Accessing attributes of a list of objects in python

Question

I have a list of objects, known as Bricks, I want to check the attributes of all objects in the list. I tried:

pos = [Brick.start_pos for Brick in self.brick_list]

to create a list of the required attribute i.e start_pos, but it populates a list with only the attribute of the last brick in the brick_list. I am using python 3.4. I need to do this for a lot of different attributes so a comprehension makes sense to me.

Brick class is mostly just a holder for data and is defined:

class Brick:
    def __init__(self, start_pos=0, current_pos=0, variety=0, moveability=0, 
                 neighbours=0, max_range=0, sides=0, rotation=0, length=0,
                 previous_pos =0):
        self.start_pos = start_pos
        self.current_pos = current_pos
        self.moveability = moveability
        self.variety = variety
        self.neighbours = neighbours
        self.max_range = max_range
        self.sides = sides
        self.rotation = rotation
        self.length = length
        self.previous_pos = previous_pos

with couple very simple functions.

How did you define the class Brick? And how did you build the list? — Reut Sharabani
– Reut Sharabani, Commented Aug 5, 2015 at 11:30
The list is held as an attribute in another class. It is built by hell.brick_list = len(lister)*[Brick()] — Elwood168
– Elwood168, Commented Aug 5, 2015 at 11:34
The fragment that you have is a 100% correct list comprehension which yields the attributes of all the bricks in the brick_list (as long as accessing that attribute doesn't modify the list as a side effect). If it isn't giving you what you expect -- the problem is probably that the list isn't what you think that it is. — John Coleman
– John Coleman, Commented Aug 5, 2015 at 11:36
@Elwood168 It appears the list contains the same Brick instance, repeated n times (n being the length of lister). — Pynchia
– Pynchia, Commented Aug 5, 2015 at 11:41

Pynchia · Accepted Answer · 2015-08-05 12:11:06Z

3

It appears the list contains the same Brick instance, repeated n times (n being the length of lister)

hell.brick_list = len(lister)*[Brick()]

instead, you might want to build the list as

hell.brick_list = [Brick() for _ in lister]

Obviously, you will need to pass the constructor all the necessary parameters, otherwise you would end-up with Bricks with identical attributes (default values).

Actually, given your second comment

for i in range (len(lister)): hell.brick_list[i].start_pos = lister[i]

the best option would be to create and initialize the list all in one go

hell.brick_list = [Brick(start_pos=pos) for pos in lister]

edited Aug 5, 2015 at 12:11

answered Aug 5, 2015 at 11:45

Pynchia

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4 Comments

SiHa Over a year ago

It's worth noting that, in the first (non-working) version, this means that in the line for i in range (len(lister)): hell.brick_list[i].start_pos = lister[i] each time you assign a value to start.pos it is assigned to all copies of the same Brick instance, because they are all the same object.

Elwood168 Over a year ago

@Pynchia. Thanks for the explanation, I did not understand that could happen. I replaced that line with what you suggested and it all worked immediately, very satisfying :)

Pynchia Over a year ago

[Brick()] is evaluated and it produces a list containing a single object instance. Then the list is multiplied by len(lister), which produces a list containing the very same elements, but repeated len(lister) times over. Only one Brick object instance is created.

Pynchia Over a year ago

@SiHa yes. And thank you for mentioning that, I had not noticed the OP's second comment, so you allowed me to improve my answer.

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