Broadcasting rule allows broadcasting 0d array to empty 1d array

提供された Python の broadcasted_shape 実装と array API のブロードキャスト仕様から始め、次に、示されている 0d から空の 1d へのケースを NumPy の動作と比較します。互換性が意図されたものかどうかを判断し、仕様上の決定または根拠を記録してください。この issue では、更新対象のプロジェクトファイルやテストは指定されていません。

Here is an implementation of the broadcasting algorithm from the spec:

import operator

def broadcasted_shape(sh1, sh2):
    if not isinstance(sh1, (tuple, list)) or not isinstance(sh2, (tuple, list)):
        raise TypeError
    shape1 = tuple(operator.index(i) for i in sh1)
    shape2 = tuple(operator.index(i) for i in sh2)
    n1 = len(shape1)
    n2 = len(shape2)
    n = max(n1, n2)
    shape = [0] * n
    i = n - 1
    while i>=0:
        _n1 = n1 - n + i
        d1 = shape1[_n1] if (_n1 >=0) else 1
        _n2 = n2 - n + i
        d2 = shape2[_n2] if (_n2 >=0) else 1
        if d1 == 1:
            shape[i] = d2
        elif d2 == 1 or d2 == d1:
            shape[i] = d1
        else:
            raise ValueError
        i = i - 1
    return tuple(shape)

With this implementation broadcasting from 0d array to empty 1d array is allowed, which is also consistent with NumPy.

I am not sure why this is a logic thing to do other than to stay compatible with NumPy.

In [1]: from broadcast import broadcasted_shape

In [2]: broadcasted_shape((1,), (0,))
Out[2]: (0,)

In [3]: broadcasted_shape(tuple(), (0,))
Out[3]: (0,)

In [4]: import numpy as np

In [5]: np.broadcast_to(np.array(0), (0,)).shape
Out[5]: (0,)

In [6]: np.broadcast_to(np.array([0]), (0,)).shape
Out[6]: (0,)

The purpose of this issue is to discuss this behavior. If this is as designed, please feel free to close.