注解中给出的两种解决方案（np.outer(a, b)和a[:, None] * b）是等效的，并且具有相似的效率：

In [25]: m, n = 300, 500
    ...:
    ...: a = np.random.randn(m)
    ...: b = np.random.randn(n)
    ...:
    ...: outer = np.outer(a, b)
    ...: broadcast = a[:, None] * b
    ...:
    ...: assert outer.shape == (m, n)
    ...: assert broadcast.shape == (m, n)
    ...: assert np.all(outer == broadcast)
    ...:
    ...: %timeit outer = np.outer(a, b)
    ...: %timeit broadcast = a[:, None] * b
118 µs ± 619 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
119 µs ± 2.46 µs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)