python - 生成连接范围的 1D​​ NumPy 数组

Question

我想生成以下数组a:

nv = np.random.randint(3, 10+1, size=(1000000,))
a = np.concatenate([np.arange(1,i+1) for i in nv])

因此，输出将类似于 -

[0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 0, 1, 2, 3, 4, 5, 0, ...]

有更好的方法吗？

Answer 1

这是使用累积求和的矢量化方法 -

def ranges(nv, start = 1):
    shifts = nv.cumsum()
    id_arr = np.ones(shifts[-1], dtype=int)
    id_arr[shifts[:-1]] = -nv[:-1]+1
    id_arr[0] = start # Skip if we know the start of ranges is 1 already
    return id_arr.cumsum()

示例运行 -

In [23]: nv
Out[23]: array([3, 2, 5, 7])

In [24]: ranges(nv, start=0)
Out[24]: array([0, 1, 2, 0, 1, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5, 6])

In [25]: ranges(nv, start=1)
Out[25]: array([1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 7])

运行时测试 -

In [62]: nv = np.random.randint(3, 10+1, size=(100000,))

In [63]: %timeit your_func(nv) # @MSeifert's solution
10 loops, best of 3: 129 ms per loop

In [64]: %timeit ranges(nv)
100 loops, best of 3: 5.54 ms per loop