[Solved] Overriding __iadd__ from the right hand side in Python

Anton Akhmerov Asks: Overriding __iadd__ from the right hand side in Python
Imagine I have a class A that is outside of my control (numpy.ndarray is the specific application), and class B within my control (scipy.sparse.coo_matrix). I want to implement in-place addition of B to A without touching class A.

Is this at all possible? Is it a bad idea? If this isn’t possible in general, is it possible with numpy arrays specifically?

For a specific example consider:

Code:
class A:
   foo = 0

   def __iadd__(self, other):
       print("Avoid calling this function.")
       return self

class B:
   def __add__(self, other):
       if isinstance(other, A):
           other.foo += 1
           return other

   __radd__ = __add__

   # Modify this class to make assertion below pass

a1, a2 = A(), A()
a1 += B()
a2 = a2 + B()
assert a1.foo == a2.foo == 1, "How to make this work"

EDIT: Actual application.

In-place addition of a sparse coordinate matrix to a dense numpy array has an efficient implementation:

Code:
from time import time

import numpy as np
from scipy import sparse

N = 1000
a = np.zeros((N, N))
b = sparse.identity(N, format="coo")

t_slow = -time()
a += b  # Want to override, converts b to array—slow!
t_slow += time()

t_fast = -time()
a[b.row, b.col] += b.data  # Desired implementation
t_fast += time()
print(f"{t_slow=:.2}s, {t_fast=:.2}s")
# t_slow=0.0017s, t_fast=0.00024s

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