numpy array of objects

Question

I’m trying to implement a simulation for a lattice model (lattice boltzmann) in Python. Each site of the lattice has a number of properties, and interact with neighboring sites according to certain rules. I figured that it might be clever to make a class with all the properties and make a grid of instances of that class. (As I’m inexperienced with Python, this might not be a good idea at all, so feel free to comment on my approach.)

Here is a toy example of what I’m doing

class site:
    def __init__(self,a,...):
        self.a = a
        .... other properties ...
    def set_a(self, new_a):
        self.a = new_a

Now I want to deal with a 2D/3D lattice (grid) of such sites so I tried to do the following (here is a 2D 3×3 grid as an example, but in simulation I would need the order of >1000x1000X1000)

lattice = np.empty( (3,3), dtype=object)
lattice[:,:] = site(3)

Now, the problem is that each lattice point refer to the same instance, for example

lattice[0,0].set_a(5)

will also set the value of lattice[0,2].a to 5. This behavior is unwanted. To avoid the problem i can loop over each grid point and assign the objects element by element, like

for i in range(3):
    for j in range(3):
        lattice[i,j] = site(a)

But is there a better way (not involving the loops) to assign objects to a multidimensional array?

Thanks

Asked By: jonalm

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Answer 1

The missing piece for you is that Python treats everything as a reference. (There are some “immutable” objects, strings and numbers and tuples, that are treated more like values.) When you do

lattice[:,:] = site(3)

you are saying “Python: make a new object site, and tell every element of lattice to point to that object.” To see that this is really the case, print the array to see that the memory addresses of the objects are all the same:

array([[<__main__.Site object at 0x1029d5610>,
        <__main__.Site object at 0x1029d5610>,
        <__main__.Site object at 0x1029d5610>],
       [<__main__.Site object at 0x1029d5610>,
        <__main__.Site object at 0x1029d5610>,
        <__main__.Site object at 0x1029d5610>],
       [<__main__.Site object at 0x1029d5610>,
        <__main__.Site object at 0x1029d5610>,
        <__main__.Site object at 0x1029d5610>]], dtype=object)

The loop way is one correct way to do it. With numpy arrays, that may be your best option; with Python lists, you could also use a list comprehension:

lattice = [ [Site(i + j) for i in range(3)] for j in range(3) ]

You can use a list comprehension with the numpy.array construction:

lattice = np.array( [ [Site(i + j) for i in range(3)] for j in range(3) ],
                    dtype=object)

Now when you print lattice, it’s

array([[<__main__.Site object at 0x1029d53d0>,
        <__main__.Site object at 0x1029d50d0>,
        <__main__.Site object at 0x1029d5390>],
       [<__main__.Site object at 0x1029d5750>,
        <__main__.Site object at 0x1029d57d0>,
        <__main__.Site object at 0x1029d5990>],
       [<__main__.Site object at 0x1029d59d0>,
        <__main__.Site object at 0x1029d5a10>,
        <__main__.Site object at 0x1029d5a50>]], dtype=object)

so you can see that every object in there is unique.

You should also note that “setter” and “getter” methods (e.g., set_a) are un-Pythonic. It’s better to set and get attributes directly, and then use the @property decorator if you REALLY need to prevent write access to an attribute.

Also note that it’s standard for Python classes to be written using CamelCase, not lowercase.

Answered By: Seth Johnson

Answer 2

You can vectorize the class’s __init__ function:

import numpy as np

class Site:
    def __init__(self, a):
        self.a = a
    def set_a(self, new_a):
        self.a = new_a

vSite = np.vectorize(Site)

init_arry = np.arange(9).reshape((3,3))

lattice = np.empty((3,3), dtype=object)
lattice[:,:] = vSite(init_arry)

This may look cleaner but has no performance advantage over your looping solution. The list comprehension answers create an intermediate python list which would cause a performance hit.

Answered By: Paul

Answer 3

I don’t know about better, but as an alternative to an explicit set of loops, you could write

lattice = np.empty( (3,3), dtype=object)
lattice.flat = [site(3) for _ in lattice.flat]

which should work whatever the shape of the lattice.

Answered By: DSM

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