使用NumPy查找条件为True的范围(Find span where condition is True using NumPy)

 想象一下，我有一个numpy数组，我需要找到条件为True的跨度/范围。 例如，我有以下数组，其中我试图找到项大于1的跨度： 
 
[0, 0, 0, 2, 2, 0, 2, 2, 2, 0]
 
 我需要找到索引（开始，停止）： 
 
(3, 5) 
(6, 9)
 
 我能够实现的最快的事情是制作一个布尔数组： 
 
truth = data > threshold
 
 然后使用numpy.argmin和numpy.argmax循环遍历数组以查找开始和结束位置。 
 
    pos = 0
    truth = container[RATIO,:] > threshold

    while pos < len(truth):
        start = numpy.argmax(truth[pos:]) + pos + offset
        end = numpy.argmin(truth[start:]) + start  + offset
        if not truth[start]:#nothing more
            break
        if start == end:#goes to the end
            end = len(truth)
        pos = end
 
 但是对于阵列中的数十亿个位置来说这已经太慢了，而且我发现的跨度通常只是连续几个位置。 有谁知道更快找到这些跨度的方法？ 

Imagine I have a numpy array and I need to find the spans/ranges where that condition is True. For example, I have the following array in which I'm trying to find spans where items are greater than 1:
 
[0, 0, 0, 2, 2, 0, 2, 2, 2, 0]
 
I would need to find indices (start, stop):
 
(3, 5) 
(6, 9)
 
The fastest thing I've been able to implement is making a boolean array of:
 
truth = data > threshold
 
and then looping through the array using numpy.argmin and numpy.argmax to find start and end positions. 
 
    pos = 0
    truth = container[RATIO,:] > threshold

    while pos < len(truth):
        start = numpy.argmax(truth[pos:]) + pos + offset
        end = numpy.argmin(truth[start:]) + start  + offset
        if not truth[start]:#nothing more
            break
        if start == end:#goes to the end
            end = len(truth)
        pos = end
 
But this has been too slow for the billions of positions in my arrays and the fact that the spans I'm finding are usually just a few positions in a row. Does anyone know a faster way to find these spans?

原文：https://stackoverflow.com/questions/17150974