How does numpy addition work?

Question

I have the following unexpected behaviour

import numpy as np
class Test:
    def __radd__(self, other):
        print(f'value: {other}')

[1,2,3] + Test()
# prints: value: [1,2,3]
np.array([1,2,3]) + Test()
# prints
# value: 1
# value: 2
# value: 3

I would expect the second addition to behave in the same way as the first one, but it doesn't. The only logical explanation I can see is that the numpy + operator somehow iterates over the arguments first, and tries to add each one of them to Test(), and the second addition (int + Test) falls back to the Test.__radd__

So

• is there any other explanation that I'm missing?
• if not, how does a + b work in numpy
• is it equivalent to np.add(a, b)?

Answer 1

This is because of NumPy 'broadcasting'.

Your explanation is pretty much correct as you can see in the docs for np.add.

Notes

Equivalent to x1 + x2 in terms of array broadcasting.

I find this makes a little more sense if you play around with NumPy and see how np.array differs from the built-in list.

Python 3.5.2 (default, Jul  6 2016, 16:37:16)
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In [1]: import numpy as np

In [2]: [1, 2, 3] + [1, 2, 3]
Out[2]: [1, 2, 3, 1, 2, 3]

In [3]: np.array([1, 2, 3]) + np.array([1, 2, 3])
Out[3]: array([2, 4, 6])

Looking at the sourcecode here for the NDArrayOperatorsMixin which implements the special methods for almost all of Python's builtin operators defined in the `operator` module. It looks like __add__, __radd__, __iadd__ are all set to the .add umath function. But I'm not sure if the actual ndarray makes use of the mix-in, I think you'd have to dig through the C code to figure out how that's handled.

Answer 2

If I expand your class a bit we may get more of an idea of when __add__ is used, and when __radd__:

class Test:
    def __radd__(self, other):
        print(f'r value: {other}, {type(other)}')
        return f'{other}'
    def __add__(self, other):
        print(f'a value: {other}, {type(other)}')
        return other+3

With a list

In [285]: [1,2,3]+Test()
r value: [1, 2, 3], <class 'list'>    # use Test.__radd__
Out[285]: '[1, 2, 3]'
In [286]: Test()+[1,2,3]             # tries to use Test.__add__
a value: [1, 2, 3], <class 'list'>
....
<ipython-input-280-cd3f564be47a> in __add__(self, other)
      5     def __add__(self, other):
      6         print(f'a value: {other}, {type(other)}')
----> 7         return other+3
      8 
TypeError: can only concatenate list (not "int") to list

With an array:

In [287]: np.arange(3)+Test()     # use Test.__radd__ with each array element
r value: 0, <class 'int'>
r value: 1, <class 'int'>
r value: 2, <class 'int'>
Out[287]: array(['0', '1', '2'], dtype=object)
In [288]: Test()+np.arange(3)
a value: [0 1 2], <class 'numpy.ndarray'>
Out[288]: array([3, 4, 5])    # use Test.__add__ on whole array

With itself, a double use of Test.__add__:

In [289]: Test()+Test()
a value: <__main__.Test object at 0x7fc33a5a7a20>, <class '__main__.Test'>
a value: 3, <class 'int'>
Out[289]: 6

As I commented it can be tricky sorting out the __add__ v __radd__ delegation, and separating that from ndarray action.

---

add with Test() second gives the same output as [287]:

In [295]: np.add(np.arange(3),Test())
r value: 0, <class 'int'>
r value: 1, <class 'int'>
r value: 2, <class 'int'>
Out[295]: array(['0', '1', '2'], dtype=object)

np.add with Test() first is different from [288] above:

In [296]: np.add(Test(),np.arange(3))
a value: 0, <class 'int'>
a value: 1, <class 'int'>
a value: 2, <class 'int'>
Out[296]: array([3, 4, 5], dtype=object)