[tdelaney]

thanks to the folks pointing out i had initialized the list incorrectly.