[1, 1, 1, 0, 0, 0, 1, 1, 0, 0]
I have a NumPy array consisting of 0's and 1's like above. How can I add all consecutive 1's like below? Any time I encounter a 0, I reset.
[1, 2, 3, 0, 0, 0, 1, 2, 0, 0]
I can do this using a for loop, but is there a vectorized solution using NumPy?
Here's a vectorized approach -
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]
Sample run -
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])
Runtime test
For the timings , I would use OP's sample input array and repeat/tile it and hopefully this should be a less opportunistic benchmark -
Small case :
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
Bigger case :
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
Nah, I want really huge case :
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