[1, 1, 1, 0, 0, 0, 1, 1, 0, 0]
我有一个由0和1组成的NumPy数组,如上所示。我该如何像下面这样添加所有连续的1? 每当我遇到0时,我就会重置。
[1, 2, 3, 0, 0, 0, 1, 2, 0, 0]
我可以使用for循环来完成这个任务,但是是否有使用NumPy的向量化解决方案?
def island_cumsum_vectorized(a):
a_ext = np.concatenate(( [0], a, [0] ))
idx = np.flatnonzero(a_ext[1:] != a_ext[:-1])
a_ext[1:][idx[1::2]] = idx[::2] - idx[1::2]
return a_ext.cumsum()[1:-1]
示例运行 -
In [91]: a = np.array([1, 1, 1, 0, 0, 0, 1, 1, 0, 0])
In [92]: island_cumsum_vectorized(a)
Out[92]: array([1, 2, 3, 0, 0, 0, 1, 2, 0, 0])
In [93]: a = np.array([0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1])
In [94]: island_cumsum_vectorized(a)
Out[94]: array([0, 1, 2, 3, 4, 0, 0, 0, 1, 2, 0, 0, 1])
运行时测试
对于计时,我将使用OP的示例输入数组并重复/平铺它,希望这可以是一个更少机会主义的基准测试
-
小案例:
In [16]: a = np.array([1, 1, 1, 0, 0, 0, 1, 1, 0, 0])
In [17]: a = np.tile(a,10) # Repeat OP's data 10 times
# @Paul Panzer's solution
In [18]: %timeit np.concatenate([np.cumsum(c) if c[0] == 1 else c for c in np.split(a, 1 + np.where(np.diff(a))[0])])
10000 loops, best of 3: 73.4 µs per loop
In [19]: %timeit island_cumsum_vectorized(a)
100000 loops, best of 3: 8.65 µs per loop
更大的机箱:
In [20]: a = np.array([1, 1, 1, 0, 0, 0, 1, 1, 0, 0])
In [21]: a = np.tile(a,1000) # Repeat OP's data 1000 times
# @Paul Panzer's solution
In [22]: %timeit np.concatenate([np.cumsum(c) if c[0] == 1 else c for c in np.split(a, 1 + np.where(np.diff(a))[0])])
100 loops, best of 3: 6.52 ms per loop
In [23]: %timeit island_cumsum_vectorized(a)
10000 loops, best of 3: 49.7 µs per loop
不,我想要一个非常大的机箱:
In [24]: a = np.array([1, 1, 1, 0, 0, 0, 1, 1, 0, 0])
In [25]: a = np.tile(a,100000) # Repeat OP's data 100000 times
# @Paul Panzer's solution
In [26]: %timeit np.concatenate([np.cumsum(c) if c[0] == 1 else c for c in np.split(a, 1 + np.where(np.diff(a))[0])])
1 loops, best of 3: 725 ms per loop
In [27]: %timeit island_cumsum_vectorized(a)
100 loops, best of 3: 7.28 ms per loop
np.concatenate([np.cumsum(c) if c[0] == 1 else c for c in np.split(a, 1 + np.where(np.diff(a))[0])])
Divakar:0.47050797403790057;0.26334572909399867;0.3771821779664606
Paul Panzer:0.6545195570215583;0.4508163968566805;0.17192376195453107
- Paul Panzer