查找numpy数组中连续索引的中断

Question

我有一个数组a的条件语句(假设它是A>10)，我得到以下布尔结果:

array([False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False,  True,  True,  True,  True,  True,  True,  True,
True,  True,  True,  True,  True,  True,  True,  True,  True,
True,  True,  True,  True,  True,  True,  True,  True,  True,
True,  True,  True,  True,  True, False, False, False, False,
False, False, False, False, False, False,  True,  True,  True,
True,  True,  True,  True,  True,  True,  True,  True,  True,
True,  True,  True,  True,  True,  True,  True,  True,  True,
True,  True,  True,  True,  True,  True,  True,  True,  True,
True, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False, False, False, False, False, False, False, False,
False, False])

现在我正在寻找值为True的索引。我得到以下数组:

array([20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36,
37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 60, 61, 62, 63,
64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80,
81, 82, 83, 84, 85, 86, 87, 88, 89, 90])

我需要做的是找到连续索引的起始索引和结束索引。例如，在上面的数组中，第一个连续索引的开始索引是20，结束索引是49。同理，第二组连续索引的起始索引为60，结束索引为90。

所以总结一下，我的输出应该是:

start_indices = array([20,60])
end_indices = array([49,90])

如何做到这一点?

Answer 1

下面是itertools中groupby和accumulate的解:

from itertools import groupby, accumulate
## input array
#a = array([False, False, ..., True, ..., False])
indices = list(accumulate(len(list(g)) for i,g in groupby(a)))
starts = indices[:len(indices)//2*2:2]
stops = [i-1 for i in indices[1::2]]

NB。

输出:

>>> starts
[20, 60]
>>> stops
[49, 90]

Answer 2

import numpy as np
# With A as the original array
changes = np.where(np.diff(A > 10))[0] # Gets the actual array out of a tuple
start = changes[::2] + 1
end = changes[1::2]