Find span where condition is True using NumPy

How's one way. First take the boolean array you have:

In [11]: aOut[11]: array([0, 0, 0, 2, 2, 0, 2, 2, 2, 0])In [12]: a1 = a > 1

Shift it one to the left (to get the next state at each index) using roll:

In [13]: a1_rshifted = np.roll(a1, 1)In [14]: starts = a1 & ~a1_rshifted  # it's True but the previous isn'tIn [15]: ends = ~a1 & a1_rshifted

Where this is non-zero is the start of each True batch (or, respectively, end batch):

In [16]: np.nonzero(starts)[0], np.nonzero(ends)[0]Out[16]: (array([3, 6]), array([5, 9]))

And zipping these together:

In [17]: zip(np.nonzero(starts)[0], np.nonzero(ends)[0])Out[17]: [(3, 5), (6, 9)]

If you have access to the scipy library:

You can use scipy.ndimage.measurements.label to identify any regions of non zero value. it returns an array where the value of each element is the id of a span or range in the original array.

You can then use scipy.ndimage.measurements.find_objects to return the slices you would need to extract those ranges. You can access the start / end values directly from those slices.

In your example:

import numpyfrom scipy.ndimage.measurements import label, find_objectsdata = numpy.array([0, 0, 0, 2, 2, 0, 2, 2, 2, 0])labels, number_of_regions = label(data)ranges = find_objects(labels)for identified_range in ranges:    print(identified_range[0].start, identified_range[0].stop)

You should see:

3 56 9

Hope this helps!